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Continuous and Discrete Time Signals and Systems

Signals and systems is a core topic for electrical and computer engineers. This

textbook presents an introduction to the fundamental concepts of continuous-

time (CT) and discrete-time (DT) signals and systems, treating them separately

in a pedagogical and self-contained manner. Emphasis is on the basic sig-

nal processing principles, with underlying concepts illustrated using practical

examples from signal processing and multimedia communications. The text is

divided into three parts. Part I presents two introductory chapters on signals and

systems. Part II covers the theories, techniques, and applications of CT signals

and systems and Part III discusses these topics for DT signals and systems, so

that the two can be taught independently or together. The focus throughout is

principally on linear time invariant systems. Accompanying the book is a CD-

ROM containing M A T L A B code for running illustrative simulations included

in the text; data files containing audio clips, images and interactive programs

used in the text, and two animations explaining the convolution operation. With

over 300 illustrations, 287 worked examples and 409 homework problems, this

textbook is an ideal introduction to the subject for undergraduates in electrical

and computer engineering. Further resources, including solutions for instruc-

tors, are available online at www.cambridge.org/9780521854559.

Mrinal Mandal is an associate professor at the Department of Electrical and

Computer Engineering, University of Alberta, Edmonton, Canada. His main

research interests include multimedia signal processing, medical image and

video analysis, image and video compression, and VLSI architectures for real-

time signal and image processing.

Amir Asif is an associate professor at the Department of Computer Science and

Engineering, York University, Toronto, Canada. His principal research areas lie

in statistical signal processing with applications in image and video processing,

multimedia communications, and bioinformatics, with particular focus on video

compression, array imaging detection, genomic signal processing, and block-

banded matrix technologies.

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Continuous and Discrete Time Signals and Systems

Mrinal Mandal University of Alberta, Edmonton, Canada

and

Amir Asif York University, Toronto, Canada

iii

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CAMBRIDGE UNIVERSITY PRESS

Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, São Paulo

Cambridge University Press

The Edinburgh Building, Cambridge CB2 8RU, UK

Published in the United States of America by Cambridge University Press, New York

www.cambridge.org

Information on this title: www.cambridge.org/9780521854559

C© Cambridge University Press 2007

This publication is in copyright. Subject to statutory exception

and to the provisions of relevant collective licensing agreements,

no reproduction of any part may take place without

the written permission of Cambridge University Press.

First published 2007

Printed in the United Kingdom at the University Press, Cambridge

A catalog record for this publication is available from the British Library

ISBN-13 978-0-521-85455-9 hardback

Cambridge University Press has no responsibility for the persistence or accuracy of URLs for

external or third-party internet websites referred to in this publication, and does not guarantee that

any content on such websites is, or will remain, accurate or appropriate.

All material contained within the CD-ROM is protected by copyright and other intellectual

property laws. The customer acquires only the right to use the CD-ROM and does not acquire any

other rights, express or implied, unless these are stated explicitly in a separate licence.

To the extent permitted by applicable law, Cambridge University Press is not liable for direct

damages or loss of any kind resulting from the use of this product or from errors or faults

contained in it, and in every case Cambridge University Press’s liability shall be limited to the

amount actually paid by the customer for the product.

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Contents

Preface page xi

Part I Introduction to signals and systems 1

1 Introduction to signals 3

1.1 Classification of signals 5

1.2 Elementary signals 25

1.3 Signal operations 35

1.4 Signal implementation with MATLAB 47

1.5 Summary 51

Problems 53

2 Introduction to systems 62

2.1 Examples of systems 63

2.2 Classification of systems 72

2.3 Interconnection of systems 90

2.4 Summary 93

Problems 94

Part II Continuous-time signals and systems 101

3 Time-domain analysis of LTIC systems 103

3.1 Representation of LTIC systems 103

3.2 Representation of signals using Dirac delta functions 112

3.3 Impulse response of a system 113

3.4 Convolution integral 116

3.5 Graphical method for evaluating the convolution integral 118

3.6 Properties of the convolution integral 125

3.7 Impulse response of LTIC systems 127

3.8 Experiments with MATLAB 131

3.9 Summary 135

Problems 137

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vi Contents

4 Signal representation using Fourier series 141

4.1 Orthogonal vector space 142

4.2 Orthogonal signal space 143

4.3 Fourier basis functions 149

4.4 Trigonometric CTFS 153

4.5 Exponential Fourier series 163

4.6 Properties of exponential CTFS 169

4.7 Existence of Fourier series 177

4.8 Application of Fourier series 179

4.9 Summary 182

Problems 184

5 Continuous-time Fourier transform 193

5.1 CTFT for aperiodic signals 193

5.2 Examples of CTFT 196

5.3 Inverse Fourier transform 209

5.4 Fourier transform of real, even, and odd functions 211

5.5 Properties of the CTFT 216

5.6 Existence of the CTFT 231

5.7 CTFT of periodic functions 233

5.8 CTFS coefficients as samples of CTFT 235

5.9 LTIC systems analysis using CTFT 237

5.10 M A T L A B exercises 246

5.11 Summary 251

Problems 253

6 Laplace transform 261

6.1 Analytical development 262

6.2 Unilateral Laplace transform 266

6.3 Inverse Laplace transform 273

6.4 Properties of the Laplace transform 276

6.5 Solution of differential equations 288

6.6 Characteristic equation, zeros, and poles 293

6.7 Properties of the ROC 295

6.8 Stable and causal LTIC systems 298

6.9 LTIC systems analysis using Laplace transform 305

6.10 Block diagram representations 307

6.11 Summary 311

Problems 313

7 Continuous-time filters 320

7.1 Filter classification 321

7.2 Non-ideal filter characteristics 324

7.3 Design of CT lowpass filters 327

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vii Contents

7.4 Frequency transformations 352

7.5 Summary 364

Problems 365

8 Case studies for CT systems 368

8.1 Amplitude modulation of baseband signals 369

8.2 Mechanical spring damper system 374

8.3 Armature-controlled dc motor 377

8.4 Immune system in humans 383

8.5 Summary 388

Problems 388

Part III Discrete-time signals and systems 391

9 Sampling and quantization 393

9.1 Ideal impulse-train sampling 395

9.2 Practical approaches to sampling 405

9.3 Quantization 410

9.4 Compact disks 413

9.5 Summary 415

Problems 416

10 Time-domain analysis of discrete-time systems systems 422

10.1 Finite-difference equation representation

of LTID systems 423

10.2 Representation of sequences using Dirac delta functions 426

10.3 Impulse response of a system 427

10.4 Convolution sum 430

10.5 Graphical method for evaluating the convolution sum 432

10.6 Periodic convolution 439

10.7 Properties of the convolution sum 448

10.8 Impulse response of LTID systems 451

10.9 Experiments with M A T L A B 455

10.10 Summary 459

Problems 460

11 Discrete-time Fourier series and transform 464

11.1 Discrete-time Fourier series 465

11.2 Fourier transform for aperiodic functions 475

11.3 Existence of the DTFT 482

11.4 DTFT of periodic functions 485

11.5 Properties of the DTFT and the DTFS 491

11.6 Frequency response of LTID systems 506

11.7 Magnitude and phase spectra 507

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viii Contents

11.8 Continuous- and discrete-time Fourier transforms 514

11.9 Summary 517

Problems 520

12 Discrete Fourier transform 525

12.1 Continuous to discrete Fourier transform 526

12.2 Discrete Fourier transform 531

12.3 Spectrum analysis using the DFT 538

12.4 Properties of the DFT 547

12.5 Convolution using the DFT 550

12.6 Fast Fourier transform 553

12.7 Summary 559

Problems 560

13 The z-transform 565

13.1 Analytical development 566

13.2 Unilateral z-transform 569

13.3 Inverse z-transform 574

13.4 Properties of the z-transform 582

13.5 Solution of difference equations 594

13.6 z-transfer function of LTID systems 596

13.7 Relationship between Laplace and z-transforms 599

13.8 Stabilty analysis in the z-domain 601

13.9 Frequency-response calculation in the z-domain 606

13.10 DTFT and the z-transform 607

13.11 Experiments with M A T L A B 609

13.12 Summary 614

Problems 616

14 Digital filters 621

14.1 Filter classification 622

14.2 FIR and IIR filters 625

14.3 Phase of a digital filter 627

14.4 Ideal versus non-ideal filters 632

14.5 Filter realization 638

14.6 FIR filters 639

14.7 IIR filters 644

14.8 Finite precision effect 651

14.9 M A T L A B examples 657

14.10 Summary 658

Problems 660

15 FIR filter design 665

15.1 Lowpass filter design using windowing method 666

15.2 Design of highpass filters using windowing 684

15.3 Design of bandpass filters using windowing 688

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ix Contents

15.4 Design of a bandstop filter using windowing 691

15.5 Optimal FIR filters 693

15.6 M A T L A B examples 700

15.7 Summary 707

Problems 709

16 IIR filter design 713

16.1 IIR filter design principles 714

16.2 Impulse invariance 715

16.3 Bilinear transformation 728

16.4 Designing highpass, bandpass, and bandstop IIR filters 734

16.5 IIR and FIR filters 737

16.6 Summary 741

Problems 742

17 Applications of digital signal processing 746

17.1 Spectral estimation 746

17.2 Digital audio 754

17.3 Audio filtering 759

17.4 Digital audio compression 765

17.5 Digital images 771

17.6 Image filtering 777

17.7 Image compression 782

17.8 Summary 789

Problems 789

Appendix A Mathematical preliminaries 793

A.1 Trigonometric identities 793

A.2 Power series 794

A.3 Series summation 794

A.4 Limits and differential calculus 795

A.5 Indefinite integrals 795

Appendix B Introduction to the complex-number system 797

B.1 Real-number system 797

B.2 Complex-number system 798

B.3 Graphical interpertation of complex numbers 801

B.4 Polar representation of complex numbers 801

B.5 Summary 805

Problems 805

Appendix C Linear constant-coefficient differential equations 806

C.1 Zero-input response 807

C.2 Zero-state response 810

C.3 Complete response 813

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x Contents

Appendix D Partial fraction expansion 814

D.1 Laplace transform 814

D.2 Continuous-time Fourier transform 822

D.3 Discrete-time Fourier transform 825

D.4 The z-transform 826

Appendix E Introduction to M A T L A B 829

E.1 Introduction 829

E.2 Entering data into M A T L A B 831

E.3 Control statements 838

E.4 Elementary matrix operations 840

E.5 Plotting functions 842

E.6 Creating M A T L A B functions 846

E.7 Summary 847

Appendix F About the CD 848

F.1 Interactive environment 848

F.2 Data 853

F.3 M A T L A B codes 854

Bibliography 858 Index 860

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Preface

The book is primarily intended for instruction in an upper-level undergraduate

or a first-year graduate course in the field of signal processing in electrical

and computer engineering. Practising engineers would find the book useful

for reference or for self study. Our main motivation in writing the book is to

deal with continuous-time (CT) and discrete-time (DT) signals and systems

separately. Many instructors have realized that covering CT and DT systems in

parallel with each other often confuses students to the extent where they are not

clear if a particular concept applies to a CT system, to a DT system, or to both.

In this book, we treat DT and CT signals and systems separately. Following

Part I, which provides an introduction to signals and systems, Part II focuses on

CT signals and systems. Since most students are familiar with the theory of CT

signals and systems from earlier courses, Part II can be taught to such students

with relative ease. For students who are new to this area, we have supplemented

the material covered in Part II with appendices, which are included at the end

of the book. Appendices A–F cover background material on complex numbers,

partial fraction expansion, differential equations, difference equations, and a

review of the basic signal processing instructions available in M A T L A B . Part

III, which covers DT signals and systems, can either be covered independently

or in conjunction with Part II.

The book focuses on linear time-invariant (LTI) systems and is organized as

follows. Chapters 1 and 2 introduce signals and systems, including their math-

ematical and graphical interpretations. In Chapter 1, we cover the classification

between CT and DT signals and we provide several practical examples in which

CT and DT signals are observed. Chapter 2 defines systems as transformations

that process the input signals and produce outputs in response to the applied

inputs. Practical examples of CT and DT systems are included in Chapter 2.

The remaining fifteen chapters of the book are divided into two parts. Part

II constitutes Chapters 3–8 of the book and focuses primarily on the theories

and applications of CT signals and systems. Part III comprises Chapters 9–17

and deals with the theories and applications of DT signals and systems. The

organization of Parts II and III is described below.

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xii Preface

Chapter 3 introduces the time-domain analysis of the linear time-invariant

continuous-time (LTIC) systems, including the convolution integral used to

evaluate the output in response to a given input signal. Chapter 4 defines the

continuous-time Fourier series (CTFS) as a frequency representation for the

CT periodic signals, and Chapter 5 generalizes the CTFS to aperiodic signals

and develops an alternative representation, referred to as the continuous-time

Fourier transform (CTFT). Not only do the CTFT and CTFS representations

provide an alternative to the convolution integral for the evaluation of the out-

put response, but also these frequency representations allow additional insights

into the behavior of the LTIC systems that are exploited later in the book to

design such systems. While the CTFT is useful for steady state analysis of

the LTIC systems, the Laplace transform, introduced in Chapter 6, is used in

control applications where transient and stability analyses are required. An

important subset of LTIC systems are frequency-selective filters, whose char-

acteristics are specified in the frequency domain. Chapter 7 presents design

techniques for several CT frequency-selective filters including the Butterworth,

Chebyshev, and elliptic filters. Finally, Chapter 8 concludes our treatment of

LTIC signals and systems by reviewing important applications of CT signal

processing.

The coverage of CT signals and systems concludes with Chapter 8 and a

course emphasizing the CT domain can be completed at this stage. In Part

III, Chapter 9 starts our consideration of DT signals and systems by providing

several practical examples in which such signals are observed directly. Most

DT sequences are, however, obtained by sampling CT signals. Chapter 9 shows

how a band-limited CT signal can be accurately represented by a DT sequence

such that no information is lost in the conversion from the CT to the DT domain.

Chapter 10 provides the time-domain analysis of linear time-invariant discrete-

time (LTID) systems, including the convolution sum used to calculate the

output of a DT system. Chapter 11 introduces the frequency representations for

DT sequences, namely the discrete-time Fourier series (DTFS) and the discrete-

time Fourier transform (DTFT). The discrete Fourier transform (DFT) samples

the DTFT representation in the frequency domain and is convenient for digital

signal processing of finite-length sequences. Chapter 12 introduces the DFT,

while Chapter 13 is devoted to a discussion of the z-transform. As for CT sys-

tems, DT systems are generally specified in the frequency domain. A particular

class of DT systems, referred to as frequency-selective digital filters, is intro-

duced in Chapter 14. Based on the length of the impulse response, digital filters

can be further classified into finite impulse response (FIR) and infinite impulse

response (IIR) filters. Chapter 15 covers the design techniques for the IIR filters,

and Chapter 16 presents the design techniques for the IIR filters. Chapter 17

concludes the book by motivating the students with several applications of

digital signal processing in audio and music, spectral analysis, and image and

video processing.

Although the book has been designed to be as self-contained as possible,

some basic prerequisites have been assumed. For example, an introductory

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xiii Preface

background in mathematics which includes trigonometry, differential calculus,

integral calculus, and complex number theory, would be helpful. A course in

electrical circuits, although not essential, would be highly useful as several

examples of electrical circuits have been used as systems to motivate the

students. For students who lack some of the required background information,

a review of the core background materials such as complex numbers, partial

fraction expansion, differential equations, and difference equations is provided

in the appendices.

The normal use of this book should be as follows. For a first course in signal

processing, at, say, sophomore or junior level, a reasonable goal is to teach

Part II, covering continuous-time (CT) signals and sysems. Part III provides the

material for a more advanced course in discrete-time (DT) signal processing. We

have also spent a great deal of time experimening with different presentations for

a single-semester signals and systems course. Typically, such a course should

include Chapters 1, 2, 3, 10, 4, 5, 11, 6, and 13 in that order. Below, we provide

course outlines for a few traditional signal processing courses. These course

outlines should be useful to an instructor teaching this type of material or using

the book for the first time.

(1) Continuous-time signals and systems: Chapters 1–8.

(2) Discrete-time signals and systems: Chapters 1, 2, 9–17.

(3) Traditional signals and systems: Chapters 1, 2, (3, 10), (4, 5, 11), 6, 13.

(4) Digital signal processing: Chapters 10–17.

(5) Transform theory: Chapters (4, 5, 11), 6, 13.

Another useful feature of the book is that the chapters are self-contained so that

they may be taught independently of each other. There is a significant difference

between reading a book and being able to apply the material to solve actual

problems of interest. An effective use of the book must include a fair coverage

of the solved examples and problem solving by motivating the students to solve

the problems included at the end of each chapter. As such, a major focus of

the book is to illustrate the basic signal processing concepts with examples.

We have included 287 worked examples, 409 supplementary problems at the

ends of the chapters, and more than 300 figures to explain the important con-

cepts. Wherever relevant, we have extensively used M A T L A B to validate our

analytical results and also to illustrate the design procedures for a variety of

problems. In most cases, the M A T L A B code is provided in the accompanying

CD, so the students can readily run the code to satisfy their curiosity. To further

enhance their understanding of the main signal processing concepts, students

are encouraged to program extensively in M A T L A B . Consequently, several

M A T L A B exercises have been included in the Problems sections.

Any suggestions or concerns regarding the book may be communicated

to the authors; email addresses are listed at http://www.cambridge.org/

9780521854559. Future updates on the book will also be available at the same

website.

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xiv Preface

A number of people have contributed in different ways, and it is a plea-

sure to acknowledge them. Anna Littlewood, Irene Pizzie, and Emily Yossarian

of Cambridge University Press contributed significantly during the production

stage of the book. Professor Tyseer Aboulnasr reviewed the complete book

and provided valuable feedback to enhance its quality. In addition, Mrinal

Mandal would like to thank Wen Chen, Meghna Singh, Saeed S. Tehrani, San-

jukta Mukhopadhayaya, and Professor Thomas Sikora for their help in the

overall preparation of the book. On behalf of Amir Asif, special thanks are due

to Professor José Moura, who introduced the fascinating field of signal pro-

cessing to him for the first time and has served as his mentor for several years.

Lastly, Mrinal Mandal thanks his parents, Iswar Chandra Mandal (late) and

Mrs Kiran Bala Mandal, and his wife Rupa, and Amir Asif thanks his parents,

Asif Mahmood (late) and Khalida Asif, his wife Sadia, and children Maaz and

Sannah for their continuous support and love over the years.

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P A R T I

Introduction to signals and systems

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C H A P T E R

1 Introduction to signals

Signals are detectable quantities used to convey information about time-varying

physical phenomena. Common examples of signals are human speech, temper-

ature, pressure, and stock prices. Electrical signals, normally expressed in the

form of voltage or current waveforms, are some of the easiest signals to generate

and process.

Mathematically, signals are modeled as functions of one or more independent

variables. Examples of independent variables used to represent signals are time,

frequency, or spatial coordinates. Before introducing the mathematical notation

used to represent signals, let us consider a few physical systems associated

with the generation of signals. Figure 1.1 illustrates some common signals and

systems encountered in different fields of engineering, with the physical sys-

tems represented in the left-hand column and the associated signals included in

the right-hand column. Figure 1.1(a) is a simple electrical circuit consisting of

three passive components: a capacitor C , an inductor L , and a resistor R. A

voltage v(t) is applied at the input of the RLC circuit, which produces an output

voltage y(t) across the capacitor. A possible waveform for y(t) is the sinusoidal

signal shown in Fig. 1.1(b). The notations v(t) and y(t) includes both the depen-

dent variable, v and y, respectively, in the two expressions, and the independent

variable t . The notation v(t) implies that the voltage v is a function of time t.

Figure 1.1(c) shows an audio recording system where the input signal is an audio

or a speech waveform. The function of the audio recording system is to convert

the audio signal into an electrical waveform, which is recorded on a magnetic

tape or a compact disc. A possible resulting waveform for the recorded electri-

cal signal is shown in Fig 1.1(d). Figure 1.1(e) shows a charge coupled device

(CCD) based digital camera where the input signal is the light emitted from a

scene. The incident light charges a CCD panel located inside the camera, thereby

storing the external scene in terms of the spatial variations of the charges on the

CCD panel. Figure 1.1(g) illustrates a thermometer that measures the ambient

temperature of its environment. Electronic thermometers typically use a thermal

resistor, known as a thermistor, whose resistance varies with temperature. The

fluctuations in the resistance are used to measure the temperature. Figure 1.1(h)

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4 Part I Introduction to signals and systems

− + −

L R1 R2

R3 C y (t)v (t)

(a)

−1 0 t

x(t) = sin(pt)

1 2−2

(b)

audio

output

(c)

0.4

audio signal waveform

n o

rm al

iz ed

a m

p li

tu d

−0.4

−0.8

0 0.2 0.4 0.6

time (s)

0.8 1.21

(d)

(e)

( f )

thermal

resistor

RcR1

Rin

voltage

temperature

conversion

+Vc

Vin

temperature

display

(g)

S M T W F SH k

21.0

22.0

20.9 21.6

22.3

20.2

23.0

(h)

Fig. 1.1. Examples of signals and systems. (a) An electrical circuit; (c) an audio recording system; (e) a

digital camera; and (g) a digital thermometer. Plots (b), (d), (f ), and (h) are output signals generated,

respectively, by the systems shown in (a), (c), (e), and (g).

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5 1 Introduction to signals

input

signal

output

signal system

Fig. 1.2. Processing of a signal

by a system.

plots the readings of the thermometer as a function of discrete time. In the

aforementioned examples of Fig. 1.1, the RLC circuit, audio recorder, CCD

camera, and thermometer represent different systems, while the information-

bearing waveforms, such as the voltage, audio, charges, and fluctuations in

resistance, represent signals. The output waveforms, for example the voltage in

the case of the electrical circuit, current for the microphone, and the fluctuations

in the resistance for the thermometer, vary with respect to only one variable

(time) and are classified as one-dimensional (1D) signals. On the other hand,

the charge distribution in the CCD panel of the camera varies spatially in two

dimensions. The independent variables are the two spatial coordinates (m, n).

The charge distribution signal is therefore classified as a two-dimensional (2D)

signal.

The examples shown in Fig. 1.1 illustrate that typically every system has one

or more signals associated with it. A system is therefore defined as an entity

that processes a set of signals (called the input signals) and produces another

set of signals (called the output signals). The voltage source in Fig. 1.1(a),

the sound in Fig. 1.1(c), the light entering the camera in Fig. 1.1(e), and the

ambient heat in Fig. 1.1(g) provide examples of the input signals. The voltage

across capacitor C in Fig. 1.1(b), the voltage generated by the microphone in

Fig. 1.1(d), the charge stored on the CCD panel of the digital camera, displayed

as an image in Fig. 1.1(f), and the voltage generated by the thermistor, used to

measure the room temperature, in Fig. 1.1(h) are examples of output signals.

Figure 1.2 shows a simplified schematic representation of a signal processing

system. The system shown processes an input signal x(t) producing an output

y(t). This model may be used to represent a range of physical processes includ-

ing electrical circuits, mechanical devices, hydraulic systems, and computer

algorithms with a single input and a single output. More complex systems have

multiple inputs and multiple outputs (MIMO).

Despite the wide scope of signals and systems, there is a set of fundamental

principles that control the operation of these systems. Understanding these basic

principles is important in order to analyze, design, and develop new systems.

The main focus of the text is to present the theories and principles used in

signals and systems. To keep the presentations simple, we focus primarily on

signals with one independent variable (usually the time variable denoted by t

or k), and systems with a single input and a single output. The theories that we

develop for single-input, single-output systems are, however, generalizable to

multidimensional signals and systems with multiple inputs and outputs.

1.1 Classification of signals

A signal is classified into several categories depending upon the criteria used

for its classification. In this section, we cover the following categories for

signals:

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6 Part I Introduction to signals and systems

(i) continuous-time and discrete-time signals;

(ii) analog and digital signals;

(iii) periodic and aperiodic (or nonperiodic) signals;

(iv) energy and power signals;

(v) deterministic and probabilistic signals;

(vi) even and odd signals.

1.1.1 Continuous-time and discrete-time signals

If a signal is defined for all values of the independent variable t , it is called

a continuous-time (CT) signal. Consider the signals shown in Figs. 1.1(b) and

(d). Since these signals vary continuously with time t and have known mag-

nitudes for all time instants, they are classified as CT signals. On the other

hand, if a signal is defined only at discrete values of time, it is called a discrete-

time (DT) signal. Figure 1.1(h) shows the output temperature of a room mea-

sured at the same hour every day for one week. No information is available

for the temperature in between the daily readings. Figure 1.1(h) is therefore

an example of a DT signal. In our notation, a CT signal is denoted by x(t)

with regular parenthesis, and a DT signal is denoted with square parenthesis as

follows:

x[kT ], k = 0, ±1, ±2, ±3, . . . ,

where T denotes the time interval between two consecutive samples. In the

example of Fig. 1.1(h), the value of T is one day. To keep the notation simple,

we denote a one-dimensional (1D) DT signal x by x[k]. Though the sampling

interval is not explicitly included in x[k], it will be incorporated if and when

required.

Note that all DT signals are not functions of time. Figure 1.1(f), for example,

shows the output of a CCD camera, where the discrete output varies spatially in

two dimensions. Here, the independent variables are denoted by (m, n), where

m and n are the discretized horizontal and vertical coordinates of the picture

element. In this case, the two-dimensional (2D) DT signal representing the

spatial charge is denoted by x[m, n].

−1 0 t

x(t) = sin(pt)

1 2−2

(a)

−4−6

−2

0 2

x[k] = sin(0.25pk)

8−8

(b)

Fig. 1.3. (a) CT sinusoidal signal

x (t ) specified in Example 1.1;

(b) DT sinusoidal signal x [k ]

obtained by discretizing x (t )

with a sampling interval

T = 0.25 s.

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7 1 Introduction to signals

Example 1.1

Consider the CT signal x(t) = sin(π t) plotted in Fig. 1.3(a) as a function of time t . Discretize the signal using a sampling interval of T = 0.25 s, and sketch the waveform of the resulting DT sequence for the range −8 ≤ k ≤ 8.

Solution

By substituting t = kT , the DT representation of the CT signal x(t) is given by

x[kT ] = sin(πk × T ) = sin(0.25πk).

For k = 0, ±1, ±2, . . . , the DT signal x[k] has the following values:

x[−8] = x(−8T ) = sin(−2π ) = 0, x[1] = x(T ) = sin(0.25π ) = 1

√ 2 ,

x[−7] = x(−7T ) = sin(−1.75π ) = 1

√ 2 , x[2] = x(2T ) = sin(0.5π ) = 1,

x[−6] = x(−6T ) = sin(−1.5π ) = 1, x[3] = x(3T ) = sin(0.75π ) = 1

√ 2 ,

x[−5] = x(−5T ) = sin(−1.25π ) = 1

√ 2 , x[4] = x(4T ) = sin(π ) = 0,

x[−4] = x(−4T ) = sin(−π ) = 0, x[5] = x(5T ) = sin(1.25π ) = − 1

√ 2 ,

x[−3] = x(−3T ) = sin(−0.75π ) = − 1

√ 2 , x[6] = x(6T ) = sin(1.5π ) = −1,

x[−2] = x(−2T ) = sin(−0.5π ) = −1, x[7] = x(7T ) = sin(1.75π ) = − 1

√ 2 ,

x[−1] = x(−T ) = sin(−0.25π ) = − 1

√ 2 , x[8] = x(8T ) = sin(2π ) = 0,

x[0] = x(0) = sin(0) = 0.

Plotted as a function of k, the waveform for the DT signal x[k] is shown in

Fig. 1.3(b), where for reference the original CT waveform is plotted with a

dotted line. We will refer to a DT plot illustrated in Fig. 1.3(b) as a bar or a

stem plot to distinguish it from the CT plot of x(t), which will be referred to as

a line plot.

Example 1.2

Consider the rectangular pulse plotted in Fig. 1.4. Mathematically, the rectan-

gular pulse is denoted by

x(t) = rect (

)

= {

1 |t | ≤ τ/2 0 |t | > τ/2.

1 x(t)

t 0.5t−0.5t

Fig. 1.4. Waveform for CT

rectangular function. It may be

noted that the rectangular

function is discontinuous at

t = ±τ /2.

From the waveform in Fig. 1.4, it is clear that x(t) is continuous in time but

has discontinuities in magnitude at time instants t = ±0.5τ . At t = 0.5τ , for example, the rectangular pulse has two values: 0 and 1. A possible way to avoid

this ambiguity in specifying the magnitude is to state the values of the signal x(t)

at t = 0.5τ− and t = 0.5τ+, i.e. immediately before and after the discontinuity. Mathematically, the time instant t = 0.5τ− is defined as t = 0.5τ − ε, where ε is an infinitely small positive number that is close to zero. Similarly, the

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8 Part I Introduction to signals and systems

time instant t = 0.5τ+ is defined as t = 0.5τ + ε. The value of the rectangular pulse at the discontinuity t = 0.5τ is, therefore, specified by x(0.5τ−) = 1 and x(0.5τ+) = 0. Likewise, the value of the rectangular pulse at its other discontinuity t = −0.5τ is specified by x(−0.5τ−) = 0 and x(−0.5τ+) = 1.

A CT signal that is continuous for all t except for a finite number of instants

is referred to as a piecewise CT signal. The value of a piecewise CT signal at the

point of discontinuity t1 can either be specified by our earlier notation, described

in the previous paragraph, or, alternatively, using the following relationship:

x(t1) = 0.5 [

x(t+1 ) + x(t − 1 )

]

. (1.1)

Equation (1.1) shows that x(±0.5τ ) = 0.5 at the points of discontinuity t = ±0.5τ . The second approach is useful in certain applications. For instance, when a piecewise CT signal is reconstructed from an infinite series (such as the

Fourier series defined later in the text), the reconstructed value at the point of

discontinuity satisfies Eq. (1.1). Discussion of piecewise CT signals is continued

in Chapter 4, where we define the CT Fourier series.

1.1.2 Analog and digital signals

A second classification of signals is based on their amplitudes. The amplitudes

of many real-world signals, such as voltage, current, temperature, and pressure,

change continuously, and these signals are called analog signals. For example,

the ambient temperature of a house is an analog number that requires an infinite

number of digits (e.g., 24.763 578. . . ) to record the readings precisely. Digital

signals, on the other hand, can only have a finite number of amplitude values.

For example, if a digital thermometer, with a resolution of 1 ◦C and a range

of [10 ◦C, 30 ◦C], is used to measure the room temperature at discrete time

instants, t = kT , then the recordings constitute a digital signal. An example of a digital signal was shown in Fig. 1.1(h), which plots the temperature readings

taken once a day for one week. This digital signal has an amplitude resolution

of 0.1 ◦C, and a sampling interval of one day.

Figure 1.5 shows an analog signal with its digital approximation. The analog

signal has a limited dynamic range between [−1, 1] but can assume any real value (rational or irrational) within this dynamic range. If the analog signal is

sampled at time instants t = kT and the magnitude of the resulting samples are quantized to a set of finite number of known values within the range [−1, 1], the resulting signal becomes a digital signal. Using the following set of eight

uniformly distributed values,

[−0.875, −0.625, −0.375, −0.125, 0.125, 0.375, 0.625, 0.875],

within the range [−1, 1], the best approximation of the analog signal is the digital signal shown with the stem plot in Fig. 1.5.

Another example of a digital signal is the music recorded on an audio com-

pact disc (CD). On a CD, the music signal is first sampled at a rate of 44 100

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9 1 Introduction to signals

1.125

0.875

0.625

0.375

0.125

−0.125

−0.375

−0.625

−0.875

−1.125 0 1 2 3 40 1 2 3 4 5 6 7 8

sampling time t = kT

si g n al

v al

u e

Fig. 1.5. Analog signal with its

digital approximation. The

waveform for the analog signal

is shown with a line plot; the

quantized digital approximation

is shown with a stem plot.

samples per second. The sampling interval T is given by 1/44 100, or 22.68

microseconds (µs). Each sample is then quantized with a 16-bit uniform quan-

tizer. In other words, a sample of the recorded music signal is approximated

from a set of uniformly distributed values that can be represented by a 16-bit

binary number. The total number of values in the discretized set is therefore

limited to 216 entries.

Digital signals may also occur naturally. For example, the price of a com-

modity is a multiple of the lowest denomination of a currency. The grades of

students on a course are also discrete, e.g. 8 out of 10, or 3.6 out of 4 on a 4-point

grade point average (GPA). The number of employees in an organization is a

non-negative integer and is also digital by nature.

1.1.3 Periodic and aperiodic signals

A CT signal x(t) is said to be periodic if it satisfies the following property:

x(t) = x(t + T0), (1.2)

at all time t and for some positive constant T0. The smallest positive value

of T0 that satisfies the periodicity condition, Eq. (1.3), is referred to as the

fundamental period of x(t).

Likewise, a DT signal x[k] is said to be periodic if it satisfies

x[k] = x[k + K0] (1.3)

at all time k and for some positive constant K0. The smallest positive value of

K0 that satisfies the periodicity condition, Eq. (1.4), is referred to as the fun-

damental period of x[k]. A signal that is not periodic is called an aperiodic or

non-periodic signal. Figure 1.6 shows examples of both periodic and aperiodic

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10 Part I Introduction to signals and systems

−2−4 0 2 4

−3

(a)

(b)

−1 0 t

1 2−2

(c)

4 8−8 −4 2

−6

−2

−4 −−5 −2 −1 21 3 4 5

3 2

−3 0

( f )(e)

(d)

Fig. 1.6. Examples of periodic

((a), (c), and (e)) and aperiodic

((b), (d), and (f)) signals. The

line plots (a) and (c) represent

CT periodic signals with

fundamental periods T0 of 4 and

2, while the stem plot (e)

represents a DT periodic signal

with fundamental period

K0 = 8.

signals. The reciprocal of the fundamental period of a signal is called the fun-

damental frequency. Mathematically, the fundamental frequency is expressed

as follows

f0 = 1

T0 , for CT signals, or f0 =

K0 , for DT signals, (1.4)

where T0 and K0 are, respectively, the fundamental periods of the CT and DT

signals. The frequency of a signal provides useful information regarding how

fast the signal changes its amplitude. The unit of frequency is cycles per second

(c/s) or hertz (Hz). Sometimes, we also use radians per second as a unit of

frequency. Since there are 2π radians (or 360◦) in one cycle, a frequency of f0 hertz is equivalent to 2π f0 radians per second. If radians per second is used as

a unit of frequency, the frequency is referred to as the angular frequency and is

given by

ω0 = 2π

T0 , for CT signals, or Ω0 =

2π

K0 , for DT signals. (1.5)

A familiar example of a periodic signal is a sinusoidal function represented

mathematically by the following expression:

x(t) = A sin(ω0t + θ ).

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11 1 Introduction to signals

The sinusoidal signal x(t) has a fundamental period T0 = 2π/ω0 as we prove next. Substituting t by t + T0 in the sinusoidal function, yields

x(t + T0) = A sin(ω0t + ω0T0 + θ ).

Since

x(t) = A sin(ω0t + θ ) = A sin(ω0t + 2mπ + θ ), for m = 0, ±1, ±2, . . . ,

the above two expressions are equal iff ω0T0 = 2mπ . Selecting m = 1, the fundamental period is given by T0 = 2π/ω0.

The sinusoidal signal x(t) can also be expressed as a function of a complex

exponential. Using the Euler identity,

ej(ω0t+θ ) = cos(ω0t + θ ) + j sin(ω0t + θ ), (1.6)

we observe that the sinusoidal signal x(t) is the imaginary component of a

complex exponential. By noting that both the imaginary and real components

of an exponential function are periodic with fundamental period T0 = 2π/ω0, it can be shown that the complex exponential x(t) = exp[j(ω0t + θ )] is also a periodic signal with the same fundamental period of T0 = 2π/ω0.

Example 1.3

(i) CT sine wave: x1(t) = sin(4π t) is a periodic signal with period T1 = 2π/4π = 1/2;

(ii) CT cosine wave: x2(t) = cos(3π t) is a periodic signal with period T2 = 2π/3π = 2/3;

(iii) CT tangent wave: x3(t) = tan(10t) is a periodic signal with period T3 = π/10;

(iv) CT complex exponential: x4(t) = e j(2t+7) is a periodic signal with period T4 = 2π/2 = π ;

(v) CT sine wave of limited duration: x6(t) = {

sin 4π t −2 ≤ t ≤ 2 0 otherwise

is an

aperiodic signal;

(vi) CT linear relationship: x7(t) = 2t + 5 is an aperiodic signal; (vii) CT real exponential: x4(t) = e−2t is an aperiodic signal.

Although all CT sinusoidals are periodic, their DT counterparts x[k] = A sin(Ω0k + θ ) may not always be periodic. In the following discussion, we derive a condition for the DT sinusoidal x[k] to be periodic.

Assuming x[k] = A sin(Ω0k + θ ) is periodic with period K0 yields

x[k + K0] = sin(Ω0(k + K0) + θ ) = sin(Ω0k + Ω0 K0) + θ ).

Since x[k] can be expressed as x[k] = sin(Ω0k + 2mπ + θ ), the value of the fundamental period is given by K0 = 2πm/�0 for m = 0, ±1, ±2, . . . Since we are dealing with DT sequences, the value of the fundamental period K0 must

be an integer. In other words, x[k] is periodic if we can find a set of values for

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12 Part I Introduction to signals and systems

m, K0 ∈ Z+, where we use the notation Z+ to denote a set of positive integer values. Based on the above discussion, we make the following proposition.

Proposition 1.1 An arbitrary DT sinusoidal sequence x[k] = A sin(Ω0k + θ ) is periodic iff Ω0/2π is a rational number.

The term rational number used in Proposition 1.1 is defined as a fraction of

two integers. Given that the DT sinusoidal sequence x[k] = A sin(Ω0k + θ ) is periodic, its fundamental period is evaluated from the relationship

Ω0

2π =

K0 (1.7)

K0 = 2π

Ω0

m. (1.8)

Proposition 1.1 can be extended to include DT complex exponential signals.

Collectively, we state the following.

(1) The fundamental period of a sinusoidal signal that satisfies Proposition 1.1

is calculated from Eq. (1.8) with m set to the smallest integer that results

in an integer value for K0.

(2) A complex exponential x[k] = A exp[j(Ω0k + θ )] must also satisfy Propo- sition 1.1 to be periodic. The fundamental period of a complex exponential

is also given by Eq. (1.8).

Example 1.4

Determine if the sinusoidal DT sequences (i)–(iv) are periodic:

(i) f [k] = sin(πk/12 + π/4); (ii) g[k] = cos(3πk/10 + θ );

(iii) h[k] = cos(0.5k + φ); (iv) p[k] = ej(7πk/8+θ ).

Solution

(i) The value of �0 in f [k] is π/12. SinceΩ0/2π = 1/24 is a rational number, the DT sequence f [k] is periodic. Using Eq. (1.8), the fundamental period of

f [k] is given by

K0 = 2π

Ω0

m = 24m.

Setting m = 1 yields the fundamental period K0 = 24. To demonstrate that f [k] is indeed a periodic signal, consider the following:

f [k + K0] = sin(π [k + K0]/12 + π/4).

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13 1 Introduction to signals

Substituting K0 = 24 in the above equation, we obtain

f [k + K0] = sin(π [k + K0]/12 + π/4) = sin(πk + 2π + π/4) = sin(πk/12 + π/4) = f [k].

(ii) The value of Ω0 in g[k] is 3π/10. Since �0/2π = 3/20 is a rational number, the DT sequence g[k] is periodic. Using Eq. (1.8), the fundamental

period of g[k] is given by

K0 = 2π

Ω0

m = 20m

3 .

Setting m = 3 yields the fundamental period K0 = 20. (iii) The value of Ω0 in h[k] is 0.5. Since Ω0/2π = 1/4π is not a rational

number, the DT sequence h[k] is not periodic.

(iv) The value of Ω0 in p[k] is 7π/8. Since Ω0/2π = 7/16 is a rational number, the DT sequence p[k] is periodic. Using Eq. (1.8), the fundamental

period of p[k] is given by

K0 = 2π

Ω0

m = 16m

7 .

Setting m = 7 yields the fundamental period K0 = 16. Example 1.3 shows that CT sinusoidal signals of the form x(t) =

sin(ω0t + θ ) are always periodic with fundamental period 2π/ω0 irrespective of the value of ω0. However, Example 1.4 shows that the DT sinusoidal sequences

are not always periodic. The DT sequences are periodic only when Ω0/2π is a

rational number. This leads to the following interesting observation.

Consider the periodic signal x(t) = sin(ω0t + θ ). Sample the signal with a sampling interval T . The DT sequence is represented as x[k] = sin(ω0kT + θ ). The DT signal will be periodic if Ω0/2π = ω0T/2π is a rational number. In other words, if you sample a CT periodic signal, the DT signal need not always

be periodic. The signal will be periodic only if you choose a sampling interval

T such that the term ω0T/2π is a rational number.

1.1.3.1 Harmonics

Consider two sinusoidal functions x(t) = sin(ω0t + θ ) and xm(t) = sin(mω0t + θ ). The fundamental angular frequencies of these two CT signals are given by ω0 and mω0 radians/s, respectively. In other words, the angular

frequency of the signal xm(t) is m times the angular frequency of the signal

x(t). In such cases, the CT signal xm(t) is referred to as the mth harmonic of

x(t). Using Eq. (1.6), it is straightforward to verify that the fundamental period

of x(t) is m times that of xm(t).

Figure 1.7 plots the waveform of a signal x(t) = sin(2π t) and its second har- monic. The fundamental period of x(t) is 1 s with a fundamental frequency of

2π radians/s. The second harmonic of x(t) is given by x2(t) = sin(4π t). Like- wise, the third harmonic of x(t) is given by x3(t) = sin(6π t). The fundamental

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14 Part I Introduction to signals and systems

−2 −1 1 2 t

x1(t) = sin(2pt)

(a)

1 2−2 −1

x2(t) = sin(4pt)

(b)

Fig. 1.7. Examples of harmonics.

(a) Waveform for the sinusoidal

signal x(t ) = sin(2π t ); (b) waveform for its second

harmonic given by

x2(t ) = sin(4π t ).

periods of the second harmonic x2(t) and third harmonics x3(t) are given by

1/2 s and 1/3 s, respectively.

Harmonics are important in signal analysis as any periodic non-sinusoidal

signal can be expressed as a linear combination of a sine wave having the same

fundamental frequency as the fundamental frequency of the original periodic

signal and the harmonics of the sine wave. This property is the basis of the

Fourier series expansion of periodic signals and will be demonstrated with

examples in later chapters.

1.1.3.2 Linear combination of two signals

Proposition 1.2 A signal g(t) that is a linear combination of two periodic sig-

nals, x1(t) with fundamental period T1 and x2(t) with fundamental period T2 as

follows:

g(t) = ax1(t) + bx2(t)

is periodic iff

T2 =

n = rational number. (1.9)

The fundamental period of g(t) is given by nT1 = mT2 provided that the values of m and n are chosen such that the greatest common divisor (gcd) between m

and n is 1.

Proposition 1.2 can also be extended to DT sequences. We illustrate the

application of Proposition 1.2 through a series of examples.

Example 1.5

Determine if the following signals are periodic. If yes, determine the funda-

mental period.

(i) g1(t) = 3 sin(4π t) + 7 cos(3π t); (ii) g2(t) = 3 sin(4π t) + 7 cos(10t).

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15 1 Introduction to signals

Solution

(i) In Example 1.3, we saw that the sinuosoidal signals sin(4π t) and cos(3π t)

are both periodic signals with fundamental periods 1/2 and 2/3, respectively.

Calculating the ratio of the two fundamental periods yields

T2 =

1/2

2/3 =

4 ,

which is a rational number. Hence, the linear combination g1(t) is a periodic

signal.

Comparing the above ratio with Eq. (1.9), we obtain m = 3 and n = 4. The fundamental period of g1(t) is given by nT1 = 4T1 = 2 s. Alternatively, the fundamental period of g1(t) can also be evaluated from mT2 = 3T2 = 2 s.

(ii) In Example 1.3, we saw that sin(4π t) and 7 cos(10t) are both periodic

signals with fundamental periods 1/2 and π/5, respectively. Calculating the

ratio of the two fundamental periods yields

T2 =

1/2

π/5 =

2π ,

which is not a rational number. Hence, the linear combination g2(t) is not a

periodic signal.

In Example 1.5, the two signals g1(t) = 3 sin(4π t) + 7 cos(3π t) and g2(t) = 3 sin(4π t) + 7 cos(10t) are almost identical since the angular frequency of the cosine terms in g1(t) is 3π = 9.426, which is fairly close to 10, the fundamental frequency for the cosine term in g2(t). Even such a minor difference can cause

one signal to be periodic and the other to be non-periodic. Since g1(t) satisfies

Proposition 1.2, it is periodic. On the other hand, signal g2(t) is not periodic

as the ratio of the fundamental periods of the two components, 3 sin(4π t) and

7 sin(10t), is 5/2π , which is not a rational number.

We can also illustrate the above result graphically. The two signals g1(t) and

g2(t) are plotted in Fig. 1.8. It is observed that g1(t) is repeating itself every two

time units, as shown in Fig. 1.8(a), where an arrowed horizontal line represents

a duration of 2 s. From Fig 1.8(b), it appears that the waveform of g2(t) is also

repetitive. Observing carefully, however, reveals that consecutive durations of

2 s in g2(t) are slightly different. For example, the amplitude of g2(t) at the two

ends of the arrowed horizontal line (of duration 2 s) are clearly different. Signal

g2(t) is not therefore a periodic waveform.

We should also note that a periodic signal by definition must strictly start at

t = −∞ and continue on forever till t approaches +∞. In practice, however, most signals are of finite duration. Therefore, we relax the periodicity condition

and consider a signal to be periodic if it repeats itself during the time it is

observed.

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16 Part I Introduction to signals and systems

−4 −3 −2 −1 0 1 2 3 4 −10

−8

−6

−4

−2

g1(t) = 3sin(4pt) + 7cos(3pt)

(a)

−4 −3 −2 −1 0 1 2 3 4 −10

−8

−6

−4

−2

g2(t) = 3sin(4pt) + 7cos(10 t)

(b)

Fig. 1.8. Signals (a) g1(t ) and (b) g2(t ) considered in Example 1.5. Signal g1(t ) is periodic with a

fundamental period of 2 s, while g2(t ) is not periodic.

1.1.4 Energy and power signals

Before presenting the conditions for classifying a signal as an energy or a power

signal, we present the formulas for calculating the energy and power in a signal.

The instantaneous power at time t = t0 of a real-valued CT signal x(t) is given by x2(t0). Similarly, the instantaneous power of a real-valued DT signal

x[k] at time instant k = k0 is given by x2[k]. If the signal is complex-valued, the expressions for the instantaneous power are modified to |x(t0)|2 or |x[k0]|2, where the symbol | · | represents the absolute value of a complex number.

The energy present in a CT or DT signal within a given time interval is given

by the following:

CT signals E(T1,T2) = T2∫

|x(t)|2dt in interval t = (T1, T2) with T2 > T1;

(1.10a)

DT sequences E[N1,N2] = N2∑

k=N1

|x[k]|2 in interval k = [N1, N2] with N2 > N1.

(1.10b)

The total energy of a CT signal is its energy calculated over the interval t = [−∞, ∞]. Likewise, the total energy of a DT signal is its energy calculated over the range k = [−∞, ∞]. The expressions for the total energy are therefore given by the following:

CT signals Ex = ∞∫

−∞

|x(t)|2dt ; (1.11a)

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17 1 Introduction to signals

DT sequences Ex = ∞∑

k=−∞ |x[k]|2. (1.11b)

Since power is defined as energy per unit time, the average power of a CT

signal x(t) over the interval t = (−∞, ∞) and of a DT signal x[k] over the range k = [−∞, ∞] are expressed as follows:

CT signals Px = lim T →∞

T/2∫

−T/2

|x(t)|2dt. (1.12)

DT sequences Px = 1

2K + 1

K∑

k=−K |x[k]|2. (1.13)

Equations (1.12) and (1.13) are simplified considerably for periodic signals.

Since a periodic signal repeats itself, the average power is calculated from one

period of the signal as follows:

CT signals Px = 1

∫

〈T0〉

|x(t)|2dt = 1

t1+T0∫

|x(t)|2dt, (1.14)

DT sequences Px = 1

∑

k=〈K0〉 |x[k]|2 =

k1+K0−1∑

k=k1

|x[k]|2, (1.15)

where t1 is an arbitrary real number and k1 is an arbitrary integer. The symbols

T0 and K0 are, respectively, the fundamental periods of the CT signal x(t) and

the DT signal x[k]. In Eq. (1.14), the duration of integration is one complete

period over the range [t1, t1 + T0], where t1 can take any arbitrary value. In other words, the lower limit of integration can have any value provided that the

upper limit is one fundamental period apart from the lower limit. To illustrate

this mathematically, we introduce the notation ∫〈T0〉 to imply that the integration is performed over a complete period T0 and is independent of the lower limit.

Likewise, while computing the average power of a DT signal x[k], the upper

and lower limits of the summation in Eq. (1.15) can take any values as long as

the duration of summation equals one fundamental period K0.

A signal x(t), or x[k], is called an energy signal if the total energy Ex has

a non-zero finite value, i.e. 0 < Ex < ∞. On the other hand, a signal is called a power signal if it has non-zero finite power, i.e. 0 < Px < ∞. Note that a signal cannot be both an energy and a power signal simultaneously. The energy

signals have zero average power whereas the power signals have infinite total

energy. Some signals, however, can be classified as neither power signals nor as

energy signals. For example, the signal e2t u(t) is a growing exponential whose

average power cannot be calculated. Such signals are generally of little interest

to us.

Most periodic signals are typically power signals. For example, the average

power of the CT sinusoidal signal, or A sin(ω0t + θ ), is given by A2/2 (see

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18 Part I Introduction to signals and systems

0−2−4 t

2 4 6−8 8 t

5 x(t)

−6 8

(a)

0−2−4 t

2 4 6−8 8 t

5 z(t)

−6 8

(b)

Fig. 1.9. CT signals for Example

1.6.

Problem 1.6). Similarly, the average power of the complex exponential signal

A exp(jω0t) is given by A 2 (see Problem 1.8).

Example 1.6

Consider the CT signals shown in Figs. 1.9(a) and (b). Calculate the instanta-

neous power, average power, and energy present in the two signals. Classify

these signals as power or energy signals.

Solution

(a) The signal x(t) can be expressed as follows:

x(t) = {

5 −2 ≤ t ≤ 2 0 otherwise.

The instantaneous power, average power, and energy of the signal are calculated

as follows:

instantaneous power Px (t) = {

25 −2 ≤ t ≤ 2 0 otherwise;

energy Ex = ∞∫

−∞

|x(t)|2dt = 2∫

−2

25 dt = 100;

average power Px = lim T →∞

T Ex = 0.

Because x(t) has finite energy (0 < Ex = 100 < ∞) it is an energy signal. (b) The signal z(t) is a periodic signal with fundamental period 8 and over

one period is expressed as follows:

z(t) = {

5 −2 ≤ t ≤ 2 0 2 < |t | ≤ 4,

with z(t + 8) = z(t). The instantaneous power, average power, and energy of the signal are calculated as follows:

instantaneous power Pz(t) = {

25 −2 ≤ t ≤ 2 0 2 < |t | ≤ 4

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19 1 Introduction to signals

and Pz(t + 8) = Pz(t);

average power Pz = 1

4∫

−4

|z(t)|2 dt = 1

2∫

−2

25 dt = 100

8 = 12.5;

energy Ez = ∞∫

−∞

|z(t)|2 dt = ∞.

Because the signal has finite power (0 < Pz = 12.5 < ∞), z(t) is a power signal.

Example 1.7

Consider the following DT sequence:

f [k] = {

e−0.5k k ≥ 0 0 k < 0.

Determine if the signal is a power or an energy signal.

Solution

The total energy of the DT sequence is calculated as follows:

E f = ∞∑

k=−∞ | f [k]|2 =

∞∑

k=0 |e−0.5k |2 =

∞∑

k=0 (e−1)k =

1 − e−1 ≈ 1.582.

Because E f is finite, the DT sequence f [k] is an energy signal.

In computing E f , we make use of the geometric progression (GP) series to

calculate the summation. The formulas for the GP series are considered in

Appendix A.3.

Example 1.8

Determine if the DT sequence g[k] = 3 cos(πk/10) is a power or an energy signal.

Solution

The DT sequence g[k] = 3 cos(πk/10) is a periodic signal with a fundamental period of 20. All periodic signals are power signals. Hence, the DT sequence

g[k] is a power signal.

Using Eq. (1.15), the average power of g[k] is given by

Pg = 1

19∑

k=0 9 cos2

( πk

)

= 9

19∑

k=0

[

1 + cos (

2πk

)]

= 9

19∑

k=0 1

︸︷︷︸

term I

+ 9

19∑

k=0 cos

( 2πk

)

︸︷︷︸

term II

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20 Part I Introduction to signals and systems

Clearly, the summation represented by term I equals 9(20)/40 = 4.5. To com- pute the summation in term II, we express the cosine as follows:

term II = 9

19∑

k=0

2 [e jπk/5 + e−jπk/5] =

19∑

k=0 (e jπ/5)k +

19∑

k=0 (e−jπ/5)k .

Using the formulas for the GP series yields

19∑

k=0 (e jπ/5)k =

1 − (e jπ/5)20

1 − (e jπ/5) =

1 − e jπ4

1 − (e jπ/5) =

1 − 1 1 − (e jπ/5)

= 0

and

19∑

k=0 (e−jπ/5)k =

1 − (e−jπ/5)20

1 − (e jπ/5) =

1 − e−jπ4

1 − (e jπ/5) =

1 − 1 1 − (e jπ/5)

= 0.

Term II, therefore, equals zero. The average power of g[k] is therefore given

Pg = 4.5 + 0 = 4.5.

In general, a periodic DT sinusoidal signal of the form x[k] − A cos (ω0k + θ ) has an average power Px = A2/2.

1.1.5 Deterministic and random signals

If the value of a signal can be predicted for all time (t or k) in advance without

any error, it is referred to as a deterministic signal. Conversely, signals whose

values cannot be predicted with complete accuracy for all time are known as

random signals.

Deterministic signals can generally be expressed in a mathematical, or graph-

ical, form. Some examples of deterministic signals are as follows.

(1) CT sinusoidal signal: x1(t) = 5 sin(20π t + 6); (2) CT exponentially decaying sinusoidal signal: x2(t) = 2e−t sin(7t);

(3) CT finite duration complex exponential signal: x3(t) = {

e j4π t |t | < 5 0 elsewhere;

(4) DT real-valued exponential sequence: x4[k] = 4e−2k ; (5) DT exponentially decaying sinusoidal sequence: x5[k] = 3e−2k×

sin

( 16πk

)

Unlike deterministic signals, random signals cannot be modeled precisely.

Random signals are generally characterized by statistical measures such as

means, standard deviations, and mean squared values. In electrical engineering,

most meaningful information-bearing signals are random signals. In a digital

communication system, for example, data are generally transmitted using a

sequence of zeros and ones. The binary signal is corrupted with interference

from other channels and additive noise from the transmission media, resulting

in a received signal that is random in nature. Another example of a random

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21 1 Introduction to signals

signal in electrical engineering is the thermal noise generated by a resistor. The

intensity of the thermal noise depends on the movement of billions of electrons

and cannot be predicted accurately.

The study of random signals is beyond the scope of this book. We therefore

restrict our discussion to deterministic signals. However, most principles and

techniques that we develop are generalizable to random signals. The readers

are advised to consult more advanced books for analysis of random signals.

1.1.6 Odd and even signals

A CT signal xe(t) is said to be an even signal if

xe(t) = xe(−t). (1.16)

Conversely, a CT signal xo(t) is said to be an odd signal if

xo(t) = −xo(−t). (1.17)

A DT signal xe[k] is said to be an even signal if

xe[k] = xe[−k]. (1.18)

Conversely, a DT signal xo[k] is said to be an odd signal if

xo[k] = −xo[−k]. (1.19)

The even signal property, Eq. (1.16) for CT signals or Eq. (1.18) for DT sig-

nals, implies that an even signal is symmetric about the vertical axis (t = 0). Likewise, the odd signal property, Eq. (1.17) for CT signals or Eq. (1.19) for

DT signals, implies that an odd signal is antisymmetric about the vertical axis

(t = 0). The symmetry characteristics of even and odd signals are illustrated in Fig. 1.10. The waveform in Fig 1.10(a) is an even signal as it is symmetric

about the y-axis and the waveform in Fig. 1.10(b) is an odd signal as it is anti-

symmetric about the y-axis. The waveforms shown in Figs. 1.6(a) and (b) are

additional examples of even signals, while the waveforms shown in Figs. 1.6(c)

and (e) are examples of odd signals.

Most practical signals are neither odd nor even. For example, the signals

shown in Figs. 1.6(d) and (f), and 1.8(a) do not exhibit any symmetry about

the y-axis. Such signals are classified in the “neither odd nor even” category.

0−2−4 t

2 4 6−8 8 t

5 xe(t)

−6 8

(a)

0−2−4 t

4 6−8 8 t

−5

xo(t)

−6 8

(b)

Fig. 1.10. Example of (a) an

even signal and (b) an odd

signal.

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22 Part I Introduction to signals and systems

Neither odd nor even signals can be expressed as a sum of even and odd signals

as follows:

x(t) = xe(t) + xo(t),

where the even component xe(t) is given by

xe(t) = 1

2 [x(t) + x(−t)], (1.20)

while the odd component xo(t) is given by

xo(t) = 1

2 [x(t) − x(−t)]. (1.21)

Example 1.9

Express the CT signal

x(t) = {

t 0 ≤ t < 1 0 elsewhere

as a combination of an even signal and an odd signal.

Solution

In order to calculate xe(t) and xo(t), we need to calculate the function x(−t), which is expressed as follows:

x(−t) = {

−t 0 ≤ −t < 1 0 elsewhere

= {

−t −1 < t ≤ 0 0 elsewhere.

Using Eq. (1.20), the even component xe(t) of x(t) is given by

xe(t) = 1

2 [x(t) + x(−t)] =



   

   

2 t 0 ≤ t < 1

− 1

2 t −1 ≤ t < 0

0 elsewhere,

while the odd component xo(t) is evaluated from Eq. (1.21) as follows:

xo(t) = 1

2 [x(t) − x(−t)] =



   

   

2 t 0 ≤ t < 1

2 t −1 ≤ t < 0

0 elsewhere.

The waveforms for the CT signal x(t) and its even and odd components are

plotted in Fig. 1.11.

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23 1 Introduction to signals

0−1 1−2 t

0.5

−0.5

xe(t)

(b)

0−1 1−2 t

0.5

−0.5

xo(t)

(c)

0−1 1−2 t

0.5

x(t)

(a)

Fig. 1.11. (a) The CT signal x(t )

for Example 1.9. (b) Even

component of x(t ). (c) Odd

component of x(t ).

1.1.6.1 Combinations of even and odd CT signals

Consider ge(t) and he(t) as two CT even signals and go(t) and ho(t) as two

CT odd signals. The following properties may be used to classify different

combinations of these four signals into the even and odd categories.

(i) Multiplication of a CT even signal with a CT odd signal results in a CT

odd signal. The CT signal x(t) = ge(t) × go(t) is therefore an odd signal. (ii) Multiplication of a CT odd signal with another CT odd signal results in a

CT even signal. The CT signal h(t) = go(t) × ho(t) is therefore an even signal.

(iii) Multiplication of two CT even signals results in another CT even signal.

The CT signal z(t) = ge(t) × he(t) is therefore an even signal. (iv) Due to its antisymmetry property, a CT odd signal is always zero at t = 0.

Therefore, go(0) = ho(0) = 0. (v) Integration of a CT odd signal within the limits [−T , T ] results in a zero

value, i.e.

T∫

−T

go(t)dt = T∫

−T

ho(t)dt = 0. (1.22)

(vi) The integral of a CT even signal within the limits [−T , T ] can be simplified as follows:

T∫

−T

ge(t)dt = 2 T∫

ge(t)dt . (1.23)

It is straightforward to prove properties (i)–(vi). Below we prove property

(vi).

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24 Part I Introduction to signals and systems

Proof of property (vi)

By expanding the left-hand side of Eq. (1.23), we obtain

T∫

−T

ge(t)dt = 0∫

−T

ge(t)dt

︸︷︷︸

integral I

+ T∫

ge(t)dt

︸︷︷︸

integral II

Substituting α = −t in integral I yields

integral I = 0∫

ge(−α)(−dα) = T∫

ge(α)dα = T∫

ge(t)dt = integral II,

which proves Eq. (1.23).

1.1.6.2 Combinations of even and odd DT signals

Properties (i)–(vi) for CT signals can be extended to DT sequences. Consider

ge[k] and he[k] as even sequences and go[k] and ho[k] are as odd sequences.

For the four DT signals, the following properties hold true.

(i) Multiplication of an even sequence with an odd sequence results in an odd

sequence. The DT sequence x[k] = ge[k] × go[k], for example, is an odd sequence.

(ii) Multiplication of two odd sequences results in an even sequence. The DT

sequence h[k] = go[k] × ho[k], for example, is an even sequence. (iii) Multiplication of two even sequences results in an even sequence. The DT

sequence z[k] = ge[k] × he[k], for example, is an even sequence. (iv) Due to its antisymmetry property, a DT odd sequence is always zero at

k = 0. Therefore, go[0] = ho[0] = 0. (v) Adding the samples of a DT odd sequence go[k] within the range [−M ,

M] is 0, i.e.

M∑

k=−M go[k] = 0 =

M∑

k=−M ho[k]. (1.24)

(vi) Adding the samples of a DT even sequence ge[k] within the range [−M , M] simplifies to

M∑

k=−M ge[k] = ge[0] + 2

M∑

k=1 ge[k]. (1.25)

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25 1 Introduction to signals

1.2 Elementary signals

In this section, we define some elementary functions that will be used frequently

to represent more complicated signals. Representing signals in terms of the

elementary functions simplifies the analysis and design of linear systems.

1.2.1 Unit step function

The CT unit step function u(t) is defined as follows:

u(t) = {

1 t ≥ 0 0 t < 0.

(1.26)

The DT unit step function u[k] is defined as follows:

u[k] = {

1 k ≥ 0 0 k < 0.

(1.27)

The waveforms for the unit step functions u(t) and u[k] are shown, respectively,

in Figs. 1.12(a) and (b). It is observed from Fig. 1.12 that the CT unit step

function u(t) is piecewise continuous with a discontinuity at t = 0. In other words, the rate of change in u(t) is infinite at t = 0. However, the DT function u[k] has no such discontinuity.

1.2.2 Rectangular pulse function

The CT rectangular pulse rect(t/τ ) is defined as follows:

rect

( t

)

{

1 |t | ≤ τ/2

0 |t | > τ/2 (1.28)

and it is plotted in Fig. 1.12(c). The DT rectangular pulse rect(k/(2N + 1)) is defined as follows:

rect

( k

2N + 1

)

= {

1 |k| ≤ N 0 |k| > N (1.29)

and it is plotted in Fig. 1.12(d).

1.2.3 Signum function

The signum (or sign) function, denoted by sgn(t), is defined as follows:

sgn(t) =







1 t > 0

0 t = 0 −1 t < 0.

(1.30)

The CT sign function sgn(t) is plotted in Fig. 1.12(e). Note that the operation

sgn(·) can be used to output the sign of the input argument. The DT signum

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26 Part I Introduction to signals and systems

slope = 1

r(t) = tu(t)

(g)

(e)

1 x(t) = u(t)

(a)

1 x(t) = rect( )

(c)

t 2 t −

t t

0 tt

−1

x(t) = sgn(t)

w0 2p 0

x(t) = sin(w0t)

(i)

w0 1− w0

x(t) = sinc(w0t)1

(k)

x[k] = rect( )2N + 1 k

r[k] = ku[k]

(h)

1 x[k] = u[k]

(b)

0−N N k

(d)

0 k

1 x[k] = sgn(k)

( f )

−1

x[k] = sin(W0k)

W0 2p

( j)

x[k] = sinc(W0k)

( l)

Fig. 1.12. CT and DT elementary functions. (a) CT and (b) DT unit step functions. (c) CT and (d) DT rectangular pulses. (e) CT and

(f) DT signum functions. (g) CT and (h) DT ramp functions. (i) CT and (j) DT sinusoidal functions. (k) CT and (l) DT sinc functions.

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27 1 Introduction to signals

function, denoted by sgn(k), is defined as follows:

sgn[k] =







1 k > 0

0 k = 0 −1 k < 0

(1.31)

and it is plotted in Fig. 1.12(f).

1.2.4 Ramp function

The CT ramp function r (t) is defined as follows:

r (t) = tu(t) = {

t t ≥ 0 0 t < 0,

(1.32)

which is plotted in Fig. 1.12(g). Similarly, the DT ramp function r [k] is defined

as follows:

r [k] = ku[k] = {

k k ≥ 0 0 k < 0,

(1.33)

which is plotted in Fig. 1.12(h).

1.2.5 Sinusoidal function

The CT sinusoid of frequency f0 (or, equivalently, an angular frequency ω0 = 2π f0) is defined as follows:

x(t) = sin(ω0t + θ ) = sin(2π f0t + θ ), (1.34)

which is plotted in Fig. 1.12(i). The DT sinusoid is defined as follows:

x[k] = sin(Ω0k + θ ) = sin(2π f0k + θ ), (1.35)

where Ω0 is the DT angular frequency. The DT sinusoid is plotted in

Fig. 1.12(j). As discussed in Section 1.1.3, a CT sinusoidal signal x(t) = sin(ω0t + θ ) is always periodic, whereas its DT counterpart x[k] = sin(Ω0k + θ ) is not necessarily periodic. The DT sinusoidal signal is periodic only if the

fraction Ω0/2π is a rational number.

1.2.6 Sinc function

The CT sinc function is defined as follows:

sinc(ω0t) = sin(πω0t)

πω0t , (1.36)

which is plotted in Fig. 1.12(k). In some text books, the sinc function is alter-

natively defined as follows:

sinc(ω0t) = sin(ω0t)

ω0t .

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28 Part I Introduction to signals and systems

In this text, we will use the definition in Eq. (1.36) for the sinc function. The

DT sinc function is defined as follows:

sinc(Ω0k) = sin(πΩ0k)

πΩ0k , (1.37)

which is plotted in Fig. 1.12(l).

1.2.7 CT exponential function

A CT exponential function, with complex frequency s = σ + jω0, is repre- sented by

x(t) = est = e(σ+jω0)t = eσ t (cos ω0t + j sin ω0t). (1.38)

The CT exponential function is, therefore, a complex-valued function with the

following real and imaginary components:

real component Re{est } = eσ t cos ω0t ; imaginary component Im{est } = eσ t sin ω0t.

Depending upon the presence or absence of the real and imaginary components,

there are two special cases of the complex exponential function.

Case 1 Imaginary component is zero (ω0 = 0) Assuming that the imaginary component ω of the complex frequency s is zero,

the exponential function takes the following form:

x(t) = eσ t ,

which is referred to as a real-valued exponential function. Figure 1.13 shows the

real-valued exponential functions for different values of σ . When the value of σ

is negative (σ < 0) then the exponential function decays with increasing time t .

x(t) = est, s < 0

(a)

x(t) = est, s = 0

(b)

x(t) = est, s > 0

(c)

Fig. 1.13. Special cases of

real-valued CT exponential

function x(t ) = exp(σ t ). (a) Decaying exponential with

σ < 0. (b) Constant with

σ = 0. (c) Rising exponential with σ > 0.

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29 1 Introduction to signals

2p w0 Re{e jwt} = cos(w0t)

(a)

2p w0 Im{e jwt} = sin(w0t)

(b)

0 0

Fig. 1.14. CT complex-valued

exponential function

x(t ) = exp( jω0t ). (a) Real component; (b) imaginary

component.

The exponential function for σ < 0 is referred to as a decaying exponential

function and is shown in Fig. 1.13(a). For σ = 0, the exponential function has a constant value, as shown in Fig. 1.13(b). For positive values of σ (σ > 0),

the exponential function increases with time t and is referred to as a rising

exponential function. The rising exponential function is shown in Fig. 1.13(c).

Case 2 Real component is zero (σ = 0) When the real component σ of the complex frequency s is zero, the exponential

function is represented by

x(t) = e jω0t = cos ω0t + j sin ω0t.

In other words, the real and imaginary parts of the complex exponential are

pure sinusoids. Figure 1.14 shows the real and imaginary parts of the complex

exponential function.

Example 1.10

Plot the real and imaginary components of the exponential function x(t) = exp[( j4π − 0.5)t] for −4 ≤ t ≤ 4.

Solution

The CT exponential function is expressed as follows:

x(t) = e(j4π−0.5)t = e−0.5t × e j4π t .

The real and imaginary components of x(t) are expressed as follows:

real component Re{(t)} = e−0.5t cos(4π t); imaginary component Im{(t)} = e−0.5t sin(4π t).

To plot the real component, we multiply the waveform of a cosine function

with ω0 = 4π , as shown in Fig. 1.14(a), by a decaying exponential exp(−0.5t). The resulting plot is shown in Fig. 1.15(a). Similarly, the imaginary component

is plotted by multiplying the waveform of a sine function with ω0 = 4π , as shown in Fig. 1.14(b), by a decaying exponential exp(−0.5t). The resulting plot is shown in Fig. 1.15(b).

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30 Part I Introduction to signals and systems

−1−2−3−4

−6

−4

0 1 2 3 4 −8

−2

0 1 2 3 4

−8

−6

−4

−2

0 1 2 3 4−1−2−4 −3

(a) (b)

Fig. 1.15. Exponential function x (t ) = exp[( j4π − 0.5)t ]. (a) Real component; (b) imaginary component.

1.2.8 DT exponential function

The DT complex exponential function with radian frequency Ω0 is defined as

follows:

x[k] = e(σ+j�0)k = eσ t (cosΩ0k + j sinΩ0k.) (1.39)

As an example of the DT complex exponential function, we consider x[k] = exp(j0.2π − 0.05k), which is plotted in Fig. 1.16, where plot (a) shows the real component and plot (b) shows the imaginary part of the complex signal.

Case 1 Imaginary component is zero (Ω0 = 0). The signal takes the following form:

x[k] = eσk

when the imaginary componentΩ0 of the DT complex frequency is zero. Similar

to CT exponential functions, the DT exponential functions can be classified as

rising, decaying, and constant-valued exponentials depending upon the value

of σ .

Case 2 Real component is zero (σ = 0). The DT exponential function takes the following form:

x[k] = e jω0k = cos ω0k + j sin ω0k.

Recall that a complex-valued exponential is periodic iff Ω0/2π is a rational

number. An alternative representation of the DT complex exponential function

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31 1 Introduction to signals

−30 −20 −10 0 10 20 30 −6

−4

−2

(a)

−30 −20 −10 0 10 20 30 −6

−4

−2

(b)

Fig. 1.16. DT complex

exponential function x[k ] = exp( j0.2πk – 0.05k). (a) Real

component; (b) imaginary

component.

is obtained by expanding

x[k] = (

e(σ+j�0) )k = γ k, (1.40)

where γ = (σ + jΩ0) is a complex number. Equation (1.40) is more compact than Eq. (1.39).

1.2.9 Causal exponential function

In practical signal processing applications, input signals start at time t = 0. Signals that start at t = 0 are referred to as causal signals. The causal exponential function is given by

x(t) = est u(t) = {

est t ≥ 0 0 t < 0,

(1.41)

where we have used the unit step function to incorporate causality in the com-

plex exponential functions. Similarly, the causal implementation of the DT

exponential function is defined as follows:

x[k] = esku[k] = {

esk k ≥ 0 0 k < 0.

(1.42)

The same concept can be extended to derive causal implementations of sinu-

soidal and other non-causal signals.

Example 1.11

Plot the DT causal exponential function x[k] = e( j0.2π–0.05)ku[k].

Solution

The real and imaginary components of the non-causal signal e(j0.2π–0.05)k are

plotted in Fig. 1.16. To plot its causal implementation, we multiply e(j0.2π–0.05)k

by the unit step function u[k]. This implies that the causal implementation will

be zero for k < 0. The real and imaginary components of the resulting function

are plotted in Fig. 1.17.

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32 Part I Introduction to signals and systems

−30 −20 −10 0 10 20 30 −6

−4

−2

k −30 −20 −10 0 10 20 30

−6

−4

−2

(a) (b)

Fig. 1.17. Causal DT complex exponential function x[k ] = exp( j0.2πk – 0.05k)u[k ]. (a) Real component; (b) imaginary component.

1.2.10 CT unit impulse function

The unit impulse function δ(t), also known as the Dirac delta function† or

simply the delta function, is defined in terms of two properties as follows:

(1) amplitude δ(t) = 0, t = 0; (1.43a)

(2) area enclosed

∞∫

−∞

δ(t)dt = 1. (1.43b)

Direct visualization of a unit impulse function in the CT domain is difficult.

One way to visualize a CT impulse function is to let it evolve from a rectangular

function. Consider a tall narrow rectangle with width ε and height 1/ε, as shown

in Fig. 1.18(a), such that the area enclosed by the rectangular function equals

one. Next, we decrease the width and increase the height at the same rate such

that the resulting rectangular functions have areas = 1. As the width ε → 0, the rectangular function converges to the CT impulse function δ(t) with an

infinite amplitude at t = 0. However, the area enclosed by CT impulse function is finite and equals one. The impulse function is illustrated in our plots by an

arrow pointing vertically upwards; see Fig. 1.18(b). The height of the arrow

corresponds to the area enclosed by the CT impulse function.

Properties of impulse function (i) The impulse function is an even function, i.e. δ(t) = δ(−t).

(ii) Integrating a unit impulse function results in one, provided that the limits

of integration enclose the origin of the impulse. Mathematically,

T∫

−T

Aδ(t − t0)dt = {

A for −T < t0 < T 0 elsewhere.

(1.44)

† The unit impulse function was introduced by Paul Adrien Maurice Dirac (1902–1984), a British

electrical engineer turned theoretical physicist.

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33 1 Introduction to signals

d(t)

0.5e−0.5e

area = 1

1/e

(a) (b)

Fig. 1.18. Impulse function δ(t ).

(a) Generating the impulse

function δ(t ) from a rectangular

pulse. (b) Notation used to

represent an impulse function.

(iii) The scaled and time-shifted version δ(at + b) of the unit impulse function is given by

δ(at + b) = 1

a δ

(

t + b

)

. (1.45)

(iv) When an arbitrary function φ(t) is multiplied by a shifted impulse function,

the product is given by

φ(t)δ(t − t0) = φ(t0)δ(t − t0). (1.46)

In other words, multiplication of a CT function and an impulse function

produces an impulse function, which has an area equal to the value of the

CT function at the location of the impulse. Combining properties (ii) and

(iv), it is straightforward to show that

∞∫

−∞

φ(t)δ(t − t0)dt = φ(t0). (1.47)

(v) The unit impulse function can be obtained by taking the derivative of the

unit step function as follows:

δ(t) = du

dt . (1.48)

(vi) Conversely, the unit step function is obtained by integrating the unit

impulse function as follows:

u(t) = t∫

−∞

δ(τ )dτ . (1.49)

Example 1.12

Simplify the following expressions:

(i) 5 − jt 7 + t2

δ(t);

(ii)

∞∫

−∞

(t + 5)δ(t − 2)dt ;

(iii)

∞∫

−∞

e j0.5πω+2δ(ω − 5)dω.

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34 Part I Introduction to signals and systems

Solution

(i) Using Eq. (1.46) yields 5 − jt 7 + t2

δ(t) = [

5 − jt 7 + t2

]

t=0 δ(t) =

7 δ(t).

(ii) Using Eq. (1.46) yields

∞∫

−∞

(t + 5)δ(t − 2)dt = ∞∫

−∞

[(t + 5)]t=2δ(t − 2)dt = 7 ∞∫

−∞

δ(t − 2)dt .

Since the integral computes the area enclosed by the unit step function, which

is one, we obtain

∞∫

−∞

(t + 5)δ(t − 2)dt = 7 ∞∫

−∞

δ(t − 2)dt = 7.

(iii) Using Eq. (1.46) yields

∞∫

−∞

ej0.5πω+2δ(ω − 5)dω = ∞∫

−∞

[ej0.5πω+2]ω=5δ(ω − 5)dω

= ej2.5π +2 ∞∫

−∞

δ(ω − 5)dω.

Since exp(j2.5π + 2) = j exp(2) and the integral equals one, we obtain ∞∫

−∞

e j0.5πω+2δ(ω − 5)dω = je2.

1.2.11 DT unit impulse function

The DT impulse function, also referred to as the Kronecker delta function or

the DT unit sample function, is defined as follows:

δ[k] = u[k] − u[k − 1] = {

1 k = 0 0 k = 0. (1.50)

Unlike the CT unit impulse function, the DT impulse function has no ambiguity

in its definition; it is well defined for all values of k. The waveform for a DT

unit impulse function is shown in Fig. 1.19.

x[k] = δ[k]

Fig. 1.19. DT unit impulse

function.

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35 1 Introduction to signals

0 1

−1 k

x[k]

0 1

−1 k

x1[k] = d[k + 1]

(b)(a)

0 1

−1 k

x2[k] = 2d[k]

(c)

0 1

−1 k

x3[k] = 3d[k − 1]

(d)

Fig. 1.20. The DT functions in

Example 1.13: (a) x[k ], (b), x[k ],

function in (a) is the sum of the

shifted DT impulse functions

shown in (b), (c), and (d).

Example 1.13

Represent the DT sequence shown in Fig. 1.20(a) as a function of time-shifted

DT unit impulse functions.

Solution

The DT signal x[k] can be represented as the summation of three functions,

x1[k], x2[k], and x3[k], as follows:

x[k] = x1[k] + x2[k] + x3[k],

where x1[k], x2[k], and x3[k] are time-shifted impulse functions,

x1[k] = δ[k + 1], x2[k] = 2δ[k], and x3[k] = 4δ[k − 1],

and are plotted in Figs. 1.20(b), (c), and (d), respectively. The DT sequence

x[k] can therefore be represented as follows:

x[k] = δ[k + 1] + 2δ[k] + 4δ[k − 1].

1.3 Signal operations

An important concept in signal and system analysis is the transformation of a

signal. In this section, we consider three elementary transformations that are

performed on a signal in the time domain. The transformations that we consider

are time shifting, time scaling, and time inversion.

1.3.1 Time shifting

The time-shifting operation delays or advances forward the input signal in time.

Consider a CT signal φ(t) obtained by shifting another signal x(t) by T time

units. The time-shifted signal φ(t) is expressed as follows:

φ(t) = x(t + T ).

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36 Part I Introduction to signals and systems

0−2−4 t

2 4 6−8 8 t

2 x(t)

−6 8

(a)

0−2−4 t

2 4 6−8 8 t

2 x(t − 3)

−6 8

(b)

0−2−4 t

2 4 6−8 8 t

2 x(t + 3)

−6 8

(c)

Fig. 1.21. Time shifting of a CT

signal. (a) Original CT signal

x(t ). (b) Time-delayed version

x(t − 3) of the CT signal x(t ). and (c) Time-advanced version

x(t + 3) of the CT signal x(t ).

In other words, a signal time-shifted by T is obtained by substituting t in x(t) by

(t + T ). If T < 0, then the signal x(t) is delayed in the time domain. Graphically this is equivalent to shifting the origin of the signal x(t) towards the right-hand

side by duration T along the t-axis. On the other hand, if T > 0, then the

signal x(t) is advanced forward in time. The plot of the time-advanced signal

is obtained by shifting x(t) towards the left-hand side by duration T along the

t-axis.

Figure 1.21(a) shows a CT signal x(t) and the corresponding two time-shifted

signals x(t − 3) and x(t + 3). Since x(t − 3) is a delayed version of x(t), the waveform of x(t − 3) is identical to that of x(t), except for a shift of three time units towards the right-hand side. Similarly, x(t + 3) is a time-advanced version of x(t). The waveform of x(t + 3) is identical to that of x(t) except for a shift of three time units towards the left-hand side.

The theory of the CT time-shifting operation can also be extended to DT

sequences. When a DT signal x[k] is shifted by m time units, the delayed signal

φ[k] is expressed as follows:

φ[k] = x[k + m].

If m < 0, the signal is said to be delayed in time. To obtain the time-delayed

signal φ[k], the origin of the signal x[k] is shifted towards the right-hand side

along the k-axis by m time units. On the other hand, if m > 0, the signal

is advanced forward in time. The time-advanced signal φ[k] is obtained by

shifting x[k] towards the left-hand side along the k-axis by m time units.

Figure 1.22 shows a DT signal x[k] and the corresponding two time-shifted

signals x[k − 4] and x[k + 4]. The waveforms of x[k − 4] and x[k + 4] are identical to that of x[k]. The time-delayed signal x[k− 4] is obtained by shifting x[k] towards the right-hand side by four time units. The time-advanced signal

x[k + 4] is obtained by shifting x[k] towards the left-hand side by four time units.

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37 1 Introduction to signals

−2−4 2 4 6−8 8−6 80

x[k]

2 3

(a)

−2−4 2 4 6−8 8−6 80 k

x[k − 4]

1 2

(b)

−2−4 2 4 6−8 8−6 80 k

x[k + 4]

1 2

(c)

Fig. 1.22. Time shifting of a DT

signal. (a) Original DT signal

x[k ]. (b) Time-delayed version

x[k − 4] of the DT signal x[k ]. (c) Time-advanced version

x[k + 4] of the DT signal x[k ].

Example 1.14

Consider the signal x(t) = e−t u(t). Determine and plot the time-shifted versions x(t − 4) and x(t + 2).

Solution

The signal x(t) can be expressed as follows:

x(t) = e−t u(t) = {

e−t t ≥ 0 0 elsewhere,

(1.51)

and is shown in Fig. 1.23(a). To determine the expression for x(t − 4), we substitute t by (t − 4) in Eq. (1.51). The resulting expression is given by

x(t − 4) = {

e−(t−4) (t − 4) ≥ 0 0 elsewhere

= {

e−(t−4) t ≥ 4 0 elsewhere.

The function x(t − 4) is plotted in Fig. 1.23(b). Similarly, we can calculate the expression for x(t + 2) by substituting t by

(t + 2) in Eq. (1.51). The resulting expression is given by

x(t + 2) = {

e−(t+2) (t + 2) ≥ 0 0 elsewhere

= {

e−(t+2) t ≥ −2 0 elsewhere.

The function x(t + 2) is plotted in Fig. 1.23(c). From Fig. 1.23, we observe that the waveform for x(t − 4) can be obtained directly from x(t) by shifting the waveform of x(t) by four time units towards the right-hand side. Similarly, the

waveform for x(t + 2) can be obtained from x(t) by shifting the waveform of x(t) by two time units towards the left-hand side. This is the result expected

from our previous discussion.

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38 Part I Introduction to signals and systems

−4 −2 0 2 4 6 8 10

0.25

0.5

0.75

1.25

(a)

−4 −2 0 2 4 6 8 10

0.25

0.5

0.75

1.25

(b)

−4 −2 0 2 4 6 8 10

0.25

0.5

0.75

1.25

(c)

Fig. 1.23. Time shifting of the

CT signal in Example 1.14.

(a) Original CT signal x(t ).

(b) Time-delayed version

x(t − 4) of the CT signal x(t ). (c) Time-advanced version

x(t + 2) of the CT signal x(t ).

Example 1.15

Consider the signal x[k] defined as follows:

x[k] = {

0.2k 0 ≤ k ≤ 5 0 elsewhere.

(1.52)

Determine and plot signals p[k] = x[k − 2] and q[k] = x[k + 2].

Solution

The signal x[k] is plotted in Fig. 1.24(a). To calculate the expression for p[k],

substitute k = m− 2 in Eq. (1.52). The resulting equation is given by

x[m − 2] = {

0.2(m − 2) 0 ≤ (m − 2) ≤ 5 0 elsewhere.

By changing the independent variable from m to k and simplifying, we obtain

p[k] = x[k − 2] = {

0.2(k − 2) 2 ≤ k ≤ 7 0 elsewhere.

The non-zero values of p[k] for −2 ≤ k ≤ 7, are shown in Table 1.1, and the stem plot p[k] is plotted in Fig. 1.24(b). To calculate the expression for q[k],

substitute k = m + 2 in Eq. (1.52). The resulting equation is as follows:

x[m + 2] = {

0.2(m + 2) 0 ≤ (m + 2) ≤ 5 0 elsewhere.

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−4 −2 0 2 4 6 8 10

0.25

0.5

0.75

1.25

(a)

k −4 −2 0 2 4 6 8 10

0.25

0.5

0.75

1.25

(b)

−4 −2 0 2 4 6 8 10

0.25

0.5

0.75

1.25

(c)

Fig. 1.24. Time shifting of the

DT sequence in Example 1.15.

(a) Original DT sequence x[k ].

(b) Time-delayed version

x[k − 2] of x[k ]. (c) Time-advanced version

x[k + 2] of x[k ].

Table 1.1. Values of the signals p[k ] and q [k ]

k −2 −1 0 1 2 3 4 5 6 7 p[k] 0 0 0 0 0 0.2 0.4 0.6 0.8 1

q[k] 0 0.2 0.4 0.6 0.8 1 0 0 0 0

By changing the independent variable from m to k and simplifying, we

obtain

q[k] = x[k + 2] = {

0.2(k + 2) −2 ≤ k ≤ 3 0 elsewhere.

Values of q[k], for −2 ≤ k ≤ 7, are shown in Table 1.1, and the stem plot for q[k] is plotted in Fig. 1.24(c).

As in Example 1.14, we observe that the waveform for p[k] = x[k − 2] can be obtained directly by shifting the waveform of x[k] towards the right-hand

side by two time units. Similarly, the waveform for q[k] = x[k + 2] can be obtained directly by shifting the waveform of x[k] towards the left-hand side

by two time units.

1.3.2 Time scaling

The time-scaling operation compresses or expands the input signal in the time

domain. A CT signal x(t) scaled by a factor c in the time domain is denoted by

x(ct). If c > 1, the signal is compressed by a factor of c. On the other hand, if

0 < c < 1 the signal is expanded. We illustrate the concept of time scaling of

CT signals with the help of a few examples.

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40 Part I Introduction to signals and systems

−4 −2 0 2 4 6 8 10

0.25

0.5

0.75

1.25

(a)

t −4 −2 0 2 4 6 8 10

0.25

0.5

0.75

1.25

(b)

−4 −2 0 2 4 6 8 10

0.25

0.5

0.75

1.25

(c)

Fig. 1.25. Time scaling of the CT

signal in Example 1.16.

(a) Original CT signal x(t ).

(b) Time-compressed version

x(2t ) of x(t ). (c) Time-expanded

version x(0.5t ) of signal x(t ).

Example 1.16

Consider a CT signal x(t) defined as follows:

x(t) =



  

  

t + 1 −1 ≤ t ≤ 0 1 0 ≤ t ≤ 2

−t + 3 2 ≤ t ≤ 3 0 elsewhere,

(1.53)

as plotted in Fig. 1.25(a). Determine the expressions for the time-scaled signals

x(2t) and x(t/2). Sketch the two signals.

Solution

Substituting t by 2α in Eq. (1.53), we obtain

x(2α) =



  

  

2α + 1 −1 ≤ 2α ≤ 0 1 0 ≤ 2α ≤ 2

−2α + 3 2 ≤ 2α ≤ 3 0 elsewhere.

By changing the independent variable from α to t and simplifying, we obtain

x(2t) =



  

  

2t + 1 −0.5 ≤ t ≤ 0 1 0 ≤ t ≤ 1

−2t + 3 1 ≤ t ≤ 1.5 0 elsewhere,

which is plotted in Fig. 1.25(b). The waveform for x(2t) can also be obtained

directly by compressing the waveform for x(t) by a factor of 2. It is important

to note that compression is performed with respect to the y-axis such that the

values x(t) and x(2t) at t = 0 are the same for both waveforms.

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41 1 Introduction to signals

Substituting t by α/2 in Eq. (1.53), we obtain

x(α/2) =



  

  

α/2 + 1 −1 ≤ α/2 ≤ 0 1 0 ≤ α/2 ≤ 2

−α/2 + 3 2 ≤ α/2 ≤ 3 0 elsewhere.

By changing the independent variable from α to t and simplifying, we obtain

x(t/2) =



  

  

t/2 + 1 −2 ≤ t ≤ 0 1 0 ≤ t ≤ 4

−t/2 + 3 4 ≤ t ≤ 6 0 elsewhere,

which is plotted in Fig. 1.25(c). The waveform for x(0.5t) can also be obtained

directly by expanding the waveform for x(t) by a factor of 2. As for compression,

expansion is performed with respect to the y-axis such that the values x(t) and

x(t/2) at t = 0 are the same for both waveforms. A CT signal x(t) can be scaled to x(ct) for any value of c. For the DTFT,

however, the time-scaling factor c is limited to integer values. We discuss the

time scaling of the DT sequence in the following.

1.3.2.1 Decimation

If a sequence x[k] is compressed by a factor c, some data samples of x[k] are

lost. For example, if we decimate x[k] by 2, the decimated function y[k] = x[2k] retains only the alternate samples given by x[0], x[2], x[4], and so on.

Compression (referred to as decimation for DT sequences) is, therefore, an

irreversible process in the DT domain as the original sequence x[k] cannot be

recovered precisely from the decimated sequence y[k].

1.3.2.2 Interpolation

In the DT domain, expansion (also referred to as interpolation) is defined as

follows:

x (m)[k] =







[ k

]

if k is a multiple of integer m

0 otherwise.

(1.54)

The interpolated sequence x (m)[k] inserts (m − 1) zeros in between adjacent samples of the DT sequence x[k]. Interpolation of the DT sequence x[k] is a

reversible process as the original sequence x[k] can be recovered from x (m)[k].

Example 1.17

Consider the DT sequence x[k] plotted in Fig. 1.26(a). Calculate and sketch

p[k] = x[2k] and q[k] = x[k/2].

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42 Part I Introduction to signals and systems

Table 1.2. Values of the signal p[k ] for −3 ≤ k ≤ 3

k −3 −2 −1 0 1 2 3 p[k] x[−6] = 0 x[−4] = 0.2 x[−2] = 0.6 x[0] = 1 x[2] = 0.6 x[4] = 0.2 x[6] = 0

Table 1.3. Values of the signal q [k ] for −10 ≤ k ≤ 10

k −10 −9 −8 −7 −6 −5 −4 q[k] x[−5] = 0 0 x[−4] = 0.2 0 x[−3] = 0.4 0 x[−2] = 0.6

k −3 −2 −1 0 1 2 3 q[k] 0 x[−1] = 0.8 0 x[0] = 1 0 x[1] = 0.8 0

k 4 5 6 7 8 9 10

q[k] x[2] = 0.6 0 x[3] = 0.4 0 x[4] = 0.2 0 x[5] = 0

−10 −8 −6 −4 −2 0 2 4 6 8 10

0.2

0.4

0.6

0.8

1.2

(a)

k −10 −8 −6 −4 −2 0 2 4 6 8 10

0.2

0.4

0.6

0.8

1.2

(b)

−10 −8 −6 −4 −2 0 2 4 6 8 10

0.2

0.4

0.6

0.8

1.2

(c)

Fig. 1.26. Time scaling of the DT

signal in Example 1.17.

(a) Original DT sequence x[k ].

(b) Decimated version x[2k ], of

x[k ]. (c) Interpolated version

x[0.5k ] of signal x[k ].

Solution

Since x[k] is non-zero for −5 ≤ k ≤ 5, the non-zero values of the decimated sequence p[k] = x[2k] lie in the range −3 ≤ k ≤ 3. The non-zero values of p[k] are shown in Table 1.2. The waveform for p[k] is plotted in Fig. 1.26(b).

The waveform for the decimated sequence p[k] can be obtained by directly

compressing the waveform for x[k] by a factor of 2 about the y-axis. While

performing the compression, the value of x[k] at k = 0 is retained in p[k]. On both sides of the k = 0 sample, every second sample of x[k] is retained in p[k].

To determine q[k] = x[k/2], we first determine the range over which x[k/2] is non-zero. The non-zero values of q[k] = x[k/2] lie in the range −10 ≤ k ≤ 10 and are shown in Table 1.3. The waveform for q[k] is plotted in Fig. 1.26(c).

The waveform for the decimated sequence q[k] can be obtained by directly

expanding the waveform for x[k] by a factor of 2 about the y-axis. During

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43 1 Introduction to signals

Table 1.4. Values of the signal q2[k ] for −10 ≤ k ≤ k

k −10 −9 −8 −7 −6 −5 −4 q2[k] x[−5] = 0 0.1 x[−4] = 0.2 0.3 x[−3] = 0.4 0.5 x[−2] = 0.6

k −3 −2 −1 0 1 2 3 q2[k] 0.7 x[−1] = 0.8 0.9 x[0] = 1 0.9 x[1] = 0.8 0.7

k 4 5 6 7 8 9 10

q2[k] x[2] = 0.6 0.5 x[3] = 0.4 0.3 x[4] = 0.2 0.1 x[5] = 0

expansion, the value of x[k] at k = 0 is retained in q[k]. The even-numbered samples, where k is a multiple of 2, of q[k] equal x[k/2]. The odd-numbered

samples in q[k] are set to zero.

While determining the interpolated sequence x[mk], Eq. (1.54) inserts (m − 1) zeros in between adjacent samples of the DT sequence x[k], where x[k] is not

defined. Instead of inserting zeros, we can possibly interpolate the undefined

values from the neighboring samples where x[k] is defined. Using linear inter-

polation, an interpolated sequence can be obtained using the following equation:

x (m)[k]=



   

   

[ k

]

if k is a multiple of integer m

(1 − α)x [⌊

⌋]

+ α x [⌈

⌉]

otherwise,

(1.55)

where ⌊

k m

⌋

denotes the nearest integer less than or equal to (k/m), ⌈

k m

⌉

denotes

the nearest integer greater than or equal to (k/m), and α = (k mod m)/m. Note that mod is the modulo operator that calculates the remainder of the division

k/m. For m = 2, Eq. (1.55) simplifies to the following:

x (2)[k] =



  

  

[ k

]

if k is even

0.5

(

[ k − 1

]

+ x [

k + 1 2

])

if k is odd.

Although, Eq. (1.55) is useful in many applications, we will use Eq. (1.54) to

denote an interpolated sequence throughout the book unless explicitly stated

otherwise.

Example 1.18

Repeat Example 1.17 to obtain the interpolated sequence q2[k] = x[k/2] using the alternative definition given by Eq. (1.55).

Solution

The non-zero values of q2[k] = x[k/2] are shown in Table 1.4, where the val- ues of the odd-numbered samples of q2[k], highlighted with the gray back-

ground, are obtained by taking the average of the values of the two neighboring

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44 Part I Introduction to signals and systems

−10 −8 −6 −4 −2 0 2 4 6 8 10

0.2

0.4

0.6

0.8

1.2

Fig. 1.27. Interpolated version

x[0.5k ] of signal x[k ], where

unknown sample values are

interpolated.

samples at k and k − 1 obtained from x[k]. The waveform for q2[k] is plotted in Fig. 1.27.

1.3.3 Time inversion

The time inversion (also known as time reversal or reflection) operation reflects

the input signal about the vertical axis (t = 0). When a CT signal x(t) is time- reversed, the inverted signal is denoted by x(−t). Likewise, when a DT signal x[k] is time-reversed, the inverted signal is denoted by x[−k]. In the following we provide examples of time inversion in both CT and DT domains.

Example 1.19

Sketch the time-inverted version of the causal decaying exponential signal

x(t) = e−t u(t) = {

e−t t ≥ 0 0 elsewhere,

(1.56)

which is plotted in Fig. 1.28(a).

Solution

To derive the expression for the time-inverted signal x(−t), substitute t = −α in Eq. (1.56). The resulting expression is given by

x (−α) = eαu (−α) = {

eα −α ≥ 0 0 elsewhere.

Simplifying the above expression and expressing it in terms of the independent

variable t yields

x(−t) = {

et t ≤ 0 0 elsewhere.

−6 −4 −2 0 2 4 6

0.2

0.4

0.6

0.8

1.2

t −6 −4 −2 0 2 4 6

0.2

0.4

0.6

0.8

1.2

(a) (b)

Fig. 1.28. Time inversion of the

CT signal in Example 1.19.

(a) Original CT signal x(t ).

(b) Time-inverted version x(−t ).

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45 1 Introduction to signals

−8 −6 −4 −2 0 2 4 6 8

0.25

0.5

0.75

1.25

(a)

k −8 −6 −4 −2 0 2 4 6 8

0.25

0.5

0.75

1.25

(b)

Fig. 1.29. Time inversion of the

DT signal in Example 1.20.

(a) Original CT sequence x[k ].

(b) Time-inverted version x[−k ].

The time-reversed signal x(−t) is plotted in Fig. 1.28(b). Signal inversion can also be performed graphically by simply flipping the signal x(t) about the

y-axis.

Example 1.20

Sketch the time-inverted version of the following DT sequence:

x[k] =







1 −4 ≤ k ≤ −1 0.25k 0 ≤ k ≤ 4 0 elsewhere,

(1.57)

which is plotted in Fig. 1.29(a).

Solution

To derive the expression for the time-inverted signal x[−k], substitute k = −m in Eq. (1.57). The resulting expression is given by

x[−m] =







1 −4 ≤ −m ≤ −1 −0.25m 0 ≤ −m ≤ 4

0 elsewhere.

Simplifying the above expression and expressing it in terms of the independent

variable k yields

x[−m] =







1 1 ≤ m ≤ 4 −0.25m −4 ≤ −m ≤ 0

0 elsewhere.

The time-reversed signal x[−k] is plotted in Fig. 1.29(b).

1.3.4 Combined operations

In Sections 1.3.1–1.3.3, we presented three basic time-domain transformations.

In many signal processing applications, these operations are combined. An

arbitrary linear operation that combines the three transformations is expressed

as x(αt + β), where α is the time-scaling factor and β is the time-shifting factor. If α is negative, the signal is inverted along with the time-scaling and

time-shifting operations. By expressing the transformed signal as

x(αt + β) = x (

[

t + β

])

, (1.58)

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46 Part I Introduction to signals and systems

−4 −3 −2 −1 0 1 2 3 4

0.25

0.5

0.75

1.25

(a)

t −4 −3 −2 −1 0 1 2 3 4

0.25

0.5

0.75

1.25

(b)

−4 −3 −2 −1 0 1 2 3 4

0.25

0.5

0.75

1.25

(c)

t −4 −3 −2 −1 0 1 2 3 4

0.25

0.5

0.75

1.25

(d)

Fig. 1.30. Combined CT

operations defined in Example

1.21. (a) Original CT signal x(t ).

(b) Time-scaled version x(2t ).

x(−2t ) of (b). (d) Time-shifted version x(4 + 2t ) of (c).

we can plot the waveform graphically for x(αt + β) by following steps (i)–(iii) outlined below.

(i) Scale the signal x(t) by |α|. The resulting waveform represents x(|α|t). (ii) If α is negative, invert the scaled signal x(|α|t) with respect to the t = 0

axis. This step produces the waveform for x(αt).

(iii) Shift the waveform for x(αt) obtained in step (ii) by |β/α| time units. Shift towards the right-hand side if (β/α) is negative. Otherwise, shift towards

the left-hand side if (β/α) is positive. The waveform resulting from this

step represents x(αt + β), which is the required transformation.

Example 1.21

Determine x(4 − 2t), where the waveform for the CT signal x(t) is plotted in Fig. 1.30(a).

Solution

Express x(4 − 2t) = x(−2[t − 2]) and follow steps (i)–(iii) as outlined below.

(i) Compress x(t) by a factor of 2 to obtain x(2t). The resulting waveform is

shown in Fig. 1.30(b).

(ii) Time-reverse x(2t) to obtain x(−2t). The waveform for x(−2t) is shown in Fig. 1.30(c).

(iii) Shift x(−2t) towards the right-hand side by two time units to obtain x(−2[t − 2]) = x(4 − 2t). The waveform for x(4 − 2t) is plotted in Fig. 1.30(d).

Example 1.22

Sketch the waveform for x[−15 – 3k] for the DT sequence x[k] plotted in Fig. 1.31(a).

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47 1 Introduction to signals

−12 −10 −6−8 −4 −2 0 2 4 6 8 10 12

(a)

k −12 −10 −6−8 −4 −2 0 2 4 6 8 10 12

0.2 0.4

0.6

0.8 1

1.2 1.4

0.2 0.4

0.6

0.8 1

1.2 1.4

(b)

−12 −10 −6−8 −4 −2 0 2 4 6 8 10 12

(c)

k −12 −10 −6−8 −4 −2 0 2 4 6 8 10 12

0.2 0.4

0.6

0.8 1

1.2 1.4

0.2 0.4

0.6

0.8 1

1.2 1.4

(d)

Fig. 1.31. Combined DT

operations defined in Example

1.22. (a) Original DT signal x[k ].

(b) Time-scaled version x[3k ].

x[−3k ] of (b). (d) Time-shifted version x [−15 − 3k ] of (c).

Solution

Express x[−15 – 3k] = x[−3(k + 5)] and follow steps (i)–(iii) as outlined below.

(i) Compress x[k] by a factor of 3 to obtain x[3k]. The resulting waveform is

shown in Fig. 1.31(b).

(ii) Time-reverse x[3k] to obtain x[−3k]. The waveform for x[−3k] is shown in Fig. 1.31(c).

(iii) Shift x[−3k] towards the left-hand side by five time units to obtain x[−3(k + 5)] = x[−15 − 3k]. The waveform for x[−15 – 3k] is plotted in Fig. 1.31(d).

1.4 Signal implementation with MATLAB

MATLAB is used frequently to simulate signals and systems. In this section,

we present a few examples to illustrate the generation of different CT and DT

signals in MATLAB . We also show how the CT and DT signals are plotted in

MATLAB . A brief introduction to MATLAB is included in Appendix E.

Example 1.23

Generate and sketch in the same figure each of the following CT signals using

MATLAB . Do not use the “for” loops in your code. In each case, the horizontal

axis t used to sketch the CT should extend only for the range over which the

three signals are defined.

(a) x1(t) = 5 sin(2π t) cos(π t − 8) for −5 ≤ t ≤ 5; (b) x2(t) = 5e−0.2t sin (2π t) for −10 ≤ t ≤ 10; (c) x3(t) = e(j4π−0.5)t u(t) for −5 ≤ t ≤ 15.

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48 Part I Introduction to signals and systems

Solution

The MATLAB code for the generation of signals (a)–(c) is as follows:

>> %%%%%%%%%%%%

>> % Part(a) %

>> %%%%%%%%%%%%

>> clf % Clear any existing figure

>> t1 = [-5:0.001:5]; % Set the time from -5 to 5

% with a sampling

% rate of 0.001s

>> x1 = 5*sin(2*pi*t1).

*cos(pi*t1-8);

% compute function x1

>> % plot x1(t)

>> subplot(2,2,1); % select the 1st out of 4

% subplots

>> plot(t1,x1); % plot a CT signal

>> grid on; % turn on the grid

>> xlabel(‘time (t)’); % Label the x-axis as time

>> ylabel(‘5sin(2\pi t) cos(\pi t - 8)’);

% Label the y-axis

>> title(‘Part (a)’); % Insert the title

>> %%%%%%%%%%%%

>> % Part(b) %

>> %%%%%%%%%%%%

>> t2 = [-10:0.002:10]; % Set the time from -10 to

% 10 with a sampling

% rate of 0.002s

>> x2 = 5*exp(-0.2*t2).

*sin(2*pi*t2);

% compute function x2

>> % plot x2(t)

>> subplot(2,2,2); % select the 2nd out of 4

% subplots

>> plot(t2,x2); % plot a CT signal

>> grid on; % turn on the grid

>> xlabel(‘time (t)’); % Label the x-axis as time

>> ylabel(‘5exp(-0.2t)

sin(2\pi t)’); % Label the y-axis

>> title(‘Part (b)’); % Insert the title

>> %%%%%%%%%%%%

>> %Part(c)%

>> %%%%%%%%%%%%

>> t3 = [-5:0.001:15]; % Set the time from -5 to

% 15 with a sampling

% rate of 0.001s

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49 1 Introduction to signals

>> x3 = exp((j*4*pi-0.5)*t3).

*(t3>=0);

% compute function x3

>> % plot the real component

of x3(t)

>> subplot(2,2,3); % select the 3rd out of 4

% subplots

>> plot(t3,real(x3)); % plot a CT signal

>> grid on; % turn on the grid

>> xlabel(‘time (t)’) % Label the x-axis as time

>> ylabel(‘5exp[(j*4\pi-0.5) t]u(t)’);

% Label the y-axis

>> title(‘Part (c): Real

Component’);

% Insert the title

>> subplot(2,2,4); % select the 4th out of 4

% subplots

>> plot(t3,imag(x3)); % plot the imaginary

% component of a CT

% signal

>> grid on; % turn on the grid

>> xlabel(‘time (t)’); % Label the x-axis as time

>> ylabel(‘5exp[(j4\pi-0.5) t]u(t)’);

% Label the y-axis

>> title(‘Part (d): Imaginary

Component’);

% Insert the title

The resulting MATLAB plot is shown in Fig. 1.32.

Example 1.24

Repeat Example 1.23 for the following DT sequences:

(a) f1[k] = −0.92 sin(0.1πk − 3π/4) for −10 ≤ k ≤ 20; (b) f2[k] = 2.0(1.1)1.8k − 2.1(0.9)0.7k for −5 ≤ k ≤ 25; (c) f3[k] = (−0.93)kejπk/

√ 350 for 0 ≤ k ≤ 50.

Solution

The MATLAB code for the generation of signals (a)–(c) is as follows:

>> %%%%%%%%%%%%

>> % Part(a) %

>> %%%%%%%%%%%%

>> clf % clear any existing figure

>> k = [-10:20]; % set the time index from

% -10 to 20

>> f1 = -0.92 * sin(0.1*pi*k - 3*pi/4); % compute function f1

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50 Part I Introduction to signals and systems

−5 −4 −3 −2 −1 0 1 2 3 4 5 −5

−4

−3

−2

−1

time (t)

−10 −8 −6 −4 −2 0 2 4 6 8 10 −40

−30

−20

−10

time (t)

−5 0 5 10 15 −1

−0.8

−0.6

−0.4

−0.2

0.2

0.4

0.6

0.8

time (t)

real component

−5 0 5 10 15 −1

−0.8

−0.6

−0.4

−0.2

0.2

0.4

0.6

0.8

time (t)

imaginary component

5 si

n (2

p t)

co s(

p t

− 8

)

5 e

−0 .2

t s in

(2 p

e( j

4 p

- 0

.5 )t u

(t )

e( j4

p -

0 .5

)t u

(t )

(a) (b)

Fig. 1.32. MATLAB plot for

Example 1.23. (a) x1(t );

(b) x2(t ); (c) Re{x3(t )}; (d) Im{x3(t )}.

>> % plot function 1

>> subplot(2,2,1), stem(k, f1, ‘filled’), grid

>> xlabel(‘k’)

>> ylabel(‘-9.2sin(0.1\pi k-0.75\pi’) >> title(‘Part (a)’)

>> %%%%%%%%%%%%

>> % Part(b) %

>> %%%%%%%%%%%%

>> k = [-5:25];

>> f2 = 2 * 1.1.ˆ(-1.8*k) - 2.1 * 0.9.ˆ(0.7*k);

>> subplot(2,2,2), stem(k, f2, ‘filled’), grid

>> xlabel(‘k’)

>> ylabel(‘2(1.1)ˆ{-1.8k} - 2.1(0.9)ˆ0.7k’)

>> title(‘Part (b)’)

>> %%%%%%%%%%%%

>> % Part(c) %

>> %%%%%%%%%%%%

>> k = [0:50];

>> f3 = (-0.93).ˆk .* exp(j*pi*k/sqrt(350));

>> subplot(2,2,3), stem(k, real(f3), ‘filled’), grid

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51 1 Introduction to signals

(a) (b)

0 20 40 60 −1

−0.5

0.5

0 20 40 60 −1

−0.5

0.5

−10 0 10 20 −1

−0.5

0.5

−0 .9

2 s

in (0

.1 p

k −

0. 75

p )

−10 0 10 20 30 −1

2 .0

(1 .1

)− 1 .8

k −

2 .1

(0 .9

)0 .7

(− 0

.9 3

)k c

o s(

p k/

√ 3

5 0

)

(− 0

.9 3

)k c

o s(

p k/

√ 3 5 0 )

Fig. 1.33. MATLAB plot for

Example 1.24.

>> xlabel(‘k’)

>> ylabel(‘(-0.93)ˆk exp(j\pi k/(350)ˆ{0.5}’) >> title(‘Part (c) - real part’)

>> %

>> subplot(2,2,4), stem(k, imag(f3), ‘filled’), grid

>> xlabel(‘k’)

>> ylabel(‘(-0.93)ˆk exp(j\pi k/(350)ˆ{0.5}’) >> title(‘Part (d) - imaginary part’)

>> print -dtiff plot.tiff

The resulting MATLAB plots are shown in Fig. 1.33.

1.5 Summary

In this chapter, we have introduced many useful concepts related to signals

and systems, including the mathematical and graphical interpretations of signal

representation. In Section 1.1, we classified signals in six different categories:

CT versus DT signals; analog versus digital signals; periodic versus aperiodic

signals; energy versus power signals; deterministic versus probabilistic signals;

and even versus odd signals. We classified the signals based on the following

definitions.

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52 Part I Introduction to signals and systems

(1) A time-varying signal is classified as a continuous time (CT) signal if it is

defined for all values of time t . A time-varying discrete time (DT) signal is

defined for certain discrete values of time, t = kTs, where Ts is the sampling interval. In our notation, a CT signal is represented by x(t) and a DT signal

is denoted by x[k].

(2) An analog signal is a CT signal whose amplitude can take any value. A

digital signal is a DT signal that can only have a discrete set of values.

The process of converting a DT signal into a digital signal is referred to as

quantization.

(3) A periodic signal repeats itself after a known fundamental period, i.e.

x(t) = x(t + T0) for CT signals and x[k] = x[k + K0] for DT signals. Note that CT complex exponentials and sinusoidal signals are always periodic,

whereas DT complex exponentials and sinusoidal signals are periodic only

if the ratio of their DT fundamental frequency Ω0, to 2π is a rational

number.

(4) A signal is classified as an energy signal if its total energy has a non-zero

finite value. A signal is classified as a power signal if it has non-zero finite

power. An energy signal has zero average power whereas a power signal

has an infinite energy. Periodic signals are generally power signals.

(5) A deterministic signal is known precisely and can be predicted in advance

without any error. A random signal cannot be predicted with 100%

accuracy.

(6) A signal that is symmetric about the vertical axis (t = 0) is referred to as an even signal. An odd signal is antisymmetric about the vertical axis

(t = 0). Mathematically, this implies x(t) = x(−t) for the CT even signals and x(t) = −x(−t) for the CT odd signals. Likewise for the DT signals.

In Section 1.2, we introduced a set of 1D elementary signals, including rectan-

gular, sinusoidal, exponential, unit step, and impulse functions, defined both in

the DT and CT domains. We illustrated through examples how the elementary

signals can be used as building blocks for implementing more complicated sig-

nals. In Section 1.3, we presented three fundamental signal operations, namely

time shifting, scaling, and inversion that operate on the independent variable.

The time-shifting operation x(t − T ) shifts signal x(t) with respect to time. If the value of T in x(t − T ) is positive, the signal is delayed by T time units. For negative values of T , the signal is time-advanced by T time units.

The time-scaling, x(ct), operation compresses (c > 0) or expands (c < 0) sig-

nal x(t). The time-inversion operation is a special case of the time-scaling

operation with c = −1. The waveform for the time-scaled signal x(−t) is the reflection of the waveform of the original signal x(t) about the vertical axis

(t = 0). The three transformations play an important role in the analysis of lin- ear time-invariant (LTI) systems, which will be covered in Chapter 2. Finally,

in Section 1.4, we used MATLAB to generate and analyze several CT and DT

signals.

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53 1 Introduction to signals

Problems

1.1 For each of the following representations: (i) z[m, n, k],

(ii) I (x, y, z, t),

establish if the signal is a CT or a DT signal. Specify the independent

and dependent variables. Think of an information signal from a physical

process that follows the mathematical representation given in (i). Repeat

for the representation in (ii).

1.2 Sketch each of the following CT signals as a function of the independent variable t over the specified range:

(i) x1(t) = cos(3π t/4 + π/8) for −1 ≤ t ≤ 2; (ii) x2(t) = sin(−3π t/8 + π/2) for −1 ≤ t ≤ 2;

(iii) x3(t) = 5t + 3 exp(−t) for −2 ≤ t ≤ 2; (iv) x4(t) = (sin(3π t/4 + π/8))2 for −1 ≤ t ≤ 2; (v) x5(t) = cos(3π t/4) + sin(π t/2) for −2 ≤ t ≤ 3;

(vi) x6(t) = t exp(−2t) for −2 ≤ t ≤ 3.

1.3 Sketch the following DT signals as a function of the independent variable k over the specified range:

(i) x1[k] = cos(3πk/4 + π/8) for −5 ≤ k ≤ 5; (ii) x2[k] = sin(−3πk/8 + π/2) for −10 ≤ k ≤ 10;

(iii) x3[k] = 5k + 3−k for −5 ≤ k ≤ 5; (iv) x4[k] = |sin(3πk/4 + π/8)| for −6 ≤ k ≤ 10; (v) x5[k] = cos(3πk/4) + sin(πk/2) for −10 ≤ k ≤ 10;

(vi) x6[k] = k4−|k| for −10 ≤ k ≤ 10.

1.4 Prove Proposition 1.2.

1.5 Determine if the following CT signals are periodic. If yes, calculate the fundamental period T0 for the CT signals:

(i) x1(t) = sin(−5π t/8 + π/2); (ii) x2(t) = |sin(−5π t/8 + π/2)|;

(iii) x3(t) = sin(6π t/7) + 2 cos(3t/5); (iv) x4(t) = exp(j(5t + π/4)); (v) x5(t) = exp(j3π t/8) + exp(π t/86);

(vi) x6(t) = 2 cos(4π t/5)∗ sin2(16t/3); (vii) x7(t) = 1 + sin 20t + cos(30t + π/3).

1.6 Determine if the following DT signals are periodic. If yes, calculate the fundamental period N0 for the DT signals:

(i) x1[k] = 5 × (−1)k ; (ii) x2[k] = exp(j(7πk/4)) + exp(j(3k/4));

(iii) x3[k] = exp(j(7πk/4)) + exp(j(3πk/4)); (iv) x4[k] = sin(3πk/8) + cos(63πk/64);

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54 Part I Introduction to signals and systems

(v) x5[k] = exp(j(7πk/4)) + cos(4πk/7 + π ); (vi) x6[k] = sin(3πk/8) cos(63πk/64).

1.7 Determine if the following CT signals are energy or power signals or neither. Calculate the energy and power of the signals in each case:

(i) x1(t) = cos(π t) sin(3π t); (v) x5(t) =

{

cos(3π t) −3 ≤ t ≤ 3; 0 elsewhere;(ii) x2(t) = exp(−2t);

(iii) x3(t) = exp(−j2t); (iv) x4(t) = exp(−2t)u(t); (vi) x6(t) =







t 0 ≤ t ≤ 2 4 − t 2 ≤ t ≤ 4 0 elsewhere.

1.8 Repeat Problem 1.7 for the following DT sequences:

(i) x1[k] = cos (

πk

)

sin

( 3πk

)

;

(ii) x2[k] =







cos

( 3πk

)

−10 ≤ k ≤ 0

0 elsewhere;

(iii) x3[k] = (−1)k ; (iv) x4[k] = exp(j(πk/2 + π/8));

(v) x5[k] =







2k 0 ≤ k ≤ 10 1 11 ≤ k ≤ 15 0 elsewhere.

1.9 Show that the average power of the CT periodic signal x(t) = A sin(ω0t + θ ), with real-valued coefficient A, is given by A2/2.

1.10 Show that the average power of the CT signal y(t) = A1 sin(ω1t + φ1) + A2 sin(ω2t + φ2), with real-valued coefficients A1 and A2, is given by

Py =



  

  

A21 2

+ A22 2

ω1 = ω2 A21 2

+ A22 2

+ A1 A2 cos(φ1 − φ2) ω1 = ω2.

1.11 Show that the average power of the CT periodic signal x(t) = D exp[j(ω0t + θ )] is given by |D|2.

1.12 Show that the average power of the following CT signal:

x(t) = N∑

n=1 Dne

jωn t , ωp = ωr if p = r,

for 1 ≤ p, r ≤ N , is given by

Px = N∑

n=1 |Dn|2.

1.13 Calculate the average power of the periodic function shown in Fig. P1.13 and defined as

x(t)|t=(0,1] = {

1 2−2m−1 < t ≤ 2−2m 0 2−2m−2 < t ≤ 2−2m−1

m ∈ Z+ and x(t) = x(t + 1).

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55 1 Introduction to signals

−1 −0.5 0 0.5 1 1.5 2 t

0.25

0.5

0.75

1.25Fig. P1.13. The CT function x(t )

in Problem 1.13.

1.14 Determine if the following CT signals are even, odd, or neither even nor odd. In the latter case, evaluate and sketch the even and odd components

of the CT signals:

(i) x1(t) = 2 sin(2π t)[2 + cos(4π t)]; (ii) x2(t) = t2 + cos(3t);

(iii) x3(t) = exp(−3t) sin(3π t); (iv) x4(t) = t sin(5t); (v) x5(t) = tu(t);

(vi) x6(t) =



  

  

3t 0 ≤ t < 2 6 2 ≤ t < 4 3(−t + 6) 4 ≤ t ≤ 6 0 elsewhere.

1.15 Determine if the following DT signals are even, odd, or neither even nor odd. In the latter case, evaluate and sketch the even and odd components

of the DT signals:

(i) x1[k] = sin(4k) + cos(2π/k3); (ii) x2[k] = sin(πk/3000) + cos(2πk/3);

(iii) x3[k] = exp(j(7πk/4)) + cos(4πk/7 + π ); (iv) x4[k] = sin(3πk/8) cos(63πk/64);

(v) x5[k] = {

(−1)k k ≥ 0 0 k < 0.

1.16 Consider the following signal:

x(t) = 3 sin (

2π (t − T ) 5

)

Determine the values of T for which the resulting signal is (a) an even

function, and (b) an odd function of the independent variable t.

1.17 By inspecting plots (a), (b), (c), and (d) in Fig. P1.17, classify the CT waveforms as even versus odd, periodic versus aperiodic, and energy

versus power signals. If the waveform is neither even nor odd, then deter-

mine the even and odd components of the signal. For periodic signals,

determine the fundamental period. Also, compute the energy and power

present in each case.

1.18 Sketch the following CT signals: (i) x1(t) = u(t) + 2u(t − 3) − 2u(t − 6) − u(t − 9);

(ii) x2(t) = u(sin(π t));

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56 Part I Introduction to signals and systems

0−1−2 t

1 2 3−4 t

x1(t)

−3 4

(a)

0−1−2 t

1 2 3−4 t

2.5

−2.5

x2(t)

−3 4

(b)

0−1−2 t

1 2 3 4−4 t

x3(t) = e−1.5t u(t)

−3

(c)

0−3−6 t

3 6 9 12−12 t

2.5

−2.5

x4(t)

−9

(d)

Fig. P1.17. Waveforms for

Problem 1.17. (iii) x3(t) = rect(t/6) + rect(t/4) + rect(t/2); (iv) x4(t) = r (t) − r (t − 2) − 2u(t − 4); (v) x5(t) = (exp(−t) − exp(−3t))u(t);

(vi) x6(t) = 3 sgn(t) · rect(t/4) + 2δ(t + 1) − 3δ(t − 3).

1.19 (a) Sketch the following functions with respect to the time variable (if a function is complex, sketch the real and imaginary components sep-

arately). (b) Locate the frequencies of the functions in the 2D complex

plane.

(i) x1(t) = e j2π t+3; (ii) x2(t) = e j2π t+3t ;

(iii) x3(t) = e−j2π t+j3t ; (iv) x4(t) = cos(2π t + 3); (v) x5(t) = cos(2π t + 3) + sin(3π t + 2);

(vi) x6(t) = 2 + 4 cos(2π t + 3) − 7 sin(5π t + 2).

1.20 Sketch the following DT signals: (i) x1[k] = u[k] + u[k − 3] − u[k − 5] − u[k − 7];

(ii) x2[k] = ∞∑

m=0 δ[k − m];

(iii) x3[k] = (3k − 2k)u[k]; (iv) x4[k] = u[cos(πk/8)]; (v) x5[k] = ku[k];

(vi) x6[k] = |k| (u[k + 4] − u[k − 4]).

1.21 Evaluate the following expressions:

(i) 5 + 2t + t2

7 + t2 + t4 δ(t − 1);

(ii) sin(t)

2t δ(t);

(iii) ω3 − 1 ω2 + 2

δ(ω − 5).

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57 1 Introduction to signals

1.22 Evaluate the following integrals:

(i)

∞∫

−∞

(t − 1)δ(t − 5)dt ;

(ii)

6∫

−∞

(t − 1)δ(t − 5)dt ;

(iii)

∞∫

(t − 1)δ(t − 5)dt ;

(iv)

∞∫

−∞

(2t/3 − 5)δ(3t/4 − 5/6)dt ;

(v)

∞∫

−∞

exp(t − 1) sin(π (t + 5)/4)δ(1 − t)dt ;

(vi)

∞∫

−∞

[sin(3π t/4) + exp(−2t + 1)]δ(−t − 1)dt ;

(vii)

∞∫

−∞

[u(t − 6) − u(t − 10)] sin(3π t/4)δ(t − 5)dt ;

(viii)

21∫

−21

( ∞∑

m=−∞ tδ(t − 5m)

)

dt .

1.23 In Section 1.2.8, the Dirac delta function was obtained as a limiting case of

the rectangular function, i.e. δ(t) = lim ε→0

ε rect

( t

)

. Show that the Dirac

delta function can also be obtained from each of the following functions

(i.e. that Eq. (1.43) is satisfied by each of the following functions):

(i) lim ε→0

π (t2 + ε2) ;

(iii) lim ε→0

π t sin εt ;

(ii) lim ε→0

2ε

4π2t2 + ε2 ;

(iv) lim ε→0

ε √

2π exp

(

− t2

2ε2

)

1.24 Consider the following signal:

x(t) =



  

  

t + 2 −2 ≤ t ≤ −1 1 −1 ≤ t ≤ 1

−t + 2 1 < t ≤ 2 0 elsewhere.

(a) Sketch the functions: (i) x(t − 3); (ii) x(−2t − 3); (iii) x(−2t − 3); (iv) x(−0.75t − 3).

(b) Determine the analytical expressions for each of the four functions.

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58 Part I Introduction to signals and systems

0−1−2 t

1 2 3−4 t

f (t)

−3

4 5

Fig. P1.25. Waveform for

Problem 1.25.

0−1−2 t

2−4−5−6 t

f (t)

−3

Fig. P1.26. Waveform for

Problem 1.26.

0−1−2 t

1 2 3−4 t

f (t)

−3 4

Fig. P1.27. Waveform for

Problem 1.27.

1.25 Consider the function f (t) shown in Fig. P1.25. (i) Sketch the function g(t) = f (−3t + 9).

(ii) Calculate the energy and power of the signal f (t). Is it a power signal

or an energy signal?

(iii) Repeat (ii) for g(t).

1.26 Consider the function f (t) shown in Fig. P1.26. (i) Sketch the function g(t) = f (−2t + 6).

(ii) Represent the function f (t) as a summation of an even and an odd

signal. Sketch the even and odd parts.

1.27 Consider the function f (t) shown in Fig. P1.27. (i) Sketch the function g(t) = t f (t + 2) − t f (t − 2).

(ii) Sketch the function g(2t).

1.28 Consider the two DT signals

x1[k] = |k|(u[k + 4] − u[k − 4])

and

x2[k] = k(u[k + 5] − u[k − 5]).

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59 1 Introduction to signals

1.2

0.8

0.4

0 0.5 1 1.5 2 time (s)

2.5 3 3.5

−0.4

−0.8

E C

G s

ig n

Fig. P1.29. ECG pattern for

Problem 1.29.

Sketch the following signals expressed as a function of x1[k] and x2[k]:

(i) x1[k];

(ii) x2[k];

(iii) x1[3 − k]; (iv) x1[6 − 2k]; (v) x1[2k];

(vi) x2[3k];

(vii) x1[k/2];

(viii) x1[2k] + x2[3k]; (ix) x1[3 − k]x2[6 − 2k]; (x) x1[2k]x2[−k].

1.29 In most parts of the human body, a small electrical current is often pro- duced by movement of different ions. For example, in cardiac cells the

electric current is produced by the movement of sodium (Na+) and potas-

sium (K+) ions (during different phases of the heart beat, these ions enter

or leave cells). The electric potential created by these ions is known as an

ECG signal, and is used by doctors to analyze heart conditions. A typical

ECG pattern is shown in Fig. P1.29.

Assume a hypothetical case in which the ECG signal corresponding to

a normal human is available from birth to death (assume a longevity of

80 years). Classify such a signal with respect to the six criteria mentioned

in Section 1.1. Justify your answer for each criterion.

1.30 It was explained in Section 1.2 that a complicated function could be represented as a sum of elementary functions. Consider the function f (t)

in Fig. P1.26. Represent f (t) in terms of the unit step function u(t) and

the ramp function r (t).

1.31 (MATLAB exercise) Write a set of MATLAB functions that compute and plot the following CT signals. In each case, use a sampling interval of

0.001 s.

hamadalsultan

Highlight

hamadalsultan

Highlight

hamadalsultan

Highlight

hamadalsultan

Highlight

hamadalsultan

Highlight

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60 Part I Introduction to signals and systems

(i) x(t) = exp(−2t) sin(10π t) for |t | ≤ 1. (ii) A periodic signal x(t) with fundamental period T = 5. The value

over one period is given by

x(t) = 5t 0 ≤ t < 5.

Use the sawtooth function available in MATLAB to plot five

periods of x(t) over the range −10 ≤ t < 15. (iii) The unit step function u(t) over [−10, 10] using the sign function

available in MATLAB .

(iv) The rectangular pulse function rect(t)

rect

( t

)

= {

1 −5 < t < 5 0 elsewhere

using the unit step function implemented in (iii).

(v) A periodic signal x(t) with fundamental period T = 6. The value over one period is given by

x(t) = {

3 |t | ≤ 1 0 1 < |t | ≤ 3.

Use the square function available in MATLAB .

1.32 (MATLAB exercise) Write a MATLAB function mydecimate with the following format:

function [y] = mydecimate(x, M)

% MYDECIMATE: computes y[k] = x[kM]

% where

% x is a column vector containing the DT input

% signal

% M is the scaling factor greater than 1

% y is a column vector containing the DT output time

% decimated by M

In other words, mydecimate accepts an input signal x[k] and produces

the signal y[k] = x[kM].

1.33 (MATLAB exercise) Repeat Problem 1.30 for the transformation y[k] = x[k/N ]. In other words, write a MATLAB function myinterpolate

with the following format:

function [y] = myinterpolate(x, N)

% MYINTERPOLATE: computes y[k] = x[k/N]

% where

% x is a column vector containing the DT input

% signal

% N is the scaling factor greater than 1

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61 1 Introduction to signals

% y is a column vector containing the DT output

% signal time expanded by N

Use linear interpolation based on the neighboring samples to predict any

required unknown values in x[k].

1.34 (MATLAB exercise) Construct a DT signal given by

x[k] = (1 − e−0.003k) cos(πk/20) for 0 ≤ k ≤ 120.

(i) Sketch the signal using the stem function.

(ii) Using the mydecimate (Problem P1.30) and myinterpolate

(Problem P1.31) functions, transform the signal x[k] based on the

operation y[k] = x[k/5] followed by the operation z[k] = y[5k]. What is the relationship between x[k] and z[k]?

(iii) Repeat (ii) with the order of interpolation and decimation reversed.

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C H A P T E R

2 Introduction to systems

In Chapter 1, we introduced the mathematical and graphical notations for repre-

senting signals, which enabled us to illustrate the effect of linear time operations

on transforming the signals. A second important component of signal process-

ing is a system that usually abstracts a physical process. Broadly speaking, a

system is characterized by its ability to accept a set of input signals xi , for i ∈

{1, 2, . . . , m}, and to produce a set of output signals y j , for j ∈ {1, 2, . . . , n}, in

response to the input signals. In other words, a system establishes a relationship

between a set of inputs and the corresponding set of outputs.

Most physical processes are modeled by multiple-input and multiple-output

(MIMO) systems of the form illustrated in Fig. 2.1(a), where the xi (t)’s repre-

sent the CT inputs while the y j (t)’s represent the CT outputs. Such systems,

which operate on CT input signals transforming them to CT output signals,

are referred to as CT systems. Using the principle of superimposition, a linear

MIMO CT system is often approximated by a combination of several single-

input CT systems. The block diagram representing a single-input, single-output

CT system is illustrated in Fig. 2.1(b). Throughout this book, we will restrict

our discussion to the analysis and design of single-input, single-output sys-

tems, knowing that the principles derived for such systems can be generalized

to MIMO systems.

In comparison to CT systems, DT systems transform DT input signals, often

referred to as sequences, into DT output signals. Two DT systems are shown in

Fig. 2.2. In Fig. 2.2(a), the schematic of a MIMO DT system is illustrated with

a set of m input sequences, denoted by xi [k]’s, and a set of n output sequences,

denoted by y j [k]’s. A single-input, single-output DT system is illustrated in

Fig. 2.2(b). As for the CT systems, we will focus on single-input, single-output

DT systems in this book.

The relationship between the input signal and its output response of a single-

input, single-output system, may it be DT or CT, will be shown by the following

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63 2 Introduction to systems

x1 (t)

x2 (t)

xm (t)

y1 (t)

y2 (t)

yn (t)

input

signals

output

signals y(t)x(t)

system

System input

signal

output

signal

(a) (b)

Fig. 2.1. General schematics of CT systems. (a) Multiple-input, multiple-output (MIMO) CT system with m

inputs and n outputs. (b) Single-input, single-output CT system.

x1 [k] x2 [k]

xm [k]

      

y1 [k]

y2 [k]

yn [k]

      

input

signals

output

signals y[k]x[k]

system

(a) (b)

Fig. 2.2. General schematics of

DT systems. (a) Multiple-input,

multiple-output (MIMO) DT

system with m inputs and n

outputs. (b) Single-input,

single-output DT system.

notation:

CT system x(t) → y(t); (2.1)

DT system x[k] → y[k]. (2.2)

The arrow in Eq. (2.1) implies that a CT signal x(t), applied at the input of a

CT system, produces a CT output y(t). Likewise, the arrow in Eq. (2.2) implies

that a DT input signal x[k] produces a DT output signal y[k]. This chapter

focuses on the classification of CT and DT systems. Before proceeding with

the classification of systems, we consider several applications of signals and

systems in electrical networks, electronic devices, communication systems, and

mechanical systems.

The organization of Chapter 2 is as follows. In Section 2.1, we provide

several examples of CT and DT systems. We show that most CT systems can be

modeled by linear, constant-coefficient differential equations, while DT systems

can be modeled by linear, constant-coefficient difference equations. Section 2.2

introduces several classifications for CT and DT systems based on the properties

of these systems. A particularly important class of systems, referred to as linear

time-invariant (LTI) systems, consists of those that satisfy both the linearity and

time-invariance properties. Most practical structures are complex and consist

of several LTI systems. Section 2.3 presents the series, parallel, and feedback

configurations used to synthesize larger systems. Section 2.4 concludes the

chapter with a summary of the important concepts.

2.1 Examples of systems

In this section, we present examples of physical systems and derive relationships

between the input and output signals associated with these systems. For linear

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64 Part I Introduction to signals and systems

CT systems, a linear, constant-coefficient differential equation is often used to

specify the relationship between its input x(t) and output y(t). For linear DT

systems, a linear, constant-coefficient difference equation often describes the

relationship between its input x[k] and output y[k]. The relationship between

the input and output signals completely specifies the physical system. In other

words, we do not require any other information to analyze the system. Once the

input/output relationship has been determined, the schematics of Figs. 2.1(b)

or 2.2(b) can be applied to model the physical system.

2.1.1 Electrical circuit

Figure 2.3 shows a simple electrical circuit comprising of three components: a

resistor R, an inductor L , and a capacitor C . A voltage signal v(t), applied at

the input of the circuit, produces an output signal y(t) representing the voltage

across capacitor C . In order to derive a relationship between the input and

output signals in the RLC circuit, we make use of the Kirchhoff’s current law,

which states “The sum of the currents flowing into a node equals the sum of the

currents flowing out of the node.”

We apply Kirchhoff’s current law to node 1, shown in the top branch of the

RLC circuit in Fig. 2.3. The equations for the currents flowing out of node 1

along resistor R, inductor L , and capacitor C , are given by

resistor R iR = y(t) − v(t)

R (2.3a)

inductor L iL = 1

t∫

−∞

y(τ )dτ (2.3b)

capacitor C iC = C dy

dt . (2.3c)

Applying Kirchhoff’s current law to node 1 and summing up all the currents

yields

y(t) − v(t)

R +

t∫

−∞

y(τ )dτ + C dy

dt = 0, (2.4)

which reduces to a linear, constant-coefficient differential equation of the second

order, given by

d2 y

dt2 +

dt +

LC y(t) =

dt . (2.5)

In conjunction with the initial conditions, y(0) and ẏ(0), Eq. (2.5) completely

specifies the relationship between the input voltage v(t) and the output voltage

y(t) for the RLC circuit shown in Fig. 2.3.

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65 2 Introduction to systems

+ −

−

R node 1

L C y (t)v (t)

iR(t)

iL(t) iC(t)

Fig. 2.3. Electrical circuit

consisting of three passive

components: resistor R,

capacitor C, and inductor L. The

RLC circuit is an example of a CT

linear system.

2.1.2 Semiconductor diode

When a piece of an intrinsic semiconductor (silicon or germanium) is doped

such that half of the piece is of n type while the other half is of p type, a pn

junction is formed. Figure 2.4(a) shows a pn junction with a voltage v applied

across its terminals. The pn junction forms a basic diode, which is fundamental

to the operation of all solid state devices. The symbol for a semiconductor diode

is shown in Fig. 2.4(b). A diode operates under one of the two bias conditions.

It is said to be forward biased when the positive polarity of the voltage source

v is connected to the p region of the diode and the negative polarity of the

voltage source v is connected to the n region. Under the forward bias condition,

the diode allows a relatively strong current i to flow across the pn junction

according to the following relationship:

i = Is[exp(v/VT ) − 1] (2.6)

where Is denotes the reverse saturation current, which for a silicon doped diode

is a constant given by Is = 4.2 × 10−15 A, and VT is the voltage equivalent of the diode’s temperature. The voltage equivalent VT is given by

VT = kT

e . (2.7)

i p n

+ v − i

+ v −

(a) (b) (c)

Fig. 2.4. Semiconductor diode:

(a) pn junction in the forward

bias mode; (b) diode

representing the pn junction

shown in (a); (c) current–voltage

characteristics of a

semiconductor diode.

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66 Part I Introduction to signals and systems

In Eq. (2.7), the Boltzmann constant k equals 1.38 × 10−23 joules/kelvin, T is the absolute temperature measured in kelvin, and e is the negative charge

contained in an electron. The value of e is 1.6 × 10−19 coulombs. At room temperature, 300 K, the value of the voltage equivalent VT , computed using

Eq. (2.7), is found to be 0.026 V. Substituting the values of the saturation

current Is and the voltage equivalent VT , Eq. (2.6) simplifies to

i = 4.2 × 10−15[exp(v/0.026) − 1] A = 0.0042[exp(38.61v) − 1] pA, (2.8)

which describes the relationship between the forward bias voltage v and the

current i flowing through the semiconductor diode. Equation (2.8) is plotted in

the first quadrant (v > 0 and i > 0) of Fig. 2.4(c).

In the reverse bias condition, the negative polarity of the voltage source is

applied to the p region of the diode and the positive polarity is applied to the

n region. When the diode is reverse biased, the current through the diode is

negligibly small and is given by its saturation value, Is = 4.2 × 10−15 A. The current–voltage relationship of a reverse biased diode is plotted in the third

quadrant (v < 0 and i < 0) of Fig. 2.4(c), where we observe a relatively small

value of current flowing through the diode.

As illustrated in Fig. 2.4(c), the input–output relationship of a semiconductor

diode is highly non-linear. Compared to the linear electrical circuit discussed

in Section 2.1.1, such non-linear systems are more difficult to analyze and are

beyond the scope of this book.

2.1.3 Amplitude modulator

Modulation is the process used to shift the frequency content of an information-

bearing signal such that the resulting modulated signal occupies a higher fre-

quency range. Modulation is the key component in modern-day communication

systems for two main reasons. One reason is that the frequency components

of the human voice are limited to a range of around 4 kHz. If a human voice

signal is transmitted directly by propagating electromagnetic radio waves, the

communication antennas required to transmit and receive these radio signals

would be impractically long. A second reason for modulation is to allow for

simultaneous transmission of several voice signals within the same geographic

region. If two signals within the same frequency range are transmitted together,

they will interfere with each other. Modulation provides us with the means of

separating the voice signals in the frequency domain by shifting each voice

signal to a different frequency band. There are different techniques used to

modulate a signal. Here we introduce the simplest form of modulation referred

to as amplitude modulation (AM).

Consider an information-bearing signal m(t) applied as an input to an AM

system, referred to as an amplitude modulator. In communications, the input

m(t) to a modulator is called the modulating signal, while its output s(t) is

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67 2 Introduction to systems

s(t)

modulated

signal

modulator

offset for

modulation Acos(2pfct)

(1 + km(t)) +attenuator

km(t) m(t)

modulating

signal

Fig. 2.5. Amplitude modulation

(AM) system.

called the modulated signal. The steps involved in an amplitude modulator

are illustrated in Fig. 2.5, where the modulating signal m(t) is first processed

by attenuating it by a factor k and adding a dc offset such that the resulting

signal (1 + km(t)) is positive for all time t . The modulated signal is produced by multiplying the processed input signal (1 + km(t)) with a high-frequency carrier c(t) = A cos(2π fct). Multiplication by a sinusoidal wave of frequency fc shifts the frequency content of the modulating signal m(t) by an additive

factor of fc. Mathematically, the amplitude modulated signal s(t) is expressed as

follows:

s(t) = A[1 + km(t)] cos(2π fct), (2.9)

where A and fc are, respectively, the amplitude and the fundamental frequency

of the sinusoidal carrier.

It may be noted that the amplitude A and frequency fc of the carrier signal,

along with the attenuation factor k used in the modulator, are fixed; therefore,

Eq. (2.9) provides a direct relationship between the input and the output signals

of an amplitude modulator. For example, if we set the attenuation factor k to

0.2 and use the carrier signal c(t) = cos(2π × 108t), Eq. (2.9) simplifies to

s(t) = [1 + 0.2m(t)] cos(2π × 108t). (2.10)

Amplitude modulation is covered in more detail in Chapter 7.

2.1.4 Mechanical water pump

The mechanical pump shown in Fig. 2.6 is another example of a linear CT

system. Water flows into the pump through a valve V1 controlled by an electrical

circuit. A second valve V2 works mechanically as the outlet. The rate of the

outlet flow depends on the height of the water in the mechanical pump. A

higher level of water exerts more pressure on the mechanical valve V2, creating

a wider opening in the valve, thus releasing water at a faster rate. As the level

of water drops, the opening of the valve narrows, and the outlet flow of water is

reduced.

A mathematical model for the mechanical pump is derived by assuming that

the rate of flow Fin of water at the input of the pump is a function of the input

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68 Part I Introduction to signals and systems

Fin = k x(t)

Fout = ch(t) h(t)

A Fig. 2.6. Mechanical water

pumping system.

voltage x(t):

Fin = kx(t), (2.11)

where k is the linearity constant. Valve V2 is designed such that the outlet flow

rate Fout is given by

Fout = ch(t), (2.12)

where c denotes the outlet flow constant and h(t) is the height of the water

level. Denoting the total volume of the water inside the tank by V (t), the rate

of change in the volume of the stored water is dV/dt , which must be equal to

the difference between the input flow rate, Eq. (2.11), and the outlet flow rate,

Eq. (2.12). The resulting equation is as follows:

dt = Fin − Fout = kx(t) − ch(t). (2.13)

Expressing V (t) as the product of the cross-sectional area A of the water tank

and the height h(t) of the water yields

A dh

dt + ch(t) = kx(t), (2.14)

which is a first-order, constant-coefficient differential equation describing the

relationship between the input current signal x(t) and height h(t) of water in

the mechanical pump. It may be noted that the input–output relationship in

the electrical circuit, discussed in Section 2.1.1, was also a constant-coefficient

differential equation. In fact, most CT linear systems are often modeled with

linear, constant-coefficient differential equations.

2.1.5 Mechanical spring damper system

The spring damping system shown in Fig. 2.7 is another classical example of a

linear mechanical system. An application of such a mechanical damping system

is in the shock absorber installed in an automobile. Figure 2.7 models a spring

damping system where mass M, which is attached to a rigid body through a

mechanical spring with a spring constant of k, is pulled downward with force

x(t). Assuming that the vertical displacement from the initial location of mass

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69 2 Introduction to systems

M is given by y(t), the three upward forces opposing the external downward

force x(t) are given by

x(t)

ky(t)My(t) ry(t)

y(t)

x(t)

r y(t) wall

friction

k spring

constant

(a)

(b)

¨ .

Fig. 2.7. (a) Mechanical spring

damper system. (b) Free-body

diagram illustrating the

opposing forces acting on mass

M of the mechanical spring

damping system.

inertial (or accelerating) force Fi = M d2 y

dt2 ; (2.15a)

frictional (or damping) force Ff = r dy

dt ; (2.15b)

spring (or restoring) force Fs = ky(t), (2.15c)

where r is the damping constant for the medium surrounding the mass. Apply-

ing Newton’s third law of motion, the input–output relationship of the spring

damping system is given by

M d2 y

dt2 + r

dt + ky(t) = x(t), (2.16)

which is a linear, constant-coefficient second-order differential equation.

Equation (2.16) describes the relationship between the applied force x(t) and

the resulting vertical displacement y(t). As in the case of the RLC circuit,

a second-order differential equation is used to model the mechanical spring

damper system.

2.1.6 Numerical differentiation and integration

Numerical methods are widely used in calculus for finding approximate values

of derivatives and definite integrals. Here, we present examples of differentiation

and integration of a CT function x(t). The systems representing integration and

differentiator are shown in Fig. 2.8. We show that the numerical approximations

of a CT differentiator and integrator lead to finite difference equations that are

frequently used to describe DT systems.

y(t)x(t) d dt

y(t)x(t) 0

dt∫

(a)

(b)

Fig. 2.8. Schematics of (a) a

differentiator and (b) an

integrator. Finite-difference

schemes are often used to

compute the values of

derivatives and finite integrals

numerically.

To discretize a derivative over a continuous interval [0, T ], the time interval T

is divided into intervals of duration �t , resulting in the sampled values x(k�t)

for k = 0, 1, 2, . . . , K , with K given by the ratio T/�t . Using a single-step backward finite-difference scheme, the time derivative can be approximated as

follows:

∣ ∣ ∣ ∣ t=k�t

≈ x(k�t) − x((k − 1)�t)

�t , (2.17)

which yields

y(k�t) = x(k�t) − x((k − 1)�t)

�t (2.18)

or,

y(k�t) = C1(x(k�t) − x((k − 1)�t)), (2.19)

where x(k�t) is the sampled value of x(t) at t = k�t and C1 is a constant, equal

to 1/�t. The CT signal y(t) = dx/dt and represents the result of differentiation.

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70 Part I Introduction to signals and systems

Usually, the sampling interval �t in Eq. (2.19) is omitted, resulting in the

following expression:

y[k] = C1x[k] − C1x[k − 1], (2.20)

which is a finite-difference representation of the differentiator shown in

Fig. 2.8(a).

To integrate a function, we use Euler’s formula, which approximates the

integral by the following:

k�t∫

(k−1)�t

x(t)dt ≈ �t x((k − 1)�t). (2.21)

In other words, the area under x(t) within the range [(k − 1)�t, k�t] is approx-

imated by a rectangle with width �t and height x((k − 1)�t). Expressing the

integral as follows:

y(t)|t=k�t =

t∫

x(t)dt =

(k−1)�t∫

x(t)dt

︸︷︷︸

y((k−1)�t)

k�t∫

(k−1)�t

x(t)dt

︸︷︷︸

�t x((k−1)�t)

(2.22)

and simplifying, we obtain

y(k�t) = y((k − 1)�t) + �t x((k − 1)�t). (2.23)

Again, omitting the sampling interval �t in Eq. (2.23) yields

y[k] − y[k − 1] = C2x[k − 1], (2.24)

where C2 = �t . Equation (2.24) is a first-order finite-difference equation mod-

eling an integrator and can be solved iteratively to compute the integral at

discrete time instants k�t . Systems represented by finite-difference equations

of the form of Eqs. (2.20) or (2.24) are referred to as DT systems and are the

focus of our discussion in the second half of the book. In the case of DT systems,

a difference equation, along with the ancillary conditions, provides a complete

description of the DT systems.

2.1.7 Delta modulation

In a digital communication system, the information-bearing analog signal is

first transformed into a binary sequence of zeros and ones, referred to as a dig-

ital signal, which is then transferred using a digital communication technique

from a transmitter to a receiver. Compared to analog transmission, digital com-

munications operate with a lower signal-to-noise ratio (SNR) and can therefore

provide almost error-free performance over long distances. In addition, digital

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71 2 Introduction to systems

1 1 1 1 1 0 1 1 1 1 1 1 0 0 0 0 0 0 0 0

t TT

∆

t T

x(t)

(a) (b)

x(t)

Fig. 2.9. A delta modulation

(DM) system. (a) Approximation

of the information-bearing

signal x(t ) with a staircase

signal x̂(kT ), referred to as the

DM signal. (b) Binary signal

transmitted to the receiver.

communications allow for other data processing features such as error cor-

rection, data encryption, and jamming resistance, which can be exploited for

secure data transmission. In this section, we study a basic waveform coding pro-

cedure, referred to as delta modulation (DM), which is widely used to transform

an analog signal into a digital signal.

The process of DM is illustrated in Fig. 2.9, where an information-bearing

analog signal x(t) is approximated by a delta modulated signal x̂(t). The analog

signal x(t) is uniformly sampled at time instants t = kT . At each sampling instant, the sampled value x(kT) of the analog signal is compared with the

amplitude of the DM signal x̂(kT ). If the magnitude of the sampled signal

x(kT) is greater than the corresponding magnitude of the DM signal x̂(kT ),

then the DM signal is increased by a fixed amplitude, say �, at t = kT . Bit 1 is transmitted to the receiver to indicate the increase in the amplitude of the DM

signal. On the other hand, if the amplitude of the sampled signal x(kT) is less

than the magnitude of the DM signal x̂(kT ), then the DM signal is decreased by

�. Bit 0 is transmitted to the receiver to indicate the decrease in the amplitude

of the DM signal. In other words, a single bit is used at each time instant t = kT to indicate an increase or decrease in the amplitude of the information-bearing

signal.

A major advantage of DM is the simple structure of the receiver. At the

receiving end, the signal x̂(t) is reconstructed using the following simple

relationship:

x̂(kT ) = x̂((k − 1)T ) + bk�, (2.25)

where bk = 1 if bit 1 is received and bk = −1 if bit 0 is received. Solving for x̂(kT ), Eq. (2.25) is represented as follows:

x̂(kT ) = n∑

k=0 bk� + x̂(0), (2.26)

where x̂(0) represents the initial value used at t = 0 in the DM signal. Equation (2.26) implies that the DM signal x̂(kT ) is obtained by accumulating

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72 Part I Introduction to signals and systems

the values of the bk�’s. Such a DT system that accumulates the values of the

input is referred to as an accumulator. It may be noted that the receiver of a

DM system is a linear system as it can be modeled by a constant-coefficient

difference equation.

2.1.8 Digital filter

Digital images are made up of tiny “dots” obtained by sampling a two-

dimensional (2D) analog image. Each dot is referred to as a picture element, or

a pixel. A digital image, therefore, can be modeled with a 2D array, x[m, n],

where the index (m, n) refers to the spatial coordinate of a pixel with m being

the number of the row and n being the number of the column. In a monochrome

image, the value x[m, n] of a pixel indicates its intensity value. When the pixels

are placed close to each other and illuminated according to their intensity values

on the computer monitor, a continuous image is perceived by the human eye.

In digital image processing, spatial averaging is frequently used for smooth-

ing noise, lowpass filtering, and subsampling of images. In spatial averaging,

the intensity of each pixel is replaced by a weighted average of the intensities

of the pixels in the neighborhood of the reference pixel. Using a unidirectional

fourth-order neighborhood, the reference pixel x[m, n] is replaced by the spa-

tially averaged value:

y[m, n] = 1

4 (x[m, n] + x[m, n − 1] + x[m − 1, n] + x[m − 1, n − 1]),

(2.27)

where y[m, n] represents the 2D output image of the spatial averaging system.

Equation (2.27) is an example of a 2D finite-difference equation and it models

a 2D DT system with input x[m, n] and output y[m, n].

In this section, we have considered some interesting applications of signal

processing in CT and DT systems. Our goal has been to motivate the reader

to learn about the techniques and basic concepts required to investigate one

or more of these application areas. Each of the discussed areas is a subject

of considerable study. Nevertheless, certain fundamentals are central to most

applications, and many of these basic concepts will be discussed in the chapters

that follow.

2.2 Classification of systems

In the analysis or design of a system, it is desirable to classify the system

according to some generic properties that the system satisfies. In this segment

we introduce a set of basic properties that may be used to categorize a system.

For a system to possess a given property, the property must hold true for all

possible input signals that can be applied to the system. If a property holds for

some input signals but not for others, the system does not satisfy that property.

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73 2 Introduction to systems

In this section, we classify systems into six basic categories:

(i) linear and non-linear systems;

(ii) time-invariant and time-varying systems;

(iii) systems with and without memory;

(iv) causal and non-causal systems;

(v) invertible and non-invertible systems;

(vi) stable and unstable systems.

In the following discussion, we make use of the notation given in Eqs. (2.1) and

(2.2), which we repeat here:

CT system x(t) → y(t);

DT system x[k] → y[k];

to refer to output y(t) resulting from input x(t) for a CT system and to output

y[k] resulting from input x[k] for a DT system.

2.2.1 Linear and non-linear systems

A CT system with the following set of inputs and outputs:

x1(t) → y1(t) and x2(t) → y2(t)

is linear iff it satisfies the additive and the homogeneity properties described

below:

additive property x1(t) + x2(t) → y1(t) + y2(t); (2.28)

homogeneity property α x1(t) → αy1(t); (2.29)

for any arbitrary value of α and all possible combinations of inputs and out-

puts. The additive and homogeneity properties are collectively referred to as

the principle of superposition. Therefore, linear systems satisfy the principle

of superposition. Based on the principle of superposition, the properties in

Eqs. (2.28) and (2.29) can be combined into a single statement as follows. A

CT system with the following sets of inputs and outputs:

x1(t) → y1(t) and x2(t) → y2(t)

is linear iff

α x1(t) + βx2(t) → αy1(t) + βy2(t) (2.30)

for any arbitrary set of values for α and β, and for all possible combinations of

inputs and outputs.

Likewise, a DT system with

x1[k] → y1[k] and x2[k] → y2[k],

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74 Part I Introduction to signals and systems

is linear iff

α x1[k] + βx2[k] → αy1[k] + βy2[k] (2.31)

for any arbitrary set of values for α and β, and for all possible combinations of

inputs and outputs.

A consequence of the linearity property is the special case when the input x

to a linear CT or DT system is zero. Substituting α = 0 in Eq. (2.29) yields

0 · x1(t) = 0 → 0 · y1(t) = 0. (2.32)

In other words, if the input x(t) to a linear system is zero, then the output

y(t) must also be zero for all time t . This property is referred to as the zero-

input, zero-output property. Both CT and DT systems that are linear satisfy

the zero-input, zero-output property for all time t . Note that Eq. (2.32) is a

necessary condition and is not sufficient to prove linearity. Many non-linear

systems satisfy this property as well.

Example 2.1

Consider the CT systems with the following input–output relationships:

(a) differentiator y(t) = dx(t)

dt ; (2.33)

(b) exponential amplifier x(t) → ex(t); (2.34)

(d) amplifier with additive bias y(t) = 3x(t) + 5. (2.36)

Determine whether the CT systems are linear.

Solution

(a) From Eq. (2.33), it follows that

x1(t) → dx1(t)

dt = y1(t)

and

x2(t) → dx2(t)

dt = y2(t),

which yields

αx1(t) + β1x2(t) → d

dt {αx1(t) + β1x2(t)} = α

dx1(t)

dt + β

dx2(t)

dt .

Since

α dx1(t)

dt + β

dx2(t)

dt = αy1(t) + βy2(t),

the differentiator as represented by Eq. (2.33) is a linear system.

(b) From Eq. (2.34), it follows that

x1(t) → e x1(t) = y1(t)

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75 2 Introduction to systems

and

x2(t) → e x2(t) = y2(t),

giving

αx1(t) + βx2(t) → e αx1(t)+βx2(t).

Since

eαx1(t)+βx2(t) = eαx1(t) · eβx2(t) = [y1(t)] α + [y2(t)]

β �= αy1(t) + βy2(t),

the exponential amplifier represented by Eq. (2.34) is not a linear system.

x1(t) → 3x1(t) = y1(t)

and

x2(t) → 3x2(t) = y2(t),

giving

αx1(t) + βx2(t) → 3{αx1(t) + βx2(t)} = 3αx1(t) + 3βx2(t)

= αy1(t) + βy2(t).

Therefore, the amplifier of Eq. (2.35) is a linear system.

(d) From Eq. (2.36), we can write

x1(t) → 3x1(t) + 5 = y1(t)

and

x2(t) → 3x2(t) + 5 = y2(t),

giving

αx1(t) + βx2(t) → 3[αx1(t) + βx2(t)] + 5.

Since

3[αx1(t) + βx2(t)] + 5 = αy1(t) + βy2(t) − 5,

the amplifier with an additive bias as specified in Eq. (2.36) is not a linear

system.

An alternative approach to check if a system is non-linear is to apply the

zero-input, zero-output property. For system (b), if x(t) = 0, then y(t) = 1.

System (b) does not satisfy the zero-input, zero-output property, hence system

(b) is non-linear. Likewise, for system (d), if x(t) = 0 then y(t) = 5. Therefore,

system (d) is not a linear system.

If a system does not satisfy the zero-input, zero-output property, we can safely

classify the system as a non-linear system. On the other hand, if it satisfies

the zero-input, zero-output property, it can be linear or non-linear. Satisfying

the zero-input, zero-output property is not a sufficient condition to prove the

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76 Part I Introduction to signals and systems

linear

system y(t)

output

signal

x(t)

input

signal

yzi(t)

Fig. 2.10. Incrementally linear

system expressed as a linear

system with an additive offset.

linearity of a system. A CT system y(t) = x2(t) is clearly a non-linear system, yet it satisfies the zero-input, zero-output property. For the system to be linear,

it must satisfy Eq. (2.30).

Incrementally linear system In Example 2.1, we proved that the amplifier y(t) = 3x(t) represents a linear system, while the amplifier with additive bias y(t) = 3x(t) + 5 represents a non-linear system. System y(t) = 3x(t) + 5 sat- isfies a different type of linearity. For two different inputs x1(t) and x2(t), the

respective outputs of system y(t) = 3x(t) + 5 are given by

input x1(t) y1(t) = 3x1(t) + 5; input x2(t) y2(t) = 3x2(t) + 5.

Calculating the difference on both sides of the above equations yield

y2(t) − y1(t) = 3[x2(t) − x1(t)]

�y(t) = 3�x(t).

In other words, the change in the output of system y(t) = 3x(t) + 5 is linearly related to the change in the input. Such systems are called incrementally linear

systems.

An incrementally linear system can be expressed as a combination of a linear

system and an adder that adds an offset yzi(t) to the output of the linear sys-

tem. The value of offset yzi(t) is the zero-input response of the original system.

System S1, y(t) = 3x(t) + 5, for example, can be expressed as a combination of a linear system S2, y(t) = 3x(t), plus an offset given by the zero-input response of S1, which equals yzi(t) = 5. Figure 2.10 illustrates the block diagram repre- sentation of an incrementally linear system in terms of a linear system and an

additive offset yzi(t).

Example 2.2

Consider two DT systems with the following input–output relationships:

(a) differencing system y[k] = 3(x[k] − x[k − 2]); (2.37) (b) sinusoidal system y[k] = sin(x[k]). (2.38)

Determine if the DT systems are linear.

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77 2 Introduction to systems

Solution

(a) From Eq. (2.37), it follows that:

x1[k] → 3x1[k] − 3x1[k − 2] = y1[k]

and

x2[k] → 3x2[k] − 3x2[k − 2] = y2[k],

giving

αx1[k] + βx2[k] → 3αx1[k] − 3αx1[k − 2] + 3βx2[k] − 3βx2[k − 2].

Since

3αx1[k] − 3αx1[k − 2] + 3βx2[k] − 3βx2[k − 2] = αy1[k] + βy2[k],

the differencing system, Eq. (2.37), is linear.

To illustrate the linearity property graphically, we consider two DT input sig-

nals x1[k] and x2[k] shown in the two top-left subplots in Figs. 2.11(a) and (c).

The resulting outputs y1[k] and y2[k] for the two inputs applied to the differ-

encing system, Eq. (2.37), are shown in the two top-right stem subplots in

Figs. 2.11(b) and (d), respectively. A linear combination, x3[k] = x1[k] +

2x2[k], of the two inputs is shown in the bottom-left subplot in Fig. 2.11(e).

The resulting output y3[k] of the system for input signal x3[k] is shown in

the bottom-right subplot in Fig. 2.11(f). By looking at the subplots, it is clear

that the output y3[k] = y1[k] + 2y2[k]. In other words, the output y3[k] can be

determined by using the same linear combination of outputs y1[k] and y2[k] as

the linear combination used to obtain x3[k] from x1[k] and x2[k].

(b) From Eq. (2.38), it follows that:

x1[k] → sin(x1[k]) = y1[k], x2[k] → sin(x2[k]) = y2[k],

giving

αx1[k] + βx2[k] → sin(αx1[k]) + sin(βx2[k]) �= αy1[k] + βy2[k];

therefore, the sinusoidal system in Eq. (2.38) is not linear.

To illustrate graphically that system (b) indeed does not satisfy the linearity

property, we consider two input signals x1[k] and x2[k] shown, respectively, in

Figs. 2.12(a) and (c). Their corresponding outputs, y1[k] and y2[k], are shown

in Figs. 2.12(b) and (d). The output y3[k] of the system for the input signal

x3[k] = x1[k] + 2x2[k], obtained by combining x1[k] and x2[k], is shown in

Fig. 2.12(f). Comparing Fig. 2.12(f) with Figs. 2.12(b) and (d), we note that

output y3[k] �= y1[k] + 2y2[k]. To check, we select k = 4. From Fig. 2.12,

inputs x1[4] = 0 and x2[4] = 2. Using Eq. (2.38), outputs y1[4] = sin(0) = 0

and y2[4] = sin(2) = 0.91. The linear combination y1[k] + 2y2[k] of y1[k] and

y2[k] at k = 4 gives a value of 1.82. If the system is linear, we should get

y3[4] = 1.82 from the combined input x3[k] = x1[k] + 2x2[k] = 4 at k = 4.

Substituting in Eq. (2.38), we obtain y3[4] = sin(4) = −0.76. Since the value

of output y3[k] at k = 4 obtained from the linear combination of individual

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78 Part I Introduction to signals and systems

0 1 2 3 4 5 6 7 8 9 10−1 −2

2 x1[k]

(a) (b)

y1[k]

3 k

0 1 2 4 5 6 7 8 9 10−1 −2

−6

≈≈≈

y2[k]

6 7 k

0 1 2 3 4 5 8 9 10−1 −2

≈≈≈≈≈

−3

≈≈≈

−6

≈≈≈≈

x2[k]

0 1 2 3 4 5 6 7 8 9 10−1 −2

1 1

x3[k]

0 1 2 3 4 5 6 7 8 9 10−1 −2

2 22 ≈≈

0 1 2 3 4 5

6 7

8 9 10−1 −2

≈

≈≈

y3[k]

≈ 6

≈

≈≈

−12

−6

≈

≈≈

(e) ( f )

Fig. 2.11. Input–output pairs of

the linear DT system specified in

Example 2.2(a). Parts (a)–(f ) are

discussed in the text.

outputs y1[k] and y2[k] is different from the value obtained directly by applying

the combined input, we may say that the system in Fig. 2.12(b) is not linear.

The graphical result is in accordance with the mathematical proof.

Example 2.3

Consider the AM system with input–output relationship given by

s(t) = [1 + 0.2m(t)] cos(2π × 108t). (2.39)

Determine if the AM system is linear.

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79 2 Introduction to systems

y1[k]

(b)

2 x1[k]

0 1 2 3 4 5 6 7 8 9 10−1 −2

(a)

0 1 2 3 4 5 6 7 8 9 10

0.91

x2[k]

0 1 2 3 4 5 6 7 8 9 10

(c)

1 1

y2[k]

(d)

0 1 2 3 4 5 6 7 8 9 10

0.91 0.84

x3[k]

0 1 2 3 4 5 6 7 8 9 10

(e)

2 22

0.76

4 0.91 0.910.91

y3[k]

(f)

0 1 2 3 5 6 7 8 9 10

−1 −2

−1 −2 −1 −2

Fig. 2.12. Input–output pairs of

the linear DT system specified in

Example 2.2(b). Parts (a)–(f) are

discussed in the text.

Solution

From Eq. (2.39), it follows that:

m1(t) → [1 + 0.2m1(t)] cos(2π × 10 8t) = s1(t)

and

m2(t) → [1 + 0.2m2(t)] cos(2π × 10 8t) = s2(t),

giving

αm1(t) + βm2(t) → [1 + 0.2{αm1(t) + βm2(t)}] cos(2π × 10 8t)

�= αs1(t) + βs2(t).

Therefore, the AM system is not linear.

2.2.2 Time-varying and time-invariant systems

A system is said to be time-invariant (TI) if a time delay or time advance of the

input signal leads to an identical time-shift in the output signal. In other words,

except for a time-shift in the output, a TI system responds exactly the same

way no matter when the input signal is applied. We now define a TI system

formally.

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80 Part I Introduction to signals and systems

A CT system with x(t) → y(t) is time-invariant iff

x(t − t0) → y(t − t0) (2.40)

for any arbitrary time-shift t0. Likewise, a DT system with x[k] → y[k] is

time-invariant iff

x[k − k0] → y[k − k0] (2.41)

for any arbitrary discrete shift k0.

Example 2.4

Consider two CT systems represented mathematically by the following input–

output relationship:

(i) system I y(t) = sin(x(t)); (2.42)

(ii) system II y(t) = t sin(x(t)). (2.43)

Determine if systems (i) and (ii) are time-invariant.

Solution

(i) From Eq. (2.42), it follows that:

x(t) → sin(x(t)) = y(t)

and

x(t − t0) → sin(x(t − t0)) = y(t − t0).

Since sin[x(t − t0)] = y(t − t0), system I is time-invariant. We demonstrate

the time-invariance property of system I graphically in Fig. 2.13, where a time-

shifted version x(t − 1) of input x(t) produces an equal shift of one time unit

in the original output y(t) obtained from x(t).

(ii) From Eq. (2.43), it follows that:

x(t) → t sin(x(t)) = y(t).

If the time-shifted signal x(t − t0) is applied at the input of Eq. (2.43), the new

output is given by

x(t − t0) → t sin(x(t − t0)).

The shifted output y(t − t0) is given by

y(t − t0) = (t − t0) sin(x(t − t0)).

Since t sin[x(t − t0)] �= y(t − t0), system II is not time-invariant. The time-

invariance property of system II is demonstrated in Fig. 2.14, where we

observe that a right shift of one time unit in input x(t) alters the shape of the

output y(t).

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81 2 Introduction to systems

x(t)

1 2 3 4−4 −3 −2 −1

y(t)

1 2 3 4−4 −3 −2 −1

(a) (b)

(d)

y2(t)

1 2 3 4−4 −3 −2 −1

x(t − 1)

1 2 3 4−4 −3 −2 −1

(c)

Fig. 2.13. Input–output pairs of

the CT time-invariant system

specified in Example 2.4(i).

(a) Arbitrary signal x(t ).

(b) Output of system for input

signal x(t ). (c) Signal x(t − 1). (d) Output of system for input

signal x(t − 1). Note that except for a time-shift, the two output

signals are identical.

Example 2.5

Consider two DT systems with the following input–output relationships:

(i) system I y[k] = 3(x[k] − x[k − 2]); (2.44) (ii) system II y[k] = k x[k]. (2.45)

Determine if the systems are time-invariant.

Solution

(i) From Eq. (2.44), it follows that:

x[k] → 3(x[k] − x[k − 2]) = y[k]

t 0 1 2 3 4

x(t)

−4 −3 −2 −1

2 y(t)

t 0 1 2 3 4−4 −3 −2 −1

t 0

x(t − 1)

1 2 3 4−4 −3 −2 −1

t 0 1 2 3 4−4 −3 −2 −1

y2(t)

(a) (b)

Fig. 2.14. Input–output pairs of the time-varying system specified in Example 2.4(ii). (a) Arbitrary signal

x(t ). (b) Output of system for input signal x(t ). (c) Signal x(t − 1). (d) Output of system for input signal

x(t − 1). Note that the output for time-shifted input x(t − 1) is different from the output y(t ) for the

original input x(t ).

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82 Part I Introduction to signals and systems

(a) (b)

x[k]

0 1 2 3 4 5 6 7 8 9 10−1 −2

1 11 1

x[k − 2]

0 1 2 3 4 5 6 7 8 9 10−1 −2

1 11 1

y[k]

0 1 2 3 4 5 6 7 8 9 10−1 −2

y2[k]

0 1 2 3 4 5 6 7 8 9 10−1 −2

Fig. 2.15. Input–output pairs of

the DT time-varying system

specified in Example 2.5(ii). The

output y2[k ] for the time-shifted

input x2[k ] = x [k − 2] is different in shape from the

output y [k ] obtained for input

x[k ]. Therefore the system is

time-variant. Parts (a)–(d) are

discussed in the text .

and

x[k − k0] → 3(x[k − k0] − x[k − k0 − 2]) = y[k − k0].

Therefore, the system in Eq. (2.44) is a time-invariant system.

(ii) From Eq. (2.45), it follows that:

x[k] → kx[k] = y[k]

and

x[k − k0] → kx[k − k0] �= y[k − k0] = (k − k0)x[k − k0].

Therefore, system II is not time-invariant. In Fig. 2.15, we plot the outputs of

the DT system in Eq. (2.45) for input x[k], shown in Fig. 2.15(a) and a shifted

version x[k − 2] of the input, shown in Fig. 2.15(c). The resulting outputs are

plotted, respectively, in Figs. 2.15(b) and (d). As expected, the Fig. 2.15(d) is

not a delayed version of Fig. 2.15(b) since the system is time-variant.

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83 2 Introduction to systems

+ −

− y(t)v (t)

(a)

− y(t)

(b)

+ −i(t)

Fig. 2.16. (a) Passive electrical circuit comprising resistors R 1 and R 2 . (b) Active electrical circuit

comprising resistor R, inductor L, and capacitor C. Both inductor L and capacitor C are storage

components, and hence lead to a system with memory.

2.2.3 Systems with and without memory

A CT system is said to be without memory (memoryless or instantaneous) if its

output y(t) at time t = t0 depends only on the values of the applied input x(t) at the same time t = t0. On the other hand, if the response of a system at t = t0 depends on the values of the input x(t) in the past or in the future of time t = t0, it is called a dynamic system, or a system with memory. Likewise, a DT system

is said to be memoryless if its output y[k] at instant k = k0 depends only on the value of its input x[k] at the same instant k = k0. Otherwise, the DT system is said to have memory.

Example 2.6

Determine if the two electrical circuits shown in Figs. 2.16(a) and (b) are

memoryless.

Solution

The relationship between the input voltage v(t) and the output voltage y(t)

across resistor R1 in the electrical circuit of Fig. 2.16(a) is given by

y(t) = R1

R1 + R2 v(τ ). (2.46)

For time t = t0, the output y(t0) depends only on the value v(t0) of the input v(t) at t = t0. The electrical circuit shown in Fig. 2.16(a) is, therefore, a memoryless system.

The relationship between the input current i(t) and the output voltage y(t) in

Fig. 2.16(b) is given by

y(t) = 1

t∫

−∞

i(τ )dτ . (2.47)

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84 Part I Introduction to signals and systems

Table 2.1. Examples of CT and DT systems with and without memory

Continuous-time Discrete-time

Memoryless systems Systems with memory Memoryless systems Systems with memory

y(t) = 3x(t) + 5 y(t) = x(t − 5) y[k] = 3x[k] + 7 y[k] = x[k − 5] y(t) = sin{x(t)} + 5 y(t) = x(t + 2) y[k] = sin(x[k]) + 3 y[k] = x[k + 3] y(t) = ex(t) y(t) = x(2t) y[k] = ex[k] y[k] = x[2k] y(t) = x2(t) y(t) = x(t/2) y[k] = x2[k] y[k] = x[k/2]

To compute the output voltage y(t0) at time t0, we require the value of the current

source for the time range (−∞, t0], which includes the entire past. Therefore,

the electrical circuit in Fig. 2.16(b) is not a memoryless system.

In Table 2.1, we consider several examples of memoryless and dynamic systems.

The reader is encouraged to verify mathematically the classifications made in

Table 2.1.

As a side note to our discussion on memoryless systems, we consider another

class of systems with memory that require only a limited set of values of input

x(t) in t0 − T ≤ t ≤ t0 to compute the value of output y(t). Such CT systems,

whose response y(t) is completely determined from the values of input x(t) over

the most recent past T time units, are referred to as finite-memory or Markov

systems with memory of length T time units. Likewise, a DT system is called

a finite-memory or a Markov system with memory of length M if output y[k]

at k = k0 depends only on the values of input x[k] for k0 − M ≤ k ≤ k0 in the

most recent past.

2.2.4 Causal and non-causal systems

A CT system is causal if the output at time t0 depends only on the input x(t) for

t ≤ t0. Likewise, a DT system is causal if the output at time instant k0 depends

only on the input x[k] for k ≤ k0. A system that violates the causality condition is

called a non-causal (or anticipative) system. Note that all memoryless systems

are causal systems because the output at any time instant depends only on

the input at that time instant. Systems with memory can either be causal or

non-causal.

Example 2.7

(i) CT time-delay system y(t) = x(t − 2) ⇒ causal system;

(ii) CT time-forward system y(t) = x(t + 2) ⇒ non-causal system;

(iii) DT time-delay system y[k] = x[k − 2] ⇒ causal system;

(iv) DT time-advance system y[k] = x[k + 2] ⇒ non-causal system;

(v) DT linear system y[k] = x[k − 2] + x[k + 10] ⇒ non-causal

system.

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85 2 Introduction to systems

Table 2.2. Examples of causal and non-causal systems

The CT and DT systems are represented using their input–output relationships. Note that all systems in the table

have memory.

CT systems DT systems

Causal Non-causal Causal Non-causal

y(t) = x(t − 5) y(t) = x(t + 2) y[k] = 3x[k − 1] + 7 y[k] = x[k + 3] y(t) = sin{x(t − 4)} + 3 y(t) = sin{x(t + 4)} + 3 y[k] = sin(x[k − 4]) + 3 y[k] = sin(x[k + 4]) + 3 y(t) = ex(t−2) y(t) = x(2t) y[k] = ex[k−2] y[k] = x[2k] y(t) = x2(t − 2) y(t) = x(t/2) y[k] = x2[k − 5] y[k] = x[k/2] y(t) = x(t − 2) + x(t − 5) y(t) = x(t − 2) + x(t + 2) y[k] = x[k − 2] + x[k − 8] y[k] = x[k + 2] + x[k − 8]

x(t) y(t)

x(t) CT

system

inverse

system x[k]

y[k] x[k] DT

system

inverse

system

(a) (b)

Fig. 2.17. Invertible systems.

(a) Inverse of a CT system.

(b) Inverse of a DT system.

Causality is a required condition for the system to be physically realizable. A

non-causal system is a predictive system and cannot be implemented physically.

Table 2.2 presents examples of causal and non-causal systems in CT and DT

domains.

2.2.5 Invertible and non-invertible systems

A CT system is invertible if the input signal x(t) can be uniquely determined

from the output y(t) produced in response to x(t) for all time t ∈ (−∞, ∞).

Similarly, a DT system is called invertible if, given an arbitrary output response

y[k] of the system for k ∈ (−∞, ∞), the corresponding input signal x[k] can be

uniquely determined for all time k ∈ (−∞, ∞). To be invertible, two different

inputs cannot produce the same output since, in such cases, the input signal

cannot be uniquely determined from the output signal.

A direct consequence of the invertibility property is the determination of a

second system that restores the original input. A system is said to be invertible

if the input to the system can be recovered by applying the output of the original

system as input to a second system. The second system is called the inverse

of the original system. The relationship between the original system and its

inverse is shown in Fig. 2.17.

Example 2.8

Determine if the following CT systems are invertible.

(i) Incrementally linear system:

y(t) = 3x(t) + 5.

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86 Part I Introduction to signals and systems

The input–output relationship is expressed as follows:

x(t) = 1

3 [y(t) − 5].

The above expression shows that input x(t) can be uniquely determined from

the output signal y(t). Therefore, the system is invertible.

(ii) Cosine system:

y(t) = cos[x(t)].

The input–output relationship is expressed as follows:

x(t) = cos−1[y(t) − 5] + 2πm,

where m is an integer with values m = 0, ±1, ±2, . . . The above relationship shows that there are several possible values of x(t) for a given value of y(t).

Therefore, system (ii) is a non-invertible system.

(iii) Squarer:

y(t) = [x(t)]2.

The input–output relationship is expressed as follows:

x(t) = ± √

y(t).

In other words, for a given y(t) value, there are two possible values of x(t).

Because x(t) is not unique, the system is non-invertible.

(iv) Time-differencing system:

y(t) = x(t) − x(t − 2).

The input–output relationship is expressed as follows:

x(t) = y(t) + x(t − 2).

Since x(t − 2) = y(t − 2) + x(t − 4), the earlier equation can be expressed as follows:

x(t) = y(t) + y(t − 2) + x(t − 4).

By recursively substituting first the value of x(t − 4) and later for other delayed versions of x(t), the above relationship can be expressed as follows:

x(t) = ∞∑

m=0

y(t − 2m).

Using the above relationship, the input signal x(t) can be uniquely reconstructed

if y(t) is known. Therefore, the system is invertible.

(v) Integrating system I:

y(t) =

t∫

−∞

x(τ )dτ .

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87 2 Introduction to systems

Differentiating both sides of the above equation yields

x(t) = dy

dt .

The above relationship shows that for a given output signal, the corresponding

input signal can be uniquely determined. Therefore, the system is invertible.

(vi) Integrating system II:

y(t) = t∫

t−2

x(τ )dτ .

We can represent y(t) as follows:

y(t) = t∫

−∞

x(τ )dτ −

t−2∫

−∞

x(τ )dτ .

Differentiating both sides, we obtain

dt = x(t) − x(t − 2).

Following the procedure used in part (iv) and expressing the result in terms of

the input signal x(t), we obtain

x(t) = ∞∑

m=0

dy(t − 2m)

dt .

The above relationship shows that for a given output signal, the corresponding

input signal can be uniquely determined. Therefore, the system is invertible.

Example 2.9

Determine if the following DT systems are invertible.

(i) Incrementally linear system:

y[k] = 2x[k] + 7.

The input–output relationship is expressed as follows:

x[k] = 1

2 {y[k] − 7}.

The above expression shows that given an output signal, the input can be

uniquely determined. Therefore, the system is invertible.

(ii) Exponential output:

y[k] = ex[k].

The input–output relationship is expressed as follows:

x[k] = ln{y[k]}.

The above expression shows that given an output signal, the input can be

uniquely determined. Therefore, the system is invertible.

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88 Part I Introduction to signals and systems

(iii) Increasing ramped output:

y[k] = k x[k].

The input–output relationship is expressed as follows:

x[k] = 1

k y[k].

The input signal can be uniquely determined for all time instant k, except at

k = 0. Therefore, the system is not invertible. (iv) Summer:

y[k] = x[k] + x[k − 1].

Following the procedure used in Example 2.8(iv), the input signal is expressed

as an infinite sum of the output y[k] as follows:

x[k] = y[k] − y[k − 1] + y[k − 2] − y[k − 3] + − · · ·

= ∞∑

m=0

(−1)m y[k − m].

The input signal x[k] can be reconstructed if y[m] is known for all m ≤ k.

Therefore, the system is invertible.

(v) Accumulator:

y[k] = k∑

m=−∞

x[m].

We express the accumulator as follows:

y[k] = x[k] + k−1∑

m=−∞

x[m] = x[k] + y[k − 1]

x[k] = y[k] − y[k − 1].

Therefore, the system is invertible.

2.2.6 Stable and unstable systems

Before defining the stability criteria for a system, we define the bounded prop-

erty for a signal. A CT signal x(t) or a DT signal x[k] is said to be bounded in

magnitude if

CT signal |x(t)| ≤ Bx < ∞ for t ∈ (−∞, ∞); (2.48)

DT signal |x[k]| ≤ Bx < ∞ for k ∈ (−∞, ∞), (2.49)

where Bx is a finite number. Next, we define the stability criteria for CT and

DT systems.

A system is referred to as bounded-input, bounded-output (BIBO) stable if

an arbitrary bounded-input signal always produces a bounded-output signal. In

other words, if an input signal x(t) for CT systems, or x[k] for DT systems,

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89 2 Introduction to systems

satisfying either Eq. (2.48) or Eq. (2.49), is applied to a stable CT or DT system,

it is always possible to find an finite number By < ∞ such that

CT system |y(t)| ≤ By < ∞ for t ∈ (−∞, ∞); (2.50)

DT system |y[k]| ≤ By < ∞ for k ∈ (−∞, ∞). (2.51)

Example 2.10

Determine if the following CT systems are stable.

(i) Incrementally linear system:

y(t) = 50x(t) + 10. (2.52)

Assume |x(t)| ≤ Bx for all t . Based on Eq. (2.52), it follows that:

y(t) ≤ 50Bx + 10 = By for all t.

As the magnitude of y(t) does not exceed 50Bx + 10, which is a finite number,

the incrementally linear system given in Eq. (2.52) is a stable system.

(ii) Integrator:

y(t) =

t∫

−∞

x(τ )dτ . (2.53)

This system integrates the input signal from t = −∞ to t . Assume that a unit-

step function x(t) = u(t) is applied at the input of the integrator. The output of

the system is given by

y(t) = tu(t) =

{

0 t < 0

t t ≥ 0.

Signal y(t) is plotted in Fig. 2.18(b). It is observed that y(t) increases steadily

for t > 0 and that there is no upper bound of y(t). Hence, the integrator is not

a BIBO stable system.

x(t)

(a)

y(t)

(b)

Fig. 2.18. Input and output of

the unstable system in Example

2.10(ii). (a) Input x(t ) to the

system. (b) Output y(t ) of the

system. The input x(t ) is

bounded for all t , but the output

y(t ) is unbounded as t → ∞.

Example 2.11

Determine if the following DT systems are stable.

(i) y[k] = 50 sin(x[k]) + 10. (2.54)

Note that sin(x[k]) is bounded between [−1, 1] for any arbitrary choice of x[k].

The output y[k] is therefore bounded within the interval [−40, 60]. Therefore,

system (i) is stable.

(ii) y[k] = ex[k]. (2.55)

Assume |x[k]| ≤ Bx for all t . Based on Eq. (2.52), it follows that:

y[k] ≤ eBx = By for all k.

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90 Part I Introduction to signals and systems

Therefore, system (ii) is stable.

(iii) y[k] = 2∑

m=−2 x[k − m]. (2.56)

The output is expressed as follows:

y[k] = x[k − 2] + x[k − 1] + x[k] + x[k + 1] + x[k + 2].

If |x[k]| ≤ Bx for all k, then |y[k]| ≤ 5Bx for all k. Therefore, the system is stable.

(iv) y[k] = k∑

m=−∞

x[m]. (2.57)

The output is calculated by summing an infinite number of input signal values.

Hence, there is no guarantee that the output will be bounded even if all the input

values are bounded. System (iv) is, therefore, not a stable system.

2.3 Interconnection of systems

In signal processing, complex structures are formed by interconnecting simple

linear and time-invariant systems. In this section, we describe three widely used

configurations for developing complex systems.

2.3.1 Cascaded configuration

As shown in Fig. 2.19(a), a series or cascaded configuration between two sys-

tems is formed by interconnecting the output of the first system S1 to the input

of the second system S2. If the interconnected systems S1 and S2 are linear, it

is straightforward to show that the overall cascaded system is also linear. Like-

wise, if the two systems S1 and S2 are time-invariant, then the overall cascaded

system is also time-invariant. Another feature of the cascaded configuration

is that the order of the two systems S1 and S2 may be interchanged without

changing the output response of the overall system.

Example 2.12

Determine the relationship between the overall output and input signals if

the two cascaded systems in Fig. 2.19(a) are specified by the following

relationships:

(i) S1 : dw

dt + 2w(t) = x(t) with w(0) = 0

and

S2 : dy

dt + 3y(t) = w(t) with y(0) = 0;

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91 2 Introduction to systems

x(t) w(t)

y(t)S2S1

∑

x(t) y(t)

S2 y2(t)

y1(t) +

−

∑ S1x(t) y(t)

w(t)

(a)

(b) (c)

Fig. 2.19. Interconnection of

systems: (a) cascaded

configuration; (b) parallel

configuration; (c) feedback

configuration. Although these

diagrams are for CT systems, the

DT systems can be

interconnected to form the three

configurations in exactly the

same manner.

(ii) S1 : w[k] − w[k − 1] = x[k] with w[0] = 0 and

S2 : y[k] − 2y[k − 1] = w[k] with y[0] = 0.

Solution

(i) Differentiating both sides of the differential equation modeling system S2 with respect to t yields

S2 : d2 y

dt2 + 3

dt =

dt .

Multiplying the differential equation modeling system S2 by 2 and adding the

result to the above equation yields

d2 y

dt2 + 5

dt + 6y(t) =

dt + 2w(t)

︸︷︷︸

x(t)

Based on the differential equation modeling system S1, the right-hand side of

the equation equals x(t). The overall relationship of the cascaded system is,

therefore, given by

d2 y

dt2 + 5

dt + 6y(t) = x(t).

(ii) Substituting k = p − 1 in the difference equation modeling system S2 yields

S2 : y[p − 1] − 2y[p − 2] = w[p − 1], or, in terms of time index k,

S2 : y[k − 1] − 2y[k − 2] = w[k − 1]. Subtracting the above equation from the original difference equation modeling

system S2 yields

S2 : y[k] − 3y[k − 1] + 2y[k − 2] = w[k] − w[k − 1] ︸︷︷︸

x[k]

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92 Part I Introduction to signals and systems

Based on the difference equation modeling system S1, the right-hand side of

the above equation equals x[k]. The overall relationship of the cascaded system

is, therefore, given by

y[k] − 3y[k − 1] + 2y[k − 2] = x[k].

2.3.2 Parallel configuration

The parallel configuration is shown in Fig. 2.19(b), where a single input is

applied simultaneously to two systems S1 and S2. The overall output response

is obtained by adding the outputs of the individual systems. In other words, if

S1 : x(t) → y1(t) and S2 : x(t) → y2(t), then Sparallel : x(t) → y1(t) + y2(t).

As for the series configuration, the system formed by a parallel combination

of two linear systems is also linear. Similarly, if the two systems S1 and S2 are

time-invariant, then the overall parallel system is also time-invariant.

Example 2.13

Determine the relationship between the overall output and input signals if the

two parallel systems in Fig. 2.19(b) are specified by the following relationships:

(i) S1 : y1(t) = x(t) + dx

dt and S2 : y2(t) = x(t) + 3

dt + 5

d2x

dt2 ;

(ii) S1 : y1[k]= x[k] − x[k − 1] and S2 : y2[k]= x[k] − 2x[k − 1] − x[k − 2].

Solution

(i) The response of the overall system is obtained by adding the two differential

equations modeling the individual systems. The resulting expression is given

y1(t) + y2(t) = 2x(t) + 4 dx

dt + 5

d2x

dt2 .

Since y(t) = y1(t) + y2(t), the response of the overall system is given by

y(t) = 2x(t) + 4 dx

dt + 5

d2x

dt2 .

(ii) The response of the overall system is obtained by adding the two dif-

ference equations modeling the individual systems. The resulting expression is

given by

y1[k] + y2[k] = 2x[k] − 3x[k − 1] − x[k − 2].

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93 2 Introduction to systems

Since y[k] = y1[k] + y2[k], the response of the overall system is given by

y[k] = 2x[k] − 3x[k − 1] − x[k − 2].

2.3.3 Feedback configuration

The feedback configuration is shown in Fig. 2.19(c), where the output of system

S1 is fed back, processed by system S2, and then subtracted from the input signal.

Such systems are difficult to analyze in the time domain and will be considered

in Chapter 6 after the introduction of the Laplace transform.

2.4 Summary

In this chapter we presented an overview of CT and DT systems, classifying

the systems into several categories. A CT system is defined as a transformation

that operates on a CT input signal to produce a CT output signal. In contrast, a

DT system transforms a DT input signal into a DT output signal. In Section 2.1,

we presented several examples of systems used to abstract everyday physical

processes. Section 2.2 classified the systems into different categories: linear

versus non-linear systems; time-invariant versus variant systems; memoryless

versus dynamic systems; causal versus non-causal systems; invertible versus

non-invertible systems; and stable versus unstable systems. We classified the

systems based on the following definitions.

(1) A system is linear if it satisfies the principle of superposition.

(2) A system is time-invariant if a time-shift in the input signal leads to

an identical shift in the output signal without affecting the shape of the

output.

(3) A system is memoryless if its output at t = t0 depends only on the value of input at t = t0 and no other value of the input signal.

(4) A system is causal if its output at t = t0 depends on the values of the input signal in the past, t ≤ t0, and does not require any future value (t > t0) of

the input signal.

(5) A system is invertible if its input can be completely determined by observing

its output.

(6) A system is BIBO stable if all bounded inputs lead to bounded outputs.

An important subset of systems is described by those that are both linear and

time-invariant (LTI). By invoking the linearity and time-invariance properties,

such systems can be analyzed mathematically with relative ease compared with

non-linear systems. In Chapters 3–8, we will focus on linear time-invariant CT

(LTIC) systems and study the time-domain and frequency-domain techniques

used to analyze such systems. DT systems and the techniques used to analyze

them will be presented in Part III, i.e. Chapters 9–17.

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94 Part I Introduction to signals and systems

+ −

−

R1 node 1

R2 C y(t)v (t)

iR1(t)

iR2(t) iC(t)

+ −

− R L C y(t)i(t)

Fig. P2.1. RC circuit consisting of

two resistors (R 1 and R 2) and a

capacitor C .

Fig. P2.2. Resonator in an AM modulator.

m(t)

v1(t) Accos(2p fct)

RLv2(t) C

non-linear

device

Fig. P2.3. AM demodulator. The

input signal is represented by

v1(t ) = A c cos(2π fc t ) + m(t ), where A c cos(2π fc t ) is the

carrier and m(t ) is the

modulating signal.

Problems

2.1 The electrical circuit shown in Fig. P2.1 consists of two resistors R1 and R2 and a capacitor C .

(i) Determine the differential equation relating the input voltage Vin(t) to

the output voltage Vout(t).

(ii) Determine whether the system is (a) linear, (b) time-invariant;

2.2 The resonant circuit shown in Fig. P2.2 is generally used as a resonator in an amplitude modulation (AM) system.

(i) Determine the relationship between the input i(t) and the output v(t)

of the AM modulator.

(ii) Determine whether the system is (a) linear, (b) time-invariant;

2.3 Figure P2.3 shows the schematic of a square-law demodulator used in the demodulation of an AM signal. Demodulation is the process of extract-

ing the information-bearing signal from the modulated signal. The input–

output relationship of the non-linear device is approximated by (assuming

v1(t) is small)

v2(t) = c1v1(t) + c2v21 (t), where c1 and c2 are constants, and v1(t) and v2(t) are, respectively, the

input and output signals.

(i) Show that the demodulator is a non-linear device.

(ii) Determine whether the non-linear device is (a) time-invariant,

(b) memoryless, (c) invertible, and (d) stable.

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95 2 Introduction to systems

2.4 The amplitude modulation (AM) system covered in Section 2.1.3 is widely used in communications as in the AM band on radio tuner sets. Assume that

the sinusoidal tone m(t) = 2 sin(2π × 100t) is modulated by the carrier c(t) = 5 cos(2π × 106t).

(i) Determine the value of the modulation index k that will ensure (1 + km(t)) ≥ 0 for all t .

(ii) Derive the expression for the AM signal s(t) and express it in the form

of Eq. (2.10).

(iii) Using the following trigonometric relationship:

2 sin θ1 cos θ2 = sin(θ1 + θ2) + sin(θ1 − θ2),

show that the frequency of the sinusoidal tone is shifted to a higher

frequency range in the frequency domain.

2.5 Equation (2.16) describes a linear, second-order, constant-coefficient dif- ferential equation used to model a mechanical spring damper system.

(i) By expressing Eq. (2.16) in the following form:

d2 y

dt2 +

ωn

dt + ω2n y(t) =

M x(t),

determine the values of ωn and Q in terms of mass M , damping factor

r , and the spring constant k.

(ii) The variable ωn denotes the natural frequency of the spring damper

system. Show that the natural frequency ωn can be increased by

increasing the value of the spring constant k or by decreasing the

mass M .

(iii) Determine whether the system is (a) linear, (b) time-invariant,

2.6 The solution to the following linear, second-order, constant-coefficient dif- ferential equation:

d2 y

dt2 + 5

dt + 6y(t) = x(t) = 0,

with input signal x(t) = 0 and initial conditions y(0) = 3 and ẏ(0) = −7,

is given by

y(t) = [e−3t + 2e−2t ]u(t).

(i) By using the backward finite-difference scheme

∣ ∣ ∣ ∣ t=k�t

≈ y(k�t) − y((k − 1)�t)

�t

and

d2 y

dt2

∣ ∣ ∣ ∣ t=k�t

≈ y(k�t) − 2y((k − 1)�t) + y((k − 2)�t)

(�t)2

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96 Part I Introduction to signals and systems

show that the finite-difference representation of the differential equa-

tion is given by

(1 + 5�t + 6(�t)2)y[k] + (−2 − 5�t)y[k − 1] + y[k − 2] = 0.

(ii) Show that the ancillary conditions for the finite-difference scheme

are given by

y[0] = 3 and y[−1] = 3 + 7�t. (iii) By iteratively computing the finite-difference scheme for �t =

0.02 s, show that the computed result from the finite-difference equa-

tion is the same as the result of the differential equation.

2.7 Assume that the delta modulation scheme, presented in Section 2.1.7, uses the following design parameters:

sampling period T = 0.1 s and quantile interval � = 0.1 V. Sketch the output of the receiver for the following binary signal:

11111011111100000000.

Assume that the initial value x(0) of the transmitted signal x(t) at t = 0 is x(0) = 0 V.

2.8 Determine if the digital filter specified in Eq. (2.27) is an invertible system. If yes, derive the difference equation modeling the inverse system. If no,

explain why.

2.9 The following CT systems are described using their input–output relation- ships between input x(t) and output y(t). Determine if the CT systems are

(a) linear, (b) time-invariant, (c) stable, and (d) causal. For the non-linear

systems, determine if they are incrementally linear systems.

(i) y(t) = x(t − 2); (ii) y(t) = x(2t − 5);

(iii) y(t) = x(2t) − 5; (iv) y(t) = t x(t + 10);

(v) y(t) = {

2 x(t) ≥ 0

0 x(t) < 0;

(vi) y(t) =

{

0 t < 0

x(t) − x(t − 5) t ≥ 0;

(vii) y(t) = 7x2(t) + 5x(t) + 3;

(viii) y(t) = sgn(x(t));

(ix) y(t) =

t0∫

−t0

x(λ)dλ + 2x(t);

(x) y(t) =

t0∫

−∞

x(λ)dλ + dx

dt ;

(xi) d4 y

dt4 + 3

d3 y

dt3 + 5

d2 y

dt2 + 3

dt + y(t) =

d2x

dt2 + 2x(t) + 1.

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97 2 Introduction to systems

2.10 The following DT systems are described using their input–output relation- ships between input x[k] and output y[k]. Determine if the DT systems are

(a) linear, (b) time-invariant, (c) stable, and (d) causal. For the non-linear

systems, determine if they are incrementally linear systems.

y(t)

1−1

Fig. P2.11. CT output y(t ) for

Problem 2.11.

(i) y[k] = ax[k] + b; (ii) y[k] = 5x[3k − 2];

(iii) y[k] = 2x[k];

(iv) y[k] = k∑

m=−∞

x[m];

(v) y[k] = k+2∑

m=k−2

x[m] − 2|x[k]|;

(vi) y[k] + 5y[k − 1] + 9y[k − 2] + 5y[k − 3] + y[k − 4]

= 2x[k] + 4x[k − 1] + 2x[k − 2].

(vii) y[k] = 0.5x[6k − 2] + 0.5x[6k + 2].

2.11 For an LTIC system, an input x(t) produces an output y(t) as shown in Fig. P2.11. Sketch the outputs for the following set of inputs:

(i) 5x(t);

(ii) 0.5x(t − 1) + 0.5x(t + 1);

(iii) x(t + 1) − x(t − 1);

(iv) dx(t)

dt + 3x(t).

2.12 For a DT linear, time-invariant system, an input x[k] produces an output y[k] as shown in Fig. P2.12. Sketch the outputs for the following set of

inputs:

(i) 4x[k − 1];

(ii) 0.5x[k − 2] + 0.5x[k + 2];

(iii) x[k + 1] − 2x[k] + x[k − 1];

(iv) x[−k].

y[k]

1 2

−1

−2

Fig. P2.12. DT output y [k ] for

Problem 2.12.

2.13 Determine if the following CT systems are invertible. If yes, find the inverse systems.

(i) y(t) = 3x(t + 2);

(ii) y(t) =

t∫

−∞

x(τ − 10)dτ ;

(iii) y(t) = |x(t)|;

(iv) dy(t)

dt + y(t) = x(t);

(v) y(t) = cos(2πx(t)).

2.14 Determine if the following DT systems are invertible. If yes, find the inverse systems.

(i) y[k] = (k + 1)x[k + 2];

(ii) y[k] =

|k|∑

m=0

x[m + 2];

(iii) y[k] = x[k] ∞∑

m=−∞

δ[k − 2m];

(iv) y[k] = x[k + 2] + 2x[k + 1] − 6x[k] + 2x[k − 1] + x[k − 2];

(v) y[k] + 2y[k − 1] + y[k − 2] = x[k].

2.15 For an LTIC system, if x(t) → y(t), show that dx(t)

dt →

dy(t)

dt . Assume

that both x(t) and y(t) are differentiable functions.

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98 Part I Introduction to signals and systems

x(t)

0.5−0.5

y(t)

1−1

xp(t)

0.5−0.5−1.5−2.5 2.51.5

(a)

(b)

Fig. P2.16. (a) Input–output

pair for an LTI CT system.

(b) Periodic input to the LTI

system.

2.16 Figure P2.16(a) shows an input–output pair of an LTI CT system. Calcu- late the output yp(t) of the system for the periodic signal xp(t) shown in

Fig. P2.16(b).

2.17 The output h(t) of a CT LTI system in response to a unit impulse function δ(t) is referred to as the impulse response of the system. Calculate the

impulse response of the CT LTI systems defined by the following input–

output relationships:

(i) y(t) = x(t + 2) − 2x(t) + 2x(t − 2);

(ii) y(t) = t+t0∫

t−t0

x(τ − 4) dτ ;

(iii) y(t) = t∫

−∞

e−2(t−τ )x(τ − 4) dτ ;

(iv) y(t) =

∞∫

−∞

f (T − τ )x(t − τ ) dτ where f (t) is a known signal and

T is a constant.

2.18 The output h[k] of a DT LTI system in response to a unit impulse function δ[k] is shown in Fig. P2.18. Find the output for the following set of inputs:

(i) x[k] = δ[k + 1] + δ[k] + δ[k − 1];

(ii) x[k] = ∞∑

m=−∞

δ[k − 4m];

(iii) x[k] = u[k].

h[k]

1−1

−2

Fig. P2.18. Output h[k ] for

input x[k ] = δ[k ] in Problem

2.18.

2.19 A DT LTI system is described by the following difference equation:

y[k] = x[k] − 2x[k − 1] + x[k − 2].

Determine the output y[k] of the system if the input x[k] is given by

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99 2 Introduction to systems

x[k] S1 S2 y[k] x[k]

+ y[k]

(a) (b)

Fig. P2.21. (a) Series

configuration; (b) parallel

configuration.

(i) x[k] = δ[k]; (ii) x[k] = δ[k − 1] + δ[k + 1];

(iii) x[k] = {

|k| |k| ≤ 3 0 elsewhere.

2.20 A five-point running average DT system is defined by the following input– output relationship:

y[k] = 1

4∑

m=0

x[k − m].

(i) Show that the five-point running average DT system is an LTI system.

(ii) Calculate the impulse response h[k] of the system when input x[k] =

δ[k].

(iii) Compute the output y[k] of the system for −10 ≤ k ≤ 10 if the input

x[k] = u[k], where u[k] is a unit step function.

(iv) Based on your answer to (iii), calculate the impulse response h[k]

of the system using the property δ[k] = u[k] – u[k − 1]. Compare

your answer to h[k] obtained in (ii).

2.21 The series and parallel configurations of systems S1 and S2 are shown in Fig. P2.21. The two systems are specified by the following input–output

relationships:

S1 : y[k] = x[k] − 2x[k − 1] + x[k − 2];

S2 : y[k] = x[k] + x[k − 1] − 2x[k − 2].

(i) Show that S1 and S2 are LTI systems.

(ii) Calculate the input–output relationship for the series configuration

of systems S1 and S2 as shown in Fig. P2.21(a).

(iii) Calculate the input–output relationship for the parallel configuration

of systems S1 and S2 as shown in Fig. P2.21(b).

(iv) Show that the series and parallel configurations of systems S1 and

S2 are LTI systems.

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P A R T I I

Continuous-time signals and systems

101

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C H A P T E R

3 Time-domain analysis of LTIC systems

In Chapter 2, we introduced CT systems and discussed a number of basic prop-

erties used to classify such systems. An important subset of CT systems satisfies

both the linearity and time-invariance properties. Such CT systems are referred

to as linear, time-invariant, continuous-time (LTIC) systems. In this chapter,

we will develop techniques for analyzing LTIC systems. Given an input–output

representation for the system under consideration, we are primarily interested

in calculating the output y(t) of the LTIC system from the applied input x(t).

The output y(t) of an LTIC system can be evaluated analytically in the time

domain in several ways. In Section 3.1, we use a linear constant-coefficient

differential equation to model an LTIC system. In such cases, the output y(t) is

obtained by directly solving the differential equation. In Sections 3.2 and 3.3, we

define the unit impulse response h(t) as the output of an LTIC system to an unit

impulse function δ(t) applied at the input. This development leads to a second

approach for calculating the output y(t) based on convolving the applied input

x(t) with the impulse response h(t). The resulting integral is referred to as the

convolution integral and is discussed in Sections 3.4 and 3.5. The properties of

the convolution integral are covered in Section 3.6. The impulse response h(t)

provides a complete description for an LTIC system. In Sections 3.7 and 3.8,

we express the properties of an LTIC system in terms of its impulse response.

The chapter is concluded in Section 3.9.

3.1 Representation of LTIC systems

For a linear CT system, the relationship between the applied input x(t) and

output y(t) can be described using a linear differential equation of the following

form:

dn y

dtn + an−1

dn−1 y

dtn−1 + · · · + a1

dt + a0 y(t)

= bm dm x

dtm + bm−1

dm−1x

dtm−1 + · · · + b1

dt + b0x(t), (3.1)

103

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104 Part II Continuous-time signals and systems

x(t) i (t)

− w(t)

− y(t)

v(t)+ − Fig. 3.1. Series RLC circuit used

in Example 3.1.

where coefficients ak , for 0 ≤ k ≤ (n− 1), and bk , for 0 ≤ k ≤ m, are parameters characterized by the linear system. If the linear system is also time-invariant,

then the ak and bk coefficients are constants. We will use the compact notation

ẏ to denote the first derivative of y(t) with respect to t . Thus ẏ = dy/dt , ÿ = d2 y/dt2, and so on for the higher derivatives. We now consider an electrical

circuit that is modeled by a differential equation.

Example 3.1

Determine the input–output representations of the series RLC circuit shown in

Fig. 3.1 for the three outputs v(t), w(t), and y(t).

Solution

Figure 3.1 illustrates an electrical circuit consisting of three passive compo-

nents: resistor R, inductor L , and capacitor C . Applying Kirchhoff’s voltage

law, the relationship between the input voltage x(t) and the loop current i(t) is

given by

x(t) = L di

dt + Ri(t) +

t∫

−∞

i(t)dt . (3.2)

Differentiating Eq. (3.2) with respect to t yields

L d2i

dt2 + R

dt +

C i(t) =

dt . (3.3)

We consider three different outputs of the RLC circuit in the following dis-

cussion, and for each output we derive the differential equation modeling the

input–output relationship of the LTIC system.

Relationship between x(t) and v(t) The output voltage v(t) is measured across inductor L . Expressed in terms of the loop current i(t), the voltage v(t) is given

v(t) = L di

dt .

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105 3 Time-domain analysis of LTIC systems

Integrating the above equation with respect to t yields

i(t) = 1

∫

v(t)dt.

By substituting the value of i(t) into Eq. (3.3), we obtain

dt +

L v(t) +

∫

v(t)dt = dx

dt .

The above input–output relationship includes both differentiation and integra-

tion operations. The integral operator can be eliminated by calculating the

derivative of both sides of the equation with respect to t . This results in the

following equation:

d2v

dt2 +

dt +

LC v(t) =

d2x

dt2 , (3.4)

which models the input–output relationship between the input voltage x(t) and

the output voltage v(t) measured across inductor L . Equation (3.4) is a linear,

second-order differential equation with constant coefficients. In fact, it can

be shown that an LTIC system can always be modeled by a linear, constant-

coefficient differential equation with the appropriate initial conditions.

Relationship between x(t) and w(t) The output voltage w(t), measured across capacitor C, is given by

w(t) = 1

t∫

−∞

i(t)dt,

which is expressed as follows:

i(t) = C dw

dt .

Substituting the value of i(t) into Eq. (3.3) yields

LC d3w

dt3 + RC

d2w

dt2 +

dt =

dt , (3.5)

which specifies the relationship between the input voltage x(t) and the output

voltage w(t) measured across capacitor C . Equation (3.5) can be further sim-

plified by integrating both sides with respect to t . The resulting equation is

simplified to

LC d2w

dt2 + RC

dt + w(t) = x(t), (3.6)

which is a linear, second-order, constant-coefficient differential equation.

Relationship between x(t) and y(t) Finally, we measure the output voltage y(t) across resistor R. Using Ohm’s law, the output voltage y(t) is given by

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106 Part II Continuous-time signals and systems

y(t) = i(t)R. Substituting the value of i(t) = y(t)/R into Eq. (3.3) yields

d2 y

dt2 +

dt +

RC y(t) =

dt , (3.7)

which is a linear, second-order, constant-coefficient, differential equation mod-

eling the relationship between the input voltage x(t) and the output voltage y(t)

measured across resistor R.

A more compact representation for Eq. (3.1) is obtained by denoting the differ-

entiation operator d/dt by D:

Dn y + an−1Dn−1 y + · · · + a1Dy + a0 y(t) = bmDm y + bm−1Dm−1 y + · · · + b1Dy + b0x(t).

By treating D as a differential operator, we obtain

(Dn + an−1Dn−1 + · · · + a1D + a0) ︸︷︷︸

Q(D)

y(t)

= (bmDm + bm−1Dm−1 + · · · + b1D + b0) ︸︷︷︸

P(D)

x(t), (3.8)

Q(D)y(t) = P(Q)x(t), (3.9)

where Q(D) is the nth-order differential operator, P(D) is the mth-order differen-

tial operator, and the ai and bi are constants. Equation (3.9) is used extensively

to describe an LTIC system.

To compute the output of an LTIC system for a given input, we must solve the

constant-coefficient differential equation, Eq. (3.9). If the reader has little or no

background in differential equations, it will be helpful to read through Appendix

C before continuing. Appendix C reviews the direct method for solving linear,

constant-coefficient differential equations and can be used as a quick look-up

of the theory of differential equations. In the material that follows, it is assumed

that the reader has adequate background in solving linear, constant-coefficient

differential equations.

From the theory of differential equations, we know that output y(t) for

Eq. (3.9) can be expressed as a sum of two components:

y(t) = yzi(t) ︸︷︷︸

zero-input response

+ yzs(t) ︸︷︷︸

zero-state response

, (3.10)

where yzi(t) is the zero-input response of the system and yzs(t) is the zero-

state response of the system. Note that the zero-input component yzi(t) is the

response produced by the system because of the initial conditions (and not due

to any external input), and hence yzi(t) is also known as the natural response

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107 3 Time-domain analysis of LTIC systems

of the system. For example, the initial conditions may include charges stored

in a capacitor or energy stored in a mechanical spring. The zero-input response

yzi(t) is evaluated by solving a homogeneous equation obtained by setting the

input signal x(t) = 0 in Eq. (3.9). For Eq. (3.9), the homogeneous equation is given by

Q(D)y(t) = 0.

The zero-state response yzs(t) arises due to the input signal and does not depend

on the initial conditions of the system. In calculating the zero-state response,

the initial conditions of the system are assumed to be zero. The zero-state

response is also referred to as the forced response of the system since the zero-

state response is forced by the input signal. For most stable LTIC systems, the

zero-input response decays to zero as t → ∞ since the energy stored in the system decays over time and eventually becomes zero. The zero-state response,

therefore, defines the steady state value of the output.

Example 3.2

Consider the RLC series circuit shown in Fig. 3.1. Assume that the inductance

L = 0 H (i.e. the inductor does not exist in the circuit), resistance R = 5 �, and capacitance C = 1/20 F. Determine the output signal y(t) when the input voltage is given by x(t) = sin(2t) and the initial voltage y(0−) = 2 V across the resistor.

Solution

Substituting L = 0, R = 5, and C = 1/20 in Eq. (3.7) yields dy

dt + 4y(t) =

dt = 2 cos(2t). (3.11)

Zero-input response of the system Using the procedure outlined in Appendix C, we determine the characteristic equation for Eq. (3.11) as

(s + 4) = 0,

which has a root at s = −4. The zero-input response of Eq. (3.11) is given by

zero input response yzi(t) = Ae−4t ,

where A is a constant. The value of A is obtained from the initial condition

y(0−) = 2 V. Substituting y(0−) = 2 V in the above equation yields A = 2. The zero-input response is given by yzi(t) = 2e−4t .

Zero-state response of the system The zero-state response is calculated by solving Eq. (3.11) with a zero initial condition, y(0−) = 0. The homogeneous component of the zero-state response of Eq. (3.11) is similar to the zero input

response and is given by

y(h)zs (t) = Ce −4t ,

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108 Part II Continuous-time signals and systems

where C is a constant. The particular component of the zero-state response of

Eq. (3.11) for input x(t) = sin(2t) is of the following form:

y(p)zs (t) = K1 cos(2t) + K2 sin(2t).

Substituting the particular component in Eq. (3.11) gives K1 = 0.4 and K2 = 0.2. The overall zero-state response of the system is as follows:

zero state response yzs(t) = Ce−4t + 0.2 sin(2t) + 0.4 cos(2t),

with zero initial condition, i.e. yzs(t) = 0. Substituting the initial condition in the zero-state response yields C = −0.4. The total response of the system is the sum of the zero-input and zero-state responses and is given by

y(t) = 1.6e−4t + 0.2 sin(2t) + 0.4 cos(2t). (3.12)

Theorem 3.1 states the total response of a LTIC system modeled with a first-

order, constant-coefficient, linear differential equation.

Theorem 3.1 The output of a first-order differential equation,

dt + f (t)y(t) = r (t), (3.13)

resulting from input r(t) is given by

y(t) = e−p [∫

epr dt + c ]

, (3.14)

where function p is given by

p(t) = ∫

f (t)dt (3.15)

and c is a constant.

Using Theorem 3.1 to solve Eq. (3.11), we obtain p(t) = ∫ 4 dt = 4t . Substi- tuting p(t) = 4t into Eq. (3.14), we obtain

y(t) = e−4t [∫

e4t 2 cos(2t)dt + c ]

where the integral simplifies to (see Section A.5 of Appendix A)

∫

e4t cos(2t)dt = 2

22 + 42 [4e4t cos(2t) + 2e4t sin(2t)].

Based on Theorem 3.1, the output is therefore given by

y(t) = ce−4t + 0.2 sin(2t) + 0.4 cos(2t).

The value of constant c in the above equation can be computed using the initial

condition. Substituting y(0−) = 2 V gives c = 1.6. The result is, therefore, the same as the solution in Eq. (3.12) obtained by following the formal procedure

outlined in Appendix C.

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109 3 Time-domain analysis of LTIC systems

Steady state value of the output The steady state value of y(t) can be obtained by applying the limit (t → ∞) to y(t). For the differential equation (3.11), the steady state solution is therefore obtained by applying the limit to Eq. (3.12),

giving

y(t) = lim t→∞

[1.6e−4t + 0.2 sin(2t) + 0.4 cos(2t)] = 0.2 sin(2t) + 0.4 cos(2t),

y(t) = √

0.42 +0.22 sin (

2t + tan−1 (

0.4

0.2

))

= √

0.42 + 0.22 sin(2t + 63.4◦)

(3.16)

The steady state solution given by Eq. (3.16) can also be verified using results

from the circuit theory. For sinusoidal inputs, the electrical circuit in Fig.

3.1 can be reduced to an equivalent impedance circuit by replacing capaci-

tor C with a capacitive reactance of 1/(jωC) and inductor L with an induc-

tive reactance of jωL , where ω is the fundamental frequency of the input

sinusoidal signal x(t) = sin(2t). In our example, ω = 2. Figure 3.1, therefore, becomes a voltage divider circuit with the steady state value of the output y(t)

given by

y(t) = R

R + jωL + (1/jωC) x(t). (3.17)

In Example 3.2, the values of the components are set to L = 0 H, R = 5 �, and C = 1/20 F. Substituting these values into Eq. (3.17) yields

y(t) = 5

5 + (10/j) x(t) =

1 − j2 sin(2t) =

∣ ∣ ∣ ∣

1 − j2

∣ ∣ ∣ ∣ sin(2t − � (1 − j2))

= 1

√ 5

sin (

2t + tan−1(2) )

= √

0.2 sin(2t + 63.4◦),

which is the same solution as given in Eq. (3.16).

Example 3.3

Consider the electrical circuit shown in Fig. 3.1 with the values of inductance,

resistance, and capacitance set to L = 1/12 H, R = 7/12 �, and C = 1 F. The circuit is assumed to be open before t = 0, i.e. no current is initially flow- ing through the circuit. However, the capacitor has an initial charge of 5 V.

Determine

(i) the zero-input response wzi(t) of the system;

(ii) the zero-state response wzs(t) of the system; and

(iii) the overall output w(t),

when the input signal is given by x(t) = 2 exp(−t)u(t) and the output w(t) is measured across capacitor C .

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110 Part II Continuous-time signals and systems

Solution

Substituting L = 1/12 H, R = 7/12 �, and C = 1 F into Eq. (3.6) and multi- plying both sides of the equation by 12 yields

d2w

dt2 + 7

dt + 12w(t) = 12x(t), (3.18)

with initial conditions, w(0−) = 5 and ẇ(0−) = 0, and the input signal is given by x(t) = 2e−t u(t).

(i) Zero-input response of the system Based on Eq. (3.18), the characteristic equation of the LTIC system is given by

s2 + 7s + 12 = 0,

which has roots at s = −4, −3. The zero-input response is therefore given by

wzi(t) = (Ae−4t + Be−3t )u(t),

where A and B are constants. To calculate the value of the constants, we sub-

stitute the initial conditions w(0−) = 5 and ẇ(0−) = 0 in the above equation. The resulting simultaneous equations are as follows:

A + B = 5, 4A + 3B = 0,

which have the solution A = −15 and B = 20. The zero-input response is therefore given by

wzi(t) = (20e−3t − 15e−4t )u(t).

(ii) Zero-state response of the system To calculate the zero-state response of the system, the initial conditions are assumed to be zero, i.e. the capaci-

tor is assumed to be uncharged. Hence, the zero-state response wzs(t) can be

calculated by solving the following differential equation:

d2w

dt2 + 7

dt + 12w(t) = 12x(t), (3.19)

with initial conditions, w(0−) = 0 and ẇ(0−) = 0, and input x(t) = 2 exp(−t)u(t).

The homogeneous solution of Eq. (3.18) has the same form as the zero-input

response and is given by

w (h)zs (t) = C1e −4t + C2e−3t ,

where C1 and C2 are constants. The particular solution for input x(t) = 2e−t u(t) is of the form w

(p) zs (t) = K e−t u(t). Substituting the particular solution into

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111 3 Time-domain analysis of LTIC systems

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 −1

w(t)

wzi(t)

wzs(t)

Fig. 3.2. Output response of the

system considered in Example

3.3.

Eq. (3.19) and solving the resulting equation yields K = 4. The zero-state response of the system is, therefore, given by

wzs(t) = (C1e−4t + C2e−3t + 4e−t )u(t).

To compute the values of constants C1 and C2, we use the initial conditions

w(0−) = 0 and ẇ(0−) = 0. Substituting the initial conditions in wzs(t) leads to the following simultaneous equations:

C1 + C2 + 4 = 0, −4C1 − 3C2 − 4 = 0,

with solutions C1 = 8 and C2 = −12. The zero-state solution of Eq. (3.18) is, therefore, given by

wzs(t) = (8e−4t − 12e−3t + 4e−t )u(t).

(iii) Overall response of the system The overall response of the system can be obtained by summing up the zero-input and zero-state responses, and can be

expressed as

w(t) = (−7e−4t + 8e−3t + 4e−t )u(t).

The zero-input, zero-state, and overall responses of the system are plotted in

Fig. 3.2.

Section 3.1 presented the procedure for calculating the output response of a

LTIC system by directly solving its input–output relationship expressed in the

form of a differential equation. However, there is an alternative and more con-

venient approach to calculate the output based on the impulse response of a

system. This approach is developed in the following sections.

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112 Part II Continuous-time signals and systems

∆0

1/∆

d∆(t)

−5∆ −3∆ −∆ 3∆∆0 5∆ 7∆ t

x(t)

(a) (b)

Fig. 3.3. Approximation of a CT signal x (t ) by a linear combination of time-shifted unit impulse functions.

(a) Rectangular function δ�(t ) used to approximate x(t ). (b) CT signal x(t ) and its approximation x̂(t )

shown with the staircase function.

3.2 Representation of signals using Dirac delta functions

In this section we will show that any arbitrary signal x(t) can be represented as

a linear combination of time-shifted impulse functions. To illustrate our result,

we define a new function δ�(t) as follows:

δ�(t) = {

1/� 0 < t < �

0 otherwise. (3.20)

The waveform for δ�(t) is shown in Fig. 3.3(a); it resembles that of a rectangular

pulse with width � and height 1/�. To approximate x(t) as a linear combination

of δ�(t), the time axis is divided into uniform intervals of duration �. Within a

time interval of duration �, say k� < t < (k + 1)�, x(t) is approximated by a constant value x(k�)δ�(t − k�)�. Following the aforementioned procedure for the entire time axis, x(t) can be approximated as follows:

x̂(t) = · · · + x(−k�)δ�(t + k�) · � + · · · + x(−�)δ�(t + �) · � + x(0)δ�(t) · � + x(�)δ�(t − �) · � + · · · + x(k�)δ�(t − k�) · � + · · · , (3.21)

which is shown as the staircase waveform in Fig. 3.3(b). For a given value of t ,

say t = m�, only one term (k = m) on the right-hand side of Eq. (3.21) is non- zero. This is because only one of the shifted functions δ�(t − k�) corresponding to k = m is non-zero. Therefore, a more compact representation for Eq. (3.21) is obtained by using the following summation:

x̂(t) = ∞∑

k=−∞ x(k�)δ�(t − k�)�. (3.22)

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113 3 Time-domain analysis of LTIC systems

Applying the limit � → 0, x̂(t) converges to x(t), giving

x(t) = lim �→0

∞∑

K=−∞ x(k�)δ�(t − k�) [(k + 1)� − k�] , (3.23)

which is the same as

x(t) = ∞∫

−∞

x(τ )δ(t − τ )dτ. (3.24)

Equation (3.24) is very important in the analysis of CT signals. It suggests that

a CT function can be represented as a weighted superposition of time-shifted

impulse functions. We will use Eq. (3.24) to calculate the output of an LTIC

system.

The above procedure used to prove Eq. (3.24) illustrates the physical sig-

nificance of the equation. A more compact proof of Eq. (3.24), based on the

properties of the impulse function, is presented below.

Alternative proof for Eq. (3.24)

In the following discussion, we present a simpler proof of Eq. (3.24), which

uses the properties of impulse functions. We start with the right-hand side of

Eq. (3.24):

RHS = ∞∫

−∞

x(τ )δ(t − τ )dτ.

Since δ(t – τ ) = δ(τ – t),

RHS = ∞∫

−∞

x(τ )δ(τ − t)dτ .

Also, x(τ )δ(τ – t) = x(t)δ(τ – t); therefore

RHS = x(t) ∞∫

−∞

δ(τ − t)dτ ,

which equals x(t), as the area enclosed by the unit impulse function equals

unity.

3.3 Impulse response of a system

In Section 3.1, a constant-coefficient differential equation is used to specify the

input–output characteristics of an LTIC system. An alternative representation

of an LTIC system can be obtained by specifying its impulse response. In this

section, we will formally define the impulse response and illustrate how the

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114 Part II Continuous-time signals and systems

impulse response of an LTIC system can be derived directly from the differential

equation modeling the LTIC system.

Definition 3.1 The impulse response h(t) of an LTIC system is the output of the

system when a unit impulse δ(t) is applied at the input. Following the notation

introduced in Eq. (2.1), the impulse response can be expressed as

δ(t) → h(t) (3.25)

with zero initial conditions. Because the system is LTIC, it satisfies the linearity

and the time-shifting properties. If the input is a scaled and time-shifted impulse

function aδ(t − t0), the output, Eq. (3.25), of the system is also scaled by the factor of a and is time-shifted by t0, i.e.

aδ(t − t0) → ah(t − t0) (3.26)

for any arbitrary constants a and t0.

Example 3.4

Calculate the impulse response of the following systems:

(i) y(t) = x(t − 1) + 2x(t − 3); (3.27)

(ii) dy

dt + 4y(t) = 2x(t). (3.28)

Solution

(i) The impulse response of a system is the output of the system when the input

signal x(t) = δ(t). Therefore, the impulse response h(t) can be obtained by substituting y(t) by h(t) and x(t) by δ(t) in Eq. (3.27). In other words,

h(t) = δ(t − 1) + 2δ(t − 3).

(ii) For input x(t) = δ(t), the resulting output y(t) = h(t). The impulse response h(t) can therefore be obtained by solving the following differential

equation:

dt + 4h(t) = 2δ(t) (3.29)

obtained by substituting x(t) = δ(t) and y(t) = h(t) in Eq. (3.28). We will use Theorem 3.1 to compute the solution of Eq. (3.29). From Eq. (3.14), p(t) is

given by

p(t) = ∫

4 dt = 4t,

which is substituted into Eq. (3.15), giving

h(t) = e−4t [

∫

e4tδ(t)dt + c ]

= 2e−4t u(t) + ce−4t , (3.30)

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115 3 Time-domain analysis of LTIC systems

−2 −1 0 1 2 3 4 5 6 7 8 9 0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

(a) (b)

−2 −1 0 1 2 3 4 5 6 7 8 9 0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

Fig. 3.4. (a) Impulse response

h(t ) of the LTIC system specified

in Example 3.5. (b) Output y(t )

of the LTIC system for input

x(t ) = δ(t + 1) + 3δ(t − 2) + 2δ(t − 6) .

where constant c is determined from the zero initial condition. Substituting

h(t) = 0 for t = 0−, in Eq. (3.30) gives c = 0. The impulse response of the system in Eq. (3.28) is therefore given by h(t) = 2 exp(−4t)u(t).

Example 3.5

The impulse response of an LTIC system is given by h(t) = exp(−3t)u(t). Determine the output of the system for the input signal x(t) = δ(t + 1) + 3δ(t − 2) + 2δ(t − 6).

Solution

Because the system is LTIC, it satisfies the linearity and time-shifting properties.

Therefore,

δ(t + 1) → h(t + 1), 3δ(t − 2) → 3h(t − 2),

and

2δ(t − 6) → 2h(t − 6).

Applying the superposition principle, we obtain

x(t) → y(t) = h(t + 1) + 3h(t − 2) + 2h(t − 6).

The impulse response h(t) is shown in Fig. 3.4(a) with the resulting output

shown in Fig. 3.4(b).

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116 Part II Continuous-time signals and systems

3.4 Convolution integral

In Section 3.3, we computed the output y(t) of an LTIC system from its impulse

response h(t) when the input signal x(t) can be represented as a linear combi-

nation of scaled and time-shifted impulse functions. In this section, we extend

the technique to general input signals.

Following the procedure of Section 3.2, an arbitrary CT signal x(t) can be

approximated by the staircase approximation illustrated in Fig. 3.3. In terms of

Eq. (3.23), the approximated function x̂(t) is given by

x̂(t) = ∞∑

k=−∞ x(k�)δ�(t − k�)�.

Note that as � → 0, the approximated Dirac delta function δ�(t – k�) approaches δ(t – k�). Therefore,

lim �→0

δ�(t − k�) → lim �→0

h(t − k�).

Multiplying both sides by x(k�)�, we obtain

lim �→0

x(k�) δ�(t − k�) × � → lim �→0

x(k�)h(t − k�) × �. (3.31)

Applying the linearity property of the system yields

lim �→0

∞∑

k=−∞ x(k�) δ�(t − k�)� → lim

�→0

∞∑

k=−∞ x(k�)h(t − k�)�. (3.32)

As � → 0, the summations on both sides of Eq. (3.32) become integrations. Substituting k� by τ and � by dτ , we obtain the following relationship:

∞∫

−∞

x(τ )δ(t − τ )dτ → ∞∫

−∞

x(τ )h(t − τ )dτ , (3.33)

x (t) → ∞∫

−∞

x(τ )h(t − τ )dτ , (3.34)

where τ is the dummy variable that disappears as the integration with limits

is computed. The integral on the left-hand side of Eq. (3.34) is referred to

as the convolution integral and is denoted by x(t) ∗ h(t). Mathematically, the

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117 3 Time-domain analysis of LTIC systems

h(t)

LTIC system

∫ ∞

−∞ x(t)d(t − t)dt x(t)h(t − t)dt

x(t) =

∫ ∞

−∞

y(t) = x(t)∗h(t) =

Fig. 3.5. Output response of a

system to a general input x(t ).

convolution of two functions x(t) and h(t) is defined as follows:

x (t) ∗ h (t) = ∞∫

−∞

x(τ )h(t − τ )dτ . (3.35)

Combining Eqs. (3.34) and (3.35), we obtain the following:

x(t) → x(t) ∗ h(t) = ∞∫

−∞

x(τ ) h(t − τ )dτ . (3.36)

Equation (3.36) is illustrated in Fig. 3.5 and can be reiterated as follows. When

an input signal x(t) is passed through an LTIC system with impulse response

h(t), the resulting output y(t) of the system can be calculated by convolving

the input signal and the impulse response.

We now consider several examples of computing the convolution integral.

Example 3.6

Determine the output response of an LTIC system when the input signal is given

by x(t) = exp(−t)u(t) and the impulse response is h(t) = exp(−2t)u(t).

Solution

Using Eq. (3.36), the output y(t) of the LTIC system is given by

y(t) = ∞∫

−∞

e−τ u(τ ) e−2(t−τ )u(t − τ )dτ ,

which can be expressed as

y(t) = e−2t ∞∫

eτ u(t − τ )dτ .

Expressed as a function of the independent variable τ , the unit step function is

given by

u(t − τ ) = {

1 τ ≤ t 0 τ > t.

Based on the value of t , we have the following two cases for the output y(t).

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118 Part II Continuous-time signals and systems

−2 −1 0 1 2 3 4 5 6 7 8 9 0

0.1

0.2

0.3

0.4

Fig. 3.6. The output of a LTIC

system with impulse response

h(t ) = exp(−2t )u(t ) resulting from the input signal x(t ) = exp(−t )u(t ) as calculated in Example 3.6.

Case I For t < 0, the shifted unit step function u(t − τ ) = 0 within the limits of integration [0, ∞]. Therefore, y(t) = 0 for t < 0.

Case II For t ≥ 0, the shifted unit step function u(t − τ ) has two different values within the limits of integration [0, ∞]. For the range [0, t], the unit step function u(t − τ ) = 1. Otherwise, for the range [t , ∞], the unit step function is zero. The output y(t) is therefore given by

y(t) = e−2t t∫

eτ dτ = e−2t [

et − 1 ]

= e−t − e−2t , for t > 0.

Combining cases I and II, the overall output y(t) is given by

y(t) = (e−t − e−2t )u(t).

The output response of the system is plotted in Fig. 3.6.

Example 3.6 shows us how to calculate the convolution integral analytically. In

many practical situations, it is more convenient to use a graphical approach to

evaluate the convolution integral, and we consider this next.

3.5 Graphical method for evaluating the convolution integral

Given input x(t) and impulse response h(t) of the LTIC system, Eq. (3.36) can

be evaluated graphically by following steps (1) to (7) listed in Box 3.1.

Box 3.1 Steps for graphical convolution

(1) Sketch the waveform for input x(τ ) by changing the independent vari-

able from t to τ and keep the waveform for x(τ ) fixed during convolution.

(2) Sketch the waveform for the impulse response h(τ ) by changing the

independent variable from t to τ .

(3) Reflect h(τ ) about the vertical axis to obtain the time-inverted impulse

response h(−τ ).

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119 3 Time-domain analysis of LTIC systems

(4) Shift the time-inverted impulse function h(−τ ) by a selected value of “t .” The resulting function represents h(t − τ ).

(5) Multiply function x(τ ) by h(t − τ ) and plot the product function x(τ )h(t − τ ).

(6) Calculate the total area under the product function x(τ )h(t − τ ) by inte- grating it over τ = [−∞, ∞].

(7) Repeat steps 4−6 for different values of t to obtain y(t) for all time, −∞ ≤ t ≤ ∞.

Example 3.7

Repeat Example 3.6 and determine the zero-state response of the system using

the graphical convolution method.

Solution

Functions x(τ ) = exp(−τ )u(τ ), h(τ ) = exp(−2τ )u(τ ), and h(−τ ) = exp(−2τ )u(−τ ) are plotted, respectively, in Figs. 3.7(a)–(c). The function h(t − τ ) = h(−(τ − t)) is obtained by shifting h(−τ ) by time t . We consider the following two cases of t .

Case 1 For t < 0, the waveform h(t − τ ) is on the left-hand side of the vertical axis. As is apparent in Fig. 3.7(e), waveforms for h(t − τ ) and x(τ ) do not overlap. In other words, x(τ )h(t − τ ) = 0 for all τ , hence y(t) = 0.

Case 2 For t ≥ 0, we see from Fig. 3.7(f) that the non-zero parts of h(t − τ ) and x(τ ) overlap over the duration t = [0, t]. Therefore,

y (t) = t∫

e−2t+τ dτ = e−2t t∫

eτ dτ = e−2t [et − 1] = e−t − e−2t .

Combining the two cases, we obtain

y(t) = {

0 t < 0

e−t − e−2t t ≥ 0,

which is equivalent to

y(t) = (e−t − e−2t )u(t).

The output y(t) of the LTIC system is plotted in Fig. 3.7(g).

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120 Part II Continuous-time signals and systems

e−tu(t) 1

x(t)

0 t

e−2tu(t)

h(t)

e−2tu(−t) 1

h(−t)

0 t

1 e−2(t−t)u(t−t)

h(t−t)

t 0

(a) (b)

(e)

(g)

(f )

x(t), h(t−t)

case 1: t < 0

x(t), h(t−t)

0 t

case 2: t > 0

0.25

y(t)

0 0.693

Fig. 3.7. Convolution of the

input signal x(t ) with the

impulse response h(t ) in

Example 3.7. Parts (a)–(g) are

discussed in the text.

Example 3.8

The input signal x(t) = exp(−t)u(t) is applied to an LTIC system whose impulse response is given by

h(t) = {

1 − t 0 ≤ t ≤ 1 0 otherwise.

Calculate the output of the system.

Solution

In order to calculate the output of the system, we need to calculate the convo-

lution integral for the two functions x(t) and h(t). Functions x(τ ), h(τ ), and

h(−τ ) are plotted as a function of the variable τ in the top three subplots of Fig. 3.8(a)–(c). The function h(t − τ ) is obtained by shifting the time-reflected function h(−τ ) by t . Depending on the value of t , three special cases may arise.

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121 3 Time-domain analysis of LTIC systems

e−tu(t) 1

x(t)

0 t

(1−t) 1

h(t)

0 1

(a)

(b)

(1+t) 1

h(−t)

0−1

(e)

(g)

( f )

h(t−t)

(1− t+t)

(t−1) t

x(t), h(t−t)

case 1: t < 0

(t−1) t t

x(t), h(t−t)

case 2: 0 < t ≤ 1

(t−1) t

x(t), h(t−t)

case 3: t > 1

(t−1) t

Fig. 3.8. Convolution of the

input signal x(t ) with the

impulse response h(t ) in

Example 3.8. Parts (a)–(g) are

discussed in the text.

Case 1 For t < 0, we see from Fig. 3.8(e) that the non-zero parts of h(t − τ ) and x(τ ) do not overlap. In other words, output y(t) = 0 for t < 0.

Case 2 For 0 ≤ t ≤ 1, we see from Fig. 3.8(f) that the non-zero parts of h(t − τ ) and x(τ ) do overlap over the duration τ = [0, t]. Therefore,

y(t) = t∫

x(τ )h(t − τ )dτ = t∫

t−1

e−τ (1 − t + τ )dτ

= (1 − t) t∫

e−τ dτ

︸︷︷︸

integral I

+ t∫

τe−τ dτ

︸︷︷︸

integral II

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122 Part II Continuous-time signals and systems

The two integrals simplify as follows:

integral I = (1 − t) [−e−τ ] t0 = (1 − t)(1 − e −t );

integral II = [−τe−τ − e−τ ] t0 = 1 − e −t − te−t .

For 0 ≤ t ≤ 1, the output y(t) is given by

y(t) = (1 − t − e−t + te−t ) + (1 − e−t − te−t ) = (2 − t − 2e−t ).

Case 3 For t > 1, we see from Fig. 3.8(g) that the non-zero part of h(t − τ ) completely overlaps x(τ ) over the region τ = [t − 1, t]. The lower limit of the overlapping region in case 3 is different from the lower limit of the over-

lapping region in case 2; therefore, case 3 results in a different convolution

integral and is considered separately from case 2. The output y(t) for case 3 is

given by

y(t) = t∫

x(τ )h(t − τ )dτ = t∫

t−1

e−τ (1 − t + τ )dτ

= (1 − t) t∫

t−1

e−τ dτ

︸︷︷︸

integral I

+ t∫

t−1

τe−τ dτ

︸︷︷︸

integral II

The two integrals simplify as follows:

integral I = (1 − t)[−e−τ ] tt−1 = (1 − t)(e −(t−1) − e−t );

integral II = [−τe−τ − e−τ ] tt−1 = (t − 1)e −(t−1) + e−(t−1) − te−t − e−t

= te−(t−1) − te−t − e−t .

For t > 1, the output y(t) is given by

y(t) = (

e−(t−1) − te−(t−1) − e−t + te−1 )

+ (

te−(t−1) − te−t − e−t )

= (

e−(t−1) − 2e−t )

Combining the above three cases, we obtain

y(t) =







0 t < 0

(2 − t − 2e−t ) 0 ≤ t ≤ 1 (e−(t−1) − 2e−t ) t > 1,

which is plotted in Fig. 3.9.

Example 3.9

Calculate the output for the following input signal and impulse response:

x(t) = {

1.5 −2 ≤ t ≤ 3 0 otherwise

and h(t) = {

2 −1 ≤ t ≤ 2 0 otherwise.

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123 3 Time-domain analysis of LTIC systems

−2 −1 0 1 2 3 4 5 6 7 8 9 0

0.1

0.2

0.3

0.4

Fig. 3.9. Output y(t ) computed

in Example 3.8.

Solution

Functions x(τ ), h(τ ), h(−τ ), and h(t − τ ) are plotted in Figs. 3.10(a)–(d). Depending on the value of t , the convolution integral takes five different forms.

We consider these five cases below.

Case 1 (t < −3). As seen in Fig. 3.10(e), the non-zero parts of h(t − τ ) and x(τ ) do not overlap. Therefore, the output signal y(t) = 0.

Case 2 (−3 ≤ t ≤ 0). As seen in Fig. 3.10(f), the non-zero part of h(t − τ ) partially overlaps with x(τ ) within the region τ = [−2, t + 1]. The product x(τ )h(t − τ ) becomes a rectangular function in the region with an amplitude of 1.5 × 2 = 3. Therefore, the output for −3 ≤ t ≤ 0 is given by

y (t) = t+1∫

−2

3 dτ = 3(t + 3).

Case 3 (0 ≤ t ≤ 2). As seen in Fig. 3.10(g), the non-zero part of h(t − τ ) overlaps completely with x(τ ). The overlapping region is given by τ = [t − 2, t + 1]. The product x(τ )h(t − τ ) is a rectangular function with an ampli- tude of 3 in the region τ = [t − 2, t + 1]. The output for 0 ≤ t ≤ 2 is given by

y(t) = t+1∫

t−2

3 dτ = 9.

Case 4 (2 ≤ t ≤ 5). The non-zero part of h(t − τ ) overlaps partially with x(τ ) within the region τ = [t − 2, 3]. Therefore, the output for 2 ≤ t ≤ 5 is given by

y(t) = 3∫

t−2

3 dτ = 3(5 − t).

Case 5 (t ≥ 0). We see from Fig. 3.10(i) that the non-zero parts of h(t − τ ) and x(τ ) do not overlap. Therefore, the output y(t) = 0.

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124 Part II Continuous-time signals and systems

1.5

x(t)

0−2 3

(a)

h(t−t)

2.0

(t−2) (t+1) (d)

2.0 h(t)

0−1 2

(b)

2.0 h(−t)

0−2 1 (c)

(e)

2.0

(t−2) (t+1) −2 t

1.5

x(t), h(t−t)

0 3

case 1: t < −3

( f )

2.0

(t−2) (t+1) t

1.5

x(t), h(t−t)

0−2 3

case 2: −3 ≤ t < 0

(g)

2.0

(t−2) (t+1)

1.5

x(t), h(t−t)

0 3−2

case 3: 0 ≤ t < 1

(h)

2.0

(t−2) (t+1) t

1.5

x(t), h(t−t)

0−2 3

case 4: 2 ≤ t < 5

( i)

2.0

(t−2) (t+1) t

1.5

x(t), h(t−t)

0−2 3

case 5: t ≥ 5

Fig. 3.10. Convolution of the

input signal x(t ) with the

impulse response h(t ) in

Example 3.9. Parts (a)–(i) are

discussed in the text.

Combining the five cases, we obtain

y(t) =



     

     

0 t < −3 3(t + 3) −3 ≤ t ≤ 0 9 0 ≤ t ≤ 2 3(5 − t) 2 ≤ t ≤ 5 0 t > 5.

The waveform for the output response is sketched in Fig. 3.11.

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125 3 Time-domain analysis of LTIC systems

−6 −4 −2 0 2 4 6 8 0

Fig. 3.11. Output y(t ) obtained

in Example 3.9.

3.6 Properties of the convolution integral

The convolution integral has several interesting properties that can be used to

simplify the analysis of LTIC systems. Some of these properties are presented

in the following discussion.

Commutative property

x1(t) ∗ x2(t) = x2(t) ∗ x1(t). (3.37)

The commutative property states that the order of the convolution operands does

not affect the result of the convolution. In calculating the output of an LTIC

system, the impulse response and input signal can be interchanged without

affecting the output. The commutative property can be proved directly from the

definition of the convolution integral by changing the dummy variable used for

integration.

Proof

By definition,

x1(t) ∗ x2(t) = ∞∫

−∞

x1(τ )x2(t − τ )dτ .

Substituting u = t – τ gives

x1(t) ∗ x2(t) = −∞∫

∞

x1(t − u)x2(u)(−du).

By interchanging the order of the upper and lower limits, we obtain

x1(t) ∗ x2(t) = ∞∫

−∞

x1(t − u)x2(u)du = x2(t) ∗ x1(t).

Below, we list the remaining properties of convolution. Each of these properties

can be proved by following the approach used in the proof for the commutative

property. To avoid redundancy, the proofs for the remaining properties are not

included.

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126 Part II Continuous-time signals and systems

Distributive property

x1(t) ∗ [x2(t) + x3(t)] = x1(t) ∗ x2(t) + x1(t) ∗ x3(t). (3.38)

The distributive property states that convolution is a linear operation.

Associative property

x1(t) ∗ [x2(t) ∗ x3(t)] = [x1(t) ∗ x2(t)] ∗ x3(t). (3.39)

This property states that changing the order of the convolution operands does

not affect the result of the convolution integral.

Shift property If x1(t) ∗ x2(t) = g(t) then

x1(t − T1) ∗ x2(t − T2) = g(t − T1 − T2), (3.40)

for any arbitrary real constants T1 and T2. In other words, if the two operands of

the convolution integral are shifted, then the result of the convolution integral

is shifted in time by a duration that is the sum of the individual time shifts

introduced in the operands.

Duration of convolution Let the non-zero durations (or widths) of the con- volution operands x1(t) and x2(t) be denoted by T1 and T2 time units, respec-

tively. It can be shown that the non-zero duration (or width) of the convolution

x1(t) ∗ x2(t) is T1 + T2 time units.

Convolution with impulse function

x(t) ∗ δ(t − t0) = x(t − t0). (3.41)

In other words, convolving a signal with a unit impulse function whose origin

is at t = t0 shifts the signal to the origin of the unit impulse function.

Convolution with unit step function

x(t) ∗ u(t) = ∞∫

−∞

x(τ )u(t − τ )dτ = t∫

−∞

x(τ )dτ . (3.42)

Equation (3.42) states that convolving a signal x(t) with a unit step function

produces the running integral of the original signal x(t) as a function of time t .

Scaling property If y(t) = x1(t) ∗ x2(t), then y(αt) = |α|x1(αt) ∗ x2(αt). In other words, if we scale the two convolution operands x1(t) and x2(t) by a

factor of α, then the result of convolution x1(t) ∗ x2(t) is (i) scaled by α and (ii) amplified by |α| to determine y(αt).

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127 3 Time-domain analysis of LTIC systems

3.7 Impulse response of LTIC systems

In Section 2.2, we considered several properties of CT systems. Since an LTIC

system is completely specified by its impulse response, it is therefore logical to

assume that its properties are completely determined from its impulse response.

In this section, we express some of the basic properties of LTIC systems defined

in Section 2.2 in terms of the impulse response of the LTIC systems. We consider

the memorylessness, causality, stability, and invertibility properties for such

systems.

3.7.1 Memoryless LTIC systems

A CT system is said to be memoryless if its output y(t) at time t = t0 depends only on the value of the applied input signal x(t) at the same time instant

t = t0. In other words, a memoryless LTIC system typically has an input–output relationship of the form

y(t) = kx(t),

where k is a constant. Substituting x(t) = δ(t), the impulse response h(t) of a memoryless system can be obtained as follows:

h(t) = kδ(t). (3.43)

An LTIC system will be memoryless if and only if its impulse response

h(t) = 0 for t �= 0.

3.7.2 Causal LTIC systems

A CT system is said to be causal if the output at time t = t0 depends only on

the value of the applied input signal x(t) at and before the time instant t = t0.

The output of an LTIC system at time t = t0 is given by

y(t0) =

∞∫

−∞

x(τ )h(t0 − τ )dτ .

In a causal system, output y(t0) must not depend on x(τ ) for τ > t0. This

condition is only satisfied if the time-shifted and reflected impulse response

h(t0 − τ ) = 0 for τ > t0. Choosing t0 = 0, the causality condition reduces to h(−τ ) = 0 for τ > 0, which is equivalent to stating that h(τ ) = 0 for τ < 0. Below we state the causality condition explicitly.

An LTIC system will be causal if and only if its impulse response h(t) = 0 for t < 0.

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128 Part II Continuous-time signals and systems

3.7.3 Stable LTIC systems

A CT system is BIBO stable if an arbitrary bounded input signal produces a

bounded output signal. Consider a bounded signal x(t) with |x(t)| < Bx for all t , applied as input to an LTIC system with impulse response h(t). The magnitude

of output y(t) is given by

|y(t)| =

∣ ∣ ∣ ∣ ∣ ∣

∞∫

−∞

h(τ )x(t − τ )dτ

∣ ∣ ∣ ∣ ∣ ∣

Using the Schwartz inequality, we can say that the output is bounded within the

range

|y(t)| ≤ ∞∫

−∞

|h(τ )||x(t − τ )|dτ .

Since x(t) is bounded, |x(t)| < Bx , therefore the above inequality reduces to

|y(t)| ≤ Bx

∞∫

−∞

|h(τ )|dτ .

It is clear from the above expression that for the output y(t) to be bounded, i.e.

|y(t)| < ∞, the integral ∫ h(τ )dτ within the limits [−∞, ∞] should also be bounded. The stability condition can, therefore, be stated as follows.

If the impulse response h(t) of an LTIC system satisfies the following

condition:

∞∫

−∞

|h(t)|dt < ∞, (3.44)

then the LTIC system is BIBO stable.

Example 3.10

Determine if systems with the following impulse responses:

(i) h(t) = δ(t) – δ(t – 2), (ii) h(t) = 2 rect(t/2),

(iii) h(t) = 2 exp(−4t)u(t), (iv) h(t) = [1 − exp(−4t)]u(t),

are memoryless, causal, and stable.

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129 3 Time-domain analysis of LTIC systems

Solution

System (i)

Memoryless property. Since h(t) �= 0 for t �= 0, system (i) is not memoryless.

The system has a limited memory as it only requires the values of the input

signal within three time units of the time instant at which the output is

being evaluated.

Causality property. Since h(t) = 0 for t < 0, system (i) is causal.

Stability property. To verify if system (i) is stable, we compute the following

integral:

∞∫

−∞

|h(t)|dt = ∞∫

−∞

|δ(t) − δ(t − 2)|dt

≤ ∞∫

−∞

|δ(t)|dt + ∞∫

−∞

|δ(t − 2)|dt = 2 < ∞,

which shows that system (i) is stable.

System (ii)

Memoryless property. Since h(t) �= 0 for t �= 0, system (ii) is not memory- less.

Causality property. Since h(t) �= 0 for t < 0, system (ii) is not causal. Stability property. To verify if system (ii) is stable, we compute the following

integral: ∞∫

−∞

|h(t)|dt = 1∫

−1

2 dt = 4 < ∞,

which shows that system (ii) is stable.

System (iii)

Memoryless property. Since h(t) �= 0 for t �= 0, system (iii) is not memo- ryless. The memory of system (iii) is infinite, as the output at any time

instant depends on the values of the input taken over the entire past.

Causality property. Since h(t) = 0 for t < 0, system (iii) is causal. Stability property. To verify that system (iii) is stable, we solve the following

integral: ∞∫

−∞

|h(t)|dt = ∞∫

2e−4t dt = −0.5 × [e−4t ]∞0 = 0.5 < ∞,

which shows that system (iii) is stable.

System (iv)

Memoryless property. Since h(t) �= 0 for t �= 0, system (iv) is not memory- less.

Causality property. Since h(t) = 0 for t < 0, system (iv) is causal.

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130 Part II Continuous-time signals and systems

Stability property. To verify that system (iv) is stable, we solve the following

integral:

∞∫

−∞

|h(t)|dt = ∞∫

(1 − e−4t )dt = [t − 0.25e−4t ]∞0 = ∞,

which shows that system (iv) is not stable.

3.7.4 Invertible LTIC systems

Consider an LTIC system with impulse response h(t). The output y1(t) of

the system for an input signal x(t) is given by y1(t) = x (t) ∗ h(t). For the system to be invertible, we cascade a second system with impulse response

hi(t) in series with the original system. The output of the second system is

given by

y2(t) = y1(t) ∗ hi(t).

For the second system to be an inverse of the original system, output y2(t) should

be the same as x(t). Substituting y1(t) = x(t) ∗ h(t) in the above expression results in the following condition for invertibility:

x(t) = [x(t) ∗ h(t)] ∗ hi(t) = x(t) ∗ [h(t) ∗ hi(t)].

The above equation is true if and only if

h(t) ∗ hi(t) = δ(t). (3.45)

The existence of hi(t) proves that an LTIC system is invertible. At times, it is

difficult to determine the inverse system hi(t) in the time domain. In Chapter 5,

when we introduce the Fourier transform, we will revisit the topic and illustrate

how the inverse system can be evaluated with relative ease in the Fourier-

transform domain.

Example 3.11

Determine if systems with the following impulse responses:

(i) h(t) = δ(t – 2), (ii) h(t) = δ(t) − δ(t − 2),

are invertible.

Solution

(i) Since δ(t − 2) ∗ δ(t + 2) = δ(t), system (i) is invertible. The impulse response of the inverse system is given by

hi(t) = δ(t + 2).

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131 3 Time-domain analysis of LTIC systems

(ii) Assuming that the impulse response of the inverse system is hi(t), the

stability condition is expressed as

h(t) ∗ hi(t) = [δ(t) − δ(t − 2)] ∗ hi(t) = δ(t).

By applying the convolution property, Eq. (3.41), the above expression simpli-

fies to

hi(t) − hi(t − 2) = δ(t)

hi(t) = δ(t) + hi(t − 2).

The above expression can be solved iteratively. For example, hi(t − 2) is given by

hi(t − 2) = δ(t − 2) + hi(t − 4).

Substituting the value of hi(t − 2) in the earlier expression gives

hi(t) = δ(t) + δ(t − 2) + hi(t − 4),

leading to the iterative expression

hi(t) = ∞∑

m=0 δ(t − 2m).

To verify that hi(t) is indeed the impulse response of the inverse system, we

convolve h(t) with hi(t). The resulting expression is as follows:

h(t) ∗ hi(t) = [δ(t) − δ(t − 2)] ∗ ∞∑

m=0 δ(t − 2m),

which simplifies to

h(t) ∗ hi(t) = δ(t) ∗ ∞∑

m=0 δ(t − 2m) + δ(t − 2) ∗

∞∑

m=0 δ(t − 2m)

h(t) ∗ hi(t) = ∞∑

m=0 δ(t − 2m) +

∞∑

m=0 δ(t − 2 − 2m) = δ(t).

Therefore, hi(t) is indeed the impulse response of the inverse system.

3.8 Experiments with MATLAB

In this chapter, we have so far presented two approaches to calculate the output

response of an LTIC system: the differential equation method and the convolu-

tion method. Both methods can be implemented using MATLAB . However, the

convolution method is more convenient for MATLAB implementation in the

discrete-time domain and this will be presented in Chapter 8. In this section,

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132 Part II Continuous-time signals and systems

therefore, we present the method for constant-coefficient differential equations

with initial conditions.

MATLAB provides several M-files for solving differential equations with

known initial conditions. The list includesode23,ode45,ode113,ode15s,

ode23s, ode23t, and ode23tb. Each of these functions uses a finite-

difference-based scheme for discretizing a CT differential equation and iterates

the resulting DT finite-difference equation for the solution. A detailed analysis

of the implementations of these MATLAB functions is beyond the scope of the

text. Instead we will focus on the procedure for solving differential equations

with MATLAB . Since the syntax used to name these M-files is similar, we

illustrate the procedure for the function call using ode23. Any other M-file

can be used instead of ode23 by replacing ode23 with the selected M-file.

We will solve first- and second-order differential equations, and we compare

the computed values with the analytical solution derived earlier.

Example 3.12

Compute the solution y(t) for Eq. (3.11), reproduced below for convenience:

dt + 4y(t) = 2 cos(2t)u(t),

with initial condition y(0) = 2 for 0 ≤ t ≤ 15. Compare the computed solution with the analytical solution given by Eq. (3.12).

Solution

The first step towards solving Eq. (3.11) is to create an M-file containing the

differential equation. We implement a reordered version of Eq. (3.11), given by

dt = −4y(t) + 2 cos(2t)u(t),

where the derivative dy/dt is the output of the M-file based on the input y and

time t . Calling the M-file myfunc1, the format for the M-file is as follows:

function [ydot] = myfunc1(t,y)

% MYFUNC1

% Computes first derivative in (3.11) given the value of

% signal y and time t.

% Usage: ydot = myfunc1(t,y)

ydot = -4*y + 2*cos(2*t).*(t >= 0)

The above function is saved in a file named myfunc1.m and placed in a direc-

tory included within the defined paths of the MATLAB environment. To solve

the differential equation defined in myfunc1 over the interval 0 ≤ t ≤ 15, we invoke ode23 after initializing the input parameters in an M-file as

shown:

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133 3 Time-domain analysis of LTIC systems

0 2 4 6 8 10 12 14 −0.5

0.5

1.5

time

o u tp

u t

re sp

o n se

y (t

)

Fig. 3.12. Solution y(t ) for Eq.

(3.11) computed using M A T L A B.

% MATLAB program to solve Equation (3.11) in Example 3.12

tspan = [0:0.01:15]; % duration with resolution

% of 0.01s.

y0 = [2]; % initial condition

[t,y] = ode23(‘myfunc1’, tspan,y0);

% solve ODE using ode23

plot(t,y) % plot the result

xlabel(‘time’) % Label of X-axis

ylabel(‘Output Response y(t)’) % Label of Y-axis

The final plot is shown in Fig. 3.12 and is the same as the analytical solution

given by Eq. (3.12).

Example 3.13

Compute the solution for the following second-order differential equation:

ÿ(t) + 5ẏ(t) + 4y(t) = (3 cos t)u(t) with initial conditions y(0) = 2 and ẏ(0) = −5,

for 0 ≤ t ≤ 20 using MATLAB . Note that the analytical solution of this problem is presented in Appendix C (see Example C.6).

Solution

Higher-order differential equations are typically represented by a system of first-

order differential equations before their solution can be computed in MATLAB .

Assuming y2(t) to be the solution of the aforementioned differential equation,

we obtain

ÿ2(t) + 5ẏ2(t) + 4y2(t) = (3 cos t)u(t).

To reduce the second-order differential equation into a system of two first-order

differential equations, assume the following:

ẏ2(t) = y1(t). (3.46)

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134 Part II Continuous-time signals and systems

Substituting y2(t) in the original equation and rearranging the terms yields

ẏ1(t) = −5y1(t) − 4y2(t) + (3 cos t)u(t). (3.47)

Equations (3.46) and (3.47) collectively define a system of first-order differ-

ential equations that simulate the original differential equation. The coupled

system can be represented in the matrix-vector form as follows: [

ẏ1(t)

ẏ2(t)

]

= [

−5y1(t) − 4y2(t) + (3 cos t)u(t) y1(t)

]

. (3.48)

To simulate the above system, we write an M-file myfunc2 that computes

the vector of derivatives on the left-hand side of Eq. (3.48) based on the input

parameters t and vector y that contains the values of y1 and y2:

function [ydot] = myfunc2(t,y)

% The function computes first derivative of (3.48) from

% vector y and time t.

% Usage: ydot = myfunc2(t,y)

ydot(1,1) = -5*y(1) - 4*y(2)+ 3*cos(t)*(t >= 0);

ydot(2,1) = y(1);

%---end of the function----------------------

Note that the output of the above M-file is the column vector ydot corre-

sponding to Eq. (3.48). The M-file myfunc2.m should be placed in a direc-

tory included within the defined paths of the MATLAB environment. To solve

the differential equation defined in myfunc2 over the interval 0 ≤ t ≤ 20, we invoke ode23 after initializing the input parameters as given below:

% MATLAB program to solve Example 3.13

tspan = [0:0.02:20]; % duration with resolution of

% 0.02s.

y0 = [-5; 2]; % initial conditions

[t,y] = ode23(‘myfunc2’, tspan,y0);

% solve ODE using ode23

plot(t,y(:,2)) % plot the result

Note that the order of the initial conditions is reversed such that ẏ2(0) = −5 is mentioned first and y2(0) = 2 later in the initial condition vector y0. Looking at the structure of Eq. (3.48), it is clear that the top entry in the first row of ydot

corresponds to ẏ1(t), which is equal to ÿ2(t). Similarly, the entry in the second

row of ydot contains the value of ẏ2(t). The function ode23 will integrate

ydot returning the value in y. The vector y, therefore, contains the values

of ẏ2(t) in the top row and the values of y2(t) in the bottom row. The order

of the initial conditions is adjusted according to the returned values such that

ẏ2(0) = −5 is mentioned first and y2(0) = 2 later in the initial condition vector y0. The solution of the differential equation is also contained in the second

column of vector y.

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135 3 Time-domain analysis of LTIC systems

0 2 4 6 8 10 12 14 16 18 20 −1

−0.5

0.5

1.5

time

o u tp

u t

re sp

o n se

y (t

)

Fig. 3.13. Solution y(t ) for

Example 3.13 computed using

MA T L A B.

The solution y(t) is plotted in Fig. 3.13. It can be easily verified that the

plot is same as the analytical solution given by Eq. (C.38), which is reproduced

below

y(t) = 1

2 e−t +

17 e−4t +

34 cos t +

34 sin t for t ≥ 0.

3.9 Summary

In Chapter 3, we developed analytical techniques for LTIC systems. We saw

that the output signal y(t) of an LTIC system can be evaluated analytically in

the time domain using two different methods. In Section 3.1, we determined the

output of an LTIC by solving a linear, constant-coefficient differential equation.

The solution of such a differential equation can be expressed as a sum of

two components: zero-input response and zero-state response. The zero-input

response is the output produced by the LTIC system because of the initial

conditions. For stable LTIC systems, the zero-input response decays to zero

with increasing time. The zero-state response is due to the input signal. The

overall output of the LTIC system is the sum of the zero-input response and

zero-state response.

An alternative representation for determining the output of an LTIC system

is based on the impulse response of the system. In Section 3.3, we defined the

impulse response h(t) as the output of an LTIC system when a unit impulse δ(t)

is applied at the input of the system. In Section 3.4, we proved that the output y(t)

of an LTIC system can be obtained by convolving the input signal x(t) with its

impulse response h(t). The resulting convolution integral can either be solved

analytically or by using a graphical approach. The graphical approach was

illustrated through several examples in Section 3.5. The convolution integral

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136 Part II Continuous-time signals and systems

satisfies the commutative, distributive, associative, time-shifting, and scaling

properties.

(1) The commutative property states that the order of the convolution operands

does not affect the result of the convolution.

(2) The distributive property states that convolution is a linear operation with

respect to addition.

(3) The associative property is an extension of the commutative property to

more than two convolution operands. It states that changing the order

of the convolution operands does not affect the result of the convolution

integral.

(4) The time-shifting property states that if the two operands of the convolution

integral are shifted in time, then the result of the convolution integral is

shifted by a duration that is the sum of the individual time shifts introduced

in the convolution operands.

(5) The duration of the waveform produced by the convolution integral is the

sum of the durations of the convolved signals.

(6) Convolving a signal with a unit impulse function with origin at t = t0 shifts the signal to the origin of the unit impulse function.

(7) Convolving a signal with a unit step function produces the running integral

of the original signal as a function of time t .

(8) If the two convolution operands are scaled by a factor α, then the result of

the convolution of the two operands is scaled by α and amplified by |α|.

In Section 3.7, we expressed the memoryless, causality, inverse, and stability

properties of an LTIC system in terms of its impulse response.

(1) An LTIC system will be memoryless if and only if its impulse response

h(t) = 0 for t �= 0. (2) An LTIC system will be causal if and only if its impulse response h(t) = 0

for t < 0.

(3) The impulse response of the inverse of an LTIC system satisfies the property

hi(t) * h(t) = δ(t).

(4) The impulse response h(t) of a (BIBO) stable LTIC system is absolutely

integrable, i.e.

∞∫

−∞

|h(t)|dt < ∞.

Finally, in Section 3.8 we presented a few MATLAB examples for solving

constant-coefficient differential equations with initial conditions.

In Chapters 4 and 5, we will introduce the frequency representations for CT

signals and systems. Such representations provide additional tools that simplify

the analysis of LTIC systems.

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137 3 Time-domain analysis of LTIC systems

Problems

3.1 Show that a system whose input x(t) and output y(t) are related by a linear differential equation of the form

dn y

dtn + an−1

dn−1 y

dtn−1 + · · · + a1

dt + a0 y(t)

= bm dm x

dtm + bm−1

dm−1x

dtm−1 + · · · + b1

dt + b0x(t)

is linear and time-invariant if the coefficients {ar , 0 ≤ r ≤ n − 1} and {br , 0 ≤ r ≤ m} are constants.

3.2 For each of the following differential equations modeling an LTIC system, determine (a) the zero-input response, (b) the zero-state response, (c) the

overall response and (d) the steady state response of the system for the

specified input x(t) and initial conditions.

(i) ÿ(t) + 4ẏ(t) + 8y(t) = ẋ(t) + x(t) with x(t) = e−4t u(t), y(0) = 0, and ẏ(0) = 0.

(ii) ÿ(t) + 6ẏ(t) + 4y(t) = ẋ(t) + x(t) with x(t) = cos(6t)u(t), y(0) = 2, and ẏ(0) = 0.

(iii) ÿ(t) + 2ẏ(t) + y(t) = ẍ(t) with x(t) = [cos(t) + sin(2t)]u(t), y(0) = 3, and ẏ(0) = 1.

(iv) ÿ(t) + 4y(t) = 5x(t) with x(t) = 4te−t u(t), y(0) = −2, and ẏ(0) = 0.

(v) ¨ÿ (t) + 2ÿ(t) + y(t) = x(t) with x(t) = 2u(t), y(0) = ÿ(0) = ˙ÿ(0) = 0, and ẏ(0) = 1.

3.3 Find the impulse responses for the following LTIC systems character- ized by linear, constant-coefficient differential equations with zero initial

conditions.

(i) ẏ(t) = 2x(t); (ii) ẏ(t) + 6y(t) = x(t);

(iii) 2ẏ(t) + 5y(t) = ẋ(t);

(iv) ẏ(t) + 3y(t) = 2ẋ(t) + 3x(t); (v) ÿ(t) + 5ẏ(t) + 4y(t) = x(t);

(vi) ÿ(t) + 2ẏ(t) + y(t) = x(t).

3.4 The input signal x(t) = e−αt u(t) is applied to an LTIC system with impulse response h(t) = e−βt u(t).

(i) Calculate the output y(t) when α �= β.

(ii) Calculate the output y(t) when α = β.

(iii) Intuitively explain why the output signals are different in parts (i) and

(ii).

3.5 Determine the output y(t) for the following pairs of input signals x(t) and impulse responses h(t):

(i) x(t) = u(t), h(t) = u(t);

(ii) x(t) = u(−t), h(t) = u(−t);

(iii) x(t) = u(t) − 2u(t − 1) + u(t − 2), h(t) = u(t + 1) − u(t − 1);

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138 Part II Continuous-time signals and systems

(iv) x(t) = e2t u(−t), h(t) = e−3t u(t); (v) x(t) = sin(2π t)(u(t − 2) − u(t − 5)), h(t) = u(t) − u(t − 2);

(vi) x(t) = e−2|t |, h(t) = e−5|t |; (vii) x(t) = sin(t)u(t), h(t) = cos(t)u(t).

3.6 For the four CT signals shown in Figs. P3.6, determine the following convolutions:

(i) y1(t) = x(t) ∗ x(t); (ii) y2(t) = x(t) ∗ z(t);

(iii) y3(t) = x(t) ∗ w(t); (iv) y4(t) = x(t) ∗ v(t); (v) y5(t) = z(t) ∗ z(t);

(vi) y6(t) = z(t) ∗ w(t); (vii) y7(t) = z(t) ∗ v(t);

(viii) y8(t) = w(t) ∗ w(t); (ix) y9(t) = w(t) ∗ v(t); (x) y10(t) = v(t) ∗ v(t).

x(t)

1 20

−1

(i)

t −1

z(t)

−1

(ii)

w(t)

0 1−1

(1− t)(1+ t)

(iii)

v(t)

0 1−1

e−2te2t

(iv)

Fig. P3.6. CT signals for

Problem P3.6.

3.7 Show that the convolution integral satisfies the distributive, associative, and scaling properties as defined in Section 3.6.

3.8 When the unit step function, u(t), is applied as the input to an LTIC sys- tem, the output produced by the system is given by y(t) = (1 − e−t )u(t). Determine the impulse response of the system. [Hint: If x(t) → y(t) then dx/dt → dy/dt (see Problem 2.15).]

3.9 A CT signal x(t), which is non-zero only over the time interval, t = [−2, 3], is applied to an LTIC system with impulse response h(t). The output y(t) is observed to be non-zero only over the time interval t = [−5, 6]. Determine the time interval in which the impulse response h(t) of the system is possibly non-zero.

3.10 An input signal

x(t) = {

1 − t 0 ≤ t ≤ 1 0 otherwise

is applied to an LTIC system whose impulse response is given by

h(t) = e−t u(t). Using the result in Example 3.8 and the properties of the convolution integral, calculate the output of the system.

3.11 An input signal g(t) = e−(t−2)u(t − 2) is applied to an LTIC system whose impulse response is given by

r (t) = {

5 − t 4 ≤ t ≤ 5 0 otherwise.

Using the result in Example 3.8 and the properties of the convolution

integral, calculate the output of the system.

3.12 Determine whether the LTIC systems characterized by the following impulse responses are memoryless, causal, and stable. Justify your

answer. For the unstable systems, demonstrate with an example that a

bounded input signal produces an unbounded output signal.

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139 3 Time-domain analysis of LTIC systems

x(t) h1(t) y(t)

h2(t)

Σ Fig. P3.15. Feedback system for

Problem P3.15.

(i) h1(t) = δ(t) + e−5t u(t); (ii) h2(t) = e−2t u(t);

(iii) h3(t) = e−5t sin(2π t)u(t); (iv) h4(t) = e−2|t | + u(t + 1) − u(t − 1); (v) h5(t) = t[u(t + 4) − u(t − 4)];

(vi) h6(t) = sin 10t ; (vii) h7(t) = cos(5t)u(t);

(viii) h8(t) = 0.95|t |;

(ix) h9(t) =







1 −1 ≤ t < 0 −1 0 ≤ t ≤ 1

0 otherwise.

3.13 Consider the systems in Example 3.10. Analyzing the impulse responses, it was shown that the systems were not memoryless. In this problem,

calculate the input–output relationships of the systems, and from these

relationships determine if the systems are memoryless.

3.14 Determine whether the LTIC systems characterized by the following impulse responses are invertible. If yes, derive the impulse response of

the inverse systems.

(i) h1(t) = 5δ(t − 2); (ii) h2(t) = δ(t) + δ(t + 2);

(iii) h3(t) = δ(t + 1) + δ(t − 1);

(iv) h4(t) = u(t); (v) h5(t) = rect(t/8);

(vi) h6(t) = e−2t u(t).

3.15 Consider the feedback configuration of the two LTIC systems shown in Fig. P3.15. System 1 is characterized by its impulse response, h1(t) = u(t). Similarly, system 2 is characterized by its impulse response, h2(t) = u(t). Determine the expression specifying the relationship between the

input x(t) and the output y(t).

3.16 A complex exponential signal x(t) = ejω0t is applied at the input of an LTIC system with impulse response h(t). Show that the output signal is

given by

y(t) = ejω0t H (ω)|ω=ω0 ,

where H (ω) is the Fourier transform of the impulse response h(t) given

H (ω) = ∞∫

−∞

h(t)e−jωt dt.

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140 Part II Continuous-time signals and systems

3.17 A sinusoidal signal x(t) = A sin(ω0t + θ ) is applied at the input of an LTIC system with real-valued impulse response h(t). By expressing the

sinusoidal signal as the imaginary term of a complex exponential, i.e. as

jA sin(ω0t + θ ) = Im {

Aej(ω0+t) }

, A ∈ ℜ,

show that the output of the LTIC system is given by

y(t) = A|H (ω0)| sin(ω0t + θ + arg(H (ω0)),

where H (ω) is the Fourier transform of the impulse response h(t) as

defined in Problem 3.16.

Hint: If h(t) is real and x(t) → y(t), then Im{x(t)} → Im{y(t)}.

3.18 Given that the LTIC system produces the output y(t) = 5 cos(2π t) when the signal x(t) = −3 sin(2π t + π/4) is applied at its input, derive the value of the tranfer function H (ω) at ω = 2π . Hint: Use the result derived in Problem 3.17.

3.19 (a) Compute the solutions of the differential equations given in P3.2 for duration 0 ≤ t ≤ 20 using MATLAB . (b) Compare the computed solution with the analytical solution obtained in P3.2.

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C H A P T E R

4 Signal representation using Fourier series

In Chapter 3, we developed analysis techniques for LTIC systems using the

convolution integral by representing the input signal x(t) as a linear combi-

nation of time-shifted impulse functions δ(t). In Chapters 4 and 5, we will

introduce alternative representations for CT signals and LTIC systems based on

the weighted superpositions of complex exponential functions. The resulting

representations are referred to as the continuous-time Fourier series (CTFS)

and continuous-time Fourier transform (CTFT). Representing CT signals as

superpositions of complex exponentials leads to frequency-domain characteri-

zations, which provide a meaningful insight into the working of many natural

systems. For example, a human ear is sensitive to audio signals within the fre-

quency range 20 Hz to 20 kHz. Typically, a musical note occupies a much

wider frequency range. Therefore, the human ear processes frequency com-

ponents within the audible range and rejects other frequency components. In

such applications, frequency-domain analysis of signals and systems provides

a convenient means of solving for the response of LTIC systems to arbitrary

input signals.

In this chapter, we focus on periodic CT signals and introduce the CTFS used

to decompose such signals into their frequency components. Chapter 5 considers

aperiodic CT signals and develops an equivalent Fourier representation, CTFT,

for aperiodic signals. The organization of Chapter 4 is as follows. In Section 4.1,

we define two- and three-dimensional orthogonal vector spaces and use them

to motivate our introduction to orthogonal signal spaces in Section 4.2. We

show that sinusoidal and complex exponential signals form complete sets of

orthogonal functions. By selecting the sinusoidal signals as an orthogonal set of

basis functions, Sections 4.3 and 4.4 present the trigonometric CTFS for a CT

periodic signal. Section 4.5 defines the exponential representation for the CTFS

based on using the complex exponentials as the basis functions. The properties

of the exponential CTFS are presented in Section 4.6. The condition for the

existence of CTFS is described in Section 4.7. Several interesting applications

of the CTFS are presented in Section 4.8, which is followed by a summary of

the chapter in Section 4.9.

141

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142 Part II Continuous-time signals

4.1 Orthogonal vector space

From the theory of vector space, we know that an arbitrary M-dimensional

vector can be represented in terms of its M orthogonal coordinates. For example,

a two-dimensional (2D) vector �V with coordinates (vi , v j ) can be expressed as follows:

�V = vi�i + v j �j, (4.1)

where �i and �j are the two basis vectors, respectively, along the x- and y-axis. A graphical representation for the 2D vector is illustrated in Fig. 4.1(a). The two

basis vectors �i and �j have unit magnitudes and are perpendicular to each other, as described by the following two properties:

j V

vii

vj j

V = vii + vj j

(a)

V = vii + vj j + vkk

vii

vj j

vkk

(b)

Fig. 4.1. Representation of

multidimensional vectors in

Cartesian planes: (a) 2D vector;

(b) 3D vector.

orthogonality property �i · �j = |�i | | �j | cos 90◦ = 0; (4.2)

unit magnitude property

{�i · �i = |�i | |�i | cos 0◦ = 1 �j · �j = |�j | | �j | cos 0◦ = 1. (4.3)

In Eqs. (4.2) and (4.3), the operator (·) denotes the dot product between the two 2D vectors.

Similarly, an arbitrary three-dimensional (3D) vector �V , illustrated in Fig. 4.1(b), with Cartesian coordinates (vi , v j , vk), is expressed as follows:

�V = vi�i + v j �j + +vk�k, (4.4)

where �i , �j , and �k represent the three basis vectors along the x-, y-, and z-axis, respectively. All possible dot product combinations of basis vectors satisfy the

orthogonality and unit magnitude properties defined in Eqs. (4.2) and (4.3), i.e.

orthogonality property �i · �j = �i · �k = �k · �j = 0; (4.5) unit magnitude property �i · �i = �j · �j = �k · �k = 1. (4.6)

Collectively, the orthogonal and unit magnitude properties are referred to as

the orthonormal property. Given vector �V , coordinates vi , v j , and vk can be calculated directly from the dot product of vector �V with the appropriate basis vectors. In other words,

vu = �V · �u �u · �u =

| �V | |�u| cos θ �V �u |�u||�u| for u ∈ {i, j, k},

(4.7)

where θ �V �u is the angle between �V and �u. Just as an arbitrary vector can be represented as a linear combination of orthonormal basis functions, it is also

possible to express an arbitrary signal as a weighted combination of orthornor-

mal (or more generally, orthogonal) waveforms. In Section 4.2, we extend the

principles of an orthogonal vector space to an orthogonal signal space.

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143 4 Signal representation using Fourier series

4.2 Orthogonal signal space

Definition 4.1 Two non-zero signals p(t) and q(t) are said to be orthogonal

over interval t = [t1, t2] if t2∫

p(t)q∗(t)dt = t2∫

p∗(t)q(t)dt = 0, (4.8)

where the superscript ∗ denotes the complex conjugation operator. In addition to Eq. (4.8), if both signals p(t) and q(t) also satisfy the unit magnitude property:

t2∫

p(t)p∗(t)dt = t2∫

q(t)q∗(t)dt = 1, (4.9)

they are said to be orthonormal to each other over the interval t = [t1, t2].

Example 4.1

Show that

(i) functions cos(2π t) and cos(3π t) are orthogonal over interval t = [0, 1]; (ii) functions exp(j2t) and exp(j4t) are orthogonal over interval t = [0, π ];

(iii) functions cos(t) and t are orthogonal over interval t = [−1, 1].

Solution

(i) Using Eq. (4.8), we obtain

1∫

cos(2π t) cos(3π t)dt = 1 2

1∫

[cos(π t) + cos(5π t)]dt

= 1 2

[ 1

π sin(π t) + 1

5π sin(5π t)

] 1

= 0.

Therefore, the functions cos(2π t) and cos(3π t) are orthogonal over interval

t = [0, 1]. Figure 4.2 illustrates the graphical interpretation of the orthogonality con-

dition for the functions cos(2π t) and cos(3π t) within interval t = [0, 1]. Equation (4.8) implies that the area enclosed by the waveform for cos(2π t) × cos(3π t) with respect to the t-axis within the interval t = [0, 1], which is shaded in Fig. 4.2(c), is zero, which can be verified visually.

(ii) Using Eq. (4.8), we obtain

π∫

e j2t e−j4t dt = π∫

e−j2t dt = 1−2j [e −j2t ]π0 = −

2j [e−j2π − 1]π0 = 0,

implying that the functions exp(j2t) and exp(j4t) are orthogonal over interval

t = [0, π ].

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144 Part II Continuous-time signals

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2

−1

−0.5

0.5

(a)

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2

−1

−0.5

0.5

(b)

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2

−1

−0.5

0.5

(c)

Fig. 4.2. Graphical illustration of

the orthogonality condition for

the functions cos(2πt ) and

cos(3πt ) used in Example 4.1(i).

(a) Waveform for cos(2πt ).

(b) Waveform for cos(3πt ).

(iii) Using Eq. (4.8), we obtain

1∫

−1

t cos(t)dt = [t sin(t) + cos(t)] 1−1 = [1 · sin(1) + cos(1)]

− [(−1) · sin(−1) + cos(−1)] = 0,

implying that the functions cos(t) and t are orthogonal over interval t = [−1, 1]. Further, it is straightforward to verify that these functions are also orthogonal

over any interval t = [−L , L] for any real value of L .

We now extend the definition of orthogonality to a larger set of functions.

Definition 4.2 A set of N functions {p1(t), p2(t), . . . , pN (t)} is mutually

orthogonal over the interval t = [t1, t2] if t2∫

pm(t)p ∗ n(t)dt =

{

En �= 0 m = n 0 m �= n for 1 ≤ m, n ≤ N . (4.10)

In addition, if En = 1 for all n, the set is referred to as an orthonormal set.

Definition 4.3 An orthogonal set {p1(t), p2(t), . . . , pN (t)} is referred to as a

complete orthogonal set if no function q(t) exists outside the set that satisfies the

orthogonality condition, Eq. (4.6), with respect to the entries pn(t), 1 ≤ n ≤ N, of the orthogonal set . Mathematically, function q(t) does not exist if

t2∫

q(t)p∗n(t)dt �= 0 for at least one value of n ∈ {1, . . . ,N } (4.11)

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145 4 Signal representation using Fourier series

with

t2∫

q(t)q∗(t)dt �= 0. (4.12)

Definition 4.4 If an orthogonal set is complete for a certain class of orthogonal

functions within interval t = [t1, t2], then any arbitrary function x(t) can be expressed within interval t = [t1, t2] as follows:

x(t) = c1 p1(t) + c2 p2(t) + · · · + cn pn(t) + · · · + cN pN (t), (4.13)

where coefficients cn, n ∈ [1, . . . , N ], are obtained using the following expression:

cn = 1

t2∫

x(t)p∗n(t)dt. (4.14)

The constant En is calculated using Eq. (4.10). The integral Eq. (4.14) is the

continuous time equivalent of the dot product in vector space, as represented

in Eq. (4.7). The coefficient cn is sometimes referred to as the nth Fourier

coefficient of the function x(t).

Definition 4.5 A complete set of orthogonal functions {pn(t)}, 1 ≤ n ≤ N, that satisfies Eq. (4.10) is referred to as a set of basis functions.

Example 4.2

For the three CT functions shown in Fig. 4.3

(a) show that the functions form an orthogonal set of functions;

(b) determine the value of T that makes the three functions orthonormal;

x(t) = {

A for 0 ≤ t ≤ T 0 elsewhere

in terms of the orthogonal set determined in (a).

t T

f2(t)

−1

−T t

f1(t)

−T t

f3(t)

−1 −T

(a) (b) (c) Fig. 4.3. Orthogonal functions

for Example 4.2.

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146 Part II Continuous-time signals

Solution

(a) We check for the unit magnitude and the orthogonality properties for all

possible combinations of the basis vectors:

unit magnitude property

T∫

−T

|φ1(t)|2dt = T∫

−T

|φ2(t)|2dt = T∫

−T

|φ3(t)|2dt

= T∫

−T

1 dt = 2T ;

orthogonality property

T∫

−T

φ1(t)φ ∗ 2(t)dt =

T∫

−T

φ∗2(t)dt = 0,

T∫

−T

φ1(t)φ ∗ 3(t)dt =

T∫

−T

φ∗3(t)dt = 0,

and

T∫

−T

φ2(t)φ ∗ 3(t)dt =

T∫

φ∗2(t)dt − 0∫

−T

φ∗2(t)dt = 0.

In other words,

T∫

−T

φm(t)φ ∗ n (t)dt =

{

2T �= 0 m = n 0 m �= n,

for 1 ≤ m, n ≤ 3. The three functions are orthogonal to each other over the interval [−T , T ].

(b) The three functions will be orthonormal to each other:

T∫

−T

φm(t)φ ∗ n (t)dt =

{

2T = 1 m = n 0 m �= n,

which implies that T = 1/2. (c) Using Definition 4.4, the CT function x(t) can be represented as x(t) =

c1φ1(t) + c2φ2(t) + c3φ3(t) with the coefficients cn , for n = 1, 2, and 3 given by

c1 = 1

T∫

−T

x(t)φ1(t)dt = 1

T∫

A dt = A 2

c2 = 1

T∫

−T

x(t)φ2(t)dt = 1

T∫

Aφ 2(t)dt

= 1 2T

T/2∫

A dt − 1 2T

T∫

T/2

A dt = 0,

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147 4 Signal representation using Fourier series

and

c3 = 1

T∫

−T

x(t)φ3(t)dt = 1

T∫

A(−1)dt = − A

2 .

In other words, x(t) = 0.5A[φ1(t) − φ3(t)].

Example 4.3

Show that the set {1, cos(ω0t), cos(2ω0t), cos(3ω0t), . . . , sin(ω0t), sin(2ω0t),

sin(3ω0t), . . . }, consisting of all possible harmonics of sine and cosine waves

with fundamental frequency of ω0, is an orthogonal set over any interval

t = [t0, t0 + T0], with duration T0 = 2π/ω0.

Solution

It may be noted that the set {1, cos(ω0t), cos(2ω0t), cos(3ω0t), . . . , sin(ω0t),

sin(2ω0t), sin(3ω0t), . . . } contains three types of functions: 1, {cos(mω0t)},

and {sin(nω0t)} for arbitrary integers m, n ∈ Z+, where Z+ is the set of positive integers. We will consider all possible combinations of these functions.

Case 1 The following proof shows that functions {cos(mω0t), m ∈ Z+} are orthogonal to each other over interval t = [t0, t0 + T0] with T0 = 2π/ω0. Equation (4.10) yields

∫

〈T0〉

cos(mω0t) cos(nω0t)dt = t0+T0∫

cos(mω0t) cos(nω0t)dt for any arbitrary t0.

Using the trigonometric identity cos(mω0t) cos(nω0t) = (1/2)[cos((m − n)ω0t) + cos((m + n)ω0t)], the above integral reduces as follows:

∫

〈T0〉

cos(mω0t) cos(nω0t)dt =



  

  

[ sin(m − n)ω0t 2(m − n)ω0

+ sin(m + n)ω0t 2(m + n)ω0

]t0+T0

t0 m �= n

[ t

2 + sin 2mω0t

4mω0

]t0+T0

t0 m = n,

∫

〈T0〉

cos(mω0t) cos(nω0t)dt =







0 m �= n T0

2 m = n, (4.15)

for m, n ∈ Z+. Equation (4.15) demonstrates that the functions in the set {cos(mω0t), m ∈ Z+} are mutually orthogonal.

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148 Part II Continuous-time signals

Case 2 By following the procedure outlined in case 1, it is straightforward to

show that

∫

〈T0〉

sin(mω0t) sin(nω0t)dt =







0 m �= n T0

2 m = n, (4.16)

for m, n ∈ Z+. Equation (4.16) proves that the set {sin(nω0t), n ∈ Z+} contains mutually orthogonal functions over interval t = [t0, t0 + T0] with T0 = 2π/ω0.

Case 3 To verify that functions {cos(mω0t)} and {sin(nω0t)} are mutually

orthogonal, consider the following:

∫

〈T0〉

cos(mω0t) sin(nω0t)dt = t0+T0∫

cos(mω0t) sin(nω0t)dt



      

      

t0+T0∫

[sin((m + n)ω0t) − sin((m − n)ω0t)]dt m �= n

t0+T0∫

[sin(2mω0t)dt m = n



   

   

−1 2

[ cos((m + n)ω0t)

(m + n)ω0

]t0+T0

+ 1 2

[ cos((m − n)ω0t)

(m − n)ω0

]t0+T0

m �= n

−1 2

[ cos(2nω0t)

2mω0

]t0+T0

m = n

= {

0 m �= n 0 m = n, (4.17)

for m, n ∈ Z+, which proves that {cos(mω0t)} and {sin(nω0t)} are orthogonal over interval t = [t0, t0 + T0] with T0 = 2π/ω0.

Case 4 The following proof demonstrates that the function “1” is orthogonal

to cos(mω0t)} and {sin(nω0t)}: ∫

〈T0〉

1 · cos(mω0t)dt = [ sin(mω0t)

mω0

]t0+T0

= [ sin(mω0t0 + 2mπ ) − sin(mω0t0)

mω0

]

= 0 (4.18)

and ∫

〈T0〉

1 · sin(mω0t)dt = [

−cos(mω0t) mω0

]t0+T0

= − [cos(mω0t0 + 2mπ ) − cos(mω0t0)

mω0

]

= 0 (4.19)

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149 4 Signal representation using Fourier series

for m, n ∈ Z+. Combining Eqs. (4.15)–(4.19), it can be inferred that the set {1, cos(ω0t), cos(2ω0t), cos(3ω0t), . . . , sin(ω0t), sin(2ω0t), sin(3ω0t), . . . } consists

of mutually orthogonal functions. It can also be shown that this particular set is

complete over t = [t0, t0 + T0] with T0 = 2π/ω0. In other words, there exists no non-trivial function outside the set which is orthogonal to all functions in

the set over the given interval.

Example 4.4

Show that the set of complex exponential functions {exp(jnω0t), n ∈ Z} is an orthogonal set over any interval t = [t0, t0 + T0] with duration T0 = 2π/ω0. The parameter Z refers to the set of integer numbers.

Solution

Equation (4.10) yields ∫

〈T0〉

exp(jmω0t)(exp(jmω0t)) ∗dt

= t0+T0∫

exp(j(m − n)mω0t)dt =



 

 

[t]t0+T0t0 m = n[ exp(j(m − n)mω0t)

j(m − n)mω0

]t0+T0

m �= n

= {

T0 m = n 0 m �= n. (4.20)

Equation (4.14) shows that the set of functions {exp(jnω0t), n ∈ Z} is indeed mutually orthogonal over interval t = [t0, t0 + T0] with duration T0 = 2π/ω0. It can also be shown that this set is complete.

Examples 4.3 and 4.4 illustrate that the sinusoidal and complex exponential

functions form two sets of complete orthogonal functions. There are sev-

eral other orthogonal set of functions, for example the Legendre polynomi-

als (Problem 4.3), Chebyshev polynomials (Problem 4.4), and Haar functions

(Problem 4.5). We are particularly interested in sinusoidal and complex expo-

nential functions since these satisfy a special property with respect to the LTIC

systems that is not observed for any other orthogonal set of functions. In Section

4.3, we discuss this special property.

4.3 Fourier basis functions

In Example 3.2, it was observed that the output response of an RLC circuit to a

sinusoidal function was another sinusoidal function of the same frequency. The

changes observed in the input sinusoidal function were only in its amplitude

and phase. Below we illustrate that the property holds true for any LTIC system.

Further, we extend the property to complex exponential signals proving that the

output response of an LTIC system to a complex exponential function is another

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150 Part II Continuous-time signals

complex exponential with the same frequency, except for possible changes in

its amplitude and phase.

Theorem 4.1 If a complex exponential function is applied to an LTIC system

with a real-valued impulse response function, the output response of the system

is identical to the complex exponential function except for changes in amplitude

and phase. In other words,

k1e jω1t → A1k1e j(ω1t+φ1),

where A1 and φ1 are constants.

Proof

Assume that the complex exponential function x(t) = k1exp(jω1t) is applied to an LTIC system with impulse response h(t). The output of the system is given

by the convolution of the input signal x(t) and the impulse response h(t) is

given by

y(t) = ∞∫

−∞

h(τ )x(t − τ )dτ = k1e jω1t ∞∫

−∞

h(τ )e−jω1τ dτ . (4.21)

Defining

H (ω) = ∞∫

−∞

h(τ )e−jωτ dτ , (4.22)

Eq. (4.21) can be expressed as follows:

y(t) = k1e jω1t H (ω1). (4.23) From the definition in Eq. (4.22), we observe that H (ω1) is a complex-valued

constant, for a given value of ω1, such that it can be expressed as H (ω1) = A1 exp(jφ1). In other words, A1 is the magnitude of the complex constant H (ω1)

and φ1 is the phase of H (ω1). Expressing H (ω1) = A1 exp(jφ1) in Eq. (4.23), we obtain

y(t) = A1k1ej(ω1t+φ1), which proves Theorem 4.1.

Corollary 4.1 The output response of an LTIC system, characterized by a real-

valued impulse response h(t), to a sinusoidal input is another sinusoidal function

with the same frequency, except for possible changes in its amplitude and phase.

In other words,

k1 sin(ω1t) → A1k1 sin(ω1t + φ1) (4.24) and

k1 cos(ω1t) → A1k1 cos(ω1t + φ1), (4.25) where constants A1 and φ1 are the magnitude and phase of H (ω1) defined in

Eq. (4.22) with ω set to ω1.

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151 4 Signal representation using Fourier series

Proof

The proof of Corollary 4.1 follows the same lines as the proof of Theorem 4.1.

The sinusoidal signals can be expressed as real (Re) and imaginary (Im) com-

ponents of a complex exponential function as follows:

cos(ω1t) = Re{e jω1t} and sin(ω1t) = Im{e jω1t}.

Because the impulse response function is real-valued, the output y1(t) to k1 sin(ω1t) is the imaginary component of y(t) given in Eq. (4.23). In other words,

y1(t) = Im {

A1k1e j(ω1t+φ1)

}

= A1k1sin(ω1t + φ1).

Likewise, the output y2(t) to k1 cos(ω1t) is the real component of y(t) given in

Eq. (4.23). In other words,

y2(t) = Re {

A1k1e j(ω1t+φ1)

}

= A1k1cos(ω1t + φ1).

Example 4.5

Calculate the output response if signal x(t) = 2 sin(5t) is applied as an input to an LTIC system with impulse response h(t) = 2e−4t u(t).

Solution

Based on Corollary 4.1, we know that output y(t) to the sinusoidal input x(t) =

2 sin(5t) is given by

y(t) = 2A1 sin(5t + φ1),

where A1 and φ1 are the magnitude and phase of the complex constant H (ω1),

given by

H (ω) = ∞∫

−∞

h(τ )e−jωτ dτ = 2 ∞∫

e−4τ e−jωτ dτ = 2 ∞∫

e−(4+jω)τ dτ = 2 4 + jω .

The magnitude A1 and phase φ1 are given by

magnitude A1 A1 = |H (ω1)| = ∣ ∣ ∣ ∣

4 + jω

∣ ∣ ∣ ∣ ω=5

= 2√ 41

phase φ1 φ1 = <H (ω1) = < 2

4 + jω

∣ ∣ ∣ ∣ ω=5

= 0 − tan−1 (

)

= −51.34o.

The output response of the system is, therefore, given by

y(t) = 4√ 41

sin(5t − 51.34o).

As shown in Example 3.4, the LTIC system with impulse response h(t) = 2e−4t u(t) can alternatively be represented by the linear, constant-coefficient differential equation as follows:

dt + 4y(t) = 2x(t).

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152 Part II Continuous-time signals

h(t) x1(t) = k1e jw1t x1(t) = A1k1e

j(w1t + f1)

x2(t) = A1k1cos(w1t + f1)

x3(t) = A1k1sin(w1t + f1)

input signals output signals

x2(t) = k1cos(w1t)

x3(t) = k1sin(w1t)

Fig. 4.4. Output response of an

LTIC system, with a real-valued

impulse response, to sinusoidal

inputs.

Substituting x(t) = 2 sin(5t) into this equation and solving the differential equa- tion, we arrive at the same value of the output y(t) obtained using the convolution

approach.

Figure 4.4 illustrates Theorem 4.1 and Corollary 4.1 graphically. It may be

noted that this property is not observed for any other input signal but only for

the sinusoids and complex exponentials.

4.3.1 Generalization of Theorem 4.1

In the preceding discussion, we have restricted the input signal x(t) to sinusoids

or complex exponentials. In such cases, Theorem 4.1 or Corollary 4.1 simplifies

the computation of the output response of a LTIC system. In cases where the

input signal x(t) is periodic but different from a sinusoidal or complex expo-

nential function, we follow an indirect approach. We express the input signal

x(t) as a linear combination of complex exponentials:

x(t) = k1e jω1t + k2e jω2t + · · · + kN e jωN t = N∑

n=1 kne

jωn t . (4.26)

Applying Theorem 4.1 to each of the N complex exponential terms in

Eq. (4.26), the output ym(t) to the complex exponential term xm(t) = kmexp(jωm t) is given by ym(t) = Amkm exp(jωm t + φm). Using the principle of superposition, the overall output y(t) is the sum of the individual outputs and

is expressed as follows:

y(t) = A1k1e j(ω1t+φ1) + A2k2e j(ω2t+φ2) + · · · + AN kN e j(ωN t+φN )

= N∑

n=1 Ankne

j(ωn t+φn ). (4.27)

In the above discussion, we have illustrated the advantage of expressing a

periodic signal x(t) as a linear combination of complex exponentials. Such a

representation provides an alternative interpretation of the signal. This interpre-

tation is referred to as the exponential CT Fourier series (CTFS).† Alternatively,

† The Fourier series is named after Jean Baptiste Joseph Fourier (1768–1830), a French

mathematician and physicist who initiated its development and applied it to problems of heat

flow for the first time.

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153 4 Signal representation using Fourier series

an arbitrary periodic signal can also be expressed as a linear combination of

sinusoidal signals:

x(t) = a0 + ∞∑

n=1 (an cos(nω0t) + bn sin(nω0t)). (4.28)

Corollary 4.1 can then be applied to calculate the output y(t). Expressing a

periodic signal as a linear combination of sinusoidal signals leads to the trigono-

metric CTFS. The trigonometric and exponential CTFS representations of CT

periodic signals are covered in Sections 4.4 and 4.5.

4.4 Trigonometric CTFS

Definition 4.6 An arbitrary periodic function x(t) with fundamental period T0 can be expressed as follows:

x(t) = a0 + ∞∑

n=1 (an cos(nω0t) + bn sin(nω0t)), (4.29)

where ω0 = 2π/T0 is the fundamental frequency of x(t) and coefficients a0, an , and bn are referred to as the trigonometric CTFS coefficients. The coefficients

are calculated as follows:

a0 = 1

∫

〈T0〉

x(t)dt, (4.30)

an = 2

∫

〈T0〉

x(t) cos(nω0t)dt, (4.31)

and

bn = 2

∫

〈T0〉

x(t) sin(nω0t)dt . (4.32)

From Eqs. (4.29)–(4.32), it is straightforward to verify that coefficient a0 rep-

resents the average or mean value (also referred to as the dc component) of

x(t). Collectively, the cosine terms represent the even component of the zero

mean signal (x(t) – a0). Likewise, the sine terms collectively represent the odd

component of the zero mean signal (x(t) – a0).

Example 4.6

Calculate the trigonometric CTFS coefficients of the periodic signal x(t) defined

over one period T0 = 3 as follows:

x(t) = {

t + 1 −1 ≤ t ≤ 1 0 1 < t < 2.

(4.33)

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154 Part II Continuous-time signals

t −8 −6 −4 −2 0 2 4 6 8 10

x(t)

Fig. 4.5. Sawtooth periodic

waveform x(t ) considered in

Example 4.6.

Solution

The periodic signal x(t) is plotted in Fig. 4.5. Since x(t) has a fundamental

period T0 = 3, the fundamental frequency ω0 = 2π/3. Using Eq. (4.30), the dc CTFS coefficient a0 is given by

a0 = 1

∫

〈T0〉

x(t)dt = 1

1∫

−1

(t + 1)dt = 1

[ 1

2 t2 + t

] 1

−1 =

3 . (4.34)

The CTFS coefficients an are given by

an = 2

∫

〈T0〉

x(t) cos(nω0t)dt = 2

1∫

−1

(t + 1) cos(nω0t)dt

= 2

1∫

−1

t cos(nω0t) ︸︷︷︸

odd function

dt + 2

1∫

−1

cos(nω0t) ︸︷︷︸

even function

dt .

Since the integral of odd functions within the limit [−t0, t0] is zero,

1∫

−1

t cos(nω0t)dt = 0,

and the value of an is given by

an = 2

1∫

−1

cos(nω0t)dt = 4

1∫

cos(nω0t)dt = 4

[ sin(nω0t)

nω0

] 1

= 4 sin(nω0)

3nω0 .

Substituting ω0 = 2π/3, we obtain

an =



     

     

0 n = 3k√ 3

nπ n = 3k + 1

− √

nπ n = 3k + 2,

(4.35)

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155 4 Signal representation using Fourier series

for k ∈ Z . Similarly, the CTFS coefficients bn are given by

bn = 2

∫

〈T0〉

x(t) sin(nω0t)dt = 2

1∫

−1

(t + 1) sin(nω0t)dt

= 2 3

1∫

−1

t sin(nω0t) ︸︷︷︸

even function

dt + 2 3

1∫

−1

sin(nω0t) ︸︷︷︸

odd function

dt .

Since the integral of odd functions within the limits [−t0, t0] is zero, 1∫

−1

sin(nω0t)dt = 0,

and the value of bn is given by

bn = 2

1∫

−1

t sin(nω0t)dt = 4

1∫

t sin(nω0t)dt

= 4 3

[

−t cos(nω0t) nω0

+ sin(nω0t) (nω0)2

= −4 cos(nω0) 3nω0

+ 4 sin(nω0) 3(nω0)2

Substituting ω0 = 2π/3, we obtain

bn =



       

       

− 2 nπ

n = 3k

nπ + 3

√ 3

2(nπ )2 n = 3k + 1

nπ − 3

√ 3

2(nπ )2 n = 3k + 2,

(4.36)

for k ∈ Z . The periodic signal x(t) is therefore expressed as follows:

x(t) = 2 3

︸︷︷︸

xav(t)

+ ∞∑

n=1 an cos

( 2nπ

3 t

)

︸︷︷︸

Ev{x(t)−a0}

+ ∞∑

n=1 bn sin

( 2nπ

3 t

)

︸︷︷︸

Odd{x(t)−a0}

, (4.37)

where coefficients an and bn are given in Eqs. (4.35) and (4.36). Coefficient a0 represents the average value of signal x(t), referred to as xav (t). The cosine terms

collectively represent the zero-mean even component of signal x(t), denoted

by Ev{x(t) – a0}, while the sine terms collectively represent the zero-mean

odd component of x(t), denoted by Odd{x(t) – a0}. Based on the values of

the coefficients, the three components of x(t) are plotted in Fig. 4.6. It can be

verified easily that the sum of these three components will indeed produce the

original signal x(t).

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156 Part II Continuous-time signals

t −9 −6 −3 0 3 6 9

xav(t)

2/3

(a) (b)

(c)

−9 −6 −4 0 3 6 9

Ev{x(t) − a0}

1/3

2/3

Odd{x(t) − a0}

−1

−9 −6 −4 0 3 6 9 Fig. 4.6. (a) The dc, (b) even,

and (c) odd components for x(t )

shown in Fig. 4.5.

4.4.1 CTFS coefficients for symmetrical signals

If the periodic signal x(t) with angular frequency ω0 exhibits some symme-

try, then the computation of the CTFS coefficients is simplified considerably.

Below, we list the properties of the trigonometric coefficients of the CTFS for

symmetrical signals.

(1) If x(t) is zero-mean, then a0 = 0. In such cases, one does not need to calculate the dc coefficient a0.

(2) If x(t) is an even function, then bn = 0 for all n. In other words, an even signal is represented by its dc component and a linear combination of a

cosine function of frequency ω0 and its higher-order harmonics.

(3) If x(t) is an odd function, then a0 = an = 0 for all n. In other words, an odd signal can be represented by a linear combination of a sine function of

frequency ω0 and its higher-order harmonics.

(4) If x(t) is a real function, then the trigonometric CTFS coefficients a0, an ,

and bn are also real-valued for all n.

(5) If g(t) = x(t) + c (where c is a constant) then the trigonometric DTFS coefficients {ag0 , a

g n , b

g n } of function g(t) are related to the CTFS coefficients

{ax0 , axn , bxn } of x(t) as follows:

dc coefficient a g 0 = a

x 0 + c, (4.38)

coefficients an a g n = a

x n for n = 1, 2, 3, . . . , (4.39)

coefficients bn b g n = b

x n for n = 1, 2, 3, . . . (4.40)

Application of the aforementioned properties is illustrated in the following

examples.

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157 4 Signal representation using Fourier series

Example 4.7

Consider the function w(t) = Ev{x(t) − a0} shown in Fig. 4.6(b). Express w(t) as a trigonometric CTFS.

Solution

From inspection, we see w(t) is even. Therefore, bn = 0 for all n. Since w(t) is periodic with a fundamental period T0 = 3, ω0 = 2π/3. The area enclosed by one period of w(t), say t = [−1, 2], is given by 2(1/3) + 1(−2/3) = 0. Function w(t) is, therefore, zero-mean, which imples that a0 = 0.

The value of an is calculated as follows:

an = 2

1.5∫

−1.5

w(t) cos(nω0t)dt = 4

1.5∫

w(t) cos(nω0t)dt,

which simplifies to

an = 4

1∫

3 cos(nω0t)dt −

1.5∫

3 cos(nω0t)dt

= 4

[ sin(nω0t)

nω0

− 8

[ sin(nω0t)

nω0

]1.5

an = 4

sin(nω0)

nω0 −

sin(1.5nω0)

nω0 +

sin(nω0)

nω0

= 4

sin(nω0)

nω0 −

sin(1.5nω0)

nω0 .

Substituting ω0 = 2π/3, we obtain

an = 2

nπ sin

( 2nπ

)

leading to the CTFS representation

w(t) = ∞∑

n=1

nπ sin

( 2nπ

)

cos

( 2nπ

3 t

)

, (4.41)

which is same as the even component Ev{x(t) − a0} in Eq. (4.26) in Example 4.6. The CTFS coefficients an are plotted in Fig. 4.7.

From Example 4.7, we observe that a rectangular pulse train w(t) = Ev{x(t) − a0}, as shown in Fig. 4.6(b), has a CTFS representation that includes a lin-

ear combination of an infinite number of cosine functions. A question that

arises is why an infinite number of cosine functions are needed. The answer

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158 Part II Continuous-time signals

0 5 10 15 20 25 30 −0.4 −0.2

0.2

0.4

0.6

a/n

Fig. 4.7. DTFS coefficients an for the rectangular pulse in Example 4.7.

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2 −0.99

−0.66

−0.33

0.33 n = 100

n = 5

n = 20

Fig. 4.8. Rectangular pulse reconstructed with a finite number n of

DTFS coefficients an . Three different values n = 5, 20, and 100 are considered.

lies in the shape of the rectangular pulse that includes two constant values

(1/3, −2/3) separated by a discontinuity within one period. The discontinuity or the sharp transition in w(t) is accounted for by a sinusoidal function with an

infinite fundamental frequency. Generally, if a function has at least one discon-

tinuity, the CTFS representation will contain an infinite number of sinusoidal

functions.

Figure 4.7 shows the exponentially decaying value of the CTFS coefficients

an . To obtain the precise waveform w(t), an infinite number of the CTFS coef-

ficients an are needed. Because of the decaying magnitude of the CTFS coeffi-

cients, however, a fairly reasonable approximation for w(t) can be obtained by

considering only a finite number of the CTFS coefficients an . Figure 4.8 shows

the reconstruction of w(t) obtained for three different values of n. We set n = 5, 20, and 100. It is observed that w(t) provides a close approximation of w(t) for

n = 20. For n = 100, the approximated waveform is almost indistinguishable from the waveform of w(t).

4.4.2 Jump discontinuity

Figure 4.8 shows that a CT function with a discontinuity can be approximated

more accurately by including a larger number of CTFS coefficients. When

approximating CT periodic functions with a finite number of CTFS coefficients,

two errors arise because of the discontinuity. First, several ripples are observed

in the approximated function. A careful observation of Fig. 4.8 reveals that, as

more terms are added to the CTFS, the separation between the ripples becomes

narrower and the approximated function is closer to the original function. The

peak magnitude of the ripples, however, does not decrease with more CTFS

terms. The presence of ripples near the discontinuity (i.e. around t = ±1 in Fig. 4.8) is a limitation of the CTFS representation of discontinuous signals,

and is known as the Gibbs phenomenon.

Secondly, an approximation error is observed at the location of the disconti-

nuity (i.e. at t = ±1 in Fig. 4.8). With a finite number of terms, it is impossible to reconstruct precisely the edge of a discontinuity. However, it is possible to

calculate the value of the approximated function at the discontinuity. Suppose

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159 4 Signal representation using Fourier series

0.8 0.85 0.9 0.95 1 1.05 1.1 1.15 1.2 −0.99

−0.66

−0.33

0.33

0.66

n = 100

n = 5

n = 20

Fig. 4.9. Magnified sketch of Fig. 4.8 at t = 1.

t −8p −6p −4p −2p 0 2p 4p 6p 8p 10p

g(t)

3 3e−0.2t

Fig. 4.10. CT periodic signal g(t ) with fundamental period T0 = 2π considered in Example 4.8.

x(t) has a jump discontinuity at t = tj. The reconstructed value for x(tj) is given by

x̃(tj) = 1

2 [x(tj+) + x(tj−)]. (4.42)

For example, the reconstructed value of w(t) in Fig. 4.8 at t = 1 is given by

w̃(1) = 1

2 [w(1−) + w(1+)] =

[ 1

3 −

]

= − 1

6 .

Figure 4.9 is an enlargement of part of Fig. 4.8 at t = 1, where it is observed that the reconstructed signals have a value of −1/6 at t = 1.

Example 4.8

Consider the periodic signal g(t) shown in Fig. 4.10. Calculate the CTFS

coefficients.

Solution

Because T0 = 2π , the fundamental frequency ω0 = 1. The dc coefficient a0 is given by

a0 = 1

∫

〈T0〉

g(t)dt = 1

2π

2π∫

3e−0.2t dt = 3

2π

[ e−0.2t

−0.2

]2π

= 15

2π [1 − e−0.4π ] ≈ 1.7079.

The CTFS coefficients an are given by

an = 1

∫

〈T0〉

g(t) cos(nω0t)dt = 1

2π

2π∫

3e−0.2t cos(nt)dt

= 3 2π

n2 + 0.22 [e −0.2t {−0.2 cos(nt) + n sin(nt)}] 2π0

an = 3

2π

n2 + 0.22 [−0.2e −0.4π + 0.2]

= 0.3 (n2 + 0.22)π [1 − e

−0.4π ] ≈ 3.4157 1 + 25n2 .

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160 Part II Continuous-time signals

t −4 −2 0 2 4

−3

f (t)Fig. 4.11. Periodic signal f(t )

considered in Example 4.9.

Similarly, the CTFS coefficients bn are given by

bn = 1

∫

〈T0〉

g(t) sin(nω0t)dt = 1

2π

2π∫

3e−0.2t sin(nt)dt

= 3

2π

n2 + 0.22 [e−0.2t {−0.2 sin(nt) − n cos(nt)}] 2π0

bn = 3

2π

n2 + 0.22 [−ne−0.4π + n] =

(n2 + 0.22)π [1 − e−0.4π ] ≈ 17.0787n

1 + 25n2 .

The trigonometric CTFS representation of g(t) is therefore given by

g(t) = 1.7079 + ∞∑

n=1

3.4157

1 + 25n2 cos(nt) + ∞∑

n=1

17.0787

1 + 25n2 n sin(nt).

Example 4.9

Consider the periodic signal f (t) as shown in Fig. 4.11. Calculate the CTFS

coefficients.

Solution

Because T0 = 4, the fundamental frequencyω0 = π/2. Since f (t) is zero-mean, the dc coefficient a0 = 0. Also, since f (t) is an even function, bn = 0 for all n. The CTFS coefficients an are given by

an = 2

2∫

−2

f (t) cos(nω0t) ︸︷︷︸

even function

dt = 4 4

2∫

(3 − 3t) cos(nω0t)dt

= [

(3 − 3t) sin(nω0t) nω0

− 3cos(nω0t) (nω0)2

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161 4 Signal representation using Fourier series

Substituting ω0 = π/2, we obtain

an = [

(−3) sin(nπ )

0.5nπ − 3

cos(nπ )

(0.5nπ )2 + 3

(0.5nπ )2

]

= 3 [

0 − (−1)n

(0.5nπ )2 +

(0.5nπ )2

]

= 12

(nπ )2 [1 − (−1)n]

an =



 

 

0 n is even

(nπ )2 n is odd.

The CTFS representation of f (t) is given by

f (t) = ∞∑

n=1,3,5,···

(nπ )2 cos(0.5nπ t)

= 24 π2

[

cos(0.5π t) + 1 9

cos(1.5π t) + 1 25

cos(2.5π t) + · · · ]

Example 4.10

Calculate the CTFS coefficients for the following signal:

x(t) = 3 + cos (

4t + π 4

)

+ sin (

10t + π 3

)

Solution

The fundamental period of cos(4t + π/4) is given by T1 = π/2, while the fundamental period of sin(10t + π/3) is given by T2 = π/5. Since the ratio

T2 = 5

is a rational number, Proposition 1.2 states that x(t) is periodic with a fun-

damental period of π . The fundamental frequency ω0 is therefore given by

ω0 = 2π/T0 = 2. Since x(t) is a linear combination of sinusoidal functions, the CTFS coef-

ficients can be calculated directly by expanding the sine and cosine terms as

follows:

x(t) = 3 + cos(4t) cos (π

)

− sin(4t) sin (π

)

+ sin(10t) cos (π

)

+ cos(10t) sin (π

)

Substituting the values of sin(π/4), cos(π/4), sin(π/3), and cos(π/3), we obtain

x(t) = 3 + 1√ 2

cos(4t) − 1√ 2

sin(4t) + 1 2

sin(10t) + √

2 cos(10t).

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162 Part II Continuous-time signals

Comparing the above equation with the CTFS expansion,

x(t) = a0 + ∞∑

n=1 (an cos(nω0t) + bn sin(nω0t)),

with ω0 = 2, we obtain

a0 = 3, a2 = 1√ 2 , a5 =

√ 3

2 , b2 = −

1√ 2 , and b5 =

2 .

The CTFS coefficients an and bn , for values of n other than n = 0, 2, and 5, are all zeros.

Example 4.11

A periodic signal is represented by the following CTFS:

x(t) = 2 π

∞∑

m=0

2m + 1 sin(4π (2m + 1)t).

(i) From the CTFS representation, determine the fundamental period T0 of

x(t).

(ii) Comment on the symmetry properties of x(t).

(iii) Plot the function to verify if your answers to (i) and (ii) are correct.

Solution

(i) The CTFS representation is obtained by expanding the summation as follows:

x(t) = 2 π

∞∑

m=0

2m + 1 sin(4π (2m + 1)t)

= 2 π

[

sin(4π t) + 1 3

sin(12π t) + 1 5

sin(20π t) + 1 7

sin(28π t) + · · · ]

Note that the signal x(t) contains the fundamental component sin(4π t) and its

higher-order harmonics. Hence, the fundamental frequency is ω0 = 4π with the fundamental period given by T0 = 2π/4π = 1/2.

(ii) Because the CTFS contains only sine terms, x(t) must be odd based on

property (3) on page 156.

(iii) It is generally difficult to evaluate the function x(t) manually. We use a

M A T L A B function ictfs.m (provided in the accompanying CD) to calculate

x(t). The function, reconstructed using the first 1000 CTFS coefficients, is plot-

ted in Fig. 4.12 for −1 ≤ t ≤ 1. It is observed that the function is a rectangular pulse train with a fundamental period of 0.5. It is also observed that the function

is odd.

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163 4 Signal representation using Fourier series

−1 −0.75 −0.5 −0.25 0 0.25 0.5 0.75 1 −1

−0.5

0.5

Fig. 4.12. Waveform

reconstructed from the first 1000

CTFS coefficients in Example

4.11.

4.5 Exponential Fourier series

In Section 4.4, we considered the trigonometric CTFS expansion using a set

of sinusoidal terms as the basis functions. An alternative expression for the

CTFS is obtained if complex exponentials {exp(jnω0t)}, for n ∈ Z , are used as the basis functions to expand a CT periodic signal. The resulting CTFS

representation is referred to as the exponential CTFS, which is defined below.

Definition 4.7 An arbitrary periodic function x(t) with a fundamental period

T0 can be expressed as follows:

x(t) = ∞∑

m=0 Dne

jnω0t , (4.43)

where the exponential CTFS coefficients Dn are calculated as

Dn = 1

∫

〈T0〉

x(t)e−jnω0t dt, (4.44)

ω0 being the fundamental frequency given by ω0 = 2π/T0.

Equation (4.43) is known as the exponential CTFS representation of x(t). Since

the basis functions corresponding to the trigonometric and exponential CTFS

are related by Euler’s identity,

e−jnω0t = cos(nω0t) − j sin(nω0t),

it is intuitively pleasing to believe that the exponential and trigonometric CTFS

coefficients are also related to each other. The exact relationship is derived by

expanding the trigonometric CTFS series as follows:

x(t) = a0 + ∞∑

n=1 (an cos(nω0t) + bn sin(nω0t))

= a0 + ∞∑

n=1

2 (e jnω0t + e−jnω0t ) +

∞∑

n=1

2j (e jnω0t − e−jnω0t ).

Combining terms with the same exponential functions, we obtain

x(t) = a0 + 1

∞∑

n=1 (an − jbn)e jnω0t +

∞∑

n=1 (an + jbn)e−jnω0t .

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164 Part II Continuous-time signals

The second summation can be expressed as follows:

∞∑

n=1 (an + jbn)e−jnω0t =

−1∑

n=−∞ (a−n + jb−n)e jnω0t ,

which leads to the following expression:

x(t) = a0 + 1

∞∑

n=1 (an − jbn)ejnω0t +

−1∑

n=−∞ (a−n + jb−n)e jnω0t .

Comparing the above expansion with the definition of exponential CTFS,

Eq. (4.31), yields

Dn =



    

    

a0 n = 0 1

2 (an − jbn) n > 0

2 (a−n + jb−n) n < 0.

(4.45)

Example 4.12

Calculate the exponential CTFS coefficients for the periodic function g(t) shown

in Fig. 4.10.

Solution

By inspection, the fundamental period T0 = 2π , which gives the fundamental frequency ω0 = 2π/2π = 1. The exponential CTFS coefficients Dn are given by

Dn = 1

∫

〈T0〉

g(t)e−jnω0t dt = 1 2π

2π∫

3e−0.2t e−jnω0t dt = 3 2π

2π∫

e−(0.2+jnω0) t dt

Dn = − 3

2π

[ e−(0.2+jnω0)t

(0.2 + jnω0)

]2π

= 3 2π

(0.2 + jnω0) [

1 − e−(0.2+jnω0)2π ]

Substituting ω0 = 1, we obtain the following expression for the exponential CTFS coefficients:

Dn = 3

2π (0.2 + jn) [

1 − e−(0.2+jn)2π ]

= 3 2π (0.2 + jn) [1 − e

−0.4π ] ≈ 0.3416 (0.2 + jn) . (4.46)

Example 4.13

Calculate the exponential CTFS coefficients for f (t) as shown in Fig. 4.11.

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165 4 Signal representation using Fourier series

Solution

Since the fundamental period T0 = 4, the angular frequency ω0 = 2π/4 = π/2. The exponential CTFS coefficients Dn are calculated directly from the definition

as follows:

Dn = 1

∫

〈T0〉

f (t)e−jnω0t dt = 1

2∫

−2

f (t)e−jnω0t dt

= 1

2∫

−2

f (t) cos(nω0t) ︸︷︷︸

even function

dt − j 1

2∫

−2

f (t) sin(nω0t) ︸︷︷︸

odd function

dt .

Since the integration of an odd function within the limits [t0, −t0] is zero,

Dn = 1

2∫

−2

f (t) cos(nω0t)dt = 1

2∫

(3 − 3t) cos(nω0t)dt,

which simplifies to

Dn = 1

[

(3 − 3t) sin(nω0t)

nω0 − 3

cos(nω0t)

(nω0)2

= 3

[

− sin(2nω0)

nω0 −

cos(2nω0)

(nω0)2 +

(nω0)2

]

Substituting ω0 = π/2, we obtain

Dn = 3

[

− sin(nπ0)

0.5nπ −

cos(nπ )

(0.5nπ )2 +

(0.5nπ )2

]

= 6

(nπ )2 [1 − (−1)n]

Dn =







0 n is even 12

(nπ )2 n is odd.

(4.47)

In Examples 4.11 and 4.12, the exponential CTFS coefficients can also be

derived from the trigonometric CTFS coefficients calculated in Examples 4.7

and 4.8 using Eq. (4.45).

Example 4.14

Calculate the exponential Fourier series of the signal x(t) shown in Fig. 4.13.

Solution

The fundamental period T0 = T , and therefore the angular frequency ω0 = 2π/T . The exponential CTFS coefficients are given by

Dn = 1

T/2∫

−T/2

x(t)e−jnω0t dt = 1

τ/2∫

−τ/2

1 · e−jnω0t dt .

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166 Part II Continuous-time signals

x(t) 1

t 2

T− − T−T

Fig. 4.13. Periodic signal x(t ) for

Example 4.14.

From integral calculus, we know that

∫

e−jnω0t dt =







− 1

jnω0 e−jnω0t+c n �= 0

t + c n = 0. (4.48)

We consider the two cases separately.

Case I For n = 0, the exponential CTFS coefficients are given by

Dn = 1

T [t]

τ/2

−τ/2 = τ

T .

Case II For n �= 0, the exponential CTFS coefficients are given by

Dn = − 1

jnω0T [e−jnω0t ]τ/2−τ/2 =

nπ sin

(nπτ

)

Dn = τ

sin (

π nτ

)

(

π nτ

) = τ T

sinc (nτ

)

In the above derivation, the CTFS coefficients are computed separately for

n = 0 and n �= 0. However, on applying the limit n → 0 to the Dn in case II, we obtain

lim n→0

Dn = lim n→0

T sinc

(nτ

)

= τ T

lim n→0

sinc (nτ

)

= τ T

[

. .. lim x→0

sinc(mx) = 1 ]

In other words, the value of Dn for n = 0 is covered by the value of Dn for n �= 0. Therefore, combining the two cases, the CTFS coefficient for the function x(t)

is expressed as follows:

Dn = τ

sin (

π nτ

)

(

π nτ

) = τ T

sinc (nτ

)

, (4.49)

for −∞ < n < ∞. As a special case, we set τ = π/2 and T = 2π . The result- ing waveform for x(t) is shown in Fig. 4.14(a). The CTFS coefficients for the

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167 4 Signal representation using Fourier series

−2p −p 0 p 2p

x(t) 1

p− 4

p −20 −15 −10 −5 0 5 10 15 20 −0.1

0.1

0.2

0.3

(a) (b)

Fig. 4.14. Exponential CTFS

coefficients for the signal x(t )

shown in Fig. 4.13 with τ = π /2 and T = 2π . (a) Waveform for x(t ). (b) Exponential CTFS

coefficients.

special case are given by

Dn = 1

4 sinc

)

for −∞ < n < ∞. The CTFS coefficients are plotted in Fig. 4.14(b). As a side note to our discussion on exponential CTFS, we make the following

observations.

(i) The exponential CTFS provide a more compact representation compared

with the trigonometric CTFS. However, the exponential CTFS coefficients

are generally complex-valued.

(ii) For real-valued functions, the coefficients Dn and D−n are complex conju- gates of each other. This is easily verified from Eq. (4.45) and the symmetry

property (4) described in Section 4.4.

4.5.1 Fourier spectrum

The exponential CTFS coefficients provide frequency information about the

content of a signal. However, it is difficult to understand the nature of the signal

by looking at the values of the coefficients, which are generally complex-valued.

Instead, the exponential CTFS coefficients are generally plotted in terms of

their magnitude and phase. The plot of the magnitude of the exponential CTFS

coefficients |Dn| versus n (or nω0) is known as the magnitude (or amplitude) spectrum, while the plot of the phase of the exponential CTFS < Dn versus n

(or nω0) is referred to as the phase spectrum.

Example 4.15

Plot the magnitude and phase spectra of the signal g(t) considered in

Example 4.12.

Solution

From Example 4.11, we know that the exponential CTFS coefficients are given

Dn = 0.3416

0.2 + jn .

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168 Part II Continuous-time signals

Table 4.1. Magnitude and phase of Dn for a few values of n given in Example 4.15

n 0 ±1 ±2 ±3 ±4 . . . ±∞ |Dn| 1.7080 0.3350 0.1700 0.1136 0.0853 . . . 0 <Dn 0 ∓0.4372π ∓0.4683π ∓0.4788π ∓0.4841π . . . ∓0.5π

−10 −8 −6 −4 −2 0 2 4 6 8 10 0

0.5

1.5

−10 −8 −6 −4 −2 0 2 4 6 8 10

−0.5p

0.25p

0.5p

−0.25p

(a) (b)

Fig. 4.15. CTFS coefficients of

signal g(t ) shown in Fig. 4.10.

(a) Magnitude spectrum.

(b) Phase spectrum.

The magnitude and phase of the exponential CTFS coefficients are as follows:

magnitude |Dn| = 0.3416

|(0.2 + jn)| = 0.3416√ 0.04 + n2

;

phase <Dn = <3.416− <(0.2 + jn) = − tan−1(5n).

Table 4.1 shows the magnitude and phase of Dn for a few selected values of n.

The phase values are expressed in radians. The magnitude and phase spectra

are plotted in Fig. 4.15.

The magnitude of the exponential CTFS coefficients Dn indicates the strength

of the frequency component nω0 (i.e. the nth harmonic) in the signal x(t). The

phase of Dn provides additional information on how different harmonics should

be shifted and added to reconstruct x(t).

Example 4.16

Calculate and plot the amplitude and phase spectra of signal x(t) considered in

Example 4.14 for τ = π/2 and T = 2π .

Solution

The exponential DTFS coefficients are given by

Dn = τ

T sinc

(nτ

)

Substituting τ = π/2 and T = 2π , we obtain

Dn = 1

4 sinc

)

which are plotted in Fig. 4.14. Note that the coefficients are all real-valued

but periodically vary between positive and negative values. Because the CTFS

coefficients Dn do not have imaginary components, the phase corresponding to

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169 4 Signal representation using Fourier series

−25 −20 −15 −10 −5 0 5 10 15 20 25 0

0.1

0.2

−25 −20 −15 −10 −5 0 5 10 15 20 25 0

0.25p

0.5p

0.75p

(a) (b)

Fig. 4.16. (a) The amplitude and

(b) the phase spectra of the

function shown in Fig. 4.14 (see

Example 4.14). The phase

spectra are given in radians/s.

the CTFS coefficients is calculated from its sign as follows:

if Dn ≥ 0, then the associated phase <Dn = 0; if Dn < 0, then the associated phase <Dn = π or −π.

The magnitude and phase spectra are plotted in Fig. 4.16. In Fig. 4.16(a),

we observe that the magnitude spectrum is always positive, while the phase

spectrum toggles between the values of 0 and π radians/s. Note that the phase

plot is not unique since the phase of π is equivalent to the value of −π .

4.6 Properties of exponential CTFS

The exponential CTFS has several interesting properties that are useful in

the analysis of CT signals. We list the important properties in the following

discussion.

Symmetry property For real-valued periodic signals, the exponential CTFS

coefficients Dn and D−n are complex conjugates of each other.

Proof

Recall that the exponential CTFS coefficients are related to the trigonometric

CTFS coefficients by Eq. (4.45), given below

Dn = 1

2 (an − jbn) for n > 0

and

D−n = 1

2 (an + jbn) for n > 0.

For real-valued functions, property (4) of the symmetric functions in

Section 4.4.1 states that the trigonometric Fourier coefficients an and bn are

always real. Based on the aforementioned equations, the exponential CTFS

coefficients Dn and D−n are therefore complex conjugates of each other.

As a corollary to this property, consider the magnitude and phase of the expo-

nential CTFS coefficients:

|D−n| = |Dn| = 1

√

a2n + b2n (4.50)

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170 Part II Continuous-time signals

and

<D−n = −<Dn = tan−1 (

)

, (4.51)

which illustrate that the magnitude spectrum is an even function and that the

phase spectrum is an odd function. Consider the magnitude and phase spectra

of the function g(t) in Example 4.11. The spectra are shown in Fig. 4.15. It

is observed that the amplitude spectrum is even, whereas the phase spectrum

is odd, which is expected as the function g(t) is real. Consider the rectangular

pulse train in Example 4.16, whose amplitude and phase spectra are shown

in Fig. 4.16. The amplitude function is again observed to be even symmetric.

However, the phase spectrum does not seem to be odd, although the time-domain

function is real-valued. Actually, the angle π r (i.e. 180o) is equivalent to −π r (i.e. −180o); the phase values π r can be changed appropriately to satisfy the odd property.

Parseval’s theorem The power of a periodic signal x(t) can be calculated from

its exponential CTFS coefficients as follows (see Problem 1.9 in Chapter 1):

Px = 1

∫

〈T0〉

|x(t)|2dt = ∞∑

n=−∞ |Dn|2. (4.52)

For real-valued signals, |Dn| = |D−n|, which results in the following simplified formula:

Px = ∞∑

n=−∞ |Dn|2 = |D0|2 + 2

∞∑

n=1 |Dn|2. (4.53)

Example 4.17

Calculate the power of the periodic signal f (t) shown in Fig. 4.11.

Solution

It was shown in Example 4.13 that the exponential CTFS coefficients of the

signal f (t) are given by

Dn =







0 n is even 12

(nπ )2 n is odd.

Since f (t) is real-valued, using Parseval’s theorem (Eq. (4.53)) yields

P f = ∞∑

n=−∞ |Dn|2 = |D0|2 + 2

∞∑

n=0 |Dn|2 = 2

∞∑

n=1,3,5,...

( 12

n2π2

= 288 π4

∞∑

n=1,3,5,...

n4 = 288

π4 × 1.015 = 3. (4.54)

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171 4 Signal representation using Fourier series

The above value of P f can also be verified by calculating the power directly in

the time domain:

P f = 1

∫

〈T0〉

|x(t)|2dt = 2

2∫

(3 − 3t)2dt = 1

[ (3 − 3t)3

−9

= 1

[ −27 −9

− 27

−9

]

= 3. (4.55)

Linearity property The exponential CTFS coefficients of a linear combination

of two periodic signals x1(t) and x2(t), both having the same fundamental

period T0, are given by the same linear combination of the exponential CTFS

coefficients for x1(t) and x2(t). Mathematically, this implies the following:

if x1(t) CTFS

←−−→ Dn and x2(t) CTFS←−−→ En then

a1x1(t) + a2x2(t) CTFS←−−→ a1 Dn + a2 En, (4.56)

with the linearly combined signal a1x1(t) + a2x2(t) having a fundamental period of T0.

A direct application of the linearity property is the periodic signal that is a

magnitude-scaled version of the original periodic signal x(t). The exponential

CTFS coefficients of the magnitude-scaled signal are given by the following

relationship:

if x(t) CTFS←−−→ Dn then ax(t)

CTFS←−−→ aDn. (4.57)

Time-shifting property If a periodic signal x(t) is time-shifted, the ampli-

tude spectrum remains unchanged. The phase spectrum changes by an expo-

nential phase shift. Mathematically, the time-shifting property is expressed as

follows:

if x(t) CTFS←−−→ Dn then x(t − t0)

CTFS←−−→ Dne −jnω0t0 , (4.58)

where x(t − t0) represents the time-shifted signal obtained by shifting x(t) towards the right-hand side by t0. The proof of the time-shifting property follows

directly by calculating the exponential CTFS representation for the time-shifted

signal x(t − t0) from the definition.

Example 4.18

Calculate the exponential CTFS coefficients of the periodic signal s(t) shown

in Fig. 4.17.

Solution

Comparing the waveform for s(t) in Fig. 4.17 with the waveform for x(t) in

Fig. 4.14, we observe that

s(t) = x(t − π/4).

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172 Part II Continuous-time signals

s(t)

−2p 0−p p 2p 3p 2

p t

Fig. 4.17. Periodic signal s(t ) for

Example 4.18.

The two waveforms s(t) and x(t) have the same time period T0 = 2π , which gives ω0 = 1. Based on the time-shifting property, we obtain

s(t) = x (

t − π

) CTFS

←−−→ Dne−jnπ/4,

where Dn denotes the exponential CTFS coefficients of x(t). Using the value

of Dn from Example 4.14, the CTFS coefficients Sn for s(t) are given by

Sn = 1

4 sinc

)

e−jnπ/4,

for −∞ < n < ∞. From the above expression, it is clear that the magnitude |Sn| = |Dn|, but that the phase of Sn changes by an additive factor of −nπ/4.

Time reversal If a periodic signal x(t) is time-reversed, the amplitude spectrum

remains unchanged. The phase spectrum changes by an exponential phase shift.

Mathematically,

if x(t) CTFS←−−→ Dn then x(−t)

CTFS←−−→ D−n , (4.59)

which implies that if a signal is time-reversed, the CTFS coefficients of a time-

reversed signal are the time-reversed CTFS coefficients of the original signal.

Example 4.19

Calculate the exponential CTFS coefficients of the periodic signal p(t) shown

in Fig. 4.18. Represent the function as a CTFS.

Solution

From Fig. 4.18, it is observed that p(t) is a time-reversed version of s(t) plotted

in Fig. 4.17. Therefore, the exponential CTFS coefficients can be obtained by

p(t)

−3p −2p −p 0 p 2p 2 p−

t Fig. 4.18. The periodic signal

p(t ) in Example 4.19.

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173 4 Signal representation using Fourier series

applying the time-reversal property to the answer in Example 4.18. Using the

latter approach, the CTFS coefficients Pn for p(t) are given by

Pn = S−n = 1

4 sinc

( −n 4

)

e−j(−n)π/4 = 1

4 sinc

)

e jnπ/4. (4.60)

Equation (4.60) can also be obtained directly by applying the time-shifting

property (t0 = −π/2) to the waveform in Fig. 4.14(a) in Example 4.14. The function p(t) can now be represented as an exponential CTFS as follows:

p(t) = ∞∑

n=−∞ Pne

jnω0t = 1 4

∞∑

n=−∞ sinc

)

e jnπ/4e jnt

= 1 4

∞∑

n=−∞ sinc

)

e jn(t+π/4),

where the fundamental frequency ω0 is set to 1.

Time scaling If a periodic signal x(t) with period T0 is time-scaled, the CTFS

spectra are inversely time-scaled. Mathematically,

if x(t) CTFS←−−→ Dn then x

( t

) CTFS←−−→ Dan, (4.61)

where the time period of the time-scaled signal x(t/a) is given by (T0/a).

Example 4.20

Calculate the exponential CTFS coefficients of the periodic function r (t) shown

in Fig. 4.19. Represent the function as a CTFS.

Solution

From Fig. 4.19, it is observed that r (t) (with T0 = π ) is a time-scaled version of x(t) (with T0 = 2π ) plotted in Fig. 4.14. The relationship between r (t) and x(t) is given by

r (t) = 2x(2t).

Using the time-scaling and linearity properties,

if x(t) CTFS←−−→ Dn then 2x(2t)

CTFS←−−→ 2Dn/2. (4.62)

−p 0 p

r (t)

2 p

2 p−

8 p

8 p−

Fig. 4.19. Periodic signal r(t ) for

Example 4.20 obtained by

time-scaling Fig. 4.14.

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174 Part II Continuous-time signals

Using the results obtained in Example 4.14, the CTFS coefficients Rn of r (t)

are given by

Rn = 2 1

4 sinc

( n/2

)

= 1

2 sinc

)

, (4.63)

for −∞ < n < ∞. The function r (t) can now be represented as an exponential CTFS as follows:

x(t) = ∞∑

n=−∞ Rne

jnω0t = 1 2

∞∑

n=−∞ sinc

)

e j2nt ,

where the fundamental frequency ω0 is set to 2.

Differentiation and integration The exponential CTFS coefficients of the

time-differentiated and time-integrated signal are expressed in terms of the

exponential CTFS coefficients of the original signal as follows:

if x(t) CTFS←−−→ Dn then

CTFS←−−→ jnω0 Dn and ∫

x(t)dt CTFS←−−→ Dn

jnω0 .

(4.64)

It may be noted that the signal obtained by differentiating or integrating a

periodic signal x(t) over one period T0 has the same period T0 as that of the

original signal.

Example 4.21

Calculate the exponential CTFS coefficients of the periodic signal g(t) shown

in Fig. 4.20.

Solution

The function g(t) can be obtained by differentiating x(t) shown in Fig. 4.14.

Therefore, the CTFS coefficients Gn can be expressed in terms of the CTFS

coefficients Dn as follows:

Gn = jnω0 Dn with ω0 = 1. Substituting the value of

Dn = 1

4 sinc

)

t −2p −p 0 p 2p

g (t)

−3 Fig. 4.20. Periodic signal g(t ) for Example 4.21.

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175 4 Signal representation using Fourier series

yields

Gn = ( jn) 1

4 sinc

)

= jn

4 sinc

)

The function r (t) can now be represented as an exponential CTFS as follows:

r (t) = ∞∑

n=−∞ Gne

jnω0t = 1 4

∞∑

n=−∞ ( jn) sinc

)

e jnt ,

where the fundamental frequency ω0 is set to 1.

4.6.1 CTFS with different periods

In this section, we consider the variation of the CTFS when the period of a

function is changed. We use the rectangular pulse train for simplicity as its

CTFS coefficients are real-valued.

Example 4.22

Consider the periodic function x(t) in Fig. 4.13 (in Example 4.14) for the

following three cases:

(a) τ = 1 ms and T = 5 ms; (b) τ = 1 ms and T = 10 ms; (c) τ = 1 ms and T = 20 ms.

In each of the above cases, (i) determine the fundamental frequency, (ii) plot

the CTFS coefficients, and (iii) determine the higher-order harmonics absent in

the function.

Solution

It was shown in Example 4.14 that the exponential DTFS coefficients are given

Dn = τ

T sinc

(nτ

)

(a) With T = 5 ms, the fundamental frequency is f0 = 1/T = 1/5 ms = 200 Hz, while the fundamental angular frequency is ω0 = 2π f0 = 400π radians/s. The corresponding exponential CTFS coefficients are given by

Dn = 1

5 sinc

)

which are plotted in Fig. 4.21(a) using two scales on the horizontal axis. The

first scale represents the number n of the CTFS coefficients and the second scale

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176 Part II Continuous-time signals

−0.05 0

0.05

0.1

0.15

0.2

−15 −10 −5 0 5 10 15 n

f −3000 −2000 −1000 0 1000 2000 3000

−30 −20 −10 0 10 20 30 n

−3000 −2000 −1000 0 1000 2000 3000 f

−60 −40 −20 0 20 40 60−0.02

0.02

0.04

0.06

0.1

0.05

−0.05

−3000 −2000 −1000 0 1000 2000 3000 f

(a) (b)

(c)

Fig. 4.21. CTFS coefficients for

square waves with different duty

cycles. (a) τ = 1 ms; T = 5 ms. (b) τ = 1 ms; T = 10 ms; (c) τ = 1 ms; T = 20 ms.

represents the corresponding frequency f = n f0 in hertz. The CTFS coefficient for n = 0 (or f = 0 Hz) has a value of 0.2, which is the strength of the dc component in the function. The spectrum at n = 1 (or f = 200 Hz) has a value of 0.19, which is the strength of the fundamental frequency (corresponding to

200 Hz, or 400π radians/s) in the function. The spectrum at n = 2 has a value of 0.15, which is the strength of the first harmonic corresponding to a frequency

f of 400 Hz, or angular frequency ω0 of 800π radians/s in the function.

From Fig. 4.21(a), we observe that the CTFS coefficients Dn are zero

at n = ±5, ±10, ±15, . . . , which correspond to frequencies ±1000 Hz, ±2000 Hz, ±3000 Hz, . . . (i.e. n f0), respectively. In other words, the miss- ing harmonics will correspond to frequencies ±1000 Hz, ±2000 Hz, ±3000 Hz, . . . or m × 103 Hz, where m is a non-zero integer.

(b) With T set to 10 ms, the fundamental frequency f0 = 1/T = 1/10 ms = 100 Hz, while the fundamental angular frequency is given by ω0 = 2π f0 = 200π radians/s. The exponential CTFS coefficients are now given by

Dn = 1

10 sinc

( n

)

which are plotted in Fig. 4.21(b). The CTFS coefficient for n = 0 has a value of 0.1. With T = 10 ms, the harmonics corresponding to n = ±10, ±20, ±30, . . . are all equal to zero. Interestingly, the missing harmonics correspond to fre-

quencies f = n f0, which are given by ±1000, ±2000, ±3000, . . . Hz, have the same values as the frequency components missing in part (a).

(c) With T set to 20 ms, the new fundamental frequency f0 = 1/T = 1/20 ms = 50 Hz, while the fundamental angular frequency is given by

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177 4 Signal representation using Fourier series

ω0 = 2π f0 = 100π radians/s. The exponential CTFS coefficients are now given by

Dn = 1

20 sinc

( n

)

which are plotted in Fig. 4.21(c). The CTFS coefficient for n = 0 has a value of 0.05. With T = 20 ms, the harmonics corresponding to n = ±20, ±40, ±60, . . . are all equal to zero. As was the case in parts (a) and (b), the missing harmonics correspond to frequencies f of ±1000, ±2000, ±3000, . . . Hz.

For a square wave, the ratio τ/T is referred to as the duty cycle, which is

defined as the ratio between the time τ that the waveform has a high value

and the fundamental period T . Cases (a)–(c) are illustrated in Figs. 4.21(a)–(c),

where the duty cycle was reduced by keeping τ constant and increasing the value

of the fundamental period T . Alternatively, the duty cycle may be decreased

by reducing the value of τ , while maintaining the fundamental period T at a

constant value. By changing the duty cycle, we observe the following variations

in the exponential DTFS representation.

DC coefficient Since the dc coefficient represents the average value of the

waveform, the value of the dc coefficient D0 decreases as the duty cycle (τ/T )

of the square wave is reduced.

Zero crossings As the duty cycle (τ/T ) is decreased, the energy within one

period of the waveform in the time domain is concentrated over a relatively

narrower fraction of the time period. Based on the time-scaling property, the

energy in the corresponding CTFS representations is distributed over a larger

number of the CTFS coefficients. In other words, the width of the main lobe

and side lobes of the discrete sinc function increases with a reduction in the

duty cycle.

4.7 Existence of Fourier series

In Sections 4.4 and 4.5, the trigonometric and exponential CTFS representations

of a periodic signal were covered. Because the CTFS coefficients are calculated

by integration, there is a possibility that the integral may result in an infinite

value. In this case, we state that the CTFS representation does not exist. Below

we list the conditions for the existence of the CTFS representation.

Definition 4.8 The CTFS representation (trigonometric or exponential) of a

periodic function x(t) exists if all CTFS coefficients are finite and the series

converges for all n. In other words, there is no infinite value in the magnitude

spectrum of the CTFS representation.

For the CTFS representation to exist, the periodic signal x(t) must satisfy the

following three conditions.

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178 Part II Continuous-time signals

(1) Absolutely integrable. The area under one period of |x(t)| is finite, i.e. ∫

|x(t)|dt < ∞. (4.65)

(2) Bounded variation. The periodic signal x(t) has a finite number of maxima

or minima in one period.

(3) Finite discontinuities. The period x(t) has a finite number of discontinuities

in one period. In addition, each of the discontinuity has a finite value.

The above conditions are known as the Dirichlet conditions.† If these condi-

tions are satisfied, it is guaranteed that perfect reconstruction is obtained from

the CTFS coefficients except at a few isolated points where the function x(t)

is discontinuous. The first condition is also known as the weak Dirichlet con-

dition, whereas the second and third conditions are known as strong Dirichlet

conditions. Most practical signals satisfy these three conditions. Examples of

the CT functions that violate these conditions are included in the following

discussion.

Example 4.23

Determine whether the following functions satisfy the Dirichlet conditions:

(i) h(t) = tan(π t); (4.66) (ii) g(t) = sin(0.5π/t) for 0 ≤ t < 1 and g(t) = g(t + 1); (4.67)

(iii) x(t) = {

1 2−2m−1 < t ≤ 2−2m 0 2−2m−2 < t ≤ 2−2m−1 (4.68)

for m ∈ Z+, 0 ≤ t < 1, and x(t) = x(t + 1).

Solution

(i) The CT function h(t) is plotted in Fig. 4.22(a). We now proceed to determine

if h(t) satisfies the Dirichlet conditions. Condition (1) is violated because

∫

|x(t)|dt = 0.5∫

−0.5

tan(π t)dt = ∞.

This is also apparent from the waveform of tan(π t), plotted in Fig. 4.22(a),

where the waveform approaches ±∞ at each discontinuity. Condition (2) is satisfied as there are only one maximum and one minimum within a single

period of h(t). Condition (3) is violated. Although there is only one discontinuity

within a single period of h(t), the magnitude of the discontinuity is infinite.

(ii) The CT function g(t) is plotted in Fig. 4.22(b). Condition (1) is sat-

isfied as the area enclosed by |g(t)| is finite. Condition (2) is violated as an

† These conditions were derived by Johann Peter Gustav Lejeune Dirichlet (1805–1859), a

German mathematician.

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179 4 Signal representation using Fourier series

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2 −10

−5

t −1 −0.5 0 0.5 1 1.5 2 2.5 3

−1

−0.5

0.5

−1 −0.5 0 0.5 1 1.5 2 0

0.25

0.5

0.75

1.25

(a) (b)

(d)

Fig. 4.22. Functions (a) h(t ),

(b) g(t ), and (c) x (t ) in Example

4.23. These functions violate one

or more of the Dirichlet

conditions, and therefore the

CTFS representation does not

exist for these functions.

infinite number of maxima and minima exist within a single period of g(t).

Condition (3) is satisfied as there are no discontinuities within a single period

of g(t).

(iii) The CT function x(t) is plotted in Fig. 4.22(c). Condition (1) is satisfied

as the area enclosed by |x(t)| is finite. Condition (2) is violated as there are an infinite number of maxima and minima within a single period of g(t). Condition

(3) is violated as an infinite number of discontinuities exist within a single period

of g(t).

4.8 Application of Fourier series

The exponential CTFS has several interesting applications. In Section 4.7.1, we

highlight an application of the CTFS representation in calculating the sum of

an infinite series. Section 4.7.2 considers the use of the CTFS representation in

calculating the response of an LTIC system to a periodic signal. By using the

CTFS representation, we avoid the convolution integral.

4.8.1 Computing the sum of an infinite series

The following example illustrates an application of the CTFS in calculating the

sum of a series:

Example 4.24

Calculate the sum S of the following infinite series:

S = ∞∑

n=0

(2n + 1)4 = 1 + 1

34 + 1

54 + 1

74 + 1

94 + 1

114 + · · ·

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180 Part II Continuous-time signals

Solution

To compute the sum S, we consider the periodic signal f (t) shown in Fig. 4.11.

As shown in Example 4.13, the exponential CTFS coefficients of f (t) are given

Dn =







0 n is even 12

(nπ )2 n is odd.

Using Parseval’s theorem, the average power of f (t) is given by

Px = ∞∑

n=−∞ |Dn|2 = |D0|2 + 2

∞∑

n=1 |Dn|2 = 2

∞∑

n=1 n=odd

144

π4 · 1

n4 = 288

π4 S.

(4.69)

Using the time-domain approach, it was shown in Example 4.17 that the average

power of f (t) is given by

P f = 1

∞∫

−∞

|x(t)|2dt = 3. (4.70)

Combining Eqs. (4.69) and (4.70) gives (288/π4) S = 3 or

S = ∞∑

n=0

(2n + 1)4 = 3π4

288 = π

56 ≈ 1.014 7.

4.8.2 Response of an LTIC system to periodic signals

As a second application of the exponential CTFS representation, we consider the

response y(t) of an LTIC system with the impulse response h(t) to an periodic

input x(t). The system is illustrated in Fig. 4.23. Assuming that the input signal

x(t) has the fundamental period T0, the exponential CTFS representation of

x(t) is given by

LTIC

system

h(t)

periodic

input

x(t)

periodic

output

y(t)

Fig. 4.23. Response of an LTIC

system to a periodic input.

x(t) = ∞∑

m=0 Dne

jnω0t , (4.71)

where the fundamental frequency ω0 = 2π/T0. The steps involved in calculat- ing the output y(t) are as follows.

Step 1 Based on Theorem 4.3.1, the output of an LTIC system yn(t) to a

complex exponential xn(t) = Dnexp(jnω0t) is given by

yn(t) = Dn H (nω0)e jnω0t , (4.72)

where H (nω0) = H (ω), evaluated at ω = nω0. The new term H (ω) is referred

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181 4 Signal representation using Fourier series

to as the transfer function of the LTIC system and is given by

H (ω) = ∞∫

−∞

h(t)e−jωt dt. (4.73)

Step 2 Using the principle of superposition, the overall output y(t) by adding

individual outputs yn(t) is given by

y(t) = ∞∑

n=−∞ yn(t) (4.74)

y(t) = ∞∑

n=−∞ Dne

jnω0t H (ω)|ω=nω0 . (4.75)

Step 3 Based on Eq. (4.75), it is clear that the response y(t) of an LTIC system

to a periodic input x(t) is also periodic with the same fundamental period as

x(t). In addition, the exponential CTFS coefficients En of the output y(t) are

related to the CTFS coefficients Dn of the periodic input signal x(t) by the

following relationship

En = Dn H (ω)|ω=nω0 . (4.76)

Example 4.25

Calculate the exponential CTFS coefficients of the output y(t) if the square

wave x(t) illustrated in Fig. 4.14 is applied as the input to an LTIC system with

impulse response h(t) = exp(−2t)u(t).

Solution

The exponential CTFS coefficients of the square wave x(t) shown in Fig. 4.14(a)

are given by (see Example 4.14)

Dn = 1

4 sinc

)

, for −∞ < n < ∞.

The transfer function H (ω) of the LTIC is given by

H (ω) = ∞∫

−∞

h(t)e−jωt dt = ∞∫

e−(2+jω)t dt = 1 (2 + jω) . (4.77)

For ω0 = 1 radian/s, the exponential CTFS coefficients of the output y(t) are given by

En = Dn H (ω)|ω=n = 1

4 sinc

)

× 1 (2 + jn) =

sinc(n/4)

8 + j4n , (4.78)

and the output y(t) is given by

y(t) = ∞∑

n=−∞ Ene

jnω0t = ∞∑

n=−∞

sinc(n/4)

8 + j4n e jnt . (4.79)

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182 Part II Continuous-time signals

0.6

0.4

y(t)

0.2

−0.2 −8 −4 0 4 8

Fig. 4.24. Response of the LTIC

system in Example 4.25.

Using the M A T L A B function ictfs.m (provided in the accompanying CD),

y(t) is calculated and shown in Fig. 4.24. It is observed that y(t) does not have

any sharp (rising or falling) edges. This is primarily because, at high frequencies,

the gain of the system (|H (ω)|) is small. As the high-frequency components of the inputs are suppressed by the system, the sharp edges are absent at the

output.

Example 4.25 used the CTFS to calculate the output y(t) of a periodic signal

x(t). Such a method is limited to periodic input signals. In Chapter 5, we show

how the continuous-time Fourier transform (CTFT) can be used to compute the

output of the LTIC systems for both periodic and aperiodic inputs. Since the

CTFT is more inclusive than the CTFS representation, our analysis of the LTIC

systems will be based primarily on the frequency decompositions using the

CTFT. The CTFS is, however, used indirectly to compute the CTFT of periodic

signals. We shall explore the relationship between the CTFS and CTFT more

fully in Chapter 5.

4.9 Summary

In Chapter 4, we introduced frequency-domain analysis of periodic sig-

nals based on the trigonometric and exponential CTFS representations. In

Sections 4.1 and 4.2, the basis functions are defined as a complete set {pn(t)},

for 1 ≤ n ≤ N , of orthogonal functions satisfying the following orthogonality properties over interval [t1, t2]:

orthogonality property

t2∫

pm(t)p ∗ n(t)dt =

{

En �= 0 m = n 0 m �= n

for 1 ≤ m, n ≤ N ,

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183 4 Signal representation using Fourier series

for any pair of functions taken from the set {pn(t)}. Section 4.3 proves that

the complex exponentials {exp(jnω0t)}, for −∞ < n < ∞, and sinusoidal functions {sin(nω0t), 1, cos(mω0t)}, for 0 < n, m < ∞, form two complete orthogonal sets over any interval [t1, t1 + 2π/ω0] of duration T0 = 2π/ω0. We refer to ω0 as the angular frequency and to its inverse T0 = 2π/ω0 as the funda- mental period. Expressing a periodic signal x(t) as a linear combination of the

sinusoidal set of functions {sin(nω0t), 1, cos(mω0t)} leads to the trigonometric

representation of the CTFS. The trigonometric CTFS is defined as follows:

x(t) = a0 + ∞∑

n=1 (an cos(nω0t) + bn sin(nω0t)),

where ω0 = 2π/T0 is the fundamental frequency of x(t) and coefficients a0, an , and bn are referred to as the trigonometric CTFS coefficients. The coefficients

are calculated using the following formulas:

a0 = 1

∫

〈T0〉

x(t)dt,

an = 2

∫

〈T0〉

x(t) cos(nω0t)dt,

and

bn = 2

∫

〈T0〉

x(t) sin(nω0t)dt .

The trigonometric CTFS is presented in Section 4.4, while its counterpart, the

exponential CTFS, is covered in Section 4.5. The exponential CTFS is obtained

by expressing the periodic signal x(t) as a linear combination of complex expo-

nentials {exp(jnω0t)} and is given by

x(t) = ∞∑

m=−∞ Dne

jnω0t ,

where the exponential CTFS coefficients Dn are calculated using the following

expression:

Dn = 1

∫

〈T0〉

x(t)e−jnω0t dt .

The exponential CTFS has several interesting properties that are useful in the

analysis of CT signals.

(1) The linearity property states that the exponential CTFS coefficients of a

linear combination of periodic signals are given by the same linear combi-

nation of the exponential CTFS coefficients of each of the periodic signals.

(2) A time shift of t0 in the periodic signal does not affect the magnitude of the

exponential CTFS coefficients. However, the phase changes by an additive

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184 Part II Continuous-time signals

factor of ±nω0t0, the sign of the phase change depending on the direction of the shift. This property is referred to as the time-shifting property.

(3) The exponential CTFS coefficients of a time-reversed periodic signal are

the time-reversed CTFS coefficients of the original signal.

(4) If a periodic signal is time-scaled, the exponential CTFS coefficients are

inversely time-scaled.

(5) The exponential CTFS coefficients of a time-differentiated periodic signal

are obtained by multiplying the CTFS coefficients of the original signal by

a factor of jnω0.

(6) The exponential CTFS coefficients of a time-integrated periodic signal are

obtained by dividing the CTFS coefficients of the original signal by a factor

of jnω0.

(7) For real-valued periodic signals, the exponential CTFS coefficients Dn and

D−n are complex conjugates of each other.

(8) Based on Parseval’s property, the power of a periodic signal x(t) with the

fundamental period of T0 is computed directly from the exponential CTFS

coefficients as follows:

Px = 1

∫

〈T0〉

|x(t)|2dt = ∞∑

n=−∞ |Dn|2.

The plot of the magnitude |Dn| of the exponential CTFS coefficients versus the coefficient number n is referred to as the magnitude spectrum, while the plot of

the phase <Dn of the exponential CTFS coefficients versus the coefficient num-

ber n is referred to as the phase spectrum of the periodic signal x(t). Section 4.6

covers the conditions for the existence of the CTFS representations, and Section

4.7 concludes the chapter by calculating the output response y(t) of an LTIC

system to a periodic input x(t). In such cases, the output y(t) is given by

y(t) = ∞∑

n=−∞ Dne

jnω0t H (ω) |ω=nω0 ,

where the transfer function H (ω) is obtained from the impulse response h(t) of

the LTIC system as follows:

H (ω) = ∞∫

−∞

h(t)e−jωt dt.

The above expression also defines the continuous-time Fourier transform

(CTFT) for aperiodic signals, which is covered in depth in Chapter 5.

Problems

4.1 Express the following functions in terms of the orthogonal basis functions

specified in Example 4.2 and illustrated in Fig. 4.3.

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185 4 Signal representation using Fourier series

(a) x1(t) = {

A 0 ≤ t ≤ T −A −T ≤ t ≤ 0;

(b) x2(t) =



 

 

A T

2 ≤ |t | ≤ T

−A 0 ≤ |t | ≤ T 2

;



 

 

A T

2 ≤ |t | ≤ T

0 0 ≤ |t | ≤ T 2

4.2 For the functions

φ1(t) = e−2|t | and φ2(t) = 1 − K e−4|t |

determine the value of K such that the functions are orthogonal over the

interval [−∞, ∞]. 4.3 The Legendre polynomials are widely used to approximate functions. An

nth-order Legendre polynomial Pn(x) is defined as

Pn(x) = 1

n!2n dn

dxn (x2 − 1)n =

M∑

m=0 anm x

where the values of anm can be expressed as follows:

anm = n∑

m=0 n,m odd n,m even

(−1)(n−m)/2 (n + m)! 2nm!(n − m/2)!(n + m/2)!

Note that anm is non-zero only when both n and m are either odd or even.

For all other values of n and m, anm is zero. The first few orders of Legendre

polynomials are given by

P0(x) = 1; P2(x) = 1

2 (3x2 − 1);

P1(x) = x ; P3(x) = 1

2 (5x3 − 3x);

and are shown in Fig. P4.3.

The Legendre polynomials {Pn(x), n = 0, 1, 2, . . .} form a set of orthogonal functions over the interval [−1, 1] by satisfying the follow- ing property:

1∫

−1

Pm(x)Pn(x)dx =







2m + 1 m = n 0 m �= n.

Verify the above orthogonality condition for m, n = 0, 1, 2, 3. 4.4 The Chebyshev polynomials of the first kind are used as the approximation

to a least-squares fit. The nth-order polynomial Tn(x) can be expressed as

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186 Part II Continuous-time signals

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1

−1

−0.5

0.5

1 P0(x)

P1(x)

P2(x)

P3(x)

Fig. P4.3. Legendre

polynomials with order 0–3.

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1

−1

−0.5

0.5

1 T0(x)

T1(x)

T5(x)T4(x) T3(x)

T2(x)

Fig. P4.4. First few orders of

Chebyshev polynomials of the

first kind.

follows:

Tn(x) = n

⌊n/2⌋∑

k=0 (−1)k

(n − k − 1)! k!(n − 2k)!

(2x)n−2k, n = 0, 1, 2, 3, . . .

The first few Chebyshev polynomials are given by

T0(x) = 1; T3(x) = 4x3 − 3x ; T1(x) = x ; T4(x) = 8x4 − 8x2 + 1; T2(x) = 2x2 − 1; T5(x) = 16x5 − 20x3 + 5x ;

which satisfy the following relationship:

Tn+1(x) = 2xTn(x) − Tn−1(x)

and are shown in Fig. P4.4.

The Chebyshev polynomials {Tn(x), n = 0, 1, 2, . . .} form an orthog- onal set on the interval [−1, 1] with respect to the weighting function by satisfying the following:

1∫

−1

1√ 1 − x2

Tm(x)Tn(x)dx =







π m = n = 0 π/2 m = n = 1, 2, 3 0 m �= n.

Verify the above orthogonality condition for m, n = 0, 1, 2, 3, 4. 4.5 The Haar functions are very popular in signal processing and wavelet appli-

cations. These functions are generated using a scale parameter (m) and a

translation parameter (n). Let the mother Haar function (m = n = 0) be defined as follows:

H0,0(t) =





1 0 ≤ t < 0.5 −1 0.5 ≤ t ≤ 1

0 otherwise.

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187 4 Signal representation using Fourier series

H0,0(t)

0 0.5 1.0 t

H1,0(t)

0 0.5 1.0 t

H2,0(t)

0 0.5 1.0 t

H2,1(t)

0 0.5 1.0 t

H2,2(t)

0 0.5 1.0 t

H2,3(t)

0 0.5 1.0 t

H1,1(t)

0 0.5 1.0 t

Fig. P4.5. Haar functions for

m = 0, 1, and 2. The other Haar functions, at scale m and with translation n, are defined

using the mother Haar function as follows:

Hm,n(t) = H0,0(2m t − n), n = 0, 1, . . . , (2m − 1).

The Haar functions for m = 0, 1, 2 are shown in Fig. P4.5. Show that the Haar wavelet functions {Hm,n(t), m = 0, 1, 2, . . . , n =

0, 1, 2, . . . (2m − 1)} form a set of orthogonal functions over the interval [0, 1] by proving the following:

1∫

Hm,n(t)Hp,q (t)dt = {

2−m m = p, n = q 0 otherwise.

4.6 Calculate the trigonometric CTFS coefficients for the periodic functions

shown in Figs. P4.6(a)–(e).

(a) Rectangular pulse train with period 2π :

x1(t) = {

3 for 0 ≤ t < π 0 for π ≤ t < 2π.

(b) Raised square wave with period 2T :

x2(t) =



 

 

0.5 for −T

2 ≤ t < T

1 for T

2 ≤ t < 3T

2 .

x3(t) = 1 − t T

for 0 ≤ t < T .

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188 Part II Continuous-time signals

x1(t)

−2π −π 0 π 2π 3π

(a)

x2(t)

−4T −2T 0 2T 4T

(b)

x3(t)

−4T −2T 0 2T 4T

(e)

x4(t)

−4T −2T 0 2T 4T

x5(t)

t −4T −2T 0 2T 4TFig. P4.6. Periodic functions in

Problem 4.6; (a)–(e) refer to the

Problem.

(d) Sawtooth wave with period 2T :

x4(t) = 1 − ∣ ∣ ∣

∣ ∣ ∣ for −T ≤ t < T .

(e) Periodic wave with period 2T .

x5(t) = {

0 for −T ≤ t < 0 1 − 0.5 sin

(π t

)

for 0 ≤ t < T .

4.7 Calculate the trigonometric CTFS coefficients for the periodic function

shown in Fig. P4.7. Note that the function

s(t) = k=∞∑

k=−∞ δ(t − kT )

is known as the sampling function and that it is used to obtain a discrete-

time signal by sampling a continuous-time signal (see Chapter 9).

4.8 Calculate the trigonometric CTFS coefficients for the following functions:

(i) xt (t) = cos 7t + sin(15t + π/2); (ii) x2(t) = 3 + sin 2t + cos(4t + π/4);

(iii) x3(t) = 1.2 + e j2t+1 + e j(5t+2) + e−j(3t+1); (iv) x4(t) = et+1 + e j(2t+3).

4.9 Show that if x(t) is an even periodic function with period T0, the expo-

nential CTFS coefficients can be calculated by evaluating the following

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189 4 Signal representation using Fourier series

s(t)

t −4T −2T 0 2T 4T

Fig. P4.7. Periodic function (an

impulse train with period T ) in

Problem 4.7.

integral:

Dn = 2

T0/2∫

x(t) cos(nω0t)dt,

where ω0 = 2π/T0.

4.10 Show that if x(t) is an odd periodic function with period T0, the exponential

CTFS coefficients can be calculated by evaluating the following integral:

Dn = −2j T0

T0/2∫

x(t) sin(nω0t)dt,

where ω0 = 2π/T0.

4.11 For the periodic functions shown in Fig. P4.6:

(i) calculate the exponential CTFS coefficients directly using Eq. (4.44);

(ii) plot the magnitude and phase spectra.

4.12 Repeat Problem 4.11 for the function shown in Fig. P4.7.

4.13 For the periodic functions shown in Fig. P4.6, calculate the exponential

CTFS coefficients by applying Eq. (4.45) to the trigonometric CTFS coef-

ficients calculated in Problem 4.6. Compare your answers with the CTFS

coefficients obtained in Problem 4.11.

4.14 Consider the raised square wave shown in Fig. P4.6(b). Using the time-

differentiation property and the exponential CTFS coefficients calcu-

lated in Problem 4.11, calculate the exponential CTFS coefficients of an

impulse train with period T0 = 2T , with impulses located at T/2 + 2kT with k ∈ Z .

4.15 Calculate the exponential CTFS coefficients for the functions given in

Problem 4.8.

4.16 The derivative of the square wave x(t) shown in Fig. 4.14 can be expressed

in terms of two shifted impulse trains as

dx(t)

dt =

∞∑

k=−∞ δ

(

t + π 4

− 2kπ )

− δ (

t − π 4

− 2kπ )

Using the time-shifting and time-scaling properties, express the exponen-

tial CTFS coefficients Dn for the square wave in terms of the exponential

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190 Part II Continuous-time signals

CTFS coefficients En of the impulse train. Calculate the CTFS coef-

ficients of the square wave and compare with the values evaluated in

Example 4.14.

4.17 Repeat Example 4.22 with the following values of τ and T such that the

duty cycle (τ/T ) is fixed at 0.2:

(i) τ = 1 ms, T = 5 ms; (ii) τ = 2 ms, T = 10 ms;

(iii) τ = 4 ms, T = 20 ms. Discuss the changes in the CTFS representations for the above selections

of τ and T .

4.18 For the periodic functions shown in Fig. P4.6:

(i) calculate the average power in the time domain, and

(ii) calculate the average power using Parseval’s theorem. Verify your

result with that obtained in step (i).

[Hint: If you find it difficult to calculate the summation

n=∞∑

n=−∞ |Dn|2

analytically, write a MATLAB program to calculate an approximate

value of

n=∞∑

n=−∞ |Dn|2 for −1000 ≤ n ≤ 1000.]

4.19 Determine whether the periodic functions shown in Fig. P4.6 satisfy the

Dirichlet conditions and have CTFS representation.

4.20 Determine if the following functions satisfy the Dirichlet conditions and

have CTFS representation:

(i) x(t) = 1/t, t = (0, 2] and x(t) = x(t + 2); (ii) g(t) = cos(π/2t), t = (0, 1] and g(t) = g(t + 1);

(iii) h(t) = sin(ln(t)), t = (0, 1] and h(t) = h(t + 1). 4.21 Consider the periodic signal f (t) considered in Example 4.9 and shown

in Fig. 4.11. From the CTFS representation, prove the following identity:

π2

8 = 1 + 1

32 + 1

52 + 1

72 + · · · .

4.22 From the half sawtooth wave shown in Fig. P4.6(c) and its trigonometric

CTFS coefficients (calculated in Problem 4.6(c)), prove the following

identity:

4 = 1 − 1

3 + 1

5 − 1

7 + 1

9 − 1

11 + · · · .

[Hint: Evaluate the function at t = T/4.] 4.23 Using the exponential CTFS coefficients of the function shown in

Fig. P4.6(c) (calculated in Problem 4.11) and Parseval’s power theorem,

prove the following identity:

π2

6 = 1 + 1

22 + 1

32 + 1

42 + 1

52 + · · · .

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191 4 Signal representation using Fourier series

4.24 The impulse response of an LTIC system is given by

h(t) = e−2|t |.

(a) Based on Eq. (4.54), calculate the transfer function H (ω) of the LTIC

system.

(b) The plot of magnitude |H (ω)| with respect to ωgs referred to as the magnitude spectrum of the LTIC system. Plot the magnitude spectrum

of the LTIC system for the range −∞ < ω < ∞. (c) Calculate the output response y(t) of the LTIC system if the impulse

train shown in Fig. P4.7 is applied as an input to the LTIC system.

4.25 Repeat P4.24 for the following LTIC system:

h(t) = [e−2t − e−4t ]u(t),

with the raised square wave function shown in Fig. P4.6(b) applied at the

input of the LTIC system.

4.26 Repeat P4.24 for the following LTIC system:

h(t) = te−4t u(t),

with the sawtooth wave function shown in Fig. P4.6(d) applied at the input

of the LTIC system.

4.27 Consider the following periodic functions represented as CTFS:

(i) x1(t) = 7

∞∑

m=0

2m + 1 sin[8π (2m + 1)t];

(ii) x2(t) = 1.5 + ∞∑

m=0

4m + 1 cos[2π (4m + 1)t].

(a) Determine the fundamental period of x(t).

(b) Determine if x(t) is an even signal or an odd signal.

the functions in the time interval −1 ≤ t ≤ 1. [Hint: You may calcu- late x(t) for t = [−1:0.01:1]. The MA T L A B “plot” function will give a smooth interpolated plot.]

(d) From the plot in step (c), determine the period of x(t). Does it match

your answer to part (a)?

4.28 Using the M A T L A B function ictfs.m (provided in the CD), show

that the periodic function f (t) (shown in Fig. 4.10) considered in

Example 4.8, can be reconstructed from its trigonometric Fourier series

coefficients.

4.29 Using the M A T L A B function ictfs.m (provided in the CD), show that

the periodic function g(t) (shown in Fig. 4.11) considered in Example 4.9,

can be reconstructed from its trigonometric Fourier series coefficients.

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192 Part II Continuous-time signals

4.30 Using the M A T L A B function ictfs.m (provided in the CD), show that

the periodic function g(t) (shown in Fig. 4.10) considered in Example

4.12, can be reconstructed from its exponential Fourier series coefficients.

4.31 Using the M A T L A B function ictfs.m (provided in the CD), show that

the periodic function f (t) (shown in Fig. 4.11) considered in Example

4.13, can be reconstructed from its trigonometric Fourier series coeffi-

cients.

4.32 Using the M A T L A B function ictfs.m (provided in the CD), plot the

output response y(t) obtained in Problem 4.24 for T = 1 s.

4.33 Using the M A T L A B function ictfs.m (provided in the CD), plot the

output response y(t) obtained in Problem 4.25. for T = 1 s.

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C H A P T E R

5 Continuous-time Fourier transform

In Chapter 4, we introduced the frequency representations for periodic sig-

nals based on the trigonometric and exponential continuous-time Fourier series

(CTFS). The exponential CTFS is useful in calculating the output response of

a linear time-invariant (LTI) system to a periodic input signal. In this chapter,

we extend the Fourier framework to continuous-time (CT) aperiodic signals.

The resulting frequency decompositions are referred to as the continuous-time

Fourier transform (CTFT) and are used to express both aperiodic and periodic

CT signals in terms of linear combinations of complex exponential functions.

We show that the convolution in the time domain is equivalent to multiplication

in the frequency domain. The CTFT, therefore, provides an alternative analysis

technique for LTIC systems in the frequency domain.

Chapter 5 is organized as follows. Section 5.1 considers the CTFT as a

limiting case of the CTFS and formally defines the CTFT and its inverse. In

Section 5.2, we provide several examples to illustrate the steps involved in the

calculation of the CTFT for a number of elementary signals. Section 5.3 presents

the look-up table and partial fraction methods for calculating the inverse CTFT.

Section 5.4 lists the symmetry properties of the CTFT for real-valued, even, and

odd signals, while Section 5.5 lists the CTFT properties arising due to linear

transformations in the time domain. The condition for the existence of the

CTFT is derived in Section 5.6, while the relationship between the CTFT and

the CTFS for periodic signals is discussed in Sections 5.7 and 5.8. Section 5.9

applies the convolution property of the CTFT to evaluate the output response of

an LTIC system to an arbitrary CT input signal. The gain and phase responses

of LTIC systems are also defined in this section. Section 5.10 demonstrates how

M A T L A B is used to compute the CTFT, and Section 5.11 concludes the chapter.

5.1 CTFT for aperiodic signals

Consider the aperiodic signal x(t) shown in Fig. 5.1(a). In order to extend

the Fourier framework of the CTFS to aperiodic signals, we consider several

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194 Part II Continuous-time signals and systems

x(t)

L−L

−T0 0 T0

~ xT (t)

L−L

(a) (b)

Fig. 5.1. Periodic extension of a

time-limited aperiodic signal.

(a) Aperiodic signal and (b) its

periodic extension.

repetitions of x(t) uniformly spaced from each other by duration T0 such that

there is no overlap between two adjacent replicas of x(t). The resulting signal

is denoted by x̃T (t) and is shown in Fig. 5.1(b). Clearly, the new signal x̃T (t) is

periodic with the fundamental period of T0 and in the limit

lim T0→∞

x̃T (t) = x(t). (5.1)

Since x̃T (t) is a periodic signal with a fundamental frequency of ω0 = 2π /T0 radians/s, its exponential CTFS representation is expressed as follows:

x̃T (t) = ∞∑

n=−∞ D̃ne

jnω0t , (5.2)

where the exponential CTFS coefficients are given by

D̃n = 1

∫

〈T0〉

x̃T (t)e −jnω0t dt . (5.3)

The spectra of x̃T (t) are the magnitude and phase plots of the CTFS coefficients

D̃n as a function of nω0. Because n takes on integer values, the magnitude

and phase spectra of x̃T (t) consist of vertical lines separated uniformly by

ω0. Applying the limit T0 → ∞ to x̃T (t) causes the spacing ω0 = 2π/T0 in the spectral lines of the magnitude and phase spectra to decrease to zero. The

resulting spectra represent the Fourier representation of the aperiodic signal x(t)

and are continuous along the frequency (ω) axis. The CTFT for aperiodic signals

is, therefore, a continuous function of frequency ω. To derive the mathematical

definition of the CTFT, we apply the limit T0 → ∞ to Eq. (5.3). The resulting expression is as follows:

lim T0→∞

D̃n = lim T0→∞

∫

〈T0〉

x(t)e−jnω0t dt

Dn = lim T0→∞

∞∫

−∞

x(t)e−jnω0t dt since lim T0→0

x̃T (t) = x(t). (5.4)

In Eq. (5.4), the term Dn denotes the exponential CTFT coefficients of x(t).

Let us define a continuous function X (ω) (with the independent variable ω) as

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195 5 Continuous-time Fourier transform

follows:

X (ω) = ∞∫

−∞

x(t)e−jωt dt . (5.5)

In terms of X (ω), Eq. (5.4) can, therefore, be expressed as follows:

Dn = lim T0→∞

T0 X (nω0). (5.6)

Using the exponential CTFS definition, x(t) can be evaluated from the CTFS

coefficients Dn as follows:

x(t) = ∞∑

n=−∞ Dne

jnω0t = lim T0→∞

∞∑

n=−∞

T0 X (nω0)e

jnω0t . (5.7)

As T0 → ∞, the fundamental frequencyω0 approaches a small value denoted by �ω. The fundamental period T0 is therefore given by T0 = 2π /�ω. Substituting T0 = 2π /�ω as ω0 → �ω in Eq. (5.7) yields

x(t) = 1

2π lim

�ω→0

∞∑

n=−∞ X (n�ω) e jn�ωt�ω

︸︷︷︸

. (5.8)

In Eq. (5.8), consider the term A as illustrated in Fig. 5.2. In the limit �ω → 0, term A represents the area under the function X (ω)exp(jωt). Therefore Eq.

(5.8) can be rewritten as follows:

CTFT synthesis equation x(t) = 1

2π =

∞∫

−∞

X (ω)e−jωt dt, (5.9)

which is referred to as the synthesis equation for the CTFT used to express

any aperiodic signal in terms of complex exponentials, exp(jωt). The analysis

equation of the CTFT is given by Eq. (5.5), which, for convenience of reference,

is repeated below.

CTFT analysis equation X (ω) = ∞∫

−∞

x(t)e−jωt dt . (5.10)

X (w)e jwt

n∆w (n + 1)∆w∆w

A = X(nw)e jn∆wt∆wFig. 5.2. Approximation of the

term ∞∑

n=−∞ X(n�ω)

e jn�ωt �ω as the area under the

function X (ω)exp( jωt ).

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196 Part II Continuous-time signals and systems

Collectively, Eqs. (5.9) and (5.10) form the CTFT pair, which is denoted by

x(t) DTFT←−−→ X (ω). (5.11)

Alternatively, the CTFT pair may also be represented as follows:

X (ω) = ℑ{x(t)} (5.12)

x(t) = ℑ−1{X (ω)}, (5.13)

where ℑ denotes for the CTFT and ℑ−1 denotes the inverse of the CTFT. Based on Eqs. (5.10) and (5.11), we make the following observations about the CTFT.

(1) The frequency representation of a periodic signal x̃(t) is obtained by

expressing x̃(t) in terms of the CTFS. The basis function of the CTFS

consists of complex exponentials {exp(jnω0t)}, which are defined at the

fundamental frequency ω0 and its harmonics nω0. The frequency repre-

sentation of an aperiodic signal x(t) is obtained through the CTFT, where

the complex exponential exp(jnωt) is the basis function. The variable ω

in the basis function of the CTFT is a continuous variable and may have

any value within the range −∞ < ω < ∞. Unlike the CTFS, the CTFT is therefore defined for all frequencies ω.

(2) In general, the CTFT X (ω) is a complex function of the angular frequency

ω. A great deal of information is obtained by plotting the magnitude and

phase of X (ω) with respect to ω. The plots of magnitude |X (ω)| and phase <X (ω) with respect to ω are, respectively, referred to as the magnitude and

phase spectra of the aperiodic function x(t).

(3) In deriving the definition of the CTFT, we assumed that the aperiodic

function x(t) is time-limited such that x(t) = 0 for |t | > L . This is not a required condition for the existence of the CTFT. In other words, the

function x(t) may be infinitely long but its CTFT can exist.

5.2 Examples of CTFT

In Section 5.2, we calculate the forward and inverse CTFT of several well known

functions. We assume that the CTFT exists in all cases. A general condition for

the existence of the CTFT is derived in Section 5.6.

Example 5.1

Determine the CTFT of the following functions and plot the corresponding

magnitude and phase spectra:

(i) x1(t) = exp(−at)u(t), a ∈ R+; (ii) x2(t) = exp(−a|t |), a ∈ R+.

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197 5 Continuous-time Fourier transform

Table 5.1. Magnitude |X (ω)| and phase <X (ω) for the CTFT of x (t ) = exp(−3t )u(t ) in Example 5.1

ω (radians/s) −∞ −1000 −100 −10 −1 0 1 10 100 ∞ Magnitude: |X (ω)| 0 0.001 0.01 0.096 0.316 0.333 0.316 0.096 0.01 0 Phase: <X (ω) π/2 1.57 1.54 1.28 0.32 0 −0.32 −1.28 −1.54 −π/2

The notation a ∈ R+ implies that a is real-valued within the range −∞ < a < ∞.

Solution

(i) Based on the definition of the CTFT, Eq. (5.10), we obtain

X1(ω) = ℑ{e−at u(t)} = ∞∫

−∞

e−at u(t)e−jωt dt = ∞∫

e−(a+jω)t dt

= − 1

(a + jω) [

e−(a+jω)t ]∞

0 = −

(a + jω)

[

lim t→∞

e−(a+jω)t − 1 ]

where the term

lim t→∞

e−(a+jω)t = lim t→∞

e−at · lim t→∞

e−jωt = 0 · lim t→∞

e−jωt = 0.

Therefore,

X1(ω) = 1

a + jω .

The magnitude and phase of X1(ω) are given by

magnitude |X1(ω)| = ∣ ∣ ∣ ∣

a + jω

∣ ∣ ∣ ∣ =

1 √

a2 + ω2 ;

phase <X1(ω) = < 1

a + jω = <1 − <(a + jω) = −tan−1

(ω

)

Table 5.1 lists the amplitude and phase of X (ω) for several values of ω with

a = 3. The exponentially decaying function x1(t) and its magnitude and phase spectra are plotted in Fig. 5.3.

(ii) Based on the definition of the CTFT, Eq. (5.10), we obtain

X2(ω) = ℑ{e−a|t |} = ∞∫

−∞

e−a|t |e−jωt dt

= ∞∫

−∞

e−a|t | cos(ωt) ︸︷︷︸

even function

dt − j ∞∫

−∞

e−a|t | sin(ωt) ︸︷︷︸

odd function

dt.

Since the integral of an odd function with limits [−L , L] is zero, the above

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198 Part II Continuous-time signals and systems

t 0

x1(t)

w 0

X1(w)1/a

w 0

< X1(w) p/2

−p/2

(a) (b) (c)

Fig. 5.3. CTFT of the causal

decaying exponential function

x(t ) = e−at u(t ). (a) x(t ); (b) magnitude spectrum;

equation reduces to

X2(ω) = ∞∫

−∞

e−a|t | cos(ωt)dt = 2 ∞∫

e−at cos(ωt)dt

= 2

a2 + ω2 [−ae−at cos(ωt) + ωe−at sin(ωt)]∞0 =

a2 + ω2 .

Since X2(ω) is positive real-valued, the magnitude and phase of X2(ω) are given

magnitude |X2(ω)| = ∣ ∣ ∣ ∣

a2 + ω2

∣ ∣ ∣ ∣ =

a2 + ω2 .

phase <X2(ω) = 0.

The non-causal exponentially decaying function x2(t) and its magnitude and

phase spectra are plotted in Fig. 5.4.

We note from Example 5.1 that the magnitude spectrum is symmetric along

the vertical axis while the phase spectrum is symmetric about the origin. The

magnitude spectrum is, therefore, an even function of ω, while the phase spec-

trum is an odd function of ω. This is a consequence of the symmetry properties

observed by real-valued functions. The symmetry properties are discussed in

detail in Section 5.3.

Example 5.2

Calculate the CTFT of a constant function x(t) = 1.

0 t

x2(t)

ω 0

X2(w)2a

w 0

< X2(w) = 0 p/2

−p/2

(a) (b) (c)

Fig. 5.4. CTFT of the causal

decaying exponential function

x2(t ) = exp(−a|t |). (a) x2(t ); (b) Magnitude spectrum;

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199 5 Continuous-time Fourier transform

t 0

x(t) = 1 1

w 0

X(ω) = 2pd(w)2p

<X(w) = 0 |X(w)|

(a) (b)

Fig. 5.5. CTFT of a constant

function. (a) Constant function,

x(t ) = 1; (b) its CTFT, X (ω) = 2πδ(ω).

Solution

Based on the definition of the CTFT, Eq. (5.10), we obtain

X (ω) = ℑ{1} = ∞∫

−∞

e−jωt dt . (5.14)

It can be shown that (see Problem 5.10)

∞∫

−∞

e jωt dt = 2πδ(ω). (5.15)

Substituting ω by −ω on both sides of Eq. (5.15), we obtain ∞∫

−∞

e−jωt dt = 2πδ(−ω) = 2πδ(ω),

which results in

X (ω) = ∞∫

−∞

e−jωt dt = 2πδ(ω).

In other words,

1 CTFT –−→ 2πδ(ω). (5.16)

The magnitude spectrum of a constant function x(t) = 1 therefore consists of an impulse function with area 2π located at the origin, ω = 0, in the frequency domain. The magnitude spectrum is plotted in Fig. 5.5(b). The phase is zero for

all frequencies (∞ ≤ ω ≤ −∞).

Example 5.3

The CTFT of an aperiodic function g(t) is given by G(ω) = 2πδ(ω). Determine the aperiodic function g(t).

Solution

Based on the CTFT analysis equation, Eq. (5.10), we obtain

g(t) = ℑ−1{2πδ(ω)} = 1

2π

∞∫

−∞

2πδ(ω)e jωt dω = ∞∫

−∞

δ(ω)dω = 1.

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200 Part II Continuous-time signals and systems

t 0

x(t) = d(t)1

w 0

X(w) = 11

<X(w) = 0

|X(w)|

(b)(a)

Fig. 5.6. CTFT of an impulse

function. (a) Impulse function,

x(t ) = δ(t ); (b) its CTFT, X(ω) = 1.

In other words,

1 CTFT←−– 2πδ(ω). (5.17)

Combining the results in Examples 5.2 and 5.3, we obtain the CTFT pair:

1 CTFT←−−→ 2πδ(ω). (5.18)

Example 5.4

Determine the Fourier transform of the impulse function x(t) = δ(t).

Solution

Based on the definition of the CTFT, Eq. (5.10), we obtain

X (ω) = ℑ{δ(t)} = ∞∫

−∞

δ(t)e−jωt dt = ∞∫

−∞

δ(t)dt = 1.

Therefore,

δ(t) CTFT –−→ 1.

The CTFT of the impulse function located at the origin (t = 0) is a constant. The magnitude spectrum is shown in Fig. 5.6. The phase spectrum is zero for

all frequencies ω.

Example 5.5

The CTFT of an aperiodic function g(t) is given by G(ω) = 1. Determine the aperiodic function g(t).

Solution

Based on the CTFT analysis equation, Eq. (5.10), we obtain

g(t) = ℑ−1{1} = 1

2π

∞∫

−∞

1 · e jωt dt . (5.19)

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201 5 Continuous-time Fourier transform

By interchanging the role of ω and t in Eq. (5.15), we obtain

∞∫

−∞

e jωt dω = 2πδ(t).

Substituting the above relationship in Eq. (5.19) yields

g(t) = 1

2π

∞∫

−∞

e jωt dt = 1

2π × 2πδ(t) = δ(t).

Therefore,

δ(t) CTFT←−– 1. (5.20)

Combining the results derived in Examples 5.4 and 5.5, we can form the CTFT

pair:

δ(t) CTFT←−−→ 1. (5.21)

In Example 5.5, we proved that the inverse CTFT of G(ω) = 1 is given by the impulse function g(t) = δ(t). In Example 5.4, we showed the converse: that the CTFT of g(t) = δ(t) is G(ω) = 1. Likewise, in Examples 5.2 and 5.3, we established the CTFT pair,

1 CTFT←−−→ 2πδ(ω),

by computing the forward and inverse CTFT. Since the CTFT pair is unique,

it is sufficient to compute either the CTFT or its inverse. Once the CTFT is

derived, its inverse is established automatically, and vice versa. In the remaining

examples, we form the CTFT pair by deriving either the forward CTFT or its

inverse.

A second observation made from the CTFT pairs given in Eqs. (5.18) and

(5.21),

1 CTFT←−−→ 2πδ(ω) and δ(t) CTFT←−−→ 1,

is that the CTFT exhibits a duality property. The CTFT of a constant is the

impulse function, while the CTFT of an impulse function is a constant. A factor

of 2π is also introduced. We revisit the duality property in Section 5.5.

Example 5.6

Calculate the CTFT of the rectangular function f (t) shown in Fig. 5.7(a).

Solution

Based on the definition of the CTFT, Eq. (5.10), we obtain

F(ω) = ℑ {rect (t /τ )} = τ/2∫

−τ/2

1 · e−jωt dt = [

e−jωt

−jω

]τ/2

−τ/2 ,

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202 Part II Continuous-time signals and systems

1 f (t) = rect( )

2 t

t t

2 t−

0 t 2p

t 2p−

t ωt 2p

F(w) = tsinc( )

(a) (b)

Fig. 5.7. CTFT of the rectangular

function. (a) Rectangular

function; (b) its CTFT given by

the sinc function.

which simplifies to

F(ω) = − 1

jω [e−jωt ]

τ/2

−τ/2 = − 1

jω [e−jωτ/2 − e jωτ/2] = −

jω

[

−2j sin (ωτ

)]

F(ω) = 2

ω sin

(ωτ

)

= τ sinc (ωτ

2π

)

The Fourier transform F(ω) is plotted in Fig. 5.7(b). The CTFT pair for a

rectangular function is given by

rect

( t

)

CTFT←−−→ τ sinc (ωτ

2π

)

. (5.22)

Example 5.7

Determine the aperiodic function g(t) whose CTFT G(ω) is the rectangular

function shown in Fig. 5.8(a).

Solution

From Fig. 5.8(a), we observe that

G(ω) = {

1 |ω| ≤ W 0 |ω| > W.

Based on the CTFT analysis equation, Eq. (5.10), we obtain

g(t) = ℑ−1 {

rect ( ω

)}

= 1

2π

W∫

−W

1 · e jωt dω = 1

2π

[ e jωt

−W , (5.23)

which simplifies to

g(t) = 1

j2π t [e jW t − e−jW t ] =

j2π t [2j sin(W t)] =

sin(W t)

π t

= W

π sinc

( W

π t

)

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203 5 Continuous-time Fourier transform

0 0W

1 2W w

G(w) = rect ( )

−W

(a)

t ( )Wpg(t) = sinc( )Wtp

W − p

W p

(b)

Fig. 5.8. Inverse CTFT of the

rectangular function.

(a) Frequency domain

representation G(ω) = rect(ω/2W ); (b) its inverse

CTFT given by the sinc function.

The aperiodic function g(t) and its CTFT are plotted in Fig. 5.8. Example 5.7

establishes the following CTFT pair:

π sinc

( W

π t

)

CTFT←−−→ rect ( ω

)

= {

1 |ω| ≤ W 0 |ω| > W. (5.24)

Example 5.8

Determine the signal x(t) whose CTFT is a frequency-shifted impulse function

X (ω) = δ(ω – ω0).

Solution

Based on the CTFT analysis equation, Eq. (5.10), we obtain

x(t) = ℑ−1{δ(ω − ω0)} = 1

2π

∞∫

−∞

δ(ω − ω0)e−jωt dω

= 1

2π e−jω0t

∞∫

−∞

δ(ω − ω0)dω = 1

2π e−jω0t .

Example 5.8 proves the following CTFT pair:

e jω0t CTFT←−−→ 2πδ(ω − ω0). (5.25)

Substituting ω0 by −ω0 in Eq. (5.25), we obtain another CTFT pair:

e−jω0t CTFT←−−→ 2πδ(ω + ω0). (5.26)

In Examples 5.1 to 5.8, we evaluated several CTFT pairs for some elementary

time functions. Table 5.2 lists the CTFTs for additional time functions. In prac-

tice, a graphical plot of the CTFT helps to understand the frequency properties

of the function. In Table 5.3, we illustrate the frequency responses of several

functions by plotting their magnitude and phase spectra. In the plots, the magni-

tude spectra are shown as solid lines and the phase spectra are shown as dashed

lines. In certain cases, the values of the corresponding phases are zero for all

frequencies, and in these cases the phase spectra are not plotted.

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204 Part II Continuous-time signals and systems

Table 5.2. CTFT pairs for elementary CT signals

Time domain Frequency domain

CT signals x(t) = 1

2π

∞∫

−∞

X (ω)e jωt dt X (ω) = ∞∫

−∞

x(t)e−jωt dt Comments

(1) Constant 1 2πδ(ω)

(2) Impulse function δ(t) 1

(3) Unit step function u(t) πδ(ω) + 1

jω

(4) Causal decaying

exponential function

e−at u(t) 1

a + jω a > 0

(5) Two-sided decaying

exponential function

e−a|t | 2a

a2 + ω2 a > 0

(6) First-order time-rising

causal decaying

exponential function

te−at u(t) 1

(a + jω)2 a > 0

(7) N th-order time-rising

causal decaying

exponential function

tne−at u(t) n!

(a + jω)n+1 a > 0

(8) Sign function sgn(t) = {

1 t > 0

−1 t < 0 2

jω

(9) Complex exponential ejω0 t 2πδ(ω − ω0) (10) Periodic cosine function cos(ω0t) π [δ(ω − ω0) + δ(ω + ω0)]

(11) Periodic sine function sin(ω0t) π

j [δ(ω − ω0) − δ(ω + ω0)]

(12) Causal cosine function cos(ω0t)u(t) π

2 [δ(ω − ω0) + δ(ω + ω0)] +

jω

ω20 − ω2

(13) Causal sine function sin(ω0t)u(t) π

2j [δ(ω − ω0) − δ(ω + ω0)] +

ω0

ω20 − ω2

(14) Causal decaying

exponential cosine

function

e−at cos(ω0t)u(t) a + jω

(a + jω)2 + ω20 a > 0

(15) Causal decaying

exponential sine function

e−at sin(ω0t)u(t) ω0

(a + jω)2 + ω20 a > 0

(16) Rectangular function rect

( t

)

= {

1 |t | ≤ τ/2 0 |t | > τ/2 τ sinc

(ωτ

2π

)

τ �= 0

(17) Sinc function W

π sinc

( W t

)

rect ( ω

)

= {

1 |ω| ≤ W 0 |ω| > W

(18) Triangular function △ (

)

{

1 − |t | τ

|t | ≤ τ 0 otherwise

τ sinc 2(ωτ

2π

)

τ > 0

(19) Impulse train

∞∑

k=−∞ δ(t − kT0) ω0

∞∑

m=−∞ δ(ω − mω0) angular

frequency

ω0 = 2π /T0 (20) Gaussian function e−t

2/2σ 2 σ √

2πe−σ 2ω2/2

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Ta b

le 5.

3. M

ag ni

tu de

an d

ph as

e sp

ec tr

a fo

r se

le ct

ed el

em en

ta ry

C T

fu nc

tio ns

M ag

ni tu

de sp

ec tr

a ar

e sh

ow n

as lin

es an

d ph

as e

sp ec

tr a

ar e

sh ow

n as

da sh

ed lin

F u

n ct

io n

T im

e- d

o m

ai n

w av

ef o

rm M

ag n

it u

d e

an d

p h

as e

sp ec

tr a

(1 )

C o

n st

an t

x (t

) =

x( t)

= 1

X (w

) =

2 p

d( w

) 2p

< X

(w )

= 0

|X (w

(2 )

U n

it im

p u

ls e

fu n

ct io

n x

(t ) =

δ (t

)

x( t)

= d

(t )

< X

(w )

= 0

|X (w

)| w

X (w

) =

1 1

(3 )

U n

it st

ep fu

n ct

io n

x (t

) =

u (t

)

t 0

x( t)

= u

(t )

0 ω

jw X

(w )

= p

d( w

) + 1

|X (w

< X

(w )

p /2

−p /2

(4 )

D ec

ay in

g ex

p o

n en

ti al

x (t

) =

e− a

t u (t

)

t 0

x( t)

= e

−a t u

(t )

< X

(w )

p /2

−p /2

a +

jw X

(w )

= 1 |X (w

(5 )

T w

o -s

id ed

d ec

ay in

g ex

p o

n en

ti al

x (t

) =

e− a |t|

t 0

x( t)

= e

−a |t |

w 0

< X

(w )

= 0

2 /a

a 2 +w

2 X

(w )

= 2 a

|X (w

(c o

n t.

)

205

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CUUK852-Mandal & Asif May 25, 2007 20:5

Ta b

le 5.

3. (c

on t.)

F u

n ct

io n

T im

e- d

o m

ai n

w av

ef o

rm M

ag n

it u

d e

an d

p h

as e

sp ec

tr a

(6 )

F ir

st -o

rd er

ti m

e- ri

si n

g d

ec ay

in g

ex p

o n

en ti

fu n

ct io

n x

(t ) =

te −

a t u

(t )

t 0

x( t)

= t

e− a

t u (t

)

< X

(w )

−p

(a +

jw )2

X (w

) =

a 21

|X (w

(7 )

N th

-o rd

er ti

m e-

ri si

n g

d ec

ay in

g ex

p o

n en

ti al

fu n

ct io

n x

(t ) =

tn e−

a t u

(t )

x( t)

= tn

e− a

t u (t

)

< X

(w )

(a +

jw )n

+ 1

n !

X (w

) =

a n

+ 1

n !

−

(n +1

|X (w

(8 )

S ig

n fu

n ct

io n

sg n

(t ) =

{

1 t >

− 1

t <

t 0

x( t)

= s

g n

(t )

−1

< X

(w )

jw X

(w )

= 2

2p −

|X (w

(9 )

C o

m p

le x

ex p

o n

en ti

al fu

n ct

io n

x (t

) =

e jω

0 t

x( t)

= e

jw 0 t

|x (t

)| =

< x

(t )

= w

0 w

w 0

X (w

) =

2 p

d( w

−w 0 )

< X

(w )

= 0

|X (w

w 02

(1 0

) C

o si

n e

fu n

ct io

n x

(t ) =

co s(

ω 0 t)

t 0

x( t)

= c

o s (w

0 t)

w 0X

(w )

= p

[d (w

−w 0 )+

d (w

+ w

0 )]

< X

(w )

= 0

w 0

−w 0

p p

206

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CUUK852-Mandal & Asif May 25, 2007 20:5

(1 1

) S

in e

fu n

ct io

n x

(t ) =

si n

(ω 0 t)

t 0

x( t)

= s

in (w

0 t)

w 0X

(w )

= j

p [d

(w +

w 0 )−

d( w

−w 0 )]

w 0

−w 0

p p

p /2

−p /2

|X (w

< X

(w )

(1 2

) C

au sa

l co

si n

e fu

n ct

io n

x (t

) =

co s(

ω 0 t)

u (t

)

t 0

x( t)

= c

o s(

w 0 t)

u (t

)

w 0

X (w

) =

[d (w

−w 0 )+

d (w

−w 0 )]

w 0

−w 0

|X (w

p 2 jw

/w 2 0 −w

(1 3

) C

au sa

l si

n e

fu n

ct io

n x

(t ) =

si n

(ω 0 t)

u (t

)

t 0

x( t)

= s

in (w

0 t)

u (t

)

w 0

X (w

) =

[d (w

−w 0 )+

d( w

−w 0 )]

w 0

−w 0

|X (w

p 2j w

0 /w

2 0 −w

(1 4

) C

au sa

l d

ec ay

in g

ex p

o n

en ti

al co

si n

e fu

n ct

io n

x (t

) =

e− a

t co

s( ω

0 t)

u (t

)

x( t)

= e

−a t c

o s(

w 0 t)

u (t

)

(a +

jw )2

+w 02

(a +

jw )

X (w

) =

< X

(w )

p /2

−p /2

|X (w

(1 5

) C

au sa

l d

ec ay

in g

ex p

o n

en ti

al si

n e

fu n

ct io

x (t

) =

e− a

t si

n (ω

0 t)

u (t

)

0 w

(a +

jw )2

+w 02

w 0

X (w

) =

< X

(w )

−p

|X (w

)| x(

t) =

e −a

t s in

(w 0 t)

u (t

)

(c o

n t.

)

207

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Ta b

le 5.

3. (c

on t.)

F u

n ct

io n

T im

e- d

o m

ai n

w av

ef o

rm M

ag n

it u

d e

an d

p h

as e

sp ec

tr a

(1 6

) G

at e

fu n

ct io

n x

(t ) =

re ct

( t τ

)

t 0

1 tt

x( t)

= re

ct ( )

2t −

2t 0

t2 p

−

< X

(w )

w t

2 p

X (w

) = t

si n

c( )

|X (w

(1 7

) S

in c

fu n

ct io

n x

(t ) =

( W π

)

si n

( W

)

0 Wp

Wp −

W p W

p x(

t) =

( )s

in c(

)

t 0

1 2

Ww X

(w )

= re

ct (

)

−W W

< X

(w )

= 0

|X (w

(1 8

) T

ri an

g u

la r

fu n

ct io

�

( t τ

)

{

1 −

|t| τ |t|

≤ τ

0 o

th er

w is

e t

1 tt

x( t)

= ∆

( )

−t t

0 t2 p

t2 p

−

w <

X (w

) =

w t

2 p

X (w

) = t

si n

c2 (

) |X

(w )|

(1 9

) Im

p u

ls e

tr ai

n x

(t ) =

∞ ∑

k =

− ∞

δ (t

− k

T )

t 0

∞ k= −

∞ x(

t) =

∑ d(

t− kT

)

−T T

|X (w

∑∞ d(w

−

)

k =

− ∞

T T

X 2

kp 2

p (w

) = T2p

T −

2 p

T2 p

T4 p

T −

4 p

< X

(w )

= 0

(2 0

) G

au ss

ia n

fu n

ct io

n x

(t ) =

e− t2

/ 2 σ

x( t)

= e

−t 2 /2

s 2

w 0

X (w

) =

s √2

p e

−s 2 w

2 /2

< X

(w )

= 0

|X (w

s √2

208

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209 5 Continuous-time Fourier transform

5.3 Inverse Fourier transform

Evaluation of the inverse CTFT is an important step in analysis of LTIC systems.

There are three main approaches that may be taken to calculate the inverse

CTFT:

(i) using the synthesis equation;

(ii) using a look-up table;

(iii) using partial fraction expansion.

In the first approach, the inverse CTFT is calculated by solving the synthe-

sis equation, Eq. (5.9). This method was used in Examples 5.3, 5.5, 5.7, and

5.8. However, this approach is difficult. We now present the second and third

approaches. Approach (ii) is straightforward as it determines the inverse CTFT

by comparing the entries with Table 5.2. We illustrate this with an example.

Example 5.9

Using the look-up table method, calculate the inverse CTFT of the following

function:

X (ω) = 2(jω) + 24

(jω)2 + 4(jω) + 29 . (5.27)

Solution

The function X (ω) is decomposed into simpler terms, whose inverse CTFT

can be determined directly from Table 5.2. One possible decomposition is as

follows:

X (ω) = 2 2 + ( jω)

(2 + jω)2 + 52 + 4

(2 + jω)2 + 52 . (5.28)

From Entries (14) and (15) of Table 5.2, we know that

e−2t cos(5t)u(t) CTFT←−−→

2 + jω (2 + jω)2 + 52

and

e−2t sin(5t)u(t) CTFT←−−→

(2 + jω)2 + 52 .

Therefore, the inverse CTFT is calculated as follows:

x(t) = 2e−2t cos(5t)u(t) + 4e−2t sin(5t)u(t). (5.29)

5.3.1 Partial fraction expansion

The look-up table approach is simple to use once a suitable decomposition is

obtained. A major problem, however, is faced in the decomposition of the CTFT

X (ω) in terms of simpler functions whose inverse CTFTs are listed in Table 5.2.

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210 Part II Continuous-time signals and systems

We now present approach (iii), which uses the partial fraction expansion to

decompose systematically a rational function in simpler terms. Consider the

CTFT

X (ω) = N (ω)

D(ω) =

bm( jω) m + bm−1( jω)m−1 + · · · + b1( jω) + b0

( jω)n + an−1( jω)n−1 + · · · + a1( jω) + a0 , (5.30)

where the numerator is an mth-order polynomial and the denominator is an

nth-order polynomial. The partial fraction method is explained in more detail

in Appendix D (see Section D.2). The main steps are summarized as follows.

(1) Factorize D(ω) into n first-order factors and express X (ω) as follows:

X (ω) = N (ω)

( jω − p1)( jω − p2) · · · ( jω − pn) . (5.31)

(2) If there are no repeated or complex roots in D(ω), X (ω) is expressed in

terms of n partial fractions:

X (ω) = k1

( jω − p1) +

( jω − p2) + · · · +

( jω − pn) , (5.32)

where the partial fraction coefficients are calculated using the Heaviside

formula as follows:

kr = [( jω − pr )X (ω)]jω=pr , (5.33)

for 1 ≤ r ≤ n. For repeated or complex roots, the partial fraction expansion is more complicated and is discussed in Appendix D.

(3) The inverse CTFT can then be calculated as follows:

x(t) = [k1ep1t + k2ep2t + · · · + knepn t ]u(t). (5.34)

Example 5.10

Using the partial fraction method, calculate the inverse CTFT of the following

function:

X (ω) = 5(jω) + 30

(jω)3 + 17(jω)2 + 80(jω) + 100 .

Solution

In terms of jω, the roots of D(ω) = (jω)3 + 17(jω)2 + 80(jω) + 100 are given by jω = −2, −5, and −10. The partial fraction expansion of X (ω) is given by

X (ω) = 5(jω) + 30

(jω + 2)( jω + 5)( jω + 10) ≡

( jω + 2) +

( jω + 5) +

( jω + 10) ,

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211 5 Continuous-time Fourier transform

where the partial fraction coefficients are given by

k1 = ( jω + 2) 5(jω) + 30

(jω + 2)( jω + 5)( jω + 10)

∣ ∣ ∣ ∣ jω=−2

= 5(jω) + 30

(jω + 5)( jω + 10)

∣ ∣ ∣ ∣ jω=−2

= 20

(3)(8) =

6 ,

k2 = ( jω + 5) 5(jω) + 30

(jω + 2)( jω + 5)( jω + 10)

∣ ∣ ∣ ∣ jω=−5

= 5(jω) + 30

(jω + 2)( jω + 10)

∣ ∣ ∣ ∣ jω=−5

= 5

(−3)(5) = −

3 ,

and

k3 = ( jω + 10) 5(jω) + 30

(jω + 2)( jω + 5)( jω + 10)

∣ ∣ ∣ ∣ jω=−10

= 5(jω) + 30

(jω + 2)( jω + 5)

∣ ∣ ∣ ∣ jω=−10

= −20

(−8)(−5) = −

2 .

Therefore, the partial fraction expansion of X (ω) is given by

X (ω) ≡ 5

6(jω + 2) −

3(jω + 5) −

2(jω + 10) . (5.35)

Using the CTFT pairs in Table 5.2 to calculate the inverse CTFT, the function

x(t) is calculated as

x(t) =

[ 5

6 e−2t −

3 e−5t −

2 e−10t

]

u(t). (5.36)

5.4 Fourier transform of real, even, and odd functions

In Example 5.1, it was observed that the CTFT of a causal decaying exponential,

e−at u(t) CTFT←−−→

(a + jω) ,

has an even magnitude spectrum, while the phase spectrum is odd. This is

known as Hermitian symmetry and holds true for the CTFT of any real-valued

function. In this section, we consider various properties of the CTFT for real-

valued functions.

5.4.1 CTFT of real-valued functions

5.4.1.1 Hermitian symmetry property

The CTFT X (ω) of a real-valued signal x(t) satisfies the following:

X (−ω) = X∗(ω) , (5.37)

where X∗(ω) denotes the complex conjugate of X (ω).

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212 Part II Continuous-time signals and systems

Proof

By definition,

X∗(ω) = [ℑ{x(t)}]∗ =





∞∫

−∞

x(t)e−jωt dt





∗

= ∞∫

−∞

[x(t)e−jωt ]∗dt,

which simplifies to

X∗(ω) = ∞∫

−∞

x∗(t)e jωt dt .

Since x(t) is a real-valued signal, x∗(t) = x(t) and we obtain

X∗(ω) = ∞∫

−∞

x(t)e−j(−ω)t dt = X (−ω),

which completes the proof.

The Hermitian property can also be expressed in terms of: (i) the real and

imaginary components of the CTFT X (ω), and (ii) the magnitude and phase

of X (ω). These lead to alternative representations for the Hermitian property,

which are listed below.

5.4.1.2 Alternative form I for Hermitian symmetry property

The real component of the CTFT X (ω) of a real-valued signal x(t) is even,

while its imaginary component is odd. Mathematically,

Re{X (−ω)} = Re{X (ω)} and Im{X (−ω)} = −Im{X (ω)}. (5.38)

Proof

Substituting X (ω) = Re{X (ω)} + j Im{X (ω)} in the Hermitian symmetry prop- erty, Eq. (5.37), yields

Re{X (−ω)} + j Im{X (−ω)} = Re{X (ω)} − j Im{X (ω)}.

Separating the real and imaginary components in the above expression proves

the alternative form I of the Hermitian symmetry property.

5.4.1.3 Alternative form II for Hermitian symmetry property

The magnitude spectrum |X (ω)| of the CTFT X (ω) of a real-valued signal x(t) is even, while its phase spectrum <X (ω) is odd. Mathematically,

|X (−ω)| = |X (ω)| and <X (−ω) = −<X (−ω). (5.39)

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213 5 Continuous-time Fourier transform

Proof

The magnitude of the complex function X (−ω) = Re{X (−ω)} + j Im{X (−ω)} is given by

|X (−ω)| = √

(Re{X (−ω)})2 + (Im{X (−ω)})2 .

Substituting Re{X (−ω)} = Re{X (ω)} and Im{X (−ω)} = −Im{X (ω)}, obtained from the alternative form I of the Hermitian symmetry property in

the above expression, yields

|X (−ω)| = √

(Re{X (ω)})2 + (−Im{X (ω)})2 = |X (ω)|,

which proves that the magnitude spectrum |X (ω)| of a real-valued signal is even. Alternatively, consider the phase of the complex function X (−ω) = Re{X (−ω)} + j Im{X (−ω)} as given by

<X (−ω) = tan−1 (

Re{X (−ω)} Im{X (−ω)}

)

Substituting Re{X (−ω)} = Re{X (ω)} and Im{X (−ω)} = −Im{X (ω)} yields

<X (−ω) = tan−1 (

Re{X (−ω)} −Im{X (−ω)}

)

= −<X (ω),

which proves that the phase spectrum <X (ω) of a real-valued signal is odd.

Example 5.11

Consider a function g(t) whose CTFT is given by G(ω) = 1 + 2πδ(ω − ω0). Determine if g(t) is a real-valued function.

Solution

Substituting ω by −ω in the CTFT G(ω) yields

G(−ω) = 1 + 2πδ(−ω − ω0) = 1 + 2πδ(ω + ω0).

The complex conjugate of G(ω) is given by

G∗(ω) = [1 + 2πδ(ω − ω0)]∗ = 1 + 2πδ(ω − ω0).

Comparing the two expressions, it is clear that G*(ω) �= G(−ω), and therefore that g(t) is not a real-valued function. In order to verify the result, we calculate

the inverse CTFT of G(ω) as follows:

g(t) = ℑ−1{G(ω)} = ℑ−1{1+2πδ(ω − ω0)} = ℑ−1{1}+2πℑ−1{δ(ω − ω0)},

which results in

g(t) = δ(t) + e jω0t ,

g(t) = δ(t) + cos(ω0t) + j sin(ω0t),

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214 Part II Continuous-time signals and systems

verifying that g(t) is indeed not real-valued. In deriving the inverse CTFT of

G(ω), we have assumed that the CTFT satisfies the linearity property, which is

formally proved in Section 5.5.

5.4.2 CTFT of real-valued even and odd functions

A second set of symmetry properties is obtained if we assume that, in addition

to being real-valued, x(t) is an even or odd function. Before expressing these

properties, we show that the expression of the CTFT is simplified considerably

if we assume that x(t) is an even or odd function.

Using the Euler identity, the CTFT is expressed as follows:

X (ω) = ∞∫

−∞

x(t)e−jωt dt = ∞∫

−∞

x(t) cos(ωt)dt − j ∞∫

−∞

x(t) sin(ωt)dt.

Case I If x(t) is even, then x(t) cos(ωt) is also an even function, while x(t) sin(ωt) is an odd function. Therefore, the CTFT for the even-valued function

can alternatively be calculated from

X (ω) = 2 ∞∫

x(t) cos(ωt)dt. (5.40)

Case II If x(t) is odd, then x(t) sin(ωt) is an even function, while x(t) cos(ωt) is an odd function. An alternative expression for the CTFT for the odd-valued

function is given by

X (ω) = −j2 ∞∫

x(t) sin(ωt)dt. (5.41)

By combining the Hermitian property with Eqs. (5.40) and (5.41), the following

two properties are obtained.

Property 5.1 CTFT of real-valued, even functions The CTFT X (ω) of a real-

valued, even function x(t) is also real and even. In other words, Re{X (ω)} = Re{X (−ω)} and Im{X (ω)} = 0.

Property 5.2 CTFT of real-valued, odd functions The CTFT X (ω) of a real-

valued, odd function x(t) is imaginary and odd. In other words, Re{X (ω)} = 0 and Im{X (ω)} = −Im{X (−ω)}.

The proofs of Properties 5.1 and 5.2 are left as exercises for the readers. See

Problems 5.6 and 5.7. The symmetry properties of the CTFT are summarized

in Table 5.4.

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215 5 Continuous-time Fourier transform

t 0 1 2

x1(t)

−1−2

0 1 2

2 x2(t)

−2

−1−2

(a) (b)

Fig. 5.9. CT signals used in

Example 5.12. (a) x1(t );

(b) x2(t ).

Example 5.12

Calculate the Fourier transform of the functions x1(t) and x2(t) shown in

Fig. 5.9.

Solution

(a) The mathematical expression for the CT function x1(t), illustrated in

Fig. 5.9(a), is given by

x1(t) =







2|t | −1 ≤ t ≤ 1 2 1 < |t | ≤ 2 0 elsewhere.

Since x1(t) is an even function, its CTFT is calculated using Eq. (5.40) as

follows:

X1(ω) = 2 ∞∫

x(t) cos(ωt)dt = 2 1∫

(2t) cos(ωt) dt + 2 2∫

2 cos(ωt) dt,

which simplifies to

X1(ω) = 4 [

t sin(ωt)

ω + 1

cos(ωt)

ω2

+ 4 [

sin(ωt)

X1(ω) = 4 ⌊

sin(ω)

ω +

cos(ω)

ω2 −

ω2

⌋

+ 4 ⌊

sin(2ω)

ω −

sin(ω)

⌋

= 4

ω2 [ω sin(2ω) + cos(ω) − 1]. (5.42)

The above result validates the symmetry property for real-valued, even func-

tions. Property 5.1 states that the CTFT of a real-valued, even function is real

and even. This is indeed the case for X1(ω) in Eq. (5.42).

(b) The function x2(t), shown in Fig. 5.9(b), is expressed as follows:

x2(t) =



  

  

−2 −2 ≤ t ≤ 1 2t −1 ≤ t ≤ 1 2 1 < t ≤ 2 0 elsewhere.

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216 Part II Continuous-time signals and systems

Since x2(t) is an odd function, its CTFT, based on Eq. (5.41), is given by

X2(ω) = −j2 ∞∫

x(t) sin(ωt)dt = −j2 1∫

(2t) sin(ωt)dt − j2 2∫

2 sin(ωt)dt,

which simplifies to

X2(ω) = −j4 [

−t cos(ωt)

ω + 1

sin(ωt)

ω2

− j4 [

− cos(ωt)

X2(ω) = j4 [

cos(ω)

ω −

sin(ω)

ω2

]

+ j4 [

cos(2ω)

ω −

cos(ω)

]

= j 4

ω2 [ω cos(2ω) − sin(ω)]. (5.43)

The above result validates the symmetry property for real-valued odd functions.

Property 5.2 states that the CTFT of a real-valued odd function is imaginary

and odd. This is indeed the case for X2(ω) in Eq. (5.43).

5.5 Properties of the CTFT

In Section 5.4, we covered the symmetry properties of the CTFT. In this section,

we present the properties of the CTFT based on the transformations of the

signals. Given the CTFT of a CT function x(t), we are interested in calculating

the CTFT of a function produced by a linear operation on x(t) in the time

domain. The linear operations being considered include superposition, time

shifting, scaling, differentiation and integration. We also consider some basic

non-linear operations like multiplication of two CT signals, convolution in the

time and frequency domain, and Parseval’s relationship. A list of the CTFT

properties is included in Table 5.4.

5.5.1 Linearity

Often we are interested in calculating the CTFT of a signal that is a linear

combination of several elementary functions whose CTFTs are known. In such

a scenario, we use the linearity property to show that the overall CTFT is

given by the same linear combination of the individual CTFTs used in the time

domain. The linearity property is defined below.

If x1(t) and x2(t) are two CT signals with the following CTFT pairs:

x1(t) CTFT←−−→ X1(ω)

and

x2(t) CTFT←−−→ X2(ω)

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217 5 Continuous-time Fourier transform

Table 5.4. Symmetry and transformation properties of the CTFT

Transformation

Time domain Frequency domain

properties x(t) = 1

2π

∞∫

−∞

X (ω)e jωt dω X (ω) = ∞∫

−∞

x(t)e− jωt dt Comments

Linearity a1x1(t) + a2x2(t) a1 X1(ω) + a2 X2(ω) a1, a2 ∈ C

Scaling x(at) 1

|a| X

(ω

)

a ∈ ℜ, real-valued

Time shifting x(t − t0) e−jωt0 X (ω) t0 ∈ ℜ, real-valued Frequency shifting e jω0t x(t) X (ω − ω0) ω0 ∈ ℜ, real-valued

Time differentiation dn x

dtn ( jω)n X (ω) provided dx/dt exists

Time integration

t∫

−∞

x(τ )dτ X (ω)

jω + π X (0)δ(ω) provided

t∫

−∞

x(τ )dτ

exists

Frequency differentiation tn x(t) ( j)n dn X

dωn provided dX/dω exists

Duality X (t) 2πx(−ω) if x(t) CTFT←−−→ X (ω) Time convolution x1(t) ∗ x2(t) X1(ω)X2(ω) convolution in time

domain

Frequency convolution x1(t) × x2(t) 1

2π [X1(ω) ∗ X2(ω)] multiplication in time

domain

Parseval’s relationship Ex = ∞∫

−∞

|x(t)|2dt = 1

2π

∞∫

−∞

|X (ω)|2dω energy in a signal

Symmetry properties

CTFT: X (−ω) = X∗(ω)

Hermitian property x(t) is a real-valued

function

real and imaginary components {

Re{X (ω)} = Re{X (−ω)} Im{X (ω)} = −Im{X (−ω)}

real component is even;

imaginary component

is odd

magnitude and phase spectra {

|X (−ω)| = |X (ω)| <X (−ω) = −<X (ω)

magnitude spectrum is

even; phase spectrum

is odd

Even function x(t) is even X (ω) = 2 ∞∫

x(t) cos(ωt)dt simplified CTFT

expression for even

signals

Odd function x(t) is odd X (ω) = −j2 ∞∫

x(t) sin(ωt)dt simplified CTFT

expression for odd

signals

Real-valued and even

function

x(t) is even and real-valued Re{X (ω)} = Re{X (−ω)} Im{X (ω)} = 0

CTFT is real-valued and

even

Real-valued and odd

function

x(t) is odd and real-valued Re{X (ω)} = 0 Im{X (ω)} = −Im{X (−ω)}

CTFT is imaginary and

odd

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218 Part II Continuous-time signals and systems

then, for any arbitrary constants a1 and a2, the linearity property states that

a1x1(t) + a2x2(t) CTFT←−−→ a1 X1(ω) + a2 X2(ω), for a1, a2 ∈ C, (5.44)

where C denotes the set of complex numbers.

Proof

By Eq. (5.10), the CTFT of the linear combination a1x1(t) and a2x2(t) is given

ℑ{a1x1(t) + a2x2(t)} = ∞∫

−∞

[a1x1(t) + a2x2(t)]e−jωt dt

= a1

∞∫

−∞

x1(t)e −jωt dt

︸︷︷︸

X1(ω)

+ a2

∞∫

−∞

x2(t)e −jωt dt

︸︷︷︸

X2(ω)

ℑ{a1x1(t) + a2x2(t)} = a1 X1(ω) + a2 X2(ω),

which completes the proof.

The application of the linearity property is demonstrated through the following

example.

Example 5.13

Using the CTFT pairs given in Eqs. (5.25) and (5.27),

e jω0t CTFT←−−→ 2πδ(ω − ω0)

and

e−jω0t CTFT←−−→ 2πδ(ω + ω0),

calculate the CTFT of the cosine function cos(ω0t).

Solution

Using Euler’s formula,

ℑ{cos(ω0t)} = {

2 [e jω0t + e−jω0t ]

}

= 1

2 ℑ{e jω0t } +

2 ℑ{e−jω0t }.

Using the aforementioned CTFT pairs for exp(jω0t) and exp(−jω0t), we obtain

ℑ{cos(ω0t)} = π [δ(ω − ω0) + δ(ω + ω0)],

which is the same as the CTFT for the periodic cosine function in Table 5.2.

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219 5 Continuous-time Fourier transform

g(t)

0 1 2 t

Fig. 5.10. Waveform g(t ) used

in Example 5.14.

Example 5.14

Calculate the CTFT of the waveform g(t) plotted in Fig. 5.10.

Solution

By inspection, the waveform g(t) can be expressed as a linear combination of

x1(t) and x2(t) from Fig. 5.9, as follows:

g(t) = 1

2 [x1(t) + x2(t)].

Using the linearity property, the CTFT of g(t) is given by

G(ω) = 1

2 X1(ω) +

2 X2(ω).

Based on Eqs. (5.42) and (5.43), the CTFT pairs for x1(t) and x2(t) are given

X1(ω) = 4

ω2 [ω sin(2ω) + cos(ω) − 1]

and

X2(ω) = j 4

ω2 [ω cos(2ω) − sin(ω)].

The CTFT of g(t) is therefore given by

G(ω) = 2

ω2 [ω sin(2ω) + cos(ω) − 1] + j

ω2 [ω cos(2ω) − sin(ω)]

= 2

ω2 [jωe−j2ω + e−jω − 1].

5.5.2 Time scaling

In Section 1.4.1, we showed that the time-scaled version of a signal x(t) is given

by x(at). If a > 1, the signal compresses in time. If a < 1, the signal expands in

time. The time-scaling property expresses the CTFT of the time-scaled signal

x(at) in terms of the CTFT of the original signal x(t).

If x(t) CTFT←−−→ X (ω) then

x(at) CTFT←−−→

|a| X

(ω

)

, for a ∈ ℜ and a �= 0, (5.45)

where ℜ denotes the set of real values.

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220 Part II Continuous-time signals and systems

0 42

3 h(t)

Fig. 5.11. Waveform h(t ) used

in Example 5.15.

Proof

Equation (5.45) can be proved separately for the two cases a > 0 and a < 0.

Case I (a > 0). By Eq. (5.10), the CTFT of the time-scaled signal x(at) is given

ℑ{x(at)} = ∞∫

−∞

x(at)e−jωt dt .

Substituting τ = at, the above integral reduces to

ℑ{x(at)} = ∞∫

−∞

x(τ )e−jωτ/2 dτ

a =

a X

(ω

)

which proves Eq. (5.45) for a > 0. The proof for a < 0 follows the above

procedure and is left as an exercise for the reader (see Problem 5.13).

Example 5.15

To illustrate the usefulness of the time-scaling property, let us calculate the

CTFT of the function h(t) shown in Fig. 5.11.

Solution

By inspection, the waveform h(t) can be expressed as a scaled version of g(t)

illustrated in Fig. 5.10 as follows:

h(t) = 3

2 g

( t

)

= 3

2 g(0.5t).

Applying the linearity and time-scaling properties with a = 0.5, the CTFT of g(t) is given by

H (ω) = 3

[ 1

0.5 G

( ω

0.5

) ]

= 3G(2ω).

Based on the result of Example 5.14, G(ω) = (2/ω2)[jωe−j2ω + e−jω − 1], which yields

H (ω) = 3 2

(2ω)2 [ j(2ω)e−j2(2ω) + e−j(2ω) − 1] =

2ω2 [ j2ωe−j4ω + e−j2ω − 1].

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221 5 Continuous-time Fourier transform

5.5.3 Time shifting

The time-shifting operation delays or advances the reference signal in time.

Given a signal x(t), the time-shifted signal is given by x(t − t0). If the value of the shift t0 is positive, the reference signal x(t) is delayed and shifted towards

the right-hand side of the t-axis. On the other hand, if the value of the shift t0 is

negative, signal x(t) advances forward and is shifted towards the left-hand side

of the t-axis.

If x(t) CTFT←−−→ X (ω) then

g(t) = x(t − t0) CTFT←−−→ e−jωt0 X (ω) for t0 ∈ ℜ, (5.46)

where ℜ denotes the set of real values.

Proof

By Eq. (5.10), the CTFT of the time-shifted signal x(t − t0) is given by

ℑ{x(t − t0)} = ∞∫

−∞

x(t − t0)e−jωt dt

= e−jωt0 ∞∫

−∞

x(τ )e−jωτ dτ by substituting τ = (t − t0)

= e−jωt0 X (ω),

which proves the time-shifting property, Eq. (5.46).

The CTFT time-shifting property states that if a signal is shifted by t0 time

units in the time domain, the CTFT of the original signal is modified by a

multiplicative factor of exp(−jω0t). The magnitude and phase of the CTFT of the time-shifted signal g(t) = x(t − t0) are given by

magnitude |G(ω)| = |e−jωt0 X (ω)| = |e−jωt0 ||X (ω)| = |X (ω)|; (5.47) phase <G(ω) = <{e−jωt0 X (ω)} = <e−jωt0+ <X (ω) = −ωt0+ <X (ω).

(5.48)

Based on Eqs. (5.47) and (5.48), we can conclude that the time shifting does not

change the magnitude spectrum of the original signal, while the phase spectrum

is modified by an additive factor of −ωt0. In Example 5.16, we illustrate the application of the time-shifting property

by calculating the CTFT of the waveform illustrated in Fig. 5.12.

Example 5.16

Express the CTFT of the function f (t) shown in Fig. 5.12 in terms of the CTFT

of g(t) shown in Fig. 5.10.

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222 Part II Continuous-time signals and systems

f (t)5

3 7 10 130−3

Fig. 5.12. Waveform f(t ) used in

Example 5.16.

Solution

By inspection, f (t) can be expressed in terms of g(t) as

f (t) = 3

2 g

( t + 3

)

+ 5

2 g

( t − 7

)

We calculate the CTFT of each term in f (t) separately. By considering the

CTFT pair g(t) CTFT←−−→ G(ω) and applying the time-shifting property with

a = 3, we obtain

( t

)

CTFT←−−→ 3G(3ω).

Using the time-shifting property,

( t + 3

)

CTFT←−−→ 3e j3ωG(3ω) and g (

t − 7 3

)

CTFT←−−→ 3e−j7ωG(3ω).

Finally, by applying the linearity property, we obtain

2 g

( t + 3

)

+ 5

2 g

( t − 7

)

CTFT←−−→ 3

2 · 3e j3ωG(3ω) +

2 · 3e−j7ωG(3ω).

Expressed in terms of the CTFT of g(t), the CTFT F(ω) of the function f (t) is

therefore given by

F(w) = 9

2 e j3ωG(3ω) +

2 e−j7ωG(3ω).

5.5.4 Frequency shifting

In the time-shifting property, we observed the change in the CTFT when a signal

x(t) is shifted in the time domain. The frequency-shifting property addresses

the converse problem of how a signal x(t) is modified in the time domain if its

CTFT is shifted in the frequency domain.

If x(t) CTFT←−−→ X (ω) then

h(t) = e jω0t x(t) CTFT←−−→ X (ω − ω0), for ω0 ∈ ℜ, (5.49)

where ℜ denotes the set of real values. The frequency-shifting property can be proved directly from Eq. (5.10) by

considering the CTFT of the signal exp(jω0t)x(t). The proof is left as an exercise

for the reader (see Problem 5.15).

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223 5 Continuous-time Fourier transform

By calculating the magnitude and phase of the term exp(jω0t)x(t) on the

left-hand side of the CTFT pair shown in Eq. (5.49), we obtain

magnitude |h(t)| = |e jω0t x(t)| = |e jω0t ||x(t)| = |x(t)|; (5.50) phase <h(t) = <e jω0t x(t) = <e jω0t + <x(t) = ω0t + <x(t). (5.51)

In other words, frequency shifting the CTFT of a signal does not change the

amplitude |x(t)| of the signal x(t) in the time domain. The only change is in the phase <x(t) of the signal x(t), which is modified by an additive factor of ω0t .

Example 5.17

In Section 2.1.3, we considered an amplitude modulator used in the AM band of

the radio transmission to transmit an information signal m(t) to the receiver. In

terms of the information signal m(t), the amplitude-modulated signal is given

s(t) = A[1 + km(t)] cos(ω0t).

Express the CTFT of the amplitude-modulated signal s(t) in terms of the CTFT

M(ω) of the information signal m(t).

Solution

The amplitude-modulated signal is a sum of two terms: A cos(ω0t) and Akm(t)

cos (ω0t). In Example 5.13, we calculated the CTFT of the A cos(ω0t) as

A cos(ω0t) CTFT←−−→ Aπ [δ(ω − ω0) + δ(ω + ω0)].

By expanding cos(ω0t), the second term Akm(t) cos(ω0t) is expressed as fol-

lows:

Akm(t) cos(ω0t) = 1

2 Akm(t)[e jω0t + e−jω0t ].

By using the frequency-shifting property, the CTFT of the terms m(t) exp(jω0t)

and m(t) exp(−jω0t) are given by

m(t)e jω0t CTFT←−−→ M(ω − ω0) and m(t)e−jω0t

CTFT←−−→ M(ω + ω0).

By using the linearity property, the CTFT of Akm(t) cos(ω0t) is then given by

Akm(t) cos(ω0t) CTFT←−−→

2 Ak[M(ω − ω0) + M(ω + ω0)].

By adding the CTFTs of the two terms, the CTFT of the amplitude-modulated

signal is given by

s(t) CTFT←−−→ A

[

πδ(ω − ω0) + πδ(ω + ω0) + k

2 M(ω − ω0) +

2 M(ω + ω0)

]

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224 Part II Continuous-time signals and systems

5.5.5 Time differentiation

The time-differentiation property expresses the CTFT of a time-differentiated

signal dx/dt in terms of the CTFT of the original signal x(t). We state the

time-differentiation property next.

If x(t) CTFT←−−→ X (ω) then

CTFT←−−→ jωX (ω) (5.52)

provided the derivative dx/dt exists at all time t.

Proof

From the CTFT synthesis equation, Eq. (5.9), we have

x(t) = 1

2π

∞∫

−∞

X (ω)e jωt dω.

Taking the derivative with respect to t on both sides of the equation yields

dt =







2π

∞∫

−∞

X (ω)e jωt dω





 .

Interchanging the order of differentiation and integration, we obtain

dt =

2π

∞∫

−∞

X (ω) d

dt {e jωt }dω =

2π

∞∫

−∞

[ jωX (ω)]e jωt dω.

Comparing this with Eq. (5.9), we obtain

CTFT←−−→ jωX (ω).

Corollary By repeatedly applying the time differentiation property, it is

straightforward to verify that

dn x

dtn CTFT←−−→ ( jω)n X (ω).

Example 5.18

In Example 5.11, we showed that the CTFT for the periodic cosine function is

given by

cos(ω0t) CTFT←−−→ π [δ(ω − ω0) + δ(ω + ω0)].

Using the above CTFT pair, derive the CTFT for the periodic sine function

sin(ω0t).

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225 5 Continuous-time Fourier transform

Solution

Taking the derivative of the CTFT pair for the cosine function yields

dt {cos(ω0t)}

CTFT←−−→ ( jω)π [δ(ω − ω0) + δ(ω + ω0)].

By rearranging terms, we obtain

−ω0 sin(ω0t) CTFT←−−→ jπ [ω0δ(ω − ω0) − ω0δ(ω + ω0)],

which can be expressed as follows:

ω0 sin(ω0t) CTFT←−−→

j [ω0δ(ω − ω0) − ω0δ(ω + ω0)],

obtained by using the multiplicative property of the impulse function,

x(t)δ(t + t0) = x(−t0)δ(t + t0). The CTFT of the periodic sine function is therefore given by

sin(ω0t) CTFT←−−→

j [δ(ω − ω0) − δ(ω + ω0)].

5.5.6 Time integration

The time-integration property expresses the CTFT of a time-integrated signal

∫ x(t)dt in terms of the CTFT of the original signal x(t).

If x(t) CTFT←−−→ X (ω), then

t∫

−∞

x(τ )dτ CTFT←−−→

X (ω)

jω + π X (0)δ(ω). (5.53)

The proof of the time-integration property is left as an exercise for the reader

(see Problem 5.14).

Example 5.19

Given δ(t) CTFT←−−→ 1, calculate the CTFT of the unit step function u(t) using

the time-integration property.

Solution

Integrating the CTFT pair for the unit impulse function yields

t∫

−∞

δ(t)dt CTFT←−−→

jω + πδ(ω).

By noting that the left-hand side of the aforementioned CTFT pair represents

the unit step function, we obtain

u(t) CTFT←−−→

jω + πδ(ω).

The above CTFT pair can be verified from Table 5.2.

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226 Part II Continuous-time signals and systems

5.5.7 Duality

The CTFTs of a constant signal x(t) = 1 and of an impulse function x(t) = δ(t) are given by the following CTFT pairs (see Table 5.2):

and

1 CTFT←−−→ 2πδ(ω) and δ(t) CTFT←−−→ 1.

For the above examples, the CTFT exhibits symmetry across the time and

frequency domains in the sense that the CTFT of a constant x(t)=1 is an impulse function, while the CTFT of an impulse function x(t) = δ(t) is a constant. This symmetry extends to the CTFT of any arbitrary signal and is referred to as the

duality property. We formally define the duality property below.

If x(t) CTFT←−−→ X (ω), then

X (t) CTFT←−−→ 2πx(−ω) (5.54)

is also a CTFT pair.

Proof

By the definition of the inverse CTFT, Eq. (5.9), we know that

x(t) = 1

2π

∞∫

−∞

X (r )e jr t dr ,

where the dummy variable r is used instead of ω. Substituting t = −ω in the above equation yields

2πx(−ω) = ∞∫

−∞

X (r )e−jωr dr = ℑ{X (t)}.

To illustrate the application of the duality property, consider the CTFT pair

δ(t) CTFT←−−→ 1,

with x(t) = δ(t) and X (ω) = 1. By interchanging the role of the independent variables t and ω, we obtain X (t) = 1 and x(ω) = δ(ω). Using the duality property, the converse CTFT pair is given by

1 CTFT←−−→ 2πδ(−ω) = 2πδ(ω),

which is indeed the CTFT of the constant signal x(t) = 1.

Example 5.20

As stated in Eq. (5.22), the following is a CTFT pair (see Example 5.6):

rect

( t

)

CTFT←−−→ τ sinc (ωτ

2π

)

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227 5 Continuous-time Fourier transform

Calculate the CTFT of x(t) = (W/π ) sinc(W t/π ) using the duality property.

Solution

By interchanging the role of variables t and ω in the following CTFT pair:

x(t) = rect (

)

CTFT←−−→ τ sinc (ωτ

2π

)

= X (ω),

we obtain X (t) = τ sinc(tτ/2π ) and x(−ω) = rect(−ω/τ ). Using the duality property, we obtain

τ sinc

( tτ

2π

)

CTFT←−−→ 2π rect (

−ω τ

)

Substituting τ = 2W and dividing both sides of the above equation by 2π yields

π sinc

( W t

)

CTFT←−−→ rect ( ω

)

The above result was proved in Example 5.7 by deriving it directly from the

definition of the CTFT.

5.5.8 Convolution

In Section 3.4, we showed that the output response of an LTIC system is obtained

by convolving the input signal with the impulse response of the system. At times,

the resulting convolution integral is difficult to solve analytically in the time

domain. The convolution property provides us with an alternative approach,

based on the CTFT, of calculating the output response. Below we define the con-

volution property and explain its application in calculating the output response

of an LTIC system.

If x1(t) CTFT←−−→ X1(ω) and x2(t)

CTFT←−−→ X2(ω), then

x1(t) ∗ x2(t) CTFT←−−→ X1(ω)X2(ω) (5.55)

and

x1(t)x2(t) CTFT←−−→

2π [X1(ω) ∗ X2(ω)]. (5.56)

In other words, convolution between two signals in the time domain is equivalent

to the multiplication of the CTFTs of the two signals in the frequency domain.

Conversely, convolution in frequency domain is equivalent to multiplication of

the inverse CTFTs in the time domain. In the case of the frequency-domain

convolution, one has to be careful in including a normalizing factor of 1/2π .

We prove the convolution property next.

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228 Part II Continuous-time signals and systems

Proof

To prove Eq. (5.55), consider the CTFT of the convolved signal [x1(t) ∗ x2(t)]. By the definition in Eq. (5.9),

ℑ{x1(t) ∗ x2(t)} = ∞∫

−∞

{x1(t) ∗ x2(t)}e−jωt dt .

Substituting the convolution [x1(t) ∗ x2(t)] by its integral, we obtain

ℑ{x1(t) ∗ x2(t)} = ∞∫

−∞





∞∫

−∞

x1(τ )x2(t − τ )dτ



 e−jωt dt .

By changing the order of the two integrations, we obtain

ℑ{x1(t) ∗ x2(t)} = ∞∫

−∞

x1(τ )





∞∫

−∞

x2(t − τ )e−jωt dt



 dτ ,

where the inner integral is given by

∞∫

−∞

x2(t − τ )e−jωt dt = ℑ{x2(t − τ )} = X2(ω)e−jωτ .

Therefore,

ℑ{x1(t) ∗ x2(t)} = X2(ω) ∞∫

−∞

x1(τ )e −jωτ dτ = X2(ω)X1(ω).

The convolution property, Eq. (5.56), in the frequency domain can be proved

similarly by taking the inverse CTFT of [X1(ω) ∗ X2(ω)] and following the aforementioned procedure.

Equation (5.55) provides us with an alternative method to calculate the convo-

lution integral using the CTFT. Expressed in terms of the CTFT pairs

x(t) CTFT←−−→ X (ω), h(t) CTFT←−−→ H (ω), and y(t) CTFT←−−→ Y (ω),

the output signal y(t) is expressed in terms of the impulse response h(t) and

the input signal x(t) as follows:

y(t) = x(t) ∗ h(t) CTFT←−−→ Y (ω) = X (ω)H (ω),

obtained by applying the convolution property in the time domain. In other

words, the CTFT of the output signal is obtained by multiplying the CTFTs

of the input signal and the impulse response. The procedure for evaluating the

output y(t) of an LTIC system in the frequency domain, therefore, consists of

the following four steps.

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229 5 Continuous-time Fourier transform

(1) Calculate the CTFT X (ω) of the input signal x(t).

(2) Calculate the CTFT H (ω) of the impulse response h(t) of the LTIC system.

The CTFT H (ω) is referred to as the transfer function of the LTIC system.

(3) Based on the convolution property, the CTFT Y (ω) of the output y(t) is

given by Y (ω) = X (ω)H (ω). (4) Calculate the output y(t) by taking the inverse CTFT of Y (ω) obtained in

step (3).

The CTFT-based approach is convenient for three reasons. First, in most cases

we can use Table 5.2 to look up the expression of the CTFTs and their inverses.

In such cases, the CTFT-based approach is simpler to use than the time-domain

approach based on the convolution integral. In cases where the CTFTs are

difficult to evaluate analytically, they are obtained by using fast computational

techniques for calculating the Fourier transform. The CTFT-based approach,

therefore, allows the use of digital computers to calculate the output. Finally, the

CTFT-based approach provides us with a meaningful insight into the behavior of

many systems. An LTIC system is typically designed in the frequency domain.

Example 5.21

In Example 3.6, we showed that in response to the input signal x(t)= e−t u(t), the LTIC system with the impulse response h(t) = e−2t u(t) produces the following output:

y(t) = (e−t − e−2t )u(t).

We will verify the above result using the CTFT-based approach.

Solution

Based on Table 5.2, the CTFTs for the input signal and the impulse response

are as follows:

e−t u(t) CTFT←−−→

1 + jω and e−2t u(t)

CTFT←−−→ 1

2 + jω .

The CTFT of the output signal is therefore calculated as follows:

Y (ω) = ℑ{[e−t u(t)] ∗ e[−2t u(t)]} = ℑ{e−t u(t)} × ℑ{e−2t u(t)}.

Using the CTFT pair

e−at u(t) CTFT←−−→

a + jω ,

we obtain

Y (ω) = 1

1 + jω ×

2 + jω ,

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230 Part II Continuous-time signals and systems

which can be expressed in terms of the following partial fraction expansion:

Y (ω) = 1

1 + jω −

2 + jω .

Taking the inverse CTFT yields

y(t) = (e−t − e−2t )u(t)

which is identical to the result obtained in Example 3.6 by direct convolution.

5.5.9 Parseval’s energy theorem

Parseval’s theorem relates the energy of a signal in the time domain to the

energy of its CTFT in the frequency domain. It shows that the CTFT is a

lossless transform as there is no loss of energy if a signal is transformed by the

CTFT.

For an energy signal x(t), the following relationship holds true:

Ex = ∞∫

−∞

|x(t)|2dt = 1

2π

∞∫

−∞

|X (ω)|2dω. (5.57)

Proof

To prove the Parseval’s theorem, consider

∞∫

−∞

|X (ω)|2dω = ∞∫

−∞

X (ω)X∗(ω)dω.

Substituting for the CTFT X (ω) using the definition in Eq. (5.10) yields

∞∫

−∞

|X (ω)|2dω = ∞∫

−∞





∞∫

−∞

x(α)e−jωαdα









∞∫

−∞

x(β)e−jωβdβ





∗

dω,

where we have used the dummy variables α and β to differentiate between the

two CTFT integrals. Taking the conjugate of the third integral and rearranging

the order of integration, we obtain

∞∫

−∞

|X (ω)|2dω = ∞∫

−∞

x(α)

∞∫

−∞

x∗(β)





∞∫

−∞

e jω(β−α)dω



 dβ dα.

Based on Eq. (5.15), we know that

∞∫

−∞

e jω(β−α)dω = 2πδ(β − α),

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231 5 Continuous-time Fourier transform

which reduces the earlier expression to

∞∫

−∞

|X (ω)|2dω = 2π ∞∫

−∞

x(α)

∞∫

−∞

x∗(β)δ(β − α)dβ

︸︷︷︸

x∗(α)

dα

Ex = ∞∫

−∞

|x(α)|2dα = 1

2π

∞∫

−∞

|X (ω)|2dω.

Example 5.22

Calculate the energy of the CT signal x(t) = e−at u(t) in the (a) time and (b) frequency domains. Verify that Eq. (5.57) is valid by comparing the two

answers.

Solution

(a) The energy in the time domain is obtained by

Ex = ∞∫

−∞

|x(t)|2dt = ∞∫

e−2at dt = [

e−2at

−2a

]∞

= 1

2a .

(b) From Table 5.2, the CTFT of x(t) = e−at u(t) is given by

e−at u(t) CTFT←−−→

a + jω .

The energy in the frequency domain is therefore given by

Ex = 1

2π

∞∫

−∞

|X (ω)|2dω = 1

2π

∞∫

−∞

a2+ω2 dω =

2π

[ 1

a tan−1

(ω

) ]∞

−∞ =

2a .

By comparison, the results in (a) and (b) are the same.

5.6 Existence of the CTFT

The CTFT X (ω) of a function x(t) is said to exist if

|X (ω)| < ∞ for −∞ < ω < ∞. (5.58)

The above definition for the existence of the CTFT agrees with our intuition

that the amplitude of a valid function should be finite for all values of the

independent variable. A simpler condition in the time domain can be derived

by considering the inverse CTFT of X (ω) as

|X (ω)| =

∣ ∣ ∣ ∣ ∣ ∣

∞∫

−∞

x(t)e−jωt dt

∣ ∣ ∣ ∣ ∣ ∣

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232 Part II Continuous-time signals and systems

Applying the triangle inequality in the CT domain, we obtain

|X (ω)| ≤ ∞∫

−∞

|x(t)e−jωt |dt = ∞∫

−∞

|x(t)||e−jωt |dt = ∞∫

−∞

|x(t)|dt,

which leads to the following condition for the existence of the CTFT.

Condition for existence of CTFT The Fourier CTFT X (ω) of a function x(t) exists if

∞∫

−∞

|x(t)|dt < ∞. (5.59)

Equation (5.59) is a sufficient condition to verify the existence of the CTFT.

Example 5.23

Determine if the CTFTs exist for the following functions:

(i) causal decaying exponential function f (t) = exp(−at)u(t); (ii) exponential function g(t) = exp(−at);

(iii) periodic cosine waveform h(t) = cos(ω0t),

where a, ω0 ∈ ℜ+.

Solution

(i) Equation (5.59) yields

∞∫

−∞

| f (t)|dt = ∞∫

−∞

|e−at u(t)|dt = ∞∫

−∞

e−at u(t)dt = ∞∫

e−at dt

= 1

−a [e−at ]∞0 =

a < ∞.

Therefore, the CTFT exists for the causal decaying exponential function.

(ii) Equation (5.59) yields

∞∫

−∞

|g(t)|dt = ∞∫

−∞

|e−at |dt = ∞∫

−∞

e−at dt = 0∫

−∞

e−at dt

︸︷︷︸

=∞

+ ∞∫

e−at dt

︸︷︷︸

=1/a

= ∞.

Therefore, the CTFT does not exist for the exponential function.

(iii) Equation (5.59) reduces to

∞∫

−∞

|h(t)|dt = ∞∫

−∞

|cos(ω0t)|dt = ∞.

Therefore, the CTFT does not exist for the exponential function.

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233 5 Continuous-time Fourier transform

In part (iii), we proved that the CTFT does not exist for a periodic cosine

function. This appears to be in violation of Table 5.2, which lists the following

CTFT pair for the periodic cosine function:

cos(ω0t) CTFT←−−→ π [δ(ω − ω0) + δ(ω + ω0)].

Actually, the two statements do not contradict each other. The condition for the

existence of the CTFT assumes that the CTFT must be finite for all values of

ω. The above CTFT pairs indicate that the CTFT of the periodic cosine func-

tion consists of two impulses at ω = ±ω0. From the definition of the impulses, we know that the magnitudes of the two impulse functions in the aforemen-

tioned CTFT pair are infinite at ω = ±ω0, and therefore that the periodic cosine function violates the condition for the existence of the CTFT.

In Section 5.7, we show that the CTFTs of most periodic signals are derived

from the CTFS representation of such signals, not directly from the CTFT

definition. Therefore, we make an exception for periodic signals and ignore the

condition of CTFT existence for periodic signals.

5.7 CTFT of periodic functions

Consider a periodic function x(t) with a fundamental period of T0. Using the

exponential CTFS, the frequency representation of x(t) is obtained from the

following expression:

x(t) = ∞∑

n=−∞ Dne

jnω0t , (5.60)

where ω0 = 2π/T0 is the fundamental frequency of the periodic signal and Dn denotes the exponential CTFS coefficients Dn , given by

Dn = 1

∫

〈T0〉

x(t)e−jnω0t dt . (5.61)

Calculating the CTFT of both sides of Eq. (5.60), we obtain

X (ω) = ℑ{x(t)} = ℑ

{ ∞∑

n=−∞ Dne

jnω0t

}

Using the linearity property, the above expression is simplified to

X (ω) = ∞∑

n=−∞ Dnℑ{e jnω0t } = 2π

∞∑

n=−∞ Dnδ(ω − nω0).

In other words, the CTFT of a periodic function x(t) is given by

x(t) CTFT←−−→ 2π

∞∑

n=−∞ Dnδ(ω − nω0). (5.62)

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234 Part II Continuous-time signals and systems

−2p −p 0 p 2p

q(t)

3/5p

3/p

1.5 Dn

3/p

3/5p n

0 2−2−4 6 84−6−8 −1/p−1/p

6/5

3p Q(w)

w 0 2−2−4 6 84−6−8

−2−2

6/5

(b) (c)

(a)

Fig. 5.13. Alternative

representations for the periodic

function considered in Example

5.24. (a) A periodic rectangular

wavefunction q(t ), (b) CTFS

coefficients Dn for q(t ), and

Equation (5.62) provides us with an alternative method for calculating the CTFT

of periodic signals using the exponential CTFS. We illustrate the procedure in

Examples 5.24 and 5.25.

Example 5.24

Calculate the CTFT representation of the periodic waveform q(t) shown in

Fig. 5.13(a).

Solution

The waveform q(t) is a special case of the rectangular wave x(t) considered in

Example 4.14 with τ = π and T = 2π . Mathematically,

q(t) = 3x(t) with duty cycle τ/T = 1/2.

Using Eq. (4.49), the CTFS coefficients of s(t) are given by

Dn = 3

2 sinc

)

Dn = 3

2 sinc

)



       

       

2 n = 0

0 n = 2k �= 0 3

nπ n = 4k + 1

− 3

nπ n = 4k + 3.

Substituting ω0 = 1 in Eq. (5.62) results in the following expression for the CTFT:

q(t) CTFT←−−→ 2π

∞∑

n=−∞ Dnδ(ω − n).

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235 5 Continuous-time Fourier transform

0 2 4 6 8−2−4−6−8

j1.5

−j1.5

−2w0 0−4w0−6w0−8w0

H(w)

w 2w0 4w0 6w0 8w0

j3p

−j3p

(a) (b)

Fig. 5.14. Alternative

representations for the sine

wave considered in Example

5.25. (a) CTFS coefficients Dn ;

(b) CTFT representation H(ω).

The CTFS coefficients Dn and the CTFT Q(ω) of the periodic rectangular wave

are plotted in Figs. 5.13(b) and (c).

Example 5.25

Calculate the CTFT for the periodic sine wave h(t) = 3 sin(ω0t).

Solution

To obtain the CTFS representation of the periodic sine wave, we expand sin(ω0t)

using Euler’s identity. The resulting expression is as follows:

h(t) = 3 sin(ω0t) = 3

2j [e jω0t − e−jω0t ],

which yields the following values for the exponential CTFS coefficients:

Dn =







−j1.5 n = 1 j1.5 n = −1 0 otherwise.

Based on Eq. (5.62), the CTFT of a periodic sine wave is given by

H (ω) = 2π ∞∑

n=−∞ Dnδ(ω − nω0) = j3π [δ(ω + ω0) − δ(ω − ω0)].

The CTFS coefficients and the CTFT for a periodic sine wave are plotted in Fig.

5.14. The above result is the same as derived in Example 5.18, with a scaling

factor of 3.

5.8 CTFS coefficients as samples of CTFT

In Section 5.7, we presented a method of calculating the CTFT of a periodic

signal from the CTFS representation. In this section, we solve the converse

problem of calculating the CTFS coefficients from the CTFT.

Consider a time-limited aperiodic function x(t), whose CTFT X (ω) is known.

By following the procedure used in Section 5.1, we construct several repetitions

of x(t) uniformly spaced from each other with a duration of T0. The process is

illustrated in Fig. 5.1, where x(t) is the aperiodic signal plotted in Fig. 5.1(a). Its

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236 Part II Continuous-time signals and systems

periodic extension x̃T (t) is shown in Fig. 5.1(b). Using Eq. (5.3), the exponential

CTFS coefficients of the periodic extension are given by

D̃n = 1

∫

〈T0〉

x(t)e−jnω0t dt = 1

T0/2∫

−T0/2

x(t)e−jnω0t dt .

Since x̃T (t) = x(t) within the range −T0 ≤ t ≤ T0, the above expression reduces to

D̃n = 1

T0/2∫

−T0/2

x(t)e−jnω0t dt = 1

∞∫

−∞

x(t)e−jnω0t dt = 1

T0 X (ω)|ω=nω0 , (5.63)

which is the relationship between the CTFT of the aperiodic signal x(t) and the

CTFS coefficients of its periodic extension x̃T (t). In other words, we can derive

the exponential CTFS coefficients of a periodic signal with period T0 from the

CTFT using the following steps.

(1) Compute the CTFT X (ω) of the aperiodic signal x(t) obtained from one

period of x̃T (t) as

x(t) = {

x̃T (t) −T0/2 ≤ t ≤ T0/2 0 elsewhere.

(2) The exponential CTFS coefficients Dn of the periodic signal x̃T (t) are given

Dn = 1

T0 X (ω)|ω=nω0 ,

where ω0 denotes the fundamental frequency of the periodic signal x̃T (t)

and is given by ω0 = 2π/T0.

Example 5.26

Calculate the exponential CTFS coefficients of the periodic signal x̃T (t) shown

in Fig. 5.13(a).

Solution

Step 1 The aperiodic signal representing one period of x̃T (t) is given by

x(t) = 3 rect (

)

Using Table 5.2, the CTFT of the rectangular gate function is given by

3 rect

( t

)

CTFT←−−→ 3π sinc (ω

)

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237 5 Continuous-time Fourier transform

Step 2 The exponential CTFS coefficients Dn of the periodic signal x̃T (t) are obtained from Eq. (5.63) as

Dn = 1

T0 X (ω)|ω=nω0 with T0 = 2π and ω0 = 1.

The above expression simplifies to

Dn = 3

2 sinc

)

= 3

nπ sin

(nπ

)

5.9 LTIC systems analysis using CTFT

In Chapters 2 and 3, we showed that an LTIC system can be modeled either

by a linear, constant-coefficient differential equation or by its impulse response

h(t). A third representation for an LTIC system is obtained by taking the CTFT

of the impulse response:

h(t) CTFT←−−→ H (ω).

The CTFT H (ω) is referred to as the Fourier transfer function of the LTIC

system and provides meaningful insights into the behavior of the system. The

impulse response relates the output response y(t) of an LTIC system to its input

x(t) using

y(t) = h(t) ∗ x(t).

Calculating the CTFT of both sides of the equation, we obtain

Y (ω) = H (ω)X (ω), (5.64)

where Y (ω) and X (ω) are the respective CTFTs of the output response y(t) and

the input signal x(t). Equation (5.64) provides an alternative definition for the

transfer function as the ratio of the CTFT of the output response and the CTFT

of the input signal. Mathematically, the transfer function H (ω) is given by

H (ω) = Y (ω)

X (ω) . (5.65)

5.9.1 Transfer function of an LTIC system

It was mentioned in Section 3.1 that, for an LTIC system, the relationship

between the applied input x(t) and output y(t) can be described using a constant-

coefficient differential equation of the following form:

n∑

k=0 ak

dk x

dtk =

m∑

k=0 bk

dk x

dtk . (5.66)

From the time-differentiation property of the CTFT, we know that

dn x

dtn CTFT←−−→ ( jω)n X (ω).

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238 Part II Continuous-time signals and systems

Calculating the CTFT of both sides of Eq. (5.66) and applying the time-

differentiation property, we obtain

n∑

k=0 ak( jω)

kY (ω) = m∑

k=0 bk( jω)

k X (ω)

H (ω) = Y (ω)

X (ω) =

n∑

k=0 bk( jω)

m∑

k=0 ak( jω)

. (5.67)

Given one representation for an LTIC system, it is straightforward to derive

the remaining two representations based on the CTFT and its properties. We

illustrate the procedure through the following examples.

Example 5.27

Consider an LTIC system whose input–output relationship is modeled by the

following third-order differential equation:

d3 y

dt3 + 6

d2 y

dt2 + 11

dt + 6y(t) = 2

dt + 3x(t). (5.68)

Calculate the transfer function H (ω) and the impulse response h(t) for the LTIC

system.

Solution

Using the time-differentiation property for the CTFT, we know that

dn x

dtn CTFT←−−→ ( jω)n X (ω).

Taking the CTFT of both sides of Eq. (5.47) and applying the time-

differentiation property yields

( jω)3Y (ω) + 6(jω)2Y (ω) + 11(jω)Y (ω) + 6Y (ω) = 2(jω)X (ω) + 3X (ω).

Making Y (ω) common on the left-hand side of the above expression, we obtain

[( jω)3 + 6(jω)2 + 11(jω) + 6]Y (ω) = [2( jω) + 3]X (ω).

Based on Eq. (5.46), the transfer function is therefore given by

H (ω) = Y (ω)

X (ω) =

2(jω) + 3 (jω)3 + 6(jω)2 + 11(jω) + 6

. (5.69)

The impulse response h(t) is obtained by taking the inverse CTFT of Eq. (5.69).

Factorizing the denominator, Eq. (5.69) is expressed as

H (ω) = 2(jω) + 3

(1 + jω)(2 + jω)(3 + jω) ,

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239 5 Continuous-time Fourier transform

which, by partial fraction expansion, reduces to

H (ω) = 1

2(1 + jω) +

(2 + jω) −

2(3 + jω) .

Taking the inverse CTFT:

h(t) = (

2 e−t + e−2t −

2 e−3t

)

u(t). (5.70)

Equations (5.68)–(5.70) provide three equivalent representations of the LTIC

system.

Example 5.28

Consider an LTIC system with the following impulse response function:

h(t) = rect (

)

= {

1 |t | ≤ τ/2 0 |t | > τ/2. (5.71)

Calculate the transfer function H (ω) and the input–output relationship for the

LTIC system.

Solution

From Table 5.2, we obtain the following transfer function:

H (ω) = τ sinc (ωτ

2π

)

= 2

ω sin

(ωτ

)

In other words,

Y (ω)

X (ω) =

ω sin

(ωτ

)

which is expressed as

jωY (ω) = j2 sin (ωτ

)

X (ω)

jωY (ω) = e jωτ/2 X (ω) − e−jωτ/2 X (ω).

Taking the inverse CTFT of both sides, we obtain

dt = x

(

t + τ

)

− x (

t − τ

)

. (5.72)

5.9.2 Response of LTIC systems to periodic signals

In Section 4.7.2, we derived the output response of an LTIC system, shown in

Fig. 5.15, of the following periodic signal:

x(t) = ∞∑

n=−∞ Dne

jnω0t (5.73)

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240 Part II Continuous-time signals and systems

y(t) = ∞∑

n=−∞ Dne

jnω0t H (ω)|ω=nω0 , (5.74)

where H (ω) is the CTFT of the impulse response h(t) of the system and is

referred to as the transfer function of the LTIC system. Corollary 4.1 is a

special case of Eq. (5.74), where the input is a sinusoidal signal and the impulse

response h(t) is real-valued. In such cases, the output y(t) can be expressed as

follows:

k1 exp(jω0t) → A1k1 exp(jω0t + jφ1), (5.75)

k1 sin(ω0t) → A1k1 sin(ω0t + φ1), (5.76)

and

k1 cos(ω0t) → A1k1 cos(ω0t + φ1), (5.77)

where A1 and φ1 are the magnitude and phase of H (ω) evaluated at ω = ω0.

LTIC

system

h(t)

periodic

input

x(t)

periodic

output

y(t)

Fig. 5.15. Response of an LTIC

system to a periodic input.

Equations (5.73)–(5.77) can be derived directly by using the CTFT. We now

prove Eq. (5.74).

Proof

The CTFT of a periodic signal x(t) is given by

x(t) CTFT←−−→ 2π

∞∑

n=−∞ Dnδ(ω − nω0).

Using the convolution property, the output of an LTIC with transfer function

H (ω) is given by

Y (ω) = 2π ∞∑

n=−∞ Dnδ(ω − nω0)H (ω) = 2π

∞∑

n=−∞ Dnδ(ω − nω0)H (nω0).

Taking the inverse CTFT of the above equation yields

y(t) = ∞∑

n=−∞ Dnℑ−1{2πδ(ω − nω0)}H (nω0) =

∞∑

n=−∞ Dn H (nω0)e

jnω0t ,

which proves Eq. (5.74).

Example 5.29

Consider an LTIC system with impulse response given by

h(t) = 10

π sinc

( 10t

)

, (5.78)

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241 5 Continuous-time Fourier transform

0 10

1 20 w

H(w) = rect ( )

−10

(b)

( )10tp10ph(t) = sinc

5 −4p

5 −2p

10 − p

5 − p

5 4p

5 2p

10 p

p 10

5 p

(a)

Fig. 5.16. LTIC system

considered in Example 5.29.

(a) Impulse response h(t );

(b) transfer function H(ω).

sketched as a function of time t in Fig. 5.16(a). Determine the output response

of the system for the following inputs:

(i) x1(t) = sin(5t); (ii) x2(t) = sin(15t);

(iii) x3(t) = sin(8t) + sin(20t).

Solution

Calculating the CTFT of Eq. (5.78), the transfer function H (ω) is given by

H (ω) = rect ( ω

)

. (5.79)

The magnitude spectrum of the LTIC system is plotted in Fig. 5.16(b). The

phase of the LTIC system is zero for all frequencies.

(i) Input x1(t) = sin(5t). The CTFT of the input signal x1(t) is given by

X1(ω) = π

j [δ(ω − 5) − δ(ω + 5)].

The CTFT Y1(ω) of the output signal is obtained by multiplying X1(ω) by H (ω)

and is given by

Y1(ω) = X1(ω)H (ω) = π

j δ(ω − 5)H (ω) −

j δ(ω + 5)H (ω).

Using the multiplication property of the impulse function, we have

Y1(ω) = π

j δ(ω − 5)H (5) −

j δ(ω + 5)H (−5).

Since H (±5) = 1, the CTFT Y1(ω) of the output signal is given by

Y1(ω) = π

j δ(ω − 5) −

j δ(ω − 5).

Taking the inverse CTFT, the output is given by

y1(t) = sin(5t).

The CTFT Y1(ω) of the output signal can also be obtained by graphical multipli-

cation, as shown in Fig. 5.17(a), where the magnitude spectrum of the transfer

function H (ω) is shown as a dashed line. Since the magnitude of the transfer

function H (ω) is one at the location of the two impulses contained in the CTFT

of the input signal, the CTFT Y1(ω) of the output signal is identical to the CTFT

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242 Part II Continuous-time signals and systems

1 X1(w) Y1(w)

10−10 −jp

0 10−10 −jp

(a)

1 X2(w) Y2(w)

10−10

−jp

0 10−10

(b)

w 0

1 X3(w) Y3(w)

10−10 −jp −jp

jp jp jp

−jp

w 0 10−10

(c)

Fig. 5.17. Frequency

interpretation of the output

response of an LTIC system.

Response of the LTIC system

(transfer function shown

as a dashed line) to:

(a) x1(t ) = sin(5t ); (b) x2(t ) = sin(15t ); (c) x3(t ) = sin(8t ) + sin(20t ).

of the input signal. By calculating the inverse CTFT, we obtain the output as

y1(t) = x1(t) = sin(5t). (ii) Input x2(t) = sin(15t). The CTFT of the input signal x2(t) is given by

X2(ω) = π

j [δ(ω − 15) − δ(ω + 15)].

The CTFT Y2(ω) of the output signal is obtained by multiplying X1(ω) by H (ω)

and is given by

Y2(ω) = X2(ω)H (ω) = π

j δ(ω − 15)H (ω) −

j δ(ω + 15)H (ω).

Using the multiplication property of the impulse function, we have

Y1(ω) = X1(ω)H (ω) = π

j δ(ω − 15)H (15) −

j δ(ω + 15)H (−15).

Since H (±5) = 0, the CTFT Y1(ω) of the output signal is given by

Y1(ω) = 0.

Taking the inverse CTFT, the output is y2(t) = 0. As in part (i), the CTFT Y2(ω) of the output signal can be obtained by

graphical multiplication shown in Fig. 5.17(b). Since the magnitude of the

transfer function H (ω) is zero at the location of the two impulses contained in

the CTFT of the input signal, the two impulses are blocked from the output of

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243 5 Continuous-time Fourier transform

the LTIC system. The CTFT Y1(ω) of the output signal is zero, which results in

y2(t) = 0. (iii) Input x3(t) = sin(8t) + sin(20t). Taking the CTFT of the input x3(t)

yields

X3(ω) = π

j [δ(ω − 8) − δ(ω + 8)] +

j [δ(ω − 20) − δ(ω + 20)].

By following the procedure used in part (i), the CTFT Y3(ω) of the output signal

is given by

Y3(ω) = [ π

j δ(ω − 8)H (8) −

j δ(ω + 8)H (−8)

]

+ [ π

j δ(ω − 20)H (20) −

j δ(ω + 20)H (−20)

]

The input signal consists of four impulse functions with two impulses located

at ω = ±8 and two located at ω = ±20. The magnitude of the transfer function at frequencies ω = ±8 is one, therefore the two impulse functions δ(ω – 8) and δ(ω + 8) are unaffected. The magnitude of the transfer function at frequencies ω = ±20 is zero, therefore the two impulses δ(ω – 20) and δ(ω + 20) are eliminated from the output. The CTFT of the output signal therefore consists

of only two impulse functions located at (ω = ±8), and is given by

Y3(ω) = [ π

j δ(ω − 8) −

j δ(ω + 8)

]

which has the inverse CTFT of

y3(t) = sin(8t).

In signal processing, the LTIC system with h(t) = (10/π ) sinc(10t/π ) is referred to as an ideal low-pass filter since it eliminates high-frequency com-

ponents and leaves the low-frequency components unaffected. In this example,

all input frequency components with frequencies greater than ω > 10 are elim-

inated. Any input components with lower frequencies (ω < 10) appear unaf-

fected in the output of the LTIC system. The frequency (ω = 10) is referred to as the cutoff frequency of the ideal low-pass filter.

5.9.3 Response of an LTIC system to quasi-periodic signals

The response of an LTIC system to ideal periodic signals is given by Eqs.

(5.73)−(5.77). In practice, however, it is difficult to produce ideal periodic signals of infinite duration. Most practical signals start at t = 0 and are of finite duration. In this section, we calculate the output of an LTIC system for input

signals that are not completely periodic. We refer to such signals as quasi-

periodic signals.

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244 Part II Continuous-time signals and systems

Example 5.30

Consider the RC series circuit shown in Fig. 5.18. Determine the overall and

steady state values of the output of the RC series circuit if the input signal is

given by x(t) = sin(3t)u(t). Assume that the capacitor is uncharged at t = 0.

Solution

The CTFT of the input signal x(t) is given by

+ y (t) −

x(t) = sin(3t)

R = 1 MΩ

C = 0.5 µF

Fig. 5.18. RC series circuit

considered in Example 5.30.

X (ω) = π

2j [δ(ω − 3) − δ(ω + 3)] +

9 − ω2 .

From the theory of electrical circuits, the transfer function of the RC series

circuit is given by

H (ω) = 1/jωC

R + 1/jωC =

1 + jωC R .

Substituting the value of the product C R = 0.5 yields

H (ω) = 1

1 + j0.5ω .

By multiplying the CTFT of the input signal by the transfer function, the CTFT

of the output y(t) is given by

Y (ω) = {

2j [δ(ω − 3) − δ(ω + 3)] +

9 − ω2

}

× 1

1 + j0.5ω .

Solving the above expression results in the following:

Y (ω) = π

⌊ δ(ω − 3) 1 + j1.5

− δ(ω + 3) 1 − j1.5

⌋

+ 3

(9 − ω2)(1 + j0.5ω) .

Taking the inverse CTFT of the above expression (see Problem 5.10) yields the

following value for the output signal:

y(t) = 2

√ 13

sin(3t − 56◦)u(t) ︸︷︷︸

steady state value

+ 6

13 e−2t u(t)

︸︷︷︸

transient value

An alternative way of obtaining the steady state value of the output of the RC

series circuit is suggested in Corollary 4.1. Expressed in terms of the given

input, Corollary 4.1 states

sin(3t) u(t) ︸︷︷︸

x(t)

−→ A1 sin(3t + φ1) u(t) ︸︷︷︸

y(t)

where A1 and φ1 are, respectively, the magnitude and phase of the transfer

function at ω = 3. The values of A1 and φ1 are given by

A1 = |H (3)| = ∣ ∣ ∣ ∣

1 + j0.5(3)

∣ ∣ ∣ ∣ =

2 √

13 and

φ1 = <H (3) = < (

1 + j0.5(3)

)

= − tan−1(1.5) = −56◦.

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245 5 Continuous-time Fourier transform

Substituting the values of A1 and φ1 in Corollary 4.1, the steady state value of

the output is given by

yss(t) = 2

√ 13

sin(3t − 56◦)u(t).

For sinusoidal signals, Corollary 4.1 provides a simpler approach of determining

the steady state output.

5.9.4 Gain and phase responses

The Fourier transfer function H (ω) provides a complete description of the

LTIC system. In many applications, the graphical plots of |H (ω)| and < H (ω) versus frequency ω are used to analyze the characteristics of the LTIC system.

The magnitude spectrum |H (ω)| response function is also referred to as the gain response of the system, while the phase spectrum <H (ω) is referred to as

the phase response of the system. Below, we provide an example to illustrate

the procedure involved in plotting the magnitude and phase spectra. We also

introduce Bode plots, where a logarithmic scale is used for the frequency ω-axis.

Example 5.31

Consider an LTIC system with the impulse response h(t) = 1.25e−0.6t sin(0.8t)u(t). Plot the gain and phase responses of the LTIC system.

Solution

The transfer function H (ω) of the LTIC system is given by

H (ω) = ℑ{1.25e−0.6t sin(0.8t)u(t)} = 1.25 × 0.8

(0.6 + jω)2 + 0.82

= 1

1 − ω2 + j1.2ω .

The magnitude and phase spectra are as follows:

magnitude spectrum |H (ω)| = 1

√

(1 − ω2)2 + (1.2ω)2

= 1

√ 1 − 0.56ω2 + ω4

;

phase spectrum <H (ω) = − tan−1 (

1.2ω

1 − ω2

)

Figure 5.19(a) plots the magnitude spectrum and Fig. 5.19(b) plots the phase

spectrum of the LTIC system. Figure 5.19(a) illustrates that the magnitude

|H (ω)| = 1 for ω = 0. As the frequency ω increases, the magnitude |H (ω)| drops and approaches zero at very high frequencies. From Fig. 5.19(b), we

observe that the phase <H (ω) is zero at ω = 0. At high frequencies, the phase <H (ω) converges to −π radians, or −180◦.

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246 Part II Continuous-time signals and systems

0 1 2 3 4 5 0

0.2 0.4 0.6 0.8

|H (w

6 0 1 2 3 4 5

−p −0.75p

−0.5p −0.25p

< H

(w )

(a) (b)

Fig. 5.19. Magnitude and phase spectra of LTIC system with impulse response h(t ) = 1.25

e−0.6t sin(0.8t )u(t ). (a) Magnitude spectrum; (b) phase spectrum.

10−2 10−1 100 101 102 −80 −60 −40 −20

|H (w

)| (d

B )

10−2 10−1 100 101 102

−p −0.75p −0.5p

−0.25p 0

< H

(w )

(a) (b)

Fig. 5.20. Bode plots for the

LTIC system considered in

Example 5.31. (a) Magnitude

plot; (b) phase plot.

Bode plots In Bode plots, the magnitude |H (ω)| in decibels and phase <H (ω) are plotted as functions of frequency ω using a logarithmic scale. Use of a loga-

rithmic scale, with base 10, on the frequency ω-axis offers two main advantages.

(1) Compared to a linear scale, the use of a logarithmic scale allows a wider

range of frequencies to be plotted, with the lower frequencies represented

at a higher resolution.

(2) The asymptotic approximations of the magnitude and phase spectra can

easily be sketched graphically by hand.

Figure 5.20 illustrates the Bode plots of the LTIC system considered in

Example 5.31 using a logarithmic scale on the frequency axis. Figure 5.20(a)

shows the magnitude Bode plot, where the magnitude |H (ω)| is expressed in decibels (dB) as 20 log10|H (ω)| and plotted as a function of log10(ω). Figure 5.20(b) shows the phase Bode plot, where the phase <H (ω) is plotted as a

function of log10(ω).

5.10 M A T L A B exercises

In this section, we will consider two applications of M A T L A B . First, we

illustrate the procedure for calculating the CTFT of a CT signal x(t) using

M A T L A B . In our explanation, we consider an example, x(t) = 4 cos(10π t), and write the appropriate M A T L A B commands for the example at each step.

Second, we list the procedure for plotting the Bode plots in M A T L A B .

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247 5 Continuous-time Fourier transform

5.10.1 CTFT using M A T L A B

Step 1 Sampling In order to manipulate the CT signals on a digital computer, the CT signals must be discretized. This is normally achieved through a process

called sampling. In reality, sampling is followed by quantization, but because

of the high resolution supported by M A T L A B , we can neglect quantization

without any appreciable loss of accuracy, at least for our purposes here. Sam-

pling converts a CT signal x(t) into an equivalent DT signal x[k]. To prevent

any loss of information and for x[k] to be an exact representation of x(t), the

sampling rate ωs must be greater than at least twice the maximum frequency

ωmax present in the signal x(t), i.e.

ωs ≥ 2ωmax. (5.80)

This is referred to as the Nyquist criterion. We will consider sampling in depth

in Chapter 9, but the information presented above is sufficient for the following

discussion.

The CTFT of the periodic cosine signal is given by (see Table 5.2)

4 cos(10π t) CTFT←−−→ 4π [δ(ω − 10π ) + δ(ω + 10π )]; (5.81)

hence, the maximum frequency in x(t) is given by ωmax = 10π radians/s. Based on the Nyquist criterion, the lower bound for the sampling rate is given by

ωs ≥ 20π radians/s. (5.82)

We choose a sampling rate that is 20 times the Nyquist rate, i.e. ωs = 400π radians/s. The sampling interval Ts is given by

Ts = 2π

ωs = 5 ms. (5.83)

Selecting a time interval from −1 to 1 second to plot the sinusoidal wave, the number N of samples in x[k] is 401. The M A T L A B command that computes

x[k] is therefore given by

> t = -1:0.005:1; % define time instants

> x = 4*cos(10*pi*t); % samples of cosine wave

> subplot(221); plot(t,x) % for CT plot

> subplot(222); stem(t,x) % for DT plot

The subplots are plotted in Fig. 5.21(a) and (b) and provide a fairly accurate

representation of the cosine wave.

Step 2 Fast Fourier transform In M A T L A B , numeric computation of the CTFT is performed by using a fast implementation referred to as the fast

Fourier transform (FFT). At this time, we will simply name the function without

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248 Part II Continuous-time signals and systems

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 −4

−2

(a)

(c)

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 −4

−2

(b)

0 50 100 150 200 250 300 350 400 450 0

200

400

600

800

1000

(d)

−800 −600 −400 −200 0 200 400 600 800 0

10p−10p

Fig. 5.21. MA T L A B subplots for

the time and frequency domain

representations of

x(t ) = 4 cos(10π t ). (a) CT plot for x (t ); (b) DT plot for x(t );

(d) CTFT of x(t ).

worrying about its implementation. The function that evaluates FFT is fft (all lower-case letters). The M A T L A B command for calculating fft is

> y = fft(x); % fft computes CTFT

> subplot(223); plot(abs(y)); % abs calculates magnitude

The subplot of y is plotted in Fig. 5.21(c). There are two differences between

y (output of the fft function) and the CTFT pair,

4 cos(10π t) CTFT←−−→ 4π [δ(ω − 10π ) + δ(ω + 10π )].

By looking at the peak value of the magnitude spectrum |y|, we note that the magnitude is not given by 4π as the CTFT pair suggests. Also, the x-axis

represents the number of points instead of the appropriate frequency range ω.

In steps (3) and (4), we compensate for these differences without going into

the details of why the differences occur. The differences between the output of

fft and CTFT will be discussed in Chapter 11.

Step 3 Compensation Scale the magnitude of y by multiplying it by π times the sampling rate (πTs). In our example, Ts is 5 ms. The following M A T L A B

command performs the scaling:

> z = pi*0.005*y; % scale the magnitude of y

We also center z about an integer index of zero. This is accomplished by

fftshift.

> z = fftshift(z); % centre the CTFT about w = 0

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249 5 Continuous-time Fourier transform

Step 4 Frequency axis For a sequence x[k] of length N with a sampling frequency ωs, the fft function y = fft(x) produces the CTFT of x(t) at N equispaced points within the frequency interval [0, ωs]. The resolution �ω in the

frequency domain is, therefore, given by �ω = ωs/(N − 1). After centering, performed by the fftshift function, the limits of the interval are changed to [−ωs/2, ωs/2]. The M A T L A B commands to compute the appropriate values for the ω-axis are given by

> dw = 400* pi/ 400;

> w = -400* pi/2:dw:400* pi/2; % calculates frequency

% axis;

> subplot(224); plot(w,abs(z)); % magnitude spectrum

The subplot of the CTFT is plotted in Fig. 5.21(d). By inspection, it is confirmed

that it does correspond to the CTFT pair in Eq. (5.81).

The phase spectrum of the CTFT can be plotted using the angle function. For our example, the M A T L A B command to plot the angle is given by

> subplot(224); plot(w,angle(z)); % phase spectrum

The above command replaces the magnitude spectrum in subplot(224) by the phase spectrum. For the given signal, x(t) = 4 cos(10π t), the phase spectrum is zero for all frequencies ω. The M A T L A B code for calculating the

CTFT of a cosine wave is provided below in a function called myctft.

function [w,z] = myctft

% MYCTFT: computes CTFT of 4*cos(10*pi*t)

% Usage: [w,z] = myctft

% compute 4 cos(10*pi*t) in time domain

A = 4; % amplitude of cosine wave

w0 = 10*pi; % maximum frequency in signal

ws = 20*w0; % sampling rate

Ts = 2*pi/ws; % sampling interval

t = -1:Ts:1; % define time instants

x = A*cos(w0*t); % samples of cosine wave

% compute the CTFT

y = fft(x); % fft computes CTFT

z = pi*Ts*y; % scale the magnitude of y

z = fftshift(z); % centre CTFT about w = 0

% compute the frequency axis

w = -ws/2:ws/length(z):ws/2-ws/length(z);

% plots

subplot(211); plot(t,x) % CT plot of cos(w0*t)

subplot(212); plot(w,abs(z)) % CTFT plot of cos(w0*t)

% end

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250 Part II Continuous-time signals and systems

To calculate the inverse CTFT, we replace the function fft with ifft and reverse the order of the instructions. The M A T L A B code to compute the inverse

CTFT is provided in a second function called myinvctft:

function [t,x] = myinvctft(w,z)

% MYINVCTFT: computes inverse CTFT of y known at

% frequencies w

% Usage: [t,x] = myinvctft(w,z)

% compute the inverse CTFT

x = ifftshift(z);

x = ifft(x); % inverse fft

% compute the time instants

ws = w(length(w)) - w(1); % sampling rate

Ts = 2*pi/ws; % sampling interval

t = Ts*[-floor(length(w))/2:floor(length(w))/2-1];

% sampling instants% amplify signal by 1/(pi*Ts)

x = x/Ts;

% plots

subplot(211); plot(w,abs(z)) % CTFT plot of cos(w0*t)

subplot(212); plot(t,real(x)) % CT plot of cos(w0*t)

% end

5.10.2 Bode plots

M A T L A B provides the bode function to sketch the Bode plot. To illustrate the application of the bode function, consider the LTIC system of Example 5.31. The system transfer function is given by

H (ω) = 1.25 × 0.8

(0.6 + jω)2 + 0.82 .

In order to avoid a complex-valued representation, M A T L A B expresses the

Fourier transfer function in terms of the Laplace variable s = jω. In Chapter 6, we will show that the independent variable s represents the entire complex

plane and leads to the generalization of the Fourier transfer function into an

alternative transfer function, referred to as the Laplace transfer function. Sub-

stituting (s = jω) in H (ω) results in the following expression for the transfer function:

H (s) = 1

(0.6 + s)2 + 0.82 =

s2 + 1.2s + 1 .

Given H (s), the Bode plots are obtained in M A T L A B using the following

instructions:

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251 5 Continuous-time Fourier transform

> clear; % clear the MATLAB environment

> num coeff = [1]; % coefficients of the numerator

% in decreasing powers of s

> denom coeff = [1 1.2 1]; % coefficient of the denominator

% in decreasing powers of s

> sys = tf(num coeff,denom coeff);

% specify the transfer function

> bode(sys,{0.01,100}); % sketch the Bode plots

In the above set of M A T L A B instructions, we have used two new functions: tf and bode. The built-in function tf specifies the LTIC system H (s) in terms of the coefficients of the polynomials of s in the numerator and denominator.

Since the numerator N (s) = 1, the coefficients of the numerator are given by num coeff = 1. The denominator D(s) = s2 + 1.2s + 1. The coefficients of the denominator are given by denom coeff = [1 1.2 1].

The built-in function bode sketches the Bode plots. It accepts two input arguments. The first input argument sys in used to represent the LTIC system, while the second input argument {0.01,100} specifies the frequency range, 0.01 radians/s to 100 radians/s, used to sketch the Bode plots. In setting the

values for the frequency range, we use the curly parenthesis. Since the square

parenthesis [0.01,100] represents only two frequencies, ω = 0.01 and ω = 100, it will result in the wrong plots. The second argument is optional. If

unspecified, M A T L A B uses a default scheme to determine the frequency range

for the Bode plots.

5.11 Summary

In this chapter, we introduced the frequency representations for CT aperiodic

signals. These frequency decompositions are referred to as the CTFT, which

for a signal x(t) is defined by the following two equations:

CTFT synthesis equation x(t) = 1

2π

∞∫

−∞

X (ω)e jωt dω;

CTFT analysis equation X (ω) = ∞∫

−∞

x(t)e−jωt dt.

Collectively, the synthesis and analysis equations form the CTFT pair, which

is denoted by

x(t) CTFT←−−→ X (ω).

In Section 5.1, we derived the synthesis and analysis equations by expressing

the CTFT as a limiting case of the CTFS. Several important CTFT pairs were

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252 Part II Continuous-time signals and systems

calculated in Section 5.2. The results are listed in Table 5.2, and their magni-

tude and phase spectra of the CTFT are plotted in Table 5.3. In Section 5.3,

we presented the partial fraction method for calculating the inverse CTFT. In

Section 5.4, we covered the following symmetry properties of the CTFT.

(1) The CTFT X (ω) of a real-valued signal x(t) is Hermitian symmetrical, i.e.

X (ω) = X∗(−ω). Due to the Hermitian symmetry property, the magnitude spectrum |X (ω)| is an even function of ω, while the phase spectrum <X (ω) is an odd function of ω.

(2) The CTFT X (ω) of a real-valued and even signal x(t) is also real-valued

and even, i.e. Re{X (ω)} = Re{X (−ω)} and Im{X (ω)} = 0. (3) The CTFT X (ω) of a real-valued and odd signal x(t) is also pure imaginary

and odd, i.e. Re{X (ω)} = 0 and Im{X (ω)} = −Im{X (−ω)}.

Section 5.5 considered the transformation properties of the CTFT, which are

summarized as follows.

(1) The linearity property states that the CTFT of a linear combination of

aperiodic signals is given by the same linear combination of the CTFT of

the individual aperiodic signals.

(2) If an aperiodic signal is time-scaled, the CTFT is inversely time-scaled.

(3) A time shift of t0 in the aperiodic signal does not affect the magnitude of

the CTFT. However, the phase changes by an additive factor of ωt0. This

property is referred to as the time-shifting property.

(4) A frequency shift of ω0 in the aperiodic signal does not affect the magnitude

of the signal in the time domain. However, the phase of the signal in the

time domain changes by an additive factor of ωt0. This property is referred

to as the frequency-shifting property.

(5) The CTFT of a time-differentiated periodic signal is obtained by multiplying

the CTFT of the original signal by a factor of jω.

(6) The CTFT of a time-integrated periodic signal is obtained by dividing the

CTFT of the original signal by a factor of jω with a scaled impulse function

at ω = 0. (7) The duality property states that there is symmetry between the time wave-

form and its frequency-domain representation such that the two functions

in a CTFT pair are dual with respect to each other. Given an arbitrary

time-domain waveform x(t) and its CTFT waveform X (ω), for example, a

second CTFT pair exists with the time-domain representation X (t), having

the same waveform as X (ω), and the CTFT 2πx(−ω) in the frequency domain.

(8) Convolution in the time domain is equivalent to multiplication of the CTFT

in the frequency domain, and vice versa. The convolution property leads

to an alternative approach for evaluating the output response of an LTIC

system to any arbitrary input.

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253 5 Continuous-time Fourier transform

(9) The Parseval’s theorem states that the total energy in a function is the same

in the time and frequency domains. Therefore, the energy in a function can

be obtained either in the time domain by calculating the energy per unit time

and integrating it over all time, or in the frequency domain by calculating

the energy per unit frequency and integrating over all frequencies.

In Section 5.6 we derived the following condition for the existence of the CTFT

of the signalx(t):

∞∫

−∞

|x(t)|dt < ∞,

while in Sections 5.7 and 5.8 we discussed the relationship between the CTFS

and CTFT of periodic signals. In particular, the CTFT of a periodic signal x(t)

is obtained by the relationship

x(t) CTFT←−−→ 2π

∞∑

n=−∞ Dnδ(ω − nω0),

where Dn denotes the exponential CTFS coefficients and ω0 is the fundamental

frequency. Conversely, the CTFS of a periodic signal is obtained by sampling

the CTFT of one period of the periodic signal at frequencies ω = nω0. Section 5.9 showed that the three representations (linear, constant-coefficient differen-

tial equation; impulse response; and transfer function) for LTIC systems are

equivalent. Given one representation, it is straightforward to derive the remain-

ing two representations based on the CTFT and its properties. The transfer

function H (ω) plays an important role in the analysis of LTIC systems, and is

typically the preferred model for representing LTIC systems. In Section 5.10,

we concluded the chapter by showing the steps involved in computing the CTFT

of a CT signal using M A T L A B .

Problems

5.1 For each of the following CT functions, calculate the expression for the CTFT directly by using Eq. (5.10). Compare the CTFT with the corre-

sponding entry in Table 5.2 to confirm the validity of your result.

(a) x1� (

t τ

)

= (1 − |t |/τ ) [

u(t + τ ) − u(t − τ ) ]

;

(b) x2(t) = t4e−at u(t), with a ∈ ℜ+; (c) x3(t) = e−at cos(ω0t)u(t), with a, ω0 ∈ ℜ+; (d) x4(t) = e−t

2/2σ 2 , with σ ∈ ℜ.

5.2 Calculate the CTFT of the functions shown in Figs. P5.2 (a)–(e).

5.3 Three functions x1(t), x2(t), and x3(t) have an identical magnitude spectrum |X (ω)| but different phase spectra denoted, respectively, by

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254 Part II Continuous-time signals and systems

x1(t)

t t 0 p

x2(t)

0 2 T−

2 3T

(a) (b)

x5(t)

0 T

T pt

1 − 0.5sin ( )

(e)

x3(t)

x4(t)

−T 0 TT t t

Fig. P5.2. Aperiodic signals for

Problem 5.2.

<X1(ω), <X2(ω), and <X3(ω); magnitude and phase plots are shown

in Figs. P5.3(a)–(d). By representing the CTFTs as X p(ω) = |X (ω) | exp(j< Xn(ω)), for p = 1, 2, and 3, and calculating the inverse CTFT, determine the functions x1(t), x2(t), and x3(t).

5.4 Using the partial fraction method, calculate the inverse Fourier transform of the following functions:

(a) X1(ω) = (1 + jω)

(2 + jω)(3 + jω) ;

(b) X2(ω) = 2 − jω

(1 + jω)(2 + jω)(3 + jω) ;

(1 + jω)(2 + jω)2(3 + jω) ;

(d) X4(ω) = 1

(1 + jω)(2 + 2jω + ( jω)2) ;

(e) X5(ω) = 1

(1 + jω)2(2 + 2jω + ( jω)2)2 .

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255 5 Continuous-time Fourier transform

0 0−W w

|X(w)|

0.5W

−W w

<X1(w)

−0.5W

0.5W

0−W w

<X2(w)

−0.5W

p/3

−W w

<X3(w)

−p/3

(a) (b)

Fig. P5.3. Amplitude and phase

spectra of the three functions in

Problem 5.3.

5.5 Prove the following identity: ∞∫

−∞

e jωt dt = 2πδ(ω).

[Hint: Show that the integral on the left-hand side is a generalized function

that satisfies Eq. (1.47) presented in Chapter 1.]

5.6 Show that the CTFT X (ω) of a real-valued even function x(t) is also real and even. In other words, that Re{X (ω)} = Re{X (−ω)} and Im{X (ω)} = 0.

5.7 Show that the CTFT X (ω) of a real-valued odd function x(t) is imag- inary and odd. In other words, that Re{X (ω)} = 0 and Im{X (ω)} = −Im{X (−ω)}.

5.8 Using the Hermitian property, determine if the time-domain functions cor- responding to following CTFTs are real-valued or complex-valued. If a

time-domain function is real-valued, determine if it has even or odd sym-

metry.

(a) X1(ω) = 5

2 + j(ω − 5) ;

(b) X2(ω) = cos (

2ω + π

)

;

(d) X4(ω) = (3 + j2)δ(ω − 10) + (1 − j2)δ(ω + 10);

(e) X5(ω) = 1

(1 + jω)(3 + jω)2(5 + ω2) .

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256 Part II Continuous-time signals and systems

t 0

h(t)

2 4−2−4

Fig. P5.12. CT signal for

Problem 5.12.

5.9 Using Table 5.2 and the properties of the CTFT, calculate the CTFT of the following functions:

(a) x1(t) = 5 + 3 cos(10t) − 7e−2t sin(3t)u(t);

(b) x2(t) = 1

π t ;

(d) x4(t) = 5sin(3π t) sin(5π t) t2

;

(e) x4(t) = 4 sin(3π t)

t ∗

[ sin(4π t)

]

5.10 Using Table 5.2 and the linearity property, show that the CTFT of the function

x(t) = [

13 e−2t −

13 cos(3t) +

13 sin(3t)

]

u(t)

is given by

X (ω) = 6

(9 − ω2)(2 + jω)

− π

13 [(3 + j2)δ(ω − 3) + (3 − j2)δ(ω + 3)].

5.11 Prove the following time-scaling property (see Eq. (5.45)) of the CTFT:

x(at) CTFT←−−→

|a| X

(ω

)

, for a ∈ ℜ and a �= 0.

5.12 Using the time-scaling property and the results in Example 5.12, calculate the CTFT of the function h(t) shown in Fig. P5.12.

5.13 Prove the following frequency-shifting property (see Eq. (5.49)) of the CTFT:

h(t) = e jω0t x(t) CTFT←−−→ X (ω − ω0), for ω0 ∈ ℜ.

5.14 Prove the following time-integration property (see Eq. (5.53)) of the CTFT:

t∫

−∞

x(τ )dτ CTFT←−−→

X (ω)

jω + π X (0)δ(ω).

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257 5 Continuous-time Fourier transform

5.15 Assume that for the CTFT pair x(t) CTFT←−−→ X (ω), the CTFT is given by

the triangular function

X (ω) = � (ω

)







1 − |ω| 3

|ω| ≤ 3

0 elsewhere.

Using the CTFT properties (listed in Table 5.4), derive the CTFT for the

following set of functions:

(a) e−j5t x(2t); (d) x2(t);

(b) t2x(t); (e) x(t) ∗ x(t); (c) (t + 5)dx

dt ; (f) cos(ω0t)x(t) with ω0 = 3/2, 3, and 6.

5.16 Using the transform pairs in Table 5.2 and the properties of the CTFT, calculate the inverse Fourier transform of the functions in Problem 5.8.

5.17 For each of the following functions, (i) draw a rough sketch of the function, and (ii) determine if the CTFT exists by evaluating Eq. (5.59):

(a) x1(t) = e−a|t |, with a ∈ ℜ+; (b) x2(t) = e−at cos(ω0t)u(t), with a, ω0 ∈ ℜ+; (c) x3(t) = t4e−at u(t), with a ∈ ℜ+; (d) x4(t) = sin(ln(t))u(t); (e) x5(t) =

t ;

(f) x6(t) = cos ( π

)

;

(g) x7(t) = e −t2/2σ 2 , with σ ∈ ℜ.

5.18 Using the exponential CTFS representations (calculated in Problem 4.11), calculate the CTFT for the periodic signals shown in Fig. P4.6.

5.19 Determine the CTFS coefficients for the periodic functions shown in Fig. P4.6 from the CTFTs calculated in Problem 5.2.

5.20 Determine (i) the transfer function, and (ii) the impulse response for the LTIC systems whose input–output relationships are represented by the

following linear, constant-coefficient differential equations. Assume zero

initial conditions in each case.

(a) d3 y

dt3 + 6

d2 y

dt2 + 11

dt + 6y(t) = x(t).

(b) d2 y

dt2 + 3

dt + 2y(t) = x(t).

dt2 + 2

dt + y(t) = x(t).

(d) d2 y

dt2 + 6

dt + 8y(t) =

dt + 4x(t).

(e) d3 y

dt3 + 8

d2 y

dt2 + 19

dt + 12y(t) = x(t).

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258 Part II Continuous-time signals and systems

0 2

T− t

v(t)

LTIC system

+ y(t)

v(t)

C −

(a) (b)

Fig. P5.22. (a) RC circuit

system; (b) input signal.

+ y(t)

x(t)

C −

Fig. P5.23. RC circuit with

sinusoidal input signal

considered in Problem 5.23.

5.21 Consider the LTIC systems with the following input–output pairs: (a) x(t) = e−2t u(t) and y(t) = 5e−2t u(t); (b) x(t) = e−2t u(t) and y(t) = 3e−2(t−4)u(t − 4); (c) x(t) = e−2t u(t) and y(t) = t3e−2t u(t); (d) x(t) = e−2t u(t) and y(t) = e−t u(t) + e−3t u(t). For each of the above systems, determine (i) the transfer function, (ii) the

impulse response function, and (iii) the input–output relationship using

linear constant-coefficient differential equations.

5.22 Determine the transfer function of the system shown in Fig. P5.22(a). Calculate the output of the system for the input signal shown in Fig.

P5.22(b).

5.23 Using the convolution property of the CTFT, calculate the output of the system shown in Fig. P5.23 for the input signals (i) x1(t) = cos(ω0t), and (ii) x2(t) = sin(ω0t).

5.24 Sketch the gain and phase responses for the LTIC systems in Problem 5.20.

5.25 Sketch the gain and phase responses for the LTIC systems in Problem 5.21.

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259 5 Continuous-time Fourier transform

5.26 Show that if the transfer function H (ω) of a system is Hermitian symmetric (i.e. its impulse response h(t) is real-valued), the outputs of the system to

cosine and sine inputs are as follows:

cos(ω0t) Hermitian Symmetric H (ω)

−−−−−−−−−−−−→ |H (ω0)| cos(ω0t +<H (ω0))

and

sin(ω0t) Hermitian Symmetric H (ω)

−−−−−−−−−−−−→ |H (ω0)| sin(ω0t +<H (ω0)).

5.27 Using the results in Problem 5.26, verify the answers in Problem 5.23.

5.28 Using the results derived in Section 5.9.2 and the linearity property of the CTFT, calculate the output of the system shown in Fig. P5.23 for the

following input signals. Assume that R = 1 M� and C = 0.1 µF. (i) x1(t) = sin(3t);

(ii) x2(t) = cos(3t) − 5 sin(6t + 30◦); (iii) x3(t) = cos(2t) + sin(2000t); (iv) x4(t) = e j3t + e j2000t .

5.29 Suppose the CT signal

x(t) = e−t u(t)

is applied as input to a causal LTIC system modeled by the impulse

response

h(t) = e−2t u(t)

Calculate the resulting output y(t) using:

(a) direct convolution;

(b) transfer function H (ω);

5.30 The periodic signals shown in Figs. P4.6(a)–(e) are applied to the follow- ing LTIC systems:

(i) H1(ω) =







1 |ω| ≤ 4

T 0 elsewhere;

(ii) H2(ω) =







1 4

T ≤ |ω| ≤

T 0 elsewhere.

Sketch the magnitude and phase spectra of the CTFT of the resulting

outputs.

5.31 The transfer function of two LTIC systems are given by

(i) H1(ω) = 20 − jω 20 + jω

;

(ii) H2(ω) = {

1 |ω| ≥ 20 0 elsewhere.

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260 Part II Continuous-time signals and systems

(a) By sketching the magnitude spectrum of each of the LTIC systems,

comment on the frequency properties of the two systems. Classify

the two systems as a lowpass, highpass, bandpass, or an allpass fil-

ter. Recall that a lowpass filter blocks high-frequency components; a

highpass filter blocks low-frequency components; a bandpass filters

blocks frequency components within a certain band of frequencies;

while an allpass filters allows all frequency components to be passed

on to the output.

(b) Determine the impulse response for each of the two LTIC systems.

5.32 Sketch the gain and phase responses for the three LTIC systems given below:

(a) h1(t) = 2te−t u(t); (b) h2(t) = u(t); (c) h3(t) = −2δ(t) + 5e−2t u(t). For each of the three systems, show that the input signal x(t) = cos t produces the same output response. How can this result be explained?

5.33 (M A T L A B exercise) By making modifications to the myctft function listed in Section 5.10, sketch the magnitude and phase spectra of the

following signals:

(i) x1(t) = sin(5π t) for −2 ≤ t ≤ 2 with sampling rate ωs = 200π samples/s;

(ii) x2(t) = sin(8π t) + sin(20π t) for −1.25 ≤ t ≤ 1.25 with sampling rate ωs = 1000π samples/s.

5.34 (M A T L A B exercise) Compute the CTFTs of the CT functions specified in Problem 5.1. By plotting the magnitude and phase spectra, compare

your computed result with the analytical expressions listed in Tables 5.2

and 5.3.

5.35 (M A T L A B exercise) Compute the output response y(t) for Problem 5.29 by computing the CTFT for x(t) and h(t), multiplying the CTFTs and

then taking the inverse CTFT of the result.

5.36 (M A T L A B exercise) Sketch the magnitude and phase Bode plots for the LTIC systems specified in Problems 5.20 and 5.21.

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C H A P T E R

6 Laplace transform

In Chapters 4 and 5, we introduced the continuous-time Fourier series (CTFS)

for CT periodic signals and the continuous-time Fourier transform (CTFT)

for both CT periodic and aperiodic signals. These frequency representations

provide a useful tool for determining the output of an LTIC system. Unfortu-

nately, the CTFT is not defined for all aperiodic signals. In cases where the

CTFT does not exist, an alternative procedure, based on the Laplace trans-

form, is used to analyze the LTIC systems. Even for the CT signals for which

the CTFT exists, the Laplace transforms are always real-valued, rational func-

tions of the independent variable s provided that the CT functions are real. The

CTFTs are complex-valued in most cases. Therefore, using the Laplace trans-

form simplifies algebraic manipulations and leads to important flow diagram

representations of the CT systems from which the hardware implementations

of the CT systems are derived. Finally, the CTFT can only be applied to stable

LTIC systems for which the impulse response is absolutely integrable. Since

the Laplace transform exists for both stable and unstable LTIC systems, it can

be used to analyze a broader range of LTIC systems.

The difference between the CTFT and the Laplace transform lies in the choice

of the basis functions used in the two representations. The CTFT expands an

aperiodic signal as a linear combination of complex exponential functions e jωt ,

which are referred to as its basis functions. The Laplace transform uses est as

the basis functions, where the independent Laplace variable s is complex and is

given by s = σ + jω. The Laplace transform is, therefore, a generalization of the CTFT, since the independent variable s can take any value in the complex s-plane

and is not simply restricted to the imaginary jω-axis, as is the case for the CTFT.

In this chapter, we will cover the Laplace transform and its applications in the

analysis of LTIC systems. To illustrate the usefulness of the Laplace transforms

in signal processing, some real-world applications are presented in Chapter 8.

Chapter 6 is organized as follows. Section 6.1 defines the bilateral, or two-

sided, Laplace transform and provides several examples to illustrate the steps

involved in its computation. The bilateral Laplace transform is used for non-

causal and causal signals. For causal signals, the bilateral Laplace transform

simplifies to the one-sided, or unilateral, Laplace transform, which is covered in

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262 Part II Continuous-time signals and systems

Section 6.2. Section 6.3 computes the time-domain representation of a Laplace-

transformed signal, while Section 6.4 considers the properties of the Laplace

transform. Sections 6.5 to 6.9 propose several applications of the Laplace trans-

form, ranging from solving differential equations (Section 6.5), evaluating the

location of poles and zeros (Section 6.6), determining the causality and stability

of LTIC systems from their Laplace transfer functions (Sections 6.7 and 6.8),

and analyzing the outputs of LTIC systems (Section 6.9). Section 6.10 presents

the cascaded, parallel, and feedback configurations for interconnecting LTI

systems, and Section 6.11 concludes the chapter.

6.1 Analytical development

In Section 5.1, the CTFT pair, x(t) CTFT←−−→X ( jω), was defined as follows:

CTFT synthesis equation x(t) = 1

2π

∞∫

−∞

X ( jω)e jωt dω; (6.1)

CTFT analysis equation X ( jω) = ∞∫

−∞

x(t)e−jωt dt . (6.2)

In Eqs. (6.1) and (6.2), the CTFT of x(t) is expressed as X ( jω), instead of

the earlier notation X (ω), to emphasize that the CTFT is computed on the

imaginary jω-axis in the complex s-plane. For a CT signal x(t), the expression

for the bilateral Laplace transform is derived by considering the CTFT of the

modified version, x(t)e−σ t , of the signal. Based on Eq. (6.2), the CTFT of the

modified signal x(t)e−σ t is given by

ℑ{x(t)e−σ t } = ∞∫

−∞

x(t)e−σ t e−jωt dt, (6.3)

which reduces to

ℑ{x(t)e−σ t } = ∞∫

−∞

x(t)e−(σ+jω)t dt

= X (σ + jω). (6.4)

Substituting s = σ + jω in Eq. (6.4) leads to the following definition for the bilateral Laplace transform:†

Laplace analysis equation X (s) = ℑ{x(t)e−σ t } = ∞∫

−∞

x(t)e−st dt . (6.5)

† The Laplace transform was discovered originally by Leonhard Euler (1707–1783), a prolific

Swiss mathematician and physicist. However, it is named in honor of another mathematician

and astronomer, Pierre-Simon Laplace (1749–1827), who used the transform in his work on

probability theory.

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263 6 Laplace transform

To derive the synthesis equation for the bilateral Laplace transform, consider

the inverse transform of the CTFT pair, x(t)e−σ t CTFT←−−→X (σ + jω) = X (s).

Based on Eq. (6.1), we obtain

x(t)e−σ t = 1

2π

∞∫

−∞

X (s)e jωt dω. (6.6)

Multiplying both sides of Eq. (6.6) by eσ t and changing the integral variable ω

to s using the relationship s = σ + jω yields

Laplace synthesis equation x(t) = 1

2π j

σ−j∞∫

σ−j∞

X (s)est ds. (6.7)

Solving Eq. (6.7) involves the use of contour integration and is seldom used

in the computation of the inverse Laplace transform. In Section 6.3, we will

consider an alternative approach based on the partial fraction expansion to

evaluate the inverse Laplace transform. Collectively, Eqs. (6.5) and (6.7) form

the bilateral Laplace transform pair, which is denoted by

x(t) L←→ X (s). (6.8)

To illustrate the steps involved in computing the Laplace transform, we consider

the following examples.

Example 6.1

Calculate the bilateral Laplace transform of the decaying exponential function:

x(t) = e−at u(t).

Solution

Substituting x(t) = e−at u(t) in Eq. (6.5), we obtain

X (s) = ∞∫

−∞

e−at u(t)e−st dt = ∞∫

e−(s+a) t dt = − 1

(s + a) e−(s+a) t

∣ ∣ ∣ ∣

∞

At the lower limit, t → 0, e−(s+a)t = 1. At the upper limit, t → ∞, e−(s+a)t = 0 if Re{s + a} > 0 or Re{s} > −a. If Re{s} ≤ −a, then the value of e−(s+a)t is infinite at the upper limit, t → ∞. Therefore,

X (s) =







(s + a) for Re{s} > −a

undefined for Re{s} ≤ −a.

The set of values of s over which the bilateral Laplace transform is defined

is referred to as the region of convergence (ROC). Assuming a to be a real

number, the ROC is given by Re{s} > −a for the Laplace transform of the decaying exponential function, x(t) = e−at u(t). Figure 6.1 highlights the ROC by shading the appropriate area in the complex s-plane.

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264 Part II Continuous-time signals and systems

t 0

x(t) = e−atu(t)

Im{s}

(a)(a) (b)

0 Re{s}

−a

Fig. 6.1. (a) Exponential

decaying function

x(t ) = e−at u(t ); (b) its associated ROC, Re{s} > −a , over which the bilateral Laplace

transform exists.

Example 6.1 shows that the bilateral Laplace transform of the decaying expo-

nential function x(t) = e−at u(t) will converge to a finite value X (s) = 1/(s + a) within the ROC (Re{s} > −a). In other words, the bilateral Laplace transform of x(t) = e−at u(t) exists for all values of a within the specified ROC. No restric- tion is imposed on the value of a for the existence of the Laplace transform.

On the other hand, the CTFT of the decaying exponential function exists only

for a > 0. For a < 0, the exponential function x(t) = e−at u(t) is not absolutely integrable, and hence its CTFT does not exist. This is an important distinction

between the CTFT and the bilateral Laplace transform. The CTFT exists for a

limited number of absolutely integrable functions. By associating an ROC with

the bilateral Laplace transform, we can evaluate the Laplace transform for a

much larger set of functions.

Example 6.2

Calculate the bilateral Laplace transform of the non-causal exponential function

g(t) = −e−at u(−t).

Solution

Substituting g(t) = −e−at u(−t) in Eq. (6.5), we obtain

G(s) = ∞∫

−∞

−e−at u(−t)e−st dt = −

0∫

−∞

e−(s+a) t dt = 1

(s + a) e−(s+a) t

∣ ∣ ∣ ∣

−∞

At the upper limit, t → 0, e−(s+a)t = 1. At the lower limit, t → −∞, e−(s+a)t

is finite only if Re{s + a} < 0, where it equals zero. The bilateral Laplace

transform is therefore given by

G(s) =







(s + a) for Re{s} < −a

undefined for Re{s} ≥ −a.

Figure 6.2 illustrates the ROC, Re{s} < −a, for the bilateral Laplace transform of g(t) = −e−at u(−t).

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265 6 Laplace transform

t 0

x(t) = −e−atu(−t)

Im{s}

(a) (b)

0 Re{s}

−a

Fig. 6.2. (a) Non-causal

decaying function

x(t ) = −e−atu(−t ); (b) its associated ROC, Re{s} < −a , over which the bilateral Laplace

transform exists.

In Examples 6.1 and 6.2, we have proved the following Laplace transform pairs:

e−at u(t) L←→

(s + a) with ROC: Re{s} > −a

and

−e−at u(−t) L←→ 1

(s + a) with ROC: Re{s} < −a.

Although the algebraic expressions for the bilateral Laplace transforms are the

same for the two functions, the ROCs are different. This implies that a bilateral

Laplace transform is completely specified only if the algebraic expression and

the ROC are both specified. This is illustrated further in Example 6.3.

Example 6.3

Calculate the inverse Laplace transform of the function H (s) = 1/(s + a) .

Solution

From Examples 6.1 and 6.2, we know that

e−at u(t) L←→

(s + a) with ROC: Re{s} > −a

and

−e−at u(−t) L←→ 1

(s + a) with ROC: Re{s} < −a.

Therefore, the inverse bilateral Laplace transform is either h(t) = e−at u(t) or h(t) = −e−at u(−t). If we want to determine a unique inverse, we need to specify the ROC associated with the Laplace transform. If the ROC is specified

as Re{s} > −a, then the inverse Laplace transform h(t) = e−at u(t). On the other hand, if the ROC is Re{s} > −a, then h(t) = e−at u(t).

The need to specify the ROC is also evident from the synthesis equation,

Eq. (6.7), of the Laplace transform. To evaluate the inverse Laplace transform

using Eq. (6.7), a straight line, parallel to the jω-axis, corresponding to all points

s satisfying Re{s} = σ within the ROC, is used as the contour of integration. The

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266 Part II Continuous-time signals and systems

complex integral, therefore, cannot be computed without having prior knowl-

edge of the ROC.

6.2 Unilateral Laplace transform

In Section 6.1, we introduced the bilateral Laplace transform that is used to

analyze both causal and non-causal LTIC systems. In signal processing, most

physical systems and signals are causal. Applying the causality condition, the

bilateral Laplace transform reduces to a simpler version of the Laplace trans-

form. The Laplace transform for causal signals and systems is referred to as the

unilateral Laplace transform and is defined as follows:

X (s) = L{x(t)} = ∞∫

0−

x(t)e−st dt, (6.9)

where the initial conditions of the system are incorporated by the lower limit

(t = 0−). In our subsequent discussions, we will mostly use the unilateral

Laplace transform. For simplicity, we will omit the term “unilateral,” there-

fore the Laplace transform implies the unilateral Laplace transform. When we

refer to the bilateral Laplace transform, the term “bilateral” will be explicitly

stated. To clarify further the differences between the unilateral and bilateral

Laplace transform, we summarize the major points.

(1) The unilateral Laplace transform simplifies the analysis of causal LTIC sys-

tems. However, it cannot analyze non-causal systems directly. Since most

physical systems are naturally causal, we will use the unilateral Laplace

transform in our computations. The bilateral transform will be used only

to analyze non-causal systems.

(2) For causal signals and systems, the unilateral and bilateral Laplace trans-

forms are the same.

(3) The synthesis equation used for calculating the inverse of the unilateral

Laplace transform is the same as Eq. (6.7) used for evaluating the inverse

of the bilateral transform.

Example 6.4

Calculate the unilateral Laplace transform for the following functions:

(i) unit impulse function, x1(t) = δ(t);

(ii) unit step function, x2(t) = u(t);

(iii) shifted gate function,

x3(t) =

{

1 2 ≤ t ≤ 4 0 otherwise;

(iv) causal complex exponential function, x4(t) = e−jω0t u(t);

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267 6 Laplace transform

(v) causal sine function, x5(t) = sin(ω0t)u(t); (vi) causal ramp function, x6(t) = tu(t);

(vii) x7(t) =







2t 0 ≤ t ≤ 1 2 1 ≤ t ≤ 2 0 otherwise.

Solution

(i) Unit impulse function. Substituting x1(t) = δ(t) in Eq. (6.9) yields

X1(s) = ∞∫

0−

δ(t)e−st dt .

Since δ(t)e−st = δ(t), the above equation reduces to

X1(s) = ∞∫

0−

δ(t)dt = 1.

The Laplace transform for an impulse function is given by

δ(t) L←→ 1 with ROC: entire s-plane.

(ii) Unit step function. Substituting x2(t) = u(t) in Eq. (6.9) yields

X2(s) = ∞∫

0−

u(t)e−st dt .

For Re{s} > 0, the above integral reduces to

X2(s) = ∞∫

0−

e−st dt = − 1

s e−st

∣ ∣ ∣ ∣

∞

= 1.

The Laplace transform for a unit step function is given by

u(t) L←→

s with ROC: Re{s} > 0.

(iii) Shifted gate function. Substituting x3(t) in Eq. (6.9) yields

X3(s) = 4∫

e−st dt = − 1

s e−st

∣ ∣ ∣ ∣

= 1

s (e−2s − e−4s).

Clearly, the above expression for the Laplace transform is not valid for s = 0. The value of the Laplace transform for s = 0 is obtained by substituting s = 0 in Eq. (6.9). The resulting expression is given by

X3(s) = ∞∫

0−

x3(t)dt = 4∫

1 · dt = t |42 = 2, for s = 0.

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268 Part II Continuous-time signals and systems

The Laplace transform for the shifted gate function is therefore given by

X3(s) =







2 for s = 0 1

s (e−2s − e−4s) for s �= 0.

The associated ROC is the entire s-plane.

(iv) Causal complex exponential function. From Example 6.1, we know that

e−at u(t) L←→

(s + a) with ROC: Re{s} > −Re{a}.

Substituting a = jω0, we obtain

e−jω t 0 u(t)

L←→ 1

(s + jω0) with ROC: Re{s} > 0.

(v) Causal sine function. By expanding sin(ω0t) = [exp(jω0t) − exp(−jω0t)]/2j, the Laplace transform for the causal sine function is given by

X5(s) = 1

∞∫

0−

[e jω0t − e−jω0t ]e−st dt = 1

∞∫

0−

e−(s−jω0)t dt − 1

∞∫

0−

e−(s−jω0)t dt .

Both integrals are finite for Re{s ±jω0} > 0 or Re{s} > 0. The Laplace trans- form is given by

X5(s) = 1

[ 1

s − jω0

]

− 1

[ 1

s + jω0

]

= ω0

s2 + ω20 .

In other words, the Laplace transform pair is given by

sin(ω0t)u(t) L←→

ω0

s2 + ω20 with ROC: Re{s} > 0.

(vi) Causal ramp function. Substituting x6(t) = tu(t) in Eq. (6.9) yields

X6(s) = ∞∫

0−

tu(t)e−st dt = ∞∫

te−st dt = te−st

(−s)

∣ ∣ ∣ ∣

∞

− e−st

(−s)2

∣ ∣ ∣ ∣

∞

which, on simplification, leads to the following Laplace transform pair:

tu(t) L←→

s2 with ROC: Re{s} > 0 .

(vii) Substituting

x7(t) =







2t 0 ≤ t ≤ 1 2 1 ≤ t ≤ 2 0 otherwise

in Eq. (6.9) leads to the following Laplace transform:

X7(s) = 2 1∫

0−

te−st dt + 2 2∫

e−st dt = 2 [

te−st

−s −

e−st

(−s)2

] 1

0− + 2

[ e−st

−s

] 2

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269 6 Laplace transform

Clearly, the above integral is not defined for s = 0. For s �= 0, the above expres- sion reduces to

X7(s) = 2 [

e−s

−s −

e−s

(−s)2 +

(−s)2

]

+ 2 [

e−2s

−s −

e−s

−s

]

= 2

s2 [1 − e−s − se−2s].

For s = 0, the Laplace transform is given by

X7(s) = ∞∫

0−

x7(t)dt = 1∫

2t · dt + 2∫

2 · dt = t2|10 + 2t | 2 1 = 3.

The Laplace transform pair is therefore given by

X7(s) =







3 for s = 0 2

s2 [1 − e−s − se−2s] for s �= 0.

The associated ROC is the entire s-plane.

6.2.1 Relationship between Fourier and Laplace transforms

Comparing Eq. (6.2) with Eq. (6.5), the CTFT can be related to the bilateral

Laplace transform as follows:

X ( jω) = ∞∫

−∞

x(t)e−jωt dt = X (s)|s=jω. (6.10)

Since, for causal functions, the bilateral and unilateral Laplace transforms are

the same, the above relationship is also valid for the unilateral Laplace transform

for causal functions. Equation (6.8) proves that the CTFT is a special case of

the Laplace transform when s = jω, i.e. σ = 0. In other words, the CTFT is the Laplace transform computed along the imaginary jω-axis in the s-plane.

Table 6.1 lists the Laplace transforms for several commonly used functions. To

compare the results with the corresponding CTFTs, we also include the CTFTs

for the same functions in the second column of Table 6.1. When the function

is causal and its CTFT exists, it is observed that the CTFT can be obtained

from the Laplace transform by substituting s = jω. An alternative condition for the existence of the CTFT is, therefore, the inclusion of the jω-axis within the

ROC of the Laplace transform. If the ROC does not contain the jω-axis, the

substitution s = jω cannot be made and the CTFT does not exist. For example, the ROC Re{s} > 0 for the unit step function x(t) = u(t) does not contain the jω-axis. Based on our earlier reasoning, its CTFT should not exist. This appears

to be in violation with the second entry in Table 6.1, where the CTFT of the

unit step function is listed as follows:

u(t) CTFT←−−→ πδ(ω) + 1/jω.

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Table 6.1. CTFT and Laplace transform pairs for several causal CT signals

CT signals x(t) CTFT

X ( jω) = ∞∫

−∞

x(t)e−jωt dt

Laplace transform

X (s) =

∞∫

−∞

x(t)e−st dt

(1) Impulse function

x(t) = δ(t)

1 1

ROC: entire s-plane

(2) Unit step function

x(t) = u(t)

πδ(ω) + 1

jω

ROC: Re{s} > 0

(3) Causal gate function

x(t) = u(t) − u(t − a)

(1 − e−jaω)

(

πδ(ω) + 1

jω

) 1

s (1 − e−as)

ROC: Re{s} > 0

(4) Causal decaying exponential function

x(t) = e−at u(t)

a + jω

a + s

ROC: Re{s} > −a

(5) Causal ramp function

x(t) = tu(t)

does not exist 1

ROC: Re{s} > 0

(6) Higher-order causal ramp function

x(t) = tnu(t)

does not exist n!

sn+1

ROC: Re{s} > 0

(7) First-order time-rising causal decaying

exponential function

x(t) = te−at u(t)

(a + jω)2

provided a > 0.

(a + s)2

ROC: Re{s} > −a

(8) Higher-order time-rising causal

decaying exponential function

x(t) = tne−at u(t)

(a + jω)n+1

provided a > 0

(a + s)n+1

ROC: Re{s} > −a

(9) Causal cosine wave

x(t) = cos(ω0t)u(t)

π [δ(ω − ω0) + δ(ω + ω0)]

+ jω

ω20 − ω 2

ω20 + s 2

ROC: Re{s} > 0

(10) Causal sine wave

x(t) = sin(ω0t)u(t)

2j [δ(ω − ω0) − δ(ω + ω0)]

+ ω0

ω20 − ω 2

ω0

ω20 + s 2

ROC: Re{s} > 0

(11) Squared causal cosine wave

x(t) = cos2(ω0t)u(t)

2 [δ(ω) + δ(ω − 2ω0) + δ(ω + 2ω0)]

+ 1

j2ω +

jω

2 (

4ω20 − ω 2 )

(

2ω20 + s 2 )

s (

4ω20 + s 2 )

ROC: Re{s} > 0

(12) Squared causal sine wave

x(t) = sin2(ω0t)u(t)

2 [δ(ω) − δ(ω − 2ω0) − δ(ω + 2ω0)]

+ 1

j2ω −

jω

2 (

4ω20 − ω 2 )

2ω20

s (

4ω20 + s 2 )

ROC: Re{s} > 0

(13) Causal decaying exponential cosine

function

x(t) = exp(−at) cos(ω0t)u(t)

a + jω

(a + jω)2 + ω20 provided a > 0

a + s

(a + s)2 + ω20

ROC: Re{s} > −a

(14) Causal decaying exponential sine

function

x(t) = exp(−at) sin(ω0t)u(t)

ω0

(a + jω)2 + ω20 provided a > 0

ω0

(a + s)2 + ω20

ROC: Re{s} > −a

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271 6 Laplace transform

The above argument is not true because the CTFT of the unit step function

contains a discontinuity at ω = 0 due to the presence of the impulse function δ(ω). Therefore, the CTFT violates the condition for the existence of CTFT. In

such cases, the CTFT is not derived from its definition but is expressed using

the impulse function, which is not a mathematical function in the strict sense.

It is therefore natural to expect Eq. (6.10) to be invalid. Likewise, the ROC for

the Laplace transform of the sine wave, cosine wave, squared cosine wave, and

squared sine wave do not contain the jω-axis, and Eq. (6.10) is also not valid

in these cases.

6.2.2 Region of convergence

As a side note to our discussion, we observe that the Laplace transform is

guaranteed to exist at all points within the ROC. For example, consider the

causal sine wave h(t) = sin(4t)u(t). We are interested in calculating the values of the Laplace transform at two points, s1 = 2 + j3 and s2 = j3 in the complex s-plane. Since s1 lies within the ROC, Re{s} > 0, the value of the Laplace transform at s1 is given by

H (2 + j3) = 4

(2 + j3)2 + 42 =

4 + j12 − 9 + 16 =

11 + j12 ,

which, as expected, is a finite value. The point s2 = j3 lies outside the ROC. However, the Laplace transform is not necessarily infinite at s2. In fact, the

Laplace transform of the causal sine wave h(t) = sin(4t)u(t) is finite for all values of s on the imaginary axis except at s = ±j4. The value of the Laplace transform at s2 is given by

H ( j3) = 4

(j3)2 + 42 = −

5 .

Since the Laplace transform is not defined at two points (s = ±j4) on the imaginary axis, the entire imaginary axis is excluded from the ROC. In short, if

a point lies on the boundary of the ROC, it is possible that the Laplace transform

exists, though the point may not be included in the ROC.

6.2.3 Spectra for the Laplace transform

In Chapter 5, the magnitude and phase spectra of the CTFT provided mean-

ingful insights into the frequency properties of the reference function. Except

for one difference, the magnitude and phase spectra of the Laplace transform

(collectively referred to as the Laplace spectra) can be plotted in a similar way.

Since the Laplace variable s is a complex variable, the Laplace spectra are

plotted with respect to a 2D complex plane with Re{s} = σ and Im{s} = ω being the two independent axes. For the magnitude spectrum, the magnitude

of the Laplace transform is plotted along the z-axis within the ROC defined

in the 2D complex plane. Likewise, for the phase spectrum, the phase of the

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272 Part II Continuous-time signals and systems

−5 −3 −2 −1 0

Re{s} = sIm{s} = w Im{s} = w

< H

(s )

(d eg

re e s)

H (s

)

1 2 3 4 −3 −5

−90

−2 −1 0

Re{s} = s

1 2 3 4

(a) (b)

Fig. 6.3. Laplace spectra for

x(t ) = e−3t u(t ). (a) Laplace magnitude spectrum;

(b) Laplace phase spectrum.

Laplace transform is plotted along the vertical z-axis within the ROC. Both

Laplace magnitude and phase spectra are, therefore, 3D plots. To illustrate the

steps involved in plotting the magnitude and phase spectra, we consider the

following example.

Example 6.5

Plot the Laplace spectra of the decaying exponential function x(t) = e−3t u(t).

Solution

Based on entry (4) of Table 6.1, the Laplace transform of the decaying expo-

nential function is given by

X (s) = 1

(s + 3) =

(σ + jω + 3) with ROC: σ = Re(s) > −3.

The Laplace spectra are therefore given by

magnitude spectrum |X (s)| = 1

√

(σ + 3)2 + ω2 ;

phase spectrum <X (s) = −tan−1 ω

(σ + 3) .

The magnitude and phase spectra are plotted with respect to the 2D complex

s-plane in Fig. 6.3. To obtain the CTFT spectra, we can simply splice out the

2D plot along the Re{s} = σ = 0 axis from the Laplace spectra. Figure 6.4 shows the resulting CTFT magnitude and phase spectra. These are identical to

the CTFT spectra obtained directly from the CTFT and plotted in Fig. 6.4.

w 0

X(w)1/4

w 0

< X(w) p/2

−p/2

(a) (b)

Fig. 6.4. CTFT spectra for

x(t ) = e−3t u(t ) obtained by extracting the 2D plot along the

Re{s} = σ = 0 axis in Fig. 6.3. (a) CTFT magnitude spectrum;

(b) CTFT phase spectrum.

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273 6 Laplace transform

6.3 Inverse Laplace transform

Evaluation of the inverse Laplace transform is an important step in the analysis

of LTIC systems. The inverse Laplace transform can be calculated directly by

solving the complex integral in the synthesis equation, Eq. (6.7). This approach

involves contour integration, which is beyond the scope of this text. In cases

where the Laplace transform takes the following rational form:

X (s) = N (s)

D(s) =

bms m + bm−1sm−1 + bm−2sm−2 + · · · + b1s + b0

sn + an−1sn−1 + an−2sn−2 + · · · + a1s + a0 , (6.11)

an alternative approach based on the partial fraction expansion is commonly

used. The approach eliminates the need for computing Eq. (6.7) and consists

of the following steps.

(1) Calculate the roots of the characteristic equation of the rational fraction, Eq.

(6.11). The characteristic equation is obtained by equating the denominator

D(s) in Eq. (6.11) to zero, i.e.

D(s) = sn + an−1sn−1 + an−2sn−2 + · · · + a1s + a0 = 0. (6.12)

For an nth-order characteristic equation, there will be n first-order roots.

Depending on the value of the coefficients {bl}, 0 ≤ l ≤ (n − 1), roots {pr}, 1 ≤ r ≤ n, of the characteristic equation may be real-valued and/or complex-valued. Assuming that roots are real-valued and do not repeat, the

Laplace transform X (s) is represented as

X (s) = N (s)

(s − p1)(s − p2) · · · (s − pn−1)(s − pn) . (6.13)

(2) Using Heaviside’s partial fraction expansion formula, explained in

Appendix D, decompose X (s) into a summation of the first- or second-

order fractions. If no roots are repeated, X (s) is decomposed as follows:

X (s) = k1

(s − p1) +

(s − p2) + · · · +

kn−1

(s − pn−1) +

(s − pn) , (6.14)

where the coefficients {kr}, 1 ≤ r ≤ n, are obtained from

kr = [

(s − pr ) N (s)

D(s)

]

s=pr . (6.15)

If there are repeated or complex roots, X (s) takes a slightly different form.

See Appendix D for more details.

(3) From Table 6.1,

epr t u(t) L←→

(s − pr ) with ROC: Re{s} > pr .

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274 Part II Continuous-time signals and systems

Using the above transform pair, the inverse Laplace transform of X (s) is

given by

x(t) = k1ep1t u(t) + k2ep2t u(t) + · · · + kn−1epn−1t u(t) + knepn t u(t)

= n∑

r=1 kr e

pr t u(t). (6.16)

To illustrate the aforementioned procedure (steps (1) to (3)) for evaluating the

inverse Laplace transform using the partial fraction expansion, we consider the

following examples.

Example 6.6

Calculate the inverse Laplace transform of a right-sided sequence with transfer

function

G(s) = 7s − 6

(s2 − s − 6) .

Solution

The characteristic equation of G(s) is given by s2 − s − 6 = 0, which has two roots at s = 3 and −2. Using the partial fraction expansion, the Laplace transform G(s) is expressed as

G(s) = 7s − 6

(s + 2)(s − 3) ≡

(s + 2) +

(s − 3) .

The coefficients of the partial fractions k1 and k2 are given by

k1 = [

(s + 2) (7s − 6)

(s + 2)(s − 3)

]

s=−2 =

[ (7s − 6) (s − 3)

]

s=2 = 4

and

k2 = [

(s − 3) (7s − 6)

(s + 2)(s − 3)

]

s=3 =

[ (7s − 6) (s + 2)

]

s=3 = 3.

The partial fraction expansion of the Laplace transform G(s) is therefore given

G(s) = 4

(s + 2) +

(s − 3) .

Using entry (4) of Table 6.1, the inverse Laplace transform is

g(t) = (4e−2t + 3e3t )u(t).

Example 6.7

Calculate the inverse Laplace transform of right-sided sequences with the fol-

lowing transfer functions:

(i) X1(s) = s + 3

s(s + 1)(s + 2) ;

(ii) X2(s) = s + 5

s3 + 5s2 + 17s + 13 .

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275 6 Laplace transform

Solution

(i) In X1(s), the characteristic equation is already factorized. In terms of the

partial fractions, X1(s) can be expressed as follows:

X1(s) = s + 3

s(s + 1)(s + 2) ≡

s +

(s + 1) +

(s + 2) ,

where the partial fraction coefficients k1, k2, and k3 are given by

k1 = [

s (s + 3)

s(s + 1)(s + 2)

]

s=0 =

[ (s + 3)

(s + 1)(s + 2)

]

s=0 =

2 ,

k2 = [

(s + 1) (s + 3)

s(s + 1)(s + 2)

]

s=−1 =

[ (s + 3) s(s + 2)

]

s=−1 = −2,

and

k3 = [

(s + 2) (s + 3)

s(s + 1)(s + 2)

]

s=−2 =

[ (s + 3) s(s + 1)

]

s=−2 =

2 .

The partial fraction expansion of the Laplace transform X1(s) is given by

X1(s) = s + 3

s(s + 1)(s + 2) ≡

2s −

(s + 1) +

2(s + 2) ,

which leads to the following inverse Laplace transform:

x1(t) = (

2 − 2e−t +

2 e−2t

)

u(t).

(ii) The characteristic equation of X2(s) is given by

D(s) = s3 + 5s2 + 17s + 13 = 0,

which has three roots at s = −1, −2, and ±j3. The partial fraction expansion of X2(s) is given by

X2(s) = s + 5

(s + 1)(s + 2 + j3)(s + 2 − j3) ≡

(s + 1) +

k2s + k3 (s2 + 4s + 13)

The partial fraction coefficient k1 is calculated to be

k1 = [

(s + 1) (s + 5)

(s + 1)(s2 + 4s + 13)

]

s=−1 =

[ (s + 5)

(s2 + 4s + 13)

]

s=−1 =

5 .

To compute coefficients k2 and k3, we substitute k1 = 2/5 in X2(s) and expand s + 5

(s + 1)(s2 + 4s + 13) ≡

5(s + 1) +

k2s + k3 (s2 + 4s + 13)

s + 5 ≡ 0.4(s2 + 4s + 13) + (k2s + k3)(s + 1).

Comparing the coefficients of s2 on both sides of the above expression yields

k2 + 0.4 = 0 ⇒ k2 = −0.4.

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276 Part II Continuous-time signals and systems

Similarly, comparing the coefficients of s yields

k2 + k3 + 1.6 = 1 ⇒ k3 = −0.2.

The partial fraction expansion of X2(s) reduces to

X2(s) = 2

5(s + 1) − 0.2

2s + 1 (s + 2)2 + 9

which is expressed as

X2(s) = 2

5(s + 1) − 0.2

2(s + 2) (s + 2)2 + 9

+ 0.2 3

(s + 2)2 + 9 .

Based on entries (4) and (13) in Table 6.1, the inverse Laplace transform is

given by

x1(t) = (0.4e−t − 0.4e−2t cos(3t) + 0.2e−2t sin(3t))u(t).

6.4 Properties of the Laplace transform

The unilateral and bilateral Laplace transforms have several interesting prop-

erties, which are used in the analysis of signals and systems. These properties

are similar to the properties of the CTFT covered in Section 5.4. In this section,

we discuss several of these properties, including their proofs and applications,

through a series of examples. A complete listing of the properties is provided

in Table 6.2. In most cases, we prove the properties for the unilateral Laplace

transform. The proof for the bilateral Laplace transform follows along similar

lines and is not included to avoid repetition.

6.4.1 Linearity

If x1(t) and x2(t) are two arbitrary functions with the following Laplace trans-

form pairs:

x1(t) L←→ X1(s) with ROC: R1

and

x2(t) L←→ X2(s) with ROC: R2 ,

then

a1x1(t) + a2x2(t) L←→ a1 X1(s) + a2 X2(s) with ROC: at least R1 ∩ R2

(6.17)

for both unilateral and bilateral Laplace transforms.

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277 6 Laplace transform

Proof

Calculating the Laplace transform of{a1x1(t) + a1x2(t)} using Eq. (6.9) yields

L{a1x1(t) + a2x2(t)} = ∞∫

0−

{a1x1(t) + a2x2(t)}e −st dt

∞∫

0−

a1x1(t)e −st dt +

∞∫

0−

a2x2(t)e −st dt

= a1

∞∫

0−

x1(t)e −st dt + a2

∞∫

0−

x2(t)e −st dt

= a1 X1(s) + a2 X2(s),

which proves Eq. (6.17).

By definition of the ROC, the Laplace transform X1(s) is finite within the

specified region R1. Similarly, X2(s) is finite within its ROC R2. Therefore, the

linear combination a1 X1(s) + a2 X2(s) must at least be finite in region R that

represents the intersection of the two regions i.e. R = R1 ∩ R2. If there is no common region between R1 and R2, then the Laplace transform of {a1x1(t) + a1x2(t)} does not exist. Due to the cancellation of certain terms in a1 X1(s) + a2 X2(s), it is also possible that the overall ROC of the linear combination is

larger than R1 ∩ R2. To illustrate the application of the linearity property, we consider the following example.

Example 6.8

Calculate the Laplace transform of the causal function g(t) shown in Fig. 6.5.

t 0

g(t) 4

1 4−1−2−3−4 32

Fig. 6.5. Causal function g(t )

considered in Example 6.8.

Solution

The causal function g(t) is expressed as the linear combination

g(t) = 4x3(t) + 2x7(t),

where the CT functions x3(t) and x7(t) are defined in Example 6.4. Based on

the results of Example 6.4, the Laplace transforms for x3(t) and x7(t) are given

X3(s) =







2 for s = 0 1

s [e−2s − e−4s] for s �= 0

with ROC: entire s-plane

and

X7(s) =







3 for s = 0 2

s2 [1 − e−s − se−2s] for s �= 0.

with ROC: entire s-plane

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278 Part II Continuous-time signals and systems

Applying the linearity property, the Laplace transform of g(t) is given by

G(s) =







4(2) + 2(3) for s = 0 4

s (e−2s − e−4s) +

s2 [1 − e−s − se−2s] for s �= 0,

which reduces to

G(s) =







14 for s = 0 4

s2 [1 − e−s − se−4s] for s �= 0.

Note that the ROC of G(s) is the intersection of the individual regions R1 and

R2. The overall ROC R is, therefore, the entire s-plane.

6.4.2 Time scaling

If x(t) L←→ X (s) with ROC:R,then the Laplace transform of the scaled signal

x(at), where a ∈ ℜ+, a > 0, is given by

x(at) L←→

|a| X

( s

)

with ROC: a R (6.18)

for both unilateral and bilateral Laplace transforms. For the unilateral Laplace

transform, the value of a must be greater than zero. If a < 0, the scaled signal

x(at) will be non-causal such that its unilateral Laplace transform will not exist.

Proof

By Eq. (6.9), the Laplace transform of the time-scaled signal x(at) is given by

L{x(at)} = ∞∫

0−

x(at)e−st dt

= ∞∫

0−

x(τ )e−sτ/a dτ

a (by substituting τ = at)

= 1

a X

( s

)

which proves Eq. (6.18) for a > 0. To prove that the ROC of the Laplace

transform of the time-scaled signal is aR, note that the values of X (s) are finite

within region R. For X (s/a), the new region over which X (s/a) is finite will

transform to aR.

Example 6.9

Calculate the Laplace transform of the function h(t) shown in Fig. 6.6. h(t)

t 0

1 4−1−2−3−4 32

Fig. 6.6. Causal function h(t )

considered in Example 6.9.

Solution

In terms of the causal function x7(t), signal h(t) is expressed as

h(t) = 1.5x7 (

)

or h(t) = 1.5x7(0.5t).

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279 6 Laplace transform

In Example 6.4, the Laplace transform of x7(t) is given by

X7(s) =







3 for s = 0 2

s2 [1 − e−s − se−2s] for s �= 0,

with the entire s-plane as the ROC.

Using the time-scaling property the Laplace transform of h(t) is given by

h(t) = 1.5x7(0.5t) L←→

0.5 (1.5)X7

( s

0.5

)

= 3X7(2s) with ROC: 2R1,

which reduces to

H (s) =







9 for s = 0 1.5

s2 [1 − e−2s − 2se−4s] for s �= 0.

The ROC associated with H (s) is the entire s-plane.

6.4.3 Time shifting

If x(t) L←→ X (s) with ROC:R,then the Laplace transform of the time-shifted

signal is

x(t − t0) L←→ e−st0 X (s) with ROC: R (6.19)

for both unilateral and bilateral Laplace transforms. For the unilateral Laplace

transform, the value of t0 should be carefully selected such that the time-shifted

signal x(t – t0) remains causal. There is no such restriction for the bilateral

Laplace transform. Also, it may be noted that time shifting a signal does not

change the ROC of its Laplace transform.

Proof

By Eq. (6.9), the Laplace transform of the time-shifted signal x(t – t0) is given

L{x(t − t0)} = ∞∫

−∞

x(t − t0)e−st dt

= e−st0 ∞∫

−∞

x(τ )e−sτ dτ (by substituting τ = t − t0)

= e−st0 X (s),

which proves the time-shifting property, Eq. (6.19). The Laplace transform of

the time-shifted signal x(t – t0) is a product of two terms: exp(−st0) and X (s). For finite values of s and t0, the first term is always finite. Therefore, the ROC

of the Laplace transform of the time-shifted signal is the same as X (s).

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280 Part II Continuous-time signals and systems

Example 6.10

Calculate the Laplace transform of the causal function f (t) shown in Fig. 6.7.

f (t)

4 7210−1 65

Fig. 6.7. Waveform f(t ) used in

Example 6.10.

Solution

In terms of the waveform h(t) shown in Fig. 6.6, f (t) is expressed as follows:

f (t) = 2h(t − 3).

In Example 6.9, the Laplace transform of h(t) is given by

H (s) =







9 for s = 0 1.5

s2 [1 − e−2s − 2se−4s] for s �= 0,

with the entire s-plane as the ROC. Using the time-shifting property, the Laplace

transform of f (t) is

f (t) = 2h(t − 3) L←→ 2e−3s H (s) with ROC: R,

which results in the following Laplace transform for f (t):

H (s) =







[18e−3s]s=0 = 18 for s = 0 3

s2 [e−3s − e−5s − 2se−7s] for s �= 0,

with the entire s-plane as the ROC.

6.4.4 Shifting in the s-domain

If x(t) L←→ X (s) with ROC: R, then the Laplace transform of

es0t x(t) L←→ X (s − s0) with ROC: R + Re{s0} (6.20)

for both unilateral and bilateral Laplace transforms. Shifting a signal in the

complex s-domain by s0 causes the ROC to shift by Re{s0}. Although the

amount of shift s0 can be complex, the shift in the ROC is always a real number.

In other words, the ROC is always shifted along the horizontal axis, irrespective

of the value of the imaginary component in s0.

The shifting property can be proved directly from Eq. (6.9) by considering

the CTFT of the signal exp(s0t)x(t). The proof is left as an exercise for the

reader (see Problem 6.6).

Example 6.11

Using the Laplace transform pair

u(t) L←→

s with ROC: Re{s} > 0,

calculate the Laplace transform of (i) x1(t) = cos(ω0t)u(t) and (ii) x2(t) = sin(ω0t)u(t).

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281 6 Laplace transform

Solution

Using the above Laplace transform pair and s-shifting property, the Laplace

transforms of exp(jω0t)u(t) and exp(−jω0t) u(t) are given by

e jω0t u(t) L←→

(s − jω0) with ROC: Re{s} > 0

and

e−jω0t u(t) L←→

(s + jω0) with ROC: Re{s} > 0.

(i) To calculate the Laplace transform of x1(t) = cos(ω0t) u(t), we add the above transform pairs to obtain

e jω0t u(t) + e−jω0t u(t) L←→ 1

(s − jω0) +

(s + jω0) with ROC: Re{s} > 0,

which reduces to

cos(ω0t)u(t) L←→

s2 + ω20 with ROC: Re{s} > 0.

(ii) To evaluate the Laplace transform of x2(t) = sin(ω0t)u(t), we take the difference of the above transform pairs to obtain

e jω0t u(t) − e−jω0t u(t) L←→ 1

(s − jω0) −

(s + jω0) with ROC: Re{s} > 0 ,

which simplifies to

sin(ω0t)u(t) L←→

ω0

s2 + ω20 with ROC: Re{s} > 0.

6.4.5 Time differentiation

If x(t) L←→ X (s) with ROC:R, then the Laplace transform of

L←→ s X (s) − x(0−) with ROC: R. (6.21)

Note that if the function x(t) is causal, x(0−) = 0.

Proof

By Eq. (6.9), the Laplace transform of the derivative dx/dt is given by

{ dx

}

= ∞∫

0−

dt e−st dt .

Applying integration by parts on the right-hand side of the equation yields

{ dx

}

= x(t)e−st ︸︷︷︸

∣ ∣ ∣ ∣ ∣ ∣

∞

0−

− (−s) ∞∫

0−

x(t)e−st dt.

︸︷︷︸

X (s)

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Considering the first term, denoted by A, we note that for the upper limit,

t → ∞, the value of A is zero due to the decaying exponential term. For the

lower limit, t → 0−, A equals x(0−). The above equation therefore reduces to

{ dx

}

= x(0−) + s X (s).

Corollary 6.1 By repeatedly applying the differentiation property n times, it is

straightforward to prove that

dn x

dtn L←→ sn X (s) − sn−1x(0−) − · · · − sx (n−2)(0−)

− x (n−1)(0−) with ROC: R, (6.22)

where x (k) denotes the kth derivative of x(t), i.e. x (k) = dk x /dtk .

Example 6.12

Based on the Laplace transform pair

u(t) L←→

s with ROC: Re{s} > 0,

calculate the Laplace transform for the impulse function x(t) = δ(t).

Solution

Based on entry (2) of Table 6.1, we know that

u(t) L←→

s with ROC: Re{s} > 0.

Using the time-differentiation property, the Laplace transform of the first deriva-

tive of u(t) is given by

du(t)

L←→ s · 1

s + u(t)|t=0− with ROC: Re{s} > 0.

Knowing that du/dt = δ(t) and u(0−) = 0, we obtain

δ(t) L←→ 1 with ROC: Re{s} > 0.

6.4.6 Time integration

If x(t) L←→ X (s) with ROC:R, then

unilateral Laplace transform

t∫

0−

x(τ )dτ L←→

X (s)

with ROC: R ∩ Re{s}>0; (6.23)

bilateral Laplace transform

t∫

−∞

x(τ )dτ L←→

X (s)

s +

0−∫

−∞

x(τ )dτ

with ROC: R ∩ Re{s} > 0. (6.24)

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283 6 Laplace transform

Table 6.2. Properties of the Laplace transform

The corresponding properties of the CTFT are also listed in the table for comparison

CTFT

X ( jω) = ∞∫

−∞

x(t)e−jωt dt

Laplace transform

X (s) =

∞∫

−∞

x(t)e−st dt

Properties in the time domain

Linearity

a1x1(t) + a2x2(t)

a1 X1(ω) + a2 X2(ω) a1 X1(s) + a2 X2(s)

ROC: at least R1 ∩ R2

Time scaling

x(at)

|a| X

(ω

) 1

|a| X

( s

)

with ROC: a R

Time shifting

x(t − t0) e−jω0t X (ω) e−st0 X (s)

with ROC: R

Frequency/s-domain shifting

x(t)e jω0t or x(t)es0t X (ω − ω0) X (s − s0)

with ROC: R + Re{s0} Time differentiation

dx/dt

jωX (ω) s X (s) − x(0−) with ROC: R

Time integration t∫

−∞

x(τ )dτ

X (ω)

jω + π X (0)δ(ω)

X (s)

s with ROC: R ∩ Re{s} > 0

Frequency/s-domain

differentiation

(−t)x(t)

−jdX/dω dX/ds

Duality

X (t)

2πx(ω) not applicable

Time convolution

x1(t) ∗ x2(t) X1(ω)X2(ω) X1(s)X2(s)

ROC includes R1 ∩ R2

Frequency/s-domain convolution

x1(t)x2(t)

2π X1(ω) ∗ X2(ω)

2π X1(s) ∗ X2(s)

ROC includes

R1 ∩ R2

Parseval’s relationship

∞∫

−∞

|x(t)|2dt = 1

2π

∞∫

−∞

|X (ω)|2dω not applicable

Initial value

x(0+) if it exists

2π

∞∫

−∞

X (ω)dω lim s→∞

sX (s)

provided s = ∞ is included in the ROC of sX(s)

Final value

x(∞) if it exists not applicable lim

s→0 sX (s)

provided s = 0 is included in the ROC of sX(s)

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284 Part II Continuous-time signals and systems

The proof of the time-integration property is left as an exercise for the reader

(see Problem 6.7).

Example 6.13

Given the Laplace transform pair

cos(ω0t)u(t) L←→

(s2 + ω20) with ROC: Re{s} > 0,

derive the unilateral Laplace transform of sin(ω0t)u(t).

Solution

By applying the time-integration property to the aforementioned unilateral

Laplace transform pair yields t∫

0−

cos(ω0τ )u(τ )dτ L←→

s (

s2 + ω20 ) with ROC: Re{s} > 0,

where the left-hand side of the transform pair is given by t∫

0−

cos(ω0τ )u(τ )dτ = t∫

cos(ω0τ )dτ = sin(ω0τ )

ω0

∣ ∣ ∣ ∣

= 1

ω0 sin(ω0t).

Substituting the value of the integral in the transform pair, we obtain

sin(ω0t)u(t) L←→

ω0

(s2 + ω20) with ROC: Re{s} > 0,

6.4.7 Time and s-plane convolution

If x1(t) and x2(t) are two arbitrary functions with the following Laplace trans-

form pairs:

x1(t) L←→ X1(s) with ROC: R1 and x2(t)

L←→ X2(s) with ROC: R2, then the convolution property states that

time convolution x1(t) ∗ x2(t) L←→ X1(s)X2(s)

containing at least ROC: R1 ∩ R2; (6.25)

s-plane convolution x1(t)x2(t) L←→

2π j [X1(s) ∗ X2(s)]

containing at least ROC: R1 ∩ R2. (6.26)

The convolution property is valid for both unilateral (for causal signals) and

bilateral (for non-causal signals), Laplace transforms. The overall ROC of the

convolved signals may be larger than the intersection of regions R1 and R2 because of possible cancellation of poles in the products.

Proof

Consider the Laplace transform of x1(t) ∗ x2(t):

L{x1(t) ∗ x2(t)} = ∞∫

0−

[x1(t) ∗ x2(t)]e−st dt = ∞∫

0−





∞∫

−∞

x1(τ )x2(t − τ )dτ



 e−st dt .

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285 6 Laplace transform

Interchanging the order of integration, we get

L{x1(t) ∗ x2(t)} = ∞∫

−∞

x1(τ )





∞∫

0−

x2(t − τ )e −st dt



 dτ.

By noting that the inner integration ∫ x2(t − τ ) exp(−st)dt = X2(s) exp(−sτ ),

the above integral simplifies to

L{x1(t) ∗ x2(t)} = X2(s)

∞∫

−∞

x1(τ )e −sτ dτ = X2(s)X1(s),

which proves Eq. (6.25). The s-plane convolution property may be proved in a

similar fashion.

Like the CTFT convolution property discussed in Section 5.5.8, the Laplace

time-convolution property provides us with an alternative approach to cal-

culate the output y(t) when a CT signal x(t) is applied at the input of an

LTIC system with the impulse response h(t). In Chapter 3, we proved that

the zero-state output response y(t) is obtained by convolving the input signal

x(t) with the impulse response h(t), i.e. y(t) = h(t) ∗ x(t). Using the time-

convolution property, the Laplace transform Y (s) of the resulting output y(t) is

given by

y(t) = x(t) ∗ h(t) L←→ Y (s) = X (s)H (s),

where X (s) and H (s) are the Laplace transforms of the input signal x(t) and

the impulse response h(t) of the LTIC systems. In other words, the Laplace

transform of the output signal is obtained by multiplying the Laplace transforms

of the input signal and the impulse response. The procedure for calculating the

output y(t) of an LTI system in the complex s-domain, therefore, consists of

the following four steps.

(1) Calculate the Laplace transform X (s) of the input signal x(t). If the input

signal and the impulse response are both causal functions, then the unilateral

Laplace transform is used. If either of the two functions is non-causal, the

bilateral Laplace transform must be used.

(2) Calculate the Laplace transform H (s) of the impulse response h(t) of the

LTIC system. The Laplace transform H (s) is referred to as the Laplace

transfer function of the LTIC system and provides a meaningful insight

into the behavior of the system.

(3) Based on the convolution property, the Laplace transform Y (s) of the output

response y(t) is given by the product of the Laplace transforms of the input

signal and the impulse response of the LTIC systems. Mathematically, this

implies that Y (s) = X (s)H (s). (4) Calculate the output response y(t) in the time domain by taking the inverse

Laplace transform of Y (s) obtained in step (3).

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286 Part II Continuous-time signals and systems

Since the Laplace-transform-based approach for calculating the output response

of an LTIC system does not involve integration, it is preferred over the time-

domain approaches.

Example 6.14

In Example 3.6, we showed that in response to the input signal x(t) = e−t u(t), the LTIC system with the impulse response h(t) = e−2t u(t) produces the following output:

y(t) = (e−t − e−2t )u(t).

Example 5.21 derived the result using the CTFT. We now derive the result using

the Laplace transform.

Solution

Since the input signal and impulse response are both causal functions, we

take the unilateral Laplace transform of both signals. Based on Table 6.1, the

resulting transform pairs are given by

x(t) = e−t u(t) L←→ X (s) = 1

(s + 1) with ROC: Re{s} > −1

and

h(t) = e−2t u(t) L←→ X (s) = 1

(s + 2) with ROC: Re{s} > −2.

Based on the time-convolution property, the Laplace transform Y (s) of the

resulting output y(t) is given by

y(t) = h(t) ∗ x(t) L←→ Y (s) = 1

(s + 1)(s + 2) with ROC: Re{s} > −1,

where the ROC of the Laplace transform of the output is obtained by taking the

intersection of the regions Re{s} > −1 and Re{s} > −2, associated with the applied input and the impulse response. Using partial fraction expansion, Y (s)

may be expressed as follows:

Y (s) = 1

(s + 1) ︸︷︷︸

ROC : Re{s}>−1

− 1

(s + 2) ︸︷︷︸

ROC : Re{s}>−2

Taking the inverse Laplace transform of the individual terms on the right-hand

side of this equation yields

y(t) = (e−t − e−2t )u(t),

which is the same as the result produced by direct convolution and the approach

based on the CTFT time-convolution property.

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6.4.8 Initial- and final-value theorems

If x(t) L←→ X (s) with ROC:R, then

initial-value theorem x(0+) = lim t→0+

x(t) = lim s→∞

s X (s) provided x(0+) exists;

(6.27)

final-value theorem x(∞) = lim t→∞

x(t) = lim s→0

s X (s) provided x(∞) exists.

(6.28)

The initial-value theorem is valid only for the unilateral Laplace transform

as it requires the reference signal x(t) to be zero for t < 0. In addition, x(t)

should not contain an impulse function or any other higher-order discontinuities

at t = 0. The second constraint is required to ensure a unique value of x(t) at t = 0+. However, the final-value theorem may be used with either the unilateral or bilateral Laplace transform. The proof of these theorems is left as an exercise

for the reader (see Problems 6.8 and 6.9).

Example 6.15

Calculate the initial and final values of the functions x1(t), x2(t), and x3(t),

whose Laplace transforms are specified below:

(i) X1(s) = s + 3

s(s + 1)(s + 2) with ROC R1: Re{s} > 0;

(ii) X2(s) = s + 5

s3 + 5s2 + 17s + 13 with ROC R2: Re{s} > −1;

(iii) X3(s) = 5

s2 + 25 with ROC R3: Re{s} > 0.

Solution

(i) Applying the initial-value theorem, Eq. (6.27), to X1(s), we obtain

x1(0 +) = lim

t→0+ x1(t) = lim

s→∞ s X1(s) = lim

s→∞

s(s + 3) s(s + 1)(s + 2)

= lim s→∞

(s + 3) (s + 1)(s + 2)

= 0.

Applying the final-value theorem, Eq. (6.28), to X1(s) yields

x1(∞) = lim t→∞

x1(t) = lim s→0

s X1(s) = lim s→0

s(s + 3) s(s + 1)(s + 2)

= lim s→0

(s + 3) (s + 1)(s + 2)

= 1.5.

These initial and final values of x(t) can be verified from the following inverse

Laplace transform of X1(s) derived in Example 6.7(i):

x1(t) = (1.5 − 2e−t + 0.5e−2t )u(t).

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288 Part II Continuous-time signals and systems

(ii) Applying the initial-value theorem, Eq. (6.27), to X2(s), we obtain

x2(0 +) = lim

t→0+ x2(t) = lim

s→∞ s X2(s) = lim

s→∞

s(s + 5)

s3 + 5s2 + 17s + 13

= lim s→∞

6s = 0.

Applying the final-value theorem, Eq. (6.28), to X2(s) yields

x2(∞) = lim t→∞

x2(t) = lim s→0

s X2(s) = lim s→0

s(s + 5)

s3 + 5s2 + 17s + 13 = 0.

The initial and final values of x(t) can be verified from the following inverse

Laplace transform of X1(s) derived in Example 6.7(ii):

x1(t) = (0.4e −t − 0.4e−2t cos(3t) + 0.2e−2t sin(3t))u(t).

(iii) Applying the initial-value theorem, Eq. (6.27), to X3(s), we obtain

x3(0 +) = lim

t→0+ x3(t) = lim

s→∞ s X3(s) = lim

s→∞

s2 + 25 = lim

s→∞

2s = 0.

Applying the final-value theorem, Eq. (6.28), to X3(s) yields

x3(∞) = lim t→∞

x3(t) = lim s→0

s X3(s) = lim s→0

s2 + 25 = 0.

To confirm the initial and final values obtained in (iii), we determine these values

directly from the inverse transform of X3(s) = 5/(s 2 + 25). From Table 6.1, the

inverse Laplace transform of X3(s) is given by x3(t) = sin(5t)u(t). Substituting

t = 0+, the initial value x3(0 +) = 0, which verifies the value determined from

the initial-value theorem. Applying the limit t → ∞ to x3(t), the final value

of x3(t) cannot be determined due to the oscillatory behavior of the sinusoidal

wave. As a result, the final-value theorem provides an erroneous answer. The

discrepancy between the result obtained from the final-value theorem and the

actual value x3(∞) occurs because the point s = 0 is not included in the ROC

of sX3(s) R3: Re{s} > 0. As such, the expression for the Laplace transform

sX3(s) is not valid for s = 0. In such cases, the final-value theorem cannot be

used to determine the value of the function as t → ∞. Similarly, the point

s = ∞ must be present within the ROC of sX3(s) to apply the initial-value

theorem.

6.5 Solution of differential equations

An important application of the Laplace transform is to solve linear, constant-

coefficient differential equations. In Section 3.1, we used a time-domain

approach to obtain the zero-input, zero-state, and overall solution of differ-

ential equations. In this section, we discuss an alternative approach based on

the Laplace transform. We illustrate the steps involved in the Laplace-transform-

based approach through Examples 6.16 and 6.17.

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Example 6.16

In Example 3.2, we calculated the output voltage y(t) across resistor R = 5 � of an RC series circuit, which is modeled by the linear, constant-coefficient

differential equation

dt + 4y(t) =

dt (6.29)

for an initial condition y(0−) = 2 V and a sinusoidal voltage x(t) = sin(2t)u(t) applied at the input of the RC circuit. Repeat Example 3.2 using the Laplace-

transform-based approach.

Solution

Overall response To compute the overall response of the RC circuit, we take the Laplace transform of each term on both sides of Eq. (6.29). The Laplace

transform X (s) of the input signal x(t) is given by

X (s) = L{x(t)} = L{sin(2t)u(t)} = 2

s2 + 4 .

Using the time-differentiation property,

{ dx

}

= s X (s) − x(0−) = 2s

s2 + 4 .

Expressed in terms of the Laplace transform pair, y(t) L←→ Y (s), the transform

of the first derivative of y(t) is given by

{ dy

}

= sY (s) − y(0−) = sY (s) − 2.

Taking the Laplace transform of Eq. (6.29) and substituting the above values

yields

[sY (s) − 2] + 4Y (s) = 2s

s2 + 4 . (6.30)

Rearranging and collecting the terms corresponding to Y (s) on the left-hand

side of the equation results in the following:

[s + 4]Y (s) = 2 + 2s

s2 + 4 or

Y (s) = 2s2 + 2s + 8

(s + 4)(s2 + 4) ≡

(s + 4) +

Bs + C (s2 + 4)

, (6.31)

where Eq. (6.31) is obtained by the partial fraction expansion. The partial

fraction coefficient A is given by

A = [

(s + 4) 2s2 + 2s + 8

(s + 4)(s2 + 4)

]

s=−4 =

20 = 1.6.

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290 Part II Continuous-time signals and systems

To obtain the values of the partial fraction coefficients B and C , we multiply

both sides of Eq. (6.31) by (s + 4) (s2 + 4) and substitute A = 1.6. The resulting expression is as follows:

2s2 + 2s + 8 = A(s2 + 4) + (s + 4)(Bs + C) = (A + B)s2 + (4B + C)s + 4(A + C) = (1.6 + B)s2 + (4B + C)s + 4(1.6 + C).

Comparing the coefficients of s2, we obtain 1.6 + B = 2, or B = 0.4. Similarly, comparing the coefficients of s gives 4B + C = 2, or C = 0.4. The expression for Y (s) is, therefore, given by

Y (s) = 1.6

(s + 4) +

0.4s + 0.4 (s2 + 4)

= 1.6

(s + 4) + 0.4

(s2 + 4) + 0.2

(s2 + 4) ,

which has the following inverse Laplace transform:

y(t) = [1.6e−4t + 0.4 cos(2t) + 0.2 sin(2t)]u(t).

The aforementioned value of the overall output signal is same as the solution

derived in Eq. (3.10) using the time-domain approach. We now proceed with

the calculation of the zero-input response yzi(t) and zero-state response yzs(t).

Zero-input response To obtain the zero-input response yzi(t), we assume that the value of input x(t) = 0 in Eq. (6.29), i.e.

dyzi

dt + 4yzi(t) = 0.

Taking the Laplace transform of the above equation and substituting:

{ dyzi

}

= sYzi(s) − yzi(0−) = sYzi(s) − 2,

gives

[s + 4]Yzi(s) = 2,

which reduces to

Yzi(s) = 2

s + 4 .

Taking the inverse Laplace transform results in the following expression for the

zero-input response:

yzi(t) = 2e−4t u(t),

which is same as the result derived in Example 3.2.

Zero-state response To obtain the zero-state response, we assume that the initial condition yzs(0

−) = 0. This changes the value of the Laplace transform of the first derivative of y(t) as follows:

{ dyzs

}

= sYzs(s) − yzs(0−) = sYzs(s).

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291 6 Laplace transform

Taking the Laplace transform of Eq. (6.29) yields

(s + 4)Yzs(s) = 2s

s2 + 4 .

Using the partial fraction expansion, the above equation is expressed as follows:

Yzs(s) = 2s

(s + 4)(s2 + 4) ≡ −

0.4

(s + 4) +

0.4s + 0.4 (s2 + 4)

Taking the inverse Laplace transform, the zero-state response is given by

y(t) = [−0.4e−4t + 0.4 cos(2t) + 0.2 sin(2t)]u(t),

which is same as the result derived in Example 3.2.

We also know from Chapter 3 that the overall response y(t) is the sum of

the zero-input response yzi(t) and the zero-state response yzs(t). This is easily

verifiable for the above results.

Example 6.17

In Example 3.3, the following differential equation

d2w

dt2 + 7

dt + 12w(t) = 12x(t) (6.32)

was used to model the RLC series circuit shown in Fig. 3.1. Determine the

zero-input, zero-state, and overall response of the system produced by the input

x(t) = 2e−t u(t) given the initial conditions, w(0−) = 5 V and ẇ(0−) = 0.

Solution

Overall response The Laplace transforms of the individual terms in Eq. (6.32) are given by

X (s) = L{x(t)} = L{2e−t u(t)} = 2

s + 1 ,

W (s) = L{w(t)},

{ dw

}

= sW (s) − w(0−) = sW (s) − 5,

and

{ d2w

dt2

}

= s2W (s) − sw(0−) − ẇ(0−) = s2W (s) − 5s.

Taking the Laplace transform of both sides of Eq. (6.32) and substituting the

above values yields

[s2W (s) − 5s] + 7[sW (s) − 5] + 12W (s) = 24

s + 1 or

[s2 + 7s + 12]W (s) = 5s + 35 + 24

s + 1 =

5s2 + 40s + 59 s + 1

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292 Part II Continuous-time signals and systems

which reduces to

W (s) = 5s2 + 40s + 59

(s + 1) (s2 + 7s + 12) =

5s2 + 40s + 59 (s + 1)(s + 3)(s + 4)

Taking the partial fraction expansion, we obtain

5s2 + 40s + 59 (s + 1)(s + 3)(s + 4)

≡ k1

(s + 1) +

(s + 3) +

(s + 4) ,

where the partial fraction coefficients are given by

k1 = [

(s + 1) 5s2 + 40s + 59

(s + 1)(s + 3)(s + 4)

]

s=−1 =

5 − 40 + 59 (2)(3)

= 4,

k2 = [

(s + 3) 5s2 + 40s + 59

(s + 1)(s + 3)(s + 4)

]

s=−3 =

45 − 120 + 59 (−2)(1)

= 8,

and

k3 = [

(s + 4) 5s2 + 40s + 59

(s + 1)(s + 3)(s + 4)

]

s=−4 =

80 − 160 + 59 (−3)(−1)

= −7.

Substituting the values of the partial fraction coefficients k1, k2, and k3, we

obtain

W (s) ≡ 4

(s + 1) +

(s + 3) −

(s + 4) .

Calculating the inverse Laplace transform of both sides, we obtain the output

signal as follows:

w(t) ≡ [4e−t + 8e−3t − 7e−4t ]u(t).

Zero-input response To calculate the zero-input output, the input signal is assumed to be zero. Equation (6.32) reduces to

d2wzi

dt2 + 7

dwzi

dt + 12wzi(t) = 0, (6.33)

with initial conditions w(0−) = 5 and ẇ(0−) = 0. Calculating the Laplace transform of Eq. 6.33 yields

[s2Wzi(s) − 5s] + 7[sWzi(s) − 5] + 12Wzi(s) = 0

Wzi(s) = 5s + 35

s2 + 7s + 12 .

Using the partial fraction expansion, the above equation is expressed as follows:

Wzi(s) = 5s + 35

s2 + 7s + 12 ≡

s + 3 −

s + 4 .

Taking the inverse Laplace transform, the zero-input response is given by

wzi(t) ≡ [20e−3t − 15e−4t ]u(t).

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Zero-state response To calculate the zero-state output, the initial conditions are assumed to be zero, i.e. w(0−) = 0 and ẇ(0−) = 0. Taking the Laplace transform Eq. (6.31) and applying zero initial conditions yields

s2Wzs(s) + 7sWzs(s) + 12Wzs(s) = 24

s + 1 or

Wzs(s) = 24

(s + 1)(s2 + 7s + 12) .

Using the partial fraction expansion, we obtain

Wzs(s) = 24

(s + 1)(s + 3)(s + 4) ≡

(s + 1) −

(s + 3) +

(s + 4) .

Taking the inverse Laplace transform, the zero-state response of the system is

given by

wzs(t) ≡ [4e−t − 12e−3t + 8e−4t ]u(t).

The overall, zero-input, and zero-state responses calculated in the Laplace

domain are the same as the results computed in Example 3.3 using the time-

domain approach

A direct consequence of solving a linear, constant-coefficient differential equa-

tion is the evaluation of the Laplace transfer function H (s) for the LTIC system.

The Laplace transfer function is defined as the ratio of the Laplace transform

Y (s) of the output signal y(t) to the Laplace transform X (s) of the input signal

x(t). Mathematically,

H (s) = Y (s)

X (s) , (6.34)

which is obtained by taking the Laplace transform of the differential equation

and solving for H (s), as defined in Eq. (6.34). The above procedure provides

an algebraic expression for the Laplace transfer function. Its ROC is obtained

by observing whether the LTIC is causal or non-causal. Given the algebraic

expression and the ROC, the inverse Laplace transform of the Laplace transfer

function H (s) leads to the impulse response h(t) of the LTIC system. The

Laplace transfer function is also useful for analyzing the stability of the LTIC

systems, which is considered in Sections 6.6 and 6.7.

6.6 Characteristic equation, zeros, and poles

In this section, we will define the key concepts related to the stability of LTIC

systems. Although these concepts can be applied to general LTIC systems, we

will assume a system with a rational transfer function H (s) of the following

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294 Part II Continuous-time signals and systems

form:

X (s) = N (s)

D(s) =

bms m + bm−1sm−1 + bm−2sm−2 + · · · + b1s + b0

sn + an−1sn−1 + an−2sn−2 + · · · + a1s + a0 . (6.35)

Characteristic equation The characteristic equation for the transfer function in Eq. (6.35) is defined as follows:

D(s) = sn + an−1sn−1 + an−2sn−2 + · · · + a1s + a0 = 0. (6.36)

It will be shown later that the characteristic equation determines the behavior

of the system, including its stability and possible modes of the output response.

In other words, it characterizes the system very well.

Zeros The zeros of the transfer function H (s) of an LTIC system are the finite locations in the complex s-plane where |H (s)| = 0. For the transfer function in Eq. (6.35), the location of the zeros can be obtained by solving the following

equation:

N (s) = bmsm + bm−1sm−1 + bm−2sm−2 + · · · + b1s + b0 = 0. (6.37)

Since N (s) is an mth-order polynomial, it will have m roots leading to m zeros

for transfer function H (s).

Poles The poles of the transfer function H (s) of an LTIC system are the loca- tions in the complex s-plane where |H (s)| has an infinite value. At these loca- tions, the Laplace magnitude spectrum takes the form of poles (due to the infinite

value), and this is the reason the term “pole” is used to denote such locations.

The poles corresponding to the transfer function in Eq. (6.35) can be obtained

by solving the characteristic equation, Eq. (6.36).

Because D(s) is an nth-order polynomial, it will have n roots leading to n

poles. In order to calculate the zeros and poles, a transfer function is factorized

and typically represented as follows:

H (s) = N (s)

D(s) =

bm(s − z1)(s − z2) · · · (s − zm) (s − p1)(s − p2) · · · (s − pn)

. (6.38)

Note that a transfer function H (s) must be finite within its ROC. On the other

hand, the magnitude of the transfer function H (s) is infinite at the location of a

pole. Therefore, the ROC of a system must not include any pole. However, an

ROC may contain any number of zeros.

Example 6.18

Determine the poles and zeros of the following LTIC systems:

(i) H1(s) = (s + 4)(s + 5)

s2(s + 2)(s − 2) ;

(ii) H2(s) = (s + 4)

s3 + 5s2 + 17s + 13 ;

(iii) H3(s) = 1

es + 10 .

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295 6 Laplace transform

Re{s} 0

Im{s}

−2−4−6−8 2 4 6 8

double poles

at s = 0

Re{s} 0

Im{s}

−2−4−6−8 2 4 8

xx xx x

−j3 6

Re{s} 0

Im{s}

−2−4−6−8 2 4 8

jpx

−jpx

x j3p

x −j3p

(a)

(c)

(b)

Fig. 6.8. Locations of zeros and

poles of LTIC systems specified

in Example 6.18. The ROCs for

causal LTIC systems are

highlighted by the shaded

regions. Parts (a)–(c) correspond

to parts (i)–(iii) of Example 6.18.

Solution

(i) The zeros are the roots of the quadratic equation (s + 4)(s + 5) = 0, which are given by s = −4, −5. The poles are the roots of the fourth-order equation s2(s + 2)(s − 2) = 0, and are given by s = 0, 0, −2, 2. Figure 6.8(a) plots the location of poles and zeros in the complex s-plane. The poles are denoted by

the “×” symbols, while the zeros are denoted by the “◦” symbols. (ii) The zeros are the roots of the equation s + 4 = 0, which are given by

s = −4. The poles are the roots of the third-order equation s3 + 5s2 + 17s +

13 = 0, and are given by s = −1, −2 ± j3. Figure 6.8(b) plots the location of

poles and zeros in the complex s-plane.

(iii) Since the numerator is a constant, there is no zero for the LTIC system.

The poles are the roots of the characteristic equation es + 0.1 = 0. Following

the procedure shown in Appendix B, it can be shown that there are an infinite

number of roots for the equation es + 0.1 = 0. The locations of the poles are

given by

s = ln 0.1 + j(2m + 1)π ≈ −2.3 + j(2m + 1)π.

The poles are plotted in Fig. 6.8(c).

6.7 Properties of the ROC

In Section 3.7.2, we showed that the impulse response h(t) of a causal LTIC

system satisfies the following condition:

h(t) = 0 for t < 0.

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296 Part II Continuous-time signals and systems

For such right-sided functions, it is straightforward to show that the ROC R of

its transfer function H (s) will be of the form Re{s} > σ0, containing the right side of the s-plane. Consider, for example, the bilateral Laplace transform pairs

e−at u(t) L←→

s + a with ROC: Re{s} > −a

and

−e−at u(−t) L←→ 1

s + a with ROC: Re{s} < −a.

The first function e−at u(t) is right sided and its ROC: Re{s} > −a occupies the right side of the s-plane. On the other hand, the second function −e−at u(−t) is left sided, and its ROC: Re{s} < −a occupies the left side of the s-plane. Based on the above observations and the Laplace transform pairs listed in Table 6.1,

we state the following properties for the ROC.

Property 1 The ROC consists of 2D strips that are parallel to the imaginary jω-axis.

Property 2 For a right-sided function, the ROC takes the form Re{s} > σ0 and consists of the right side of the complex s-plane.

Property 3 For a left-sided function, the ROC takes the form Re{s} < σ0 and consists of the most of the left side of the complex s-plane.

Property 4 For a finite duration function, the ROC consists of the entire s-plane except for the possible deletion of the point s = 0.

Property 5 For a double-sided function, the ROC takes the form σ1 < Re{s} < σ2 and is a confined strip within the complex s-plane.

Property 6 The ROC of a rational transfer function does not contain any pole.

Combining Property 6 with the causality constraint (Re{s} > σ0) discussed earlier in the section, we obtain the following condition for a causal LTIC

system.

Property 7 The ROC R for a right-sided LTIC system with the rational transfer function H (s) is given by R: Re{s} > Re{pr}, where pr is the location of the rightmost pole among the n poles determined using

Eq. (6.36).

Since the impulse response of a causal system is a right-sided function, the

ROC of a causal system satisfies Property 7. The converse of Property 7 leads

to Property 8 for a left-sided sequence.

Property 8 The ROC R for a left-sided function with the rational transfer func- tion H (s) is given by R: Re{s} < Re{pl} where pl is the leftmost pole among the n poles determined using Eq. (6.36).

To illustrate the application of the properties of the ROC, we consider the

following example.

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297 6 Laplace transform

Example 6.19

Consider the LTIC systems in Example 6.18(i) and (ii). Calculate the impulse

response if the specified LTIC systems are causal. Repeat for non-causal

systems.

Solution

(i) Using the partial fraction expansion, H1(s) can be expressed as follows:

H1(s) = (s + 4)(s + 5)

s2(s + 2)(s − 2) ≡

s +

s2 +

(s + 2) +

(s − 2) ,

where

k1 = [

( (s + 4)(s + 5) (s + 2)(s − 2)

)]

s=0 ≡

[ 2s + 9 s2 − 4

− (s + 4)(s + 5)

(s2 − 4)2 2s

]

s=0 = −

4 ,

k2 = [

s2 (s + 4)(s + 5)

s2(s + 2)(s − 2)

]

s=0 ≡

(4)(5)

2(−2) = −5,

k3 = [

(s + 2) (s + 4)(s + 5)

s2(s + 2)(s − 2)

]

s=−2 ≡

(2)(3)

4(−4) = −

8 ,

and

k4 = [

(s − 2) (s + 4)(s + 5)

s2(s + 2)(s − 2)

]

s=2 ≡

(6)(7)

4(4) =

8 .

Therefore,

H1(s) ≡ − 9

4s −

s2 −

8(s + 2) +

8(s − 2) .

If H1(s) represents a causal LTIC system, then its ROC, based on Property 7,

is given by Rc: Re{s} > 2. Based on the linearity property, the overall ROC Rc is only possible if the ROCs for the individual terms in H1(s) are given by

H1(s) = − 9

4s ︸︷︷︸

ROC:Re{s}>0

− 5

s2 ︸︷︷︸

ROC:Re{s}>0

− 3

8(s + 2) ︸︷︷︸

ROC:Re{s}>−2

+ 21

8(s − 2) .

︸︷︷︸

ROC:Re{s}>2

By calculating the inverse Laplace transform, the impulse response for a causal

LTIC system is obtained as follows:

h1(t) = [

− 9

4 − 5t −

8 e−2t +

8 e2t

]

u(t).

If H1(s) represents a non-causal system, then its ROC can have three different

values: Re{s} < −2; −2 < Re{s} < 0; or 0 < Re{s} < 2 in the s-plane. Select- ing Re{s} < −2 as the ROC will lead to a left-sided signal. The remaining two choices will lead to a double-sided signal. Assuming that we select the overall

ROC to be Rnc: Re{s} < −2 , the ROCs for the individual terms in H1(s) are

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298 Part II Continuous-time signals and systems

given by

H1(s) = − 9

4s ︸︷︷︸

ROC:Re{s}<0

− 5

s2 ︸︷︷︸

ROC:Re{s}<0

− 3

8(s + 2) ︸︷︷︸

ROC:Re{s}<−2

+ 21

8(s − 2) .

︸︷︷︸

ROC:Re{s}<2

Taking the inverse Laplace transform, the impulse response for a non-causal

LTIC system is given by

h1(t) = [

4 + 5t +

8 e−2t −

8 e2t

]

u(−t).

(ii) Using the partial fraction expansion, H2(s) may be expressed as follows:

H2(s) = 3

10(s + 1) −

3s − 1 10(s2 + 4s + 13)

If H2(s) represents a causal system, then its ROC is given by Rc: Re{s} > −1. The ROCs associated with the individual terms in H2(s) are given by

H2(s) = 3

10(s + 1) ︸︷︷︸

ROC:Re{s}>−1

− 3s − 1

10(s2 + 4s + 13) ︸︷︷︸

ROC:Re{s}>−2

Taking the inverse Laplace transform, the impulse response for a causal LTIC

system is given by

h2(t) = [

10 e−t −

10 e−2t cos(3t) +

30 e−2t sin(3t)

]

u(t).

If H2(s) represents a non-causal system, then several different choices of ROC

are possible. One possible choice is given by Rnc: Re{s} < −2. The ROCs associated with the individual terms in H2(s) are given by

H2(s) = 3

10(s + 1) ︸︷︷︸

ROC:Re{s}<−1

− 3s − 1

10(s2 + 4s + 13) ︸︷︷︸

ROC:Re{s}<−2

Taking the inverse Laplace transform, the impulse response for a causal LTIC

system is given by

h2(t) = [

− 3

10 e−t +

10 e−2t cos(3t) −

30 e−2t sin(3t)

]

u(−t).

6.8 Stable and causal LTIC systems

In Section 3.7.3, we showed that the impulse response h(t) of a BIBO stable

system satisfies the condition

∞∫

−∞

|h(t)|dt < ∞. (6.39)

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In this section, we derive an equivalent condition to determine the stability of

an LTIC system modeled with a rational Laplace transfer function H (s) given

in Eq. (6.35). Since we are mostly interested in causal systems, we assume

that the Laplace transfer function H (s) corresponds to a right-sided system.

The poles of a system with a transfer function as given in Eq. (6.35) can be

calculated by solving the characteristic equation, Eq. (6.36). Three types of

poles are possible. Out of the n possible poles, assume that there are L poles at

s = 0, K real poles at s = −σk , 1 ≤ k ≤ K , and M pairs of complex-conjugate poles at s = −αm ± jωm , 1 ≤ m ≤ M , such that L + K + 2M = n. In terms of its poles, the transfer function, Eq. (6.35), is given by

H (s) = N (s)

D(s) =

N (s)

sL K∏

k=1 (s + σk)

M∏

m=1

(

s2 + 2αms + (

α2m + ω 2 m

))

. (6.40)

From Table 6.1, the repeated roots at s = 0 correspond to the following term in the time domain:

n! tnu(t)

L←→ 1

sn . (6.41)

Since term tnu(t) is unbounded as t → ∞, a stable LTIC system will not contain such unstable terms. Therefore, we assume that L = 0. The partial fraction expansion of Eq. (6.40) with L = 0 results in the following expression:

H (s) = A1

(s + σ1) + · · · +

(s + σK ) +

B1s + C1 (

s2 + 2α1s + (

α21 + ω21 )) + · · ·

+ BM s + CM

(

s2 + 2αM s + (

α2M + ω2M )) , (6.42)

where {Ak , Bm , Cm} are the partial fraction coefficients. Calculating the inverse

Laplace transform of Eq. (6.42) and assuming a causal system, we obtain the

following expression for the impulse response h(t) of the LTIC system:

h(t) = K∑

k=1 Ake

−σk t u(t) ︸︷︷︸

hk (t)

+ M∑

m=1 rme

−αm t cos(ωm t + θm) u(t) ︸︷︷︸

hm (t)

, (6.43)

where we have expressed the terms with conjugate poles in the polar format.

Constants {rm , θm} are determined from the values of the partial fraction coef-

ficients {Bm , Cm} and αm .

In Eq. (6.43), we have two types of terms on the right-hand side of the

equation. Summation I consists of K real exponential functions of the type

hk(t) = Ak exp(−σk t)u(t). Depending upon the value of σk , each of these func- tions hk(t) may have a constant, decaying exponential or a rising exponential

waveform.

Summation II consists of exponentially modulated sinusoidal functions of

the type hm(t) = rm exp(−αm t) cos(ωm t + θm)u(t). The stability characteristic

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300 Part II Continuous-time signals and systems

(−s1, 2w1) (s1, 2w1)

(s1, w1)

(s1, 0)

(−s1, w1)

(−s1, 0) Re{s}

Im{s}

(0, 2w1)

(0, w1)

(0, 0)

Fig. 6.9. Nature of the shape of

the terms hk (t ) and hm (t ) for

different sets of values for

σk and αm . For real-valued

coefficients bn in D(s ), the

complex poles occur as

complex-conjugate pairs.

of the functions hm(t) included in the second summation depends upon the

value of αm . To illustrate the effect of the values of σk and αm on the stability of

the LTIC system, Fig. 6.9 plots the shape of the waveforms in the time domain

corresponding to terms hk(t) and hm(t) for different sets of values for σk and

αm . The three plots along the real axis, Re{s}, at coordinates (−σ1, 0), (0, 0), and (σ1, 0) represent the terms hk(t) in summation I. For H (s) to correspond to

a stable LTIC system, each of the terms in Eq. (6.43) should satisfy the stability

condition, Eq. (6.39). Clearly, terms hk(t) = Ak exp(−σk t)u(t) are stable if σk > 0, where terms hk(t) would correspond to decaying exponential functions.

In the three cases plotted along the real axis in Fig. 6.9, this is observed by the

impulse response hk(t) at coordinate (−σ1, 0). In other words, summation I will be stable if the value of σk in term hk(t) = Ak exp(−σk t)u(t) is positive. The real roots s = −σk , for 1 ≤ k ≤ K , must therefore lie in the left-half s-plane for summation I to be stable.

Similarly, term hm(t) = rm exp(−αm t) cos(ωm t + θm)u(t) in summation II is stable if αm > 0, where hm(t) would correspond to a decaying sinusoidal

waveform. This is evident from the remaining six coordinates selected in

Fig. 6.9. If the value of αm in term hm(t) = rm exp(−αm t) cos(ωm t + θm)u(t) is set to a negative value, corresponding to the two impulse responses hm(t)

at coordinates (α1, ω1) and (α1, 2ω1), term hm(t) corresponds to an unstable

waveform. Only when the value of αm is set to be positive, corresponding

to the waveforms at coordinates (−α1, ω1) and (−α1, 2ω1), is term hm(t) stable. This implies that the location of the complex poles s = −αm ± jωm , 1 ≤ m ≤ M , should also lie in the left-half s-plane for the LTIC system to be stable. Based on the above discussion, we state the following conditions for

the stability of the LTIC systems with causal implementation for the impulse

responses.

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301 6 Laplace transform

Property 9 A causal LTIC system with n poles {pl}, 1 ≤ l ≤ n, will be abso- lutely BIBO stable if and only if the real part of all poles are

non-zero negative numbers, i.e. if

Re{pl} < 0 for all l. (6.44)

Equation (6.44) states that a causal LTIC system will be absolutely BIBO stable

if and only if all of its poles lie in the left half of the s-plane, (i.e. to the left of the

jω-axis). In other words, a causal LTIC system will be absolutely BIBO stable

and causal if the ROC occupies the entire right half of the s-plane including the

jω-axis.

We illustrate the application of the stability condition in Eq. (6.44) in Example

6.20.

Example 6.20

In Example 6.18, we considered the following LTIC systems:

(i) H1(s) = (s + 4)(s + 5)

s2(s + 2)(s − 2) ;

(ii) H2(s) = (s + 4)

s3 + 5s2 + 17s + 13 ;

(iii) H3(s) = 1

es + 10 .

Assuming that the systems are causal, determine if the systems are BIBO stable.

Solution

(i) The LTIC system with transfer function H1(s) has four poles located at

s = −2, 0, 0, 2. Since all the poles do not lie in the left half of the s-plane, the transfer function does not represent an absolutely BIBO stable and causal

system. The impulse response of the causal implementation of the LTIC system

was calculated in Example 6.19. It can be easily verified that the time-domain

stability condition, Eq. (6.39), is not satisfied because of the rising exponential

function 21/8 exp(2t)u(t) and the ramp function 5t , which have infinite areas.

(ii) The LTIC system with transfer function H2(s) has three poles located

at s = −1, −2 ± j3. Since all the poles lie in the left-half s-plane, the transfer function represents an absolutely BIBO stable and causal system. The impulse

response of the causal implementation of the LTIC system was calculated in

Example 6.19. It can be easily verified that the time-domain stability condition,

Eq. (6.39), is satisfied as all terms are decaying exponential functions with finite

areas.

(iii) The LTIC system with transfer function H3(s) has multiple poles located

at s = −2.3 + j(2m + 1)π , for m = 0, ±1, ±2, . . . Since all the poles lie in the left-half s-plane, the transfer function represents an absolutely BIBO stable and

causal system.

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302 Part II Continuous-time signals and systems

6.8.1 Marginal stability

In our previous discussion, we considered absolutely stable and unstable sys-

tems. An absolutely stable system has all the poles in the left half of the complex

s-plane. A causal implementation of such a system is stable in the sense that

as long as the input is bounded, the system produces a bounded output. On

the contrary, an absolutely unstable system has one or more poles in the right

half of the complex s-plane. The impulse response of a causal implementation

of such a system includes a growing exponential function, making the system

unstable. An intermediate case arises when a system has unrepeated poles on

the imaginary jω-axis. The remaining poles are in the left half of the complex s-

plane. Such a system is referred to as a marginally stable system. The condition

for marginally stable system is stated below.

Property 10 An LTIC system, with K unrepeated poles sk = jωk, 1 ≤ k ≤ K , on the imaginary jω-axis and all remaining poles in the left-half s-plane, is

stable for all bounded input signals that do not include complex exponential

terms of the form exp(−jωk t), for 1 ≤ k ≤ K . If the poles on the imaginary jω-axis are repeated, then the LTIC system is unstable.

The following example demonstrates that a marginally stable system becomes

unstable if the input signal includes a complex exponential exp(−jω0t) with frequency ω0 corresponding to coordinate s = jω0 of the location of the pole of the system on the imaginary jω-axis in the complex s-plane.

Example 6.21

Consider an LTIC system with transfer function

H (s) = 25

s2 + 25 representing a marginally stable system. Determine the output of the LTIC

system for the following inputs:

(i) x1(t) = u(t); (ii) x2(t) = sin(5t)u(t).

Solution

(i) Taking the Laplace transform of the input gives X1(s) = 1/s. The Laplace transform of the output is given by

Y1(s) = H (s)X1(s) = 25

s(s2 + 25) ≡

s −

(s2 + 25) .

Taking the inverse Laplace transform gives the following value of the output in

the time domain:

y1(t) = (1 − cos(5t))u(t).

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303 6 Laplace transform

y1(t)

0.4p 0.8p 1.2p 1.6p 2.0p 2.4p 2.8p

y2(t)

12.5

−12.5

−25

0.4p 0.8p 1.2p 1.6p 2.0p 2.4p 2.8p

(a) (b)

Fig. 6.10. Waveforms of the

output signals produced by a

marginally stable system

resulting from

(a) x1(t ) = u(t ) and (b) x2(t ) = sin(5t )u(t ), as considered in Example 6.21.

As expected for a marginally stable system, the output y1(t) produced by a

bounded input x1(t) = u(t) in the above expression is bounded for all time t . Figure 6.10(a) plots the bounded output y1(t) as a function of time t .

(ii) Taking the Laplace transform of the input gives X2(s) = 5/s2 + 25. The Laplace transform of the output is given by

Y2(s) = H (s)X2(s) = 125

(s2 + 25)2 .3

Using the transform pair

2a3 (sin(at) − at cos(at))u(t) L←→

(s2 + a2)2 ,

the output y2(t) in the time domain is given by

y2(t) = 0.5(sin(5t) − 5t cos(5t))u(t).

1In part (ii), a sinusoidal signal sin(5t) = (exp(j5t) − exp(−j5t))/2j is applied at the input of a marginally stable system with poles located at s = ±j5 on the imaginary jω-axis. Note that the fundamental frequency (ω0 = 5) of the sinusoidal input is the same as the location (s = ±j5) of the poles in the com- plex s-plane. In such cases, Property 6.10 states that the resulting output y2(t)

will be unbounded. The second term −5t cos(5t)u(t) indeed makes the output unbounded. This is illustrated in Fig. 6.10(b), where y2(t) is plotted as a function

of time t .

6.8.2 Improving stability using zeros

To conclude our discussion on stability, let us consider an LTIC system with

transfer function given by

Hap(s) = (s − a − jb) (s + a − jb)

, (6.45)

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304 Part II Continuous-time signals and systems

Re{s} 0

Im{s}

−a

x jb

Fig. 6.11. Locations of poles

“×” and zeros “o” of an allpass system in the complex s-plane.

The ROC of the causal

implementation of the allpass

system is highlighted by the

shaded region.

having a pole at s = (−a + jb) and a zero at s = (a + jb). As shown in Fig. 6.11, the locations of the pole and zero are symmetric about the

imaginary jω-axis in the complex s-plane. Clearly, a causal implementa-

tion of the transfer function H (s) of the system will be stable as its ROC:

Re{s} > −a includes the imaginary jω-axis. The CTFT of the LTIC system is evaluated as

Hap( jω) = Hap(s)|s=jω = ( jω − a − jb) ( jω + a − jb)

, (6.46)

with the CTFT spectra as follows:

magnitude spectrum |Hap( jω)| = √

(−a)2 + (ω − b)2 √

(a)2 + (ω − b)2 = 1; (6.47)

phase spectrum <Hap( jω) = tan−1 (

ω − b −a

)

− tan−1 (

ω − b a

)

. (6.48)

Such a system is referred to as an allpass system, since it allows all frequencies

present in the input signal to pass through the system without any attenuation.

Of course, the phase of the input signal is affected, but in most applications we

are more concerned about the magnitude of the signal.

An allpass system specified in Eq. (6.45) is frequently used to stabilize an

unstable system. Consider an LTIC system with the transfer function

H (s) = H1(s)

(s − a − jb) , (6.49)

where the component H1(s) is assumed to have all poles in the left half of

the s-plane and is, therefore, stable. A causal implementation of the transfer

function H (s) is unstable because of the existence of the term (s − a − jb) into the denominator. This term results in a pole at s = (a + jb) and introduces instability into the system. Such a system can be made stable by cascading it

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305 6 Laplace transform

with an allpass system that has a zero at the location of the unstable pole. The

transfer function of the overall cascaded system is given by

Hoverall(s) = H (s)Hap(s) = H1(s)

(s + a − jb) , (6.50)

which is stable because the unstable pole at s = (a + jb) is canceled by the zero of the allpass system. The new pole at s = (−a + jb) lies in the left-half s-plane and satisfies the stability requirements. Note that the magnitude response

of the overall cascaded system is the product of the magnitude responses of the

unstable and allpass systems, and is given by

|Hoverall( jω)| = |H ( jω)||Hap( jω)| = |H ( jω)|, (6.51)

since |Hap( jω)| = 1. Hence, by cascading an unstable system with an allpass system, which has a zero at the location of the unstable pole, we have stabilized

the system without affecting its magnitude response. The only change in the

system is in its phase. Such a pole–zero cancelation approach is frequently

used in applications where information is contained in the magnitude of the

signal and the phase is relatively unimportant. One such application is the

amplitude modulation system described in Section 2.1.3, which is used for

radio communications.

6.9 LTIC systems analysis using Laplace transform

In Section 6.4.7, we showed that the output response of an LTIC system could

be computed using the convolution property in the complex s-plane. This elim-

inates the need to compute the computationally intense convolution integral in

the time domain. Below, we provide another example for calculating the output

using the Laplace transform. Our motivation in reintroducing this topic is to

compare the Laplace-transform-based analysis technique with the CTFT-based

approach.

Example 6.22

In Example 5.26, we determined the overall and steady state values of the output

of the RC series circuit with the CTFT transfer function

H (ω) = 1/jωC

R + 1/jωC =

1 + jωCR

and constant CR = 0.5 for the input signal x(t) = sin(3t)u(t). For simplicity, we assumed that the capacitor is uncharged at t = 0. Here we solve the problem in Example 5.26 using the Laplace transform.

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306 Part II Continuous-time signals and systems

Solution

The Laplace transform of the input signal x(t) is given by

X (s) = L{sin(3t)u(t)} = 3

s2 + 9 .

The Laplace transfer function of the RC series circuit is given by

H (s) = H (ω)| jω=s = 1

1 + sCR .

Substituting the value of the product CR = 0.5 yields

H (s) = 1

1 + 0.5 s =

s + 2 .

The Laplace transform Y (s) of the output signal is given by

Y (s) = H (s)X (s) = 6

(s + 2) (s2 + 9) ≡

13(s + 2) −

6s − 12 13(s2 + 9)

Y (s) = 6

13(s + 2) −

(s2 + 9) +

(s2 + 9) .

Taking the inverse transform leads to the following expression for the overall

output in the time domain:

y(t) = [

13 e−2t −

13 cos(3t) +

13 sin(3t)

]

u(t) = [

13 e−2t +

2 √

13 sin(3t − 56◦)

]

u(t).

The steady state value of the output is computed by applying the limit t → ∞ to the overall output:

yss(t) = lim t←∞

y(t) = 2

√ 13

sin(3t − 56◦)u(t).

In Chapters 5 and 6, we presented two frequency-domain approaches to analyze

CT signals and systems. The CTFT-based approach introduced in Chapter 5 uses

the real frequency ω, whereas the Laplace-transform-based approach uses the

complex frequency σ . Both approaches have advantages. Depending upon the

application under consideration, the appropriate transform is selected.

Comparing Example 6.22 with Example 5.26, the Laplace transform appears

to be a more convenient tool for the transient analysis. For the steady state

analysis, the Laplace transform does not seem to offer any advantage over

the CTFT. The transient analysis is very important for applications in control

systems, including process control and guided missiles. In signal processing

applications, such as audio, image, and video processing, the transients are

generally ignored. In such applications, the CTFT is sufficient to analyze the

steady state response. This is precisely why most signal processing literature

uses the CTFT, while the control systems literature uses the Laplace transform.

Important applications of the Laplace transforms such as analysis of the spring

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307 6 Laplace transform

damper system and the modeling of the human immune system are presented

in Chapter 8.

6.10 Block diagram representations

In the preceding discussion, we considered relatively elementary LTIC systems

described by linear, constant-coefficient differential equations. Most practi-

cal structures are more complex, consisting of a combination of several LTIC

systems. In this section, we analyze the cascaded, parallel, and feedback con-

figurations used to synthesize larger systems.

6.10.1 Cascaded configuration

A series or cascaded configuration between two systems is illustrated in

Fig. 6.12(a). The output of the first system H1(s) is applied as input to the

second system H2(s). Assuming that the Laplace transform of the input x(t),

applied to the first system, is given by X (s), the Laplace transform W (s) of the

output w(t) of the first system is given by

w(t) = x(t) ∗ h1(t) L←→ W (s) = X (s)H1(s). (6.52)

The resulting signal w(t) is applied as input to the second system H2(s), which

leads to the following overall output:

y(t) = w(t) ∗ h2(t) L←→ Y (s) = W (s)H2(s). (6.53)

Substituting the value of w(t) from Eq. (6.52), Eq. (6.53) reduces to

y(t) = x(t) ∗ h1(t) ∗ h2(t) L←→ Y (s) = W (s)H1(s)H2(s). (6.54)

In other words, the cascaded configuration is equivalent to a single LTIC system

with transfer function

h(t) = h1(t) ∗ h2(t) L←→ H (s) = H1(s)H2(s). (6.55)

The system H (s) equivalent to the cascaded configuration is shown in

Fig. 6.12(b).

H2(s) Y(s) W(s)

H1(s)X(s) Y(s)X(s) H(s) = H1(s)H2(s)

(a) (b)

Fig. 6.12. Cascaded

configuration for connecting

LTIC systems: (a) cascaded

connection; (b) its equivalent

single system.

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308 Part II Continuous-time signals and systems

Y(s)X(s)

H2(s)

H1(s) Y1(s)

Y2(s)

∑

Y(s)X(s) H(s) = H1(s) + H2(s)

(b)(a)

Fig. 6.13. Parallel configuration for connecting LTIC systems: (a) parallel connection; (b) its equivalent

single system.

6.10.2 Parallel configuration

The parallel configuration between two systems is illustrated in Fig. 6.13(a).

A single input x(t) is applied simultaneously to the two systems. The overall

output y(t) is obtained by adding the individual outputs y1(t) and y2(t) of the

two systems. The individual outputs of the two systems are given by

system (1) y1(t) = x(t) ∗ h1(t) L←→ Y1(s) = X (s)H1(s); (6.56)

system (2) y2(t) = x(t) ∗ h2(t) L←→ Y2(s) = X (s)H2(s). (6.57)

Combining the two outputs, the overall output y(t) is given by

y(t) = y1(t) + y2(t) L←→ Y (s) = Y1(s) + Y2(s). (6.58)

Substituting Eqs. (6.56) and (6.57) into the above equation yields

y(t) = x(t) ∗ [h1(t) + h2(t)] L←→ Y (s) = X (s)[H1(s) + H2(s)]. (6.59)

In other words, the parallel configuration is equivalent to a single LTIC system

with transfer function

h(t) = h1(t) + h2(t) L←→ H (s) = H1(s) + H2(s). (6.60)

The parallel configuration and its equivalent single-stage system are shown in

Fig. 6.13.

6.10.3 Feedback configuration

The feedback connection between two systems is shown in Fig. 6.14(a). In a

feedback system, the overall output y(t) is applied at the input of the second

system H2(s). The output w(t) of the second system is fed back into the input

of the overall system through an adder. In terms of the applied input x(t) and

w(t), the output of the adder is given by

E(s) = X (s) − W (s). (6.61)

The outputs of the two LTIC systems are given by

system (1) Y (s) = E(s)H1(s); (6.62) system (2) W (s) = Y (s)H2(s). (6.63)

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309 6 Laplace transform

Y(s)X(s)

H2(s)

H1(s) E(s)

W(s)

∑ +

−

(b)(a)

Y(s)X(s) H(s) = H1(s)

1+H1(s)H2(s)

Fig. 6.14. Feedback

configuration for connecting

LTIC systems: (a) feedback

connection; (b) its equivalent

single system.

Substituting the value of E(s) = Y (s)/H1(s) from Eq. (6.62) and W (s) from Eq. (6.63) into Eq. (6.61) yields

Y (s) = H1(s)[X (s) − H2(s)Y (s)]. (6.64)

Rearranging terms containing Y (s), we obtain

[1 + H1(s)H2(s)]Y (s) = H1(s)X (s),

which leads to the following transfer function for the feedback system:

H (s) = Y (s)

X (s) =

H1(s)

1 + H1(s)H2(s) . (6.65)

The feedback configuration and its equivalent single system are shown in

Fig. 6.14.

Example 6.23

Determine (i) the impulse response and (ii) the transfer function of the inter-

connected systems shown in Figs. 6.15(a)–(c).

Solution

(a) To calculate the overall impulse response, we proceed in the Laplace

domain. The transfer function H1(s) of the cascaded systems shown in the

lower branch of the system in Fig.6.15(a) is given by

H1(s) = L{δ(t − 1)}H (s) = e−s H (s).

The overall transfer function Ha(s) is therefore given by

Ha(s) = H (s) + H1(s) = (1 + e−s)H (s).

Taking the inverse of the above transfer function gives the impulse response:

ha(t) = h(t) + h(t − 1).

(b) The system in Fig. 6.15(b) is the feedback configuration with transfer

functions H1(s) = 1 and H2(s) = L{αδ(t – T )} = αe−T s . Substituting the val- ues of H1(s) and H2(s) into Eq. (6.65) yields

Hb(s) = 1

1 + αe−Ts .

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310 Part II Continuous-time signals and systems

∑

−

d(t−1) h(t)

h(t)

x(t) y (t)

h1(t) = d(t−1)∗h(t)

−

∑ y (t)x(t)

ad(t−T )

(a) (b)

(c)

h4(t)

h1(t) ∑

−

∑

y (t)

h2(t)

h23(t) = h2(t)−h3(t)

h3(t)

x(t) +

Fig. 6.15. Interconnections

between LTIC systems. Parts

(a)–(c) correspond to parts

(a)–(c) of Example 6.23.

Since Hb(s) is not a rational function of s, the inverse Laplace transform is

evaluated from the definition in Eq. (6.7), which involves contour integration.

box is given by

H23(s) = H2(s) − H3(s).

In terms of H23(s), the transfer function H123(s) of the top path is given by

H123(s) = H1(s)H23(s).

Substituting the value of H23(s), the above expression reduces to

H123(s) = H1(s)[H2(s) − H3(s)].

The overall transfer function of the system in Fig. 6.15(c) is given by

Hc(s) = H123(s) + H4(s)

Hc(s) = H1(s)[H2(s) − H3(s)] + H4(s).

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311 6 Laplace transform

Taking the inverse Laplace transform of the above equation leads to the follow-

ing expression for the overall impulse response:

hc(t) = h1(t) ∗ h2(t) − h1(t) ∗ h3(t) + h4(t).

6.11 Summary

In this chapter, we introduced the bilateral and unilateral Laplace transforms

used for the analysis of LTIC signals and systems. The Laplace transforms

are a generalization of the CTFT, where the independent Laplace variable,

s = σ + jω, can take any value in the complex s-plane and is not simply restricted to the jω-axis, as is the case for the CTFT. The values of s for

which the Laplace transforms converge constitute the region of convergence

(ROC) of the Laplace transforms. In Section 6.2, we derived the unilateral

Laplace transforms and the associated ROCs for a number of elementary CT

signals; these transform pairs are listed in Table 6.1. Direct computation of

the inverse Laplace transform involves contour integration, which is difficult

to compute analytically. For Laplace transforms, which take a rational form,

the inverse can be easily determined using the partial fraction approach cov-

ered in Section 6.3. The properties of the Laplace transform are covered in

Section 6.4 and listed in Table 6.2. In particular, we covered the linearity, scaling,

shifting, differentiation, integration, and convolution properties, as summarized

below.

(1) The linearity property implies that the Laplace transform of a linear com-

bination of signals is obtained by taking the same linear combination in the

complex s-domain. In other words,

a1x1(t) + a2x2(t) L←→ a1 X1(s) + a2 X2(s) with ROC: at least R1 ∩ R2.

(2) Scaling a signal by a factor of a in the time domain is equivalent to scaling

its Laplace transform by a factor of 1/a in the s-domain; i.e.

x(at) L←→

|a| X

( s

)

with ROC: a R.

(3) Shifting a signal in the time domain is equivalent to multiplication by a

complex exponential in the s-domain. Mathematically, the time-shifting

property is expressed as follows:

x(t − t0) L←→ e−st0 X (s) with ROC: R.

(4) The converse of the time-shifting property is also true. In other words,

shifting a signal in the s-domain is equivalent to multiplication by a complex

exponential in the time domain:

es0t x(t) L←→ X (s − s0) with ROC: R + Re{s0}.

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312 Part II Continuous-time signals and systems

(5) Differentiation in the time domain is equivalent to multiplication by s in the

complex s-domain. This is referred to as the time-differentiation property

and is expressed as follows:

L←→ s X (s) − x(0−) with ROC: R.

(6) Integration in the time domain is equivalent to division by s in the complex

s-domain. This is referred to as the time-integration property and is

expressed as follows:

unilateral Laplace transform

t∫

0−

x(τ )dτ L←→

X (s)

with ROC: R ∩ Re{s} > 0;

bilateral Laplace transform

t∫

−∞

x(τ )dτ L←→

X (s)

s +

0−∫

−∞

x(τ )dτ

with ROC: R ∩ Re{s} > 0.

(7) The convolution property states that convolution in the time domain is

equivalent to multiplication in the s-domain, and vice versa. Mathemati-

cally, the convolution property is stated as follows:

time convolution x1(t) ∗ x2(t) L←→ X1(s)X2(s)

containing at least ROC: R1 ∩ R2;

s-plane convolution x1(t)x2(t) L←→

2π j [X1(s) ∗ X2(s)]

containing at least ROC: R1 ∩ R2.

(8) The initial- and final-value theorems provide us with an alternative approach

for calculating the limits of a CT function x(t) as t → 0 and t → ∞ from the following expressions:

initial-value theorem x(0+) = lim t→0+

x(t) = lim s→∞

s X (s)

provided x(0+) exists;

final-value theorem x(∞) = lim t→∞

x(t) = lim s→0

s X (s)

provided x(∞) exists.

The initial-value theorem is valid for the unilateral Laplace transform, while

the final-value theorem is valid for both unilateral and bilateral transforms.

Sections 6.5 to 6.9 discussed various applications of the Laplace transform. The

time-differentiation property is used in Section 6.5 to solve linear, constant-

coefficient differential equations. Section 6.6 uses the properties of the ROC

associated with the Laplace transform with an emphasis on causal systems.

Sections 6.7 and 6.8 define the stability of the causal LTIC systems in terms

of the poles and zeros of its transfer function. The key points are summarized

below.

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313 6 Laplace transform

(1) The causal implementation of an absolutely BIBO stable system must have

all of its poles in the left half of the complex s-plane.

(2) If even a single pole lies in the right half of the s-plane, the causal imple-

mentation of the system is unstable.

(3) If no pole lies in the right half of the s-plane, but one or more first-order

poles lie on the imaginary jω-axis, the LTIC system is referred to as a

marginally stable system.

(4) An unstable system may be transformed into a stable system by cascading

the unstable system with an allpass system, which has zeros at the locations

of the unstable poles.

Section 6.9 described an analysis technique based on the Laplace transform to

calculate the output of an LTIC system. We showed that the Laplace-transform-

based analysis approach is suitable for studying the transient response of the

systems. The CTFT-based approach is appropriate for analyzing the steady state

response of the system.

Finally, Section 6.10 discussed the cascaded, parallel, and feedback config-

urations used to interconnect two LTIC systems. If two systems with impulse

responses h1(t) and h2(t) are connected, the overall impulse response and the

corresponding transfer functions are as follows:

cascaded configuration h(t) = h1(t) ∗ h2(t) L←→ H (s) = H1(s)H2(s);

parallel configuration h(t) = h1(t) + h2(t) L←→ H (s) = H1(s) + H2(s);

feedback configuration H (s) = H1(s)

1 + H1(s)H2(s) .

A practical system comprises multiple LTIC systems interconnected with a

combination of cascaded, parallel, and feedback configurations.

Problems

6.1 Using the definition in Eq. (6.5), calculate the bilateral Laplace transform and the associated ROC for the following CT functions:

(a) x(t) = e−5t u(t) + e4t u(−t); (d) x(t) = e−3|t | cos(5t); (b) x(t) = e−3|t |; (e) x(t) = e7t cos(9t)u(−t);

1 − |t | 0 ≤ |t | ≤ 1 0 otherwise.

6.2 Using Eq. (6.9), calculate the unilateral Laplace transform and the associ- ated ROC for the following CT functions:

(a) x(t) = t5u(t); (d) x(t) = e−3t cos(9t)u(t); (b) x(t) = sin(6t)u(t); (e) x(t) = t2 cos(10t)u(t);

1 − |t | 0 ≤ |t | ≤ 1 0 otherwise.

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314 Part II Continuous-time signals and systems

6.3 Using the partial fraction expansion approach, calculate the inverse Laplace transform for the following rational functions of s:

(a) X (s) = s2 + 2s + 1

(s + 1)(s2 + 5s + 6) ; ROC : Re{s} > −1;

(b) X (s) = s2 + 2s + 1

(s + 1)(s2 + 5s + 6) ; ROC : Re{s} < −3;

(s + 1)(s2 + 5s + 6) ; ROC : Re{s} > −1;

(d) X (s) = s2 + 3s − 4

(s + 1)(s2 + 5s + 6) ; ROC : Re{s} < −3;

(e) X (s) = s2 + 1

s(s + 1)(s2 + 2s + 17) ; ROC : Re{s} > 0;

(f) X (s) = s + 1

(s + 2)2(s2 + 7s + 12) ; ROC : Re{s} > −2;

(g) X (s) = s2 − 2s + 1

(s + 1)3(s2 + 16) ; ROC : Re{s} < −1.

6.4 The Laplace transforms of two CT signals x1(t) and x2(t) are given by the following expressions:

x1(t) L←→

s2 + 5s + 6 with ROC(R1) : Re{s} > −2

and

x2(t) L←→

s2 + 5s + 6 with ROC(R2) : Re{s} > −2.

Determine the Laplace transform and the associated ROC R of the com-

bined signal x1(t) + 2x2(t). Explain how the ROC R of the combined signal exceeds the intersection (R1 ∩ R2) of the individual ROCs R1 and R2.

6.5 Calculate the time-domain representation of the bilateral Laplace transform

X (s) = s2

(s2 − 1)(s2 − 4s + 5)(s2 + 4s + 5)

if the ROC R is specified as follows:

(a) R : Re{s} < −2; (b) R : −2 < Re{s} < −1; (c) R : −1 < Re{s} < 1;

(d) R : 1 < Re{s} < 2; (e) R : Re{s} > 2.

6.6 Prove the frequency-shifting property, Eq. (6.20), as stated in Section 6.4.4.

6.7 Prove the time-integration property for the unilateral and bilateral Laplace transform as stated in Section 6.4.6.

6.8 Prove the initial-value theorem, Eq. (6.27), as stated in Section 6.4.8.

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315 6 Laplace transform

6.9 Prove the final-value theorem, Eq. (6.28), as stated in Section 6.4.8.

6.10 Using the transform pairs in Table 6.1 and the properties of the Laplace transform, prove the following Laplace transform pairs:

(a) t cos(ω0t)u(t) L←→

s2 − ω20 (s2 + a2)2

;

(b) t sin(ω0t)u(t) L←→

2ω0s

(s2 + a2)2 ;

2a3 (sin(at) − at cos(at))u(t) L←→

(s2 + a2)2 .

6.11 Express the Laplace transform and the associated ROC for the following functions in terms of the Laplace transform X (s) with ROC Rx of the CT

function x(t):

(a) cos(10t)x(t);

(b) e−5t x(4t − 3);

dt [x(t − 4)];

(d) [x(t) + 2]2;

(e)

t∫

−∞

e−αs0 x(α)dα.

6.12 Using the initial- and final-value theorems, calculate the initial and final values of the causal CT functions with the following unilateral Laplace

transforms. In each case, first determine the ROC to see if the initial value

exists.

(a) X (s) = s

s2 + 7s + 1 ;

(b) X (s) = s

s2 + 5s − 4 ;

s2 − 25 ;

(d) X (s) = s2 + 2s + 1 s2 + 3s + 4

;

(e) X (s) = e−5s s2 + 4

s(s + 1)(s + 2)(s + 3) .

6.13 Solve the following initial-value differential equations using the Laplace transform method:

(a) d2 y

dt2 + 3

dt + 2y(t) = δ(t); y(0−) = ẏ(0−) = 0;

(b) d2 y

dt2 + 4

dt + 4y(t) = u(t); y(0−) = ẏ(0−) = 0;

dt2 + 6

dt + 8y(t) = te−3t u(t); y(0−) = ẏ(0−) = 1;

(d) d3 y

dt3 + 8

d2 y

dt2 + 19

dt + 12y(t) = tu(t);

y(0−) = 1; ẏ(0−) =ÿ(0−) = 0;

(e) d4 y

dt4 + 2

d2 y

dt2 + y(t) = u(t); y(0−) = ẏ(0−) = ÿ(0−) = ¨ẏ(0−) = 0.

6.14 Determine (i) the Laplace transfer function, (ii) the impulse response function, and (iii) the input–output relationship (in the form of a linear

constant-coefficient differential equation) for the causal LTIC systems

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316 Part II Continuous-time signals and systems

Re{s}

Im{s}

−2−4 −2

−4

2 4

xx Re{s}

Im{s}

−2−4 −2

−4

2 4

xxx Re{s}

Im{s}

−2−4 −2

−4

2 4

Re{s}

Im{s}

−2−4 −2

−4

2 4

4 x

double

poles

(i) (ii) (iii) (iv)

Fig. P6.17. Pole – zero plots for

Problem 6.17.

with the following input–output pairs:

(a) x(t) = 4u(t) and y(t) = tu(t) + e−2t u(t); (b) x(t) = e−2t u(t) and y(t) = 3e−2(t−4)u(t − 4); (c) x(t) = tu(t) and y(t) = [t2 − 3e−4t ]u(t); (d) x(t) = e−2t u(t) and y(t) = e−t u(t) + e−3t u(t); (e) x(t) = e−3t u(t) and y(t) = et u(−t) + e−3t u(t).

6.15 Sketch the location of the poles and zeros for the following transfer func- tions, and determine if the corresponding causal systems are stable, unsta-

ble, or marginally stable:

(a) H (s) = s2 + 1

s2 + 2s + 1 ;

(b) H (s) = 2s + 5

s2 + s − 6 ;

s2 + 9s + 18 ;

(d) H (s) = s + 2 s2 + 9

;

(e) H (s) = s2 + 3s + 2

s3 + 3s2 + 2s .

6.16 Without explicitly calculating the output, determine if the LTIC system with the transfer function

H (s) = s2 + 1

(s + 5)(s2 + 4)(s2 + 9)(s2 + 4s + 5)

produces a bounded output for the following set of inputs:

(a) x(t) = e−j2t u(t); (b) x(t) = [e−(1+j4)t + e−(2+j5)t ]u(t); (c) x(t) = [cos(t) + sin(4t)]u(t); (d) x(t) = [cos(2t) + sin(3t)]u(t); (e) x(t) = [e−(1+j2)t sin(3t)]u(t).

6.17 The pole–zero plots of four causal LTIC systems are shown in Fig. P6.17. Determine if the LTIC systems are stable. Also determine the transfer

function H (s) for each system. Assume that H (4) = 1 in all cases, and the poles and zeros are all located at integer coordinates in the s-plane.

6.18 Determine the transfer functions of all possible non-causal implementa- tions of the LTIC systems considered in Fig. P6.17. Specify which transfer

functions represent stable systems.

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317 6 Laplace transform

6.19 The inverse of an LTIC system is defined as the system that when cascaded with the original system results in an overall transfer function of unity.

Without calculating the transfer functions, determine the pole–zero plots

of the inverse systems associated with the LTIC systems whose pole–zero

plots are specified in Fig. P6.17.

6.20 An LTIC system has an impulse response h(t) with the Laplace transfer function H (s), which satisfies the following properties:

(a) the impulse response h(t) is even and real-valued;

(b) the area enclosed by the impulse response is 8, i.e.

∞∫

−∞

h(t)dt = 8;

(d) the Laplace transfer function H (s) has a complex pole at s =

0.5 exp(jπ/4).

Determine the Laplace transfer function H (s) and the associated ROC.

6.21 Consider the RLC series circuit shown in Fig. 3.1. The relationship between the input voltage x(t) and the output voltage w(t) is given by the

following differential equation:

d2w

dt2 +

dt +

L w(t) =

LC x(t).

By determining the locations of the poles of the transfer function describ-

ing the RLC series circuit, show that the causal implementation of the RLC

circuit is always stable for positive values (R > 0, L > 0, and C > 0) of

the passive components.

6.22 Given the transfer function

H (s) = s2 − s − 6

(s2 + 3s + 1)(s2 + 7s + 12)

(a) determine all possible choices for the ROC;

(b) determine the impulse response of a causal implementation of the

transfer function H (s);

function H (s);

(d) determine all possible choices of double-sided impulse responses hav-

ing the specified transfer function H (s).

(e) Which of the four impulse responses obtained in (b)–(d) are stable?

6.23 Repeat Problem 6.22 for the following transfer function:

H (s) = s2 − 5s − 84

(s2 − 2s − 35)(s2 + 9s + 20) .

6.24 For most practical applications, we are interested in implementing a causal and stable system. The causal implementations of some of the transfer

functions specified in Problem 6.15 are not stable. For each such trans-

fer function, specify an allpass system that may be cascaded in a series

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318 Part II Continuous-time signals and systems

+ −

+ (s+1)

1 (s+2)

1 (s+3)

(s+5) 1

(s+6) 1

(s+4) 1

− ∑∑x(t) y(t)

(i)

−

+ −

∑∑x(t) y(t) (s+1)

1 (s+2)

1 (s+3)

(s+5) 1

(s+4) 1

(ii)

− +

−

y(t)x(t) ∑

∑

(s+1) 1

(s+7) 1

(s+6) 1

(s+4) 1

(s+5) 1

(s+2) 1

(s+3) 1

(iii)

Fig. P6.25. Interconnected

systems specified in Problem

6.25.

configuration to the specified transfer function to make its causal imple-

mentation stable.

6.25 Determine the overall transfer function for the three interconnected sys- tems shown in Fig. P6.25.

6.26 Using the function residue available in M A T L A B toolboxes, calculate the partial fraction coefficients for the transfer functions considered in

Problem 6.3.

6.27 Using the functions tf and bode available in the M A T L A B control toolbox, plot the frequency characteristics of the systems with transfer

functions considered in Problem 6.15.

6.28 Repeat Problem 6.27 using the functionfreqs available in the M A T L A B signal toolbox.

6.29 Using the functions tf and impulse available in the M A T L A B con- trol toolbox, calculate the impulse response of the systems with transfer

functions considered in Problem 6.15.

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319 6 Laplace transform

6.30 (a) Using the M A T L A B function roots, calculate the location of poles and zeros of the following transfer functions:

(i) H1(s) = s2 − 5s − 84

s4 + 7s3 − 33s2 − 355s − 700 ;

(ii) H2(s) = s2 − 19s + 84

s4 + 7s3 − 33s2 − 355s − 700 ;

(iii) H3(s) = s3 + 20s2 + 15s + 61

s4 + 5s3 + 31s2 + 125s + 150 ;

(iv) H4(s) = s3 − 10s2 + 25s + 7

s6 + 6s5 + 42s4 + 48s3 + 288s2 + 96s + 544 ;

(v) H5(s) = s2 + 3s + 7

s3 + (6 − j7)s2 + (11 − j28)s + (6 − j21) .

(b) From the location of poles and zeros in the s-plane, determine if the

systems are (i) absolutely stable, (ii) marginally stable, or (iii) unstable.

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C H A P T E R

7 Continuous-time filters

A common requirement in signal processing is to modify the frequency contents

of a continuous-time (CT) signal in a predefined manner. In communication sys-

tems, for example, noise and interference from the neighboring channels cor-

rupt the information-bearing signal transmitted via a communication channel,

such as a telephone line. By exploiting the differences between the frequency

characteristics of the transmitted signal and the channel noise, a linear time-

invariant system (LTI) system can be designed to compensate for the distortion

introduced during the transmission. Such an LTI system is referred to as a

frequency-selective filter, which processes the received signal to eliminate the

high-frequency components introduced by the channel interference and noise

from the low-frequency components constituting the information-bearing sig-

nal. The range of frequencies eliminated from the CT signal applied at the input

of the filter is referred to as the stop band of the filter, while the range of fre-

quencies that is left relatively unaffected by the filter constitute the pass band

of the filter.

Graphic equalizers used in stereo sound systems provide another application

for the continuous-time (CT) filters. A graphic equalizer consists of a combina-

tion of CT filters, each tuned to a different band of frequencies. By selectively

amplifying or attenuating the frequencies within the operational bands of the

constituent filters, a graphic equalizer maintains sound consistency within dis-

similar acoustic environments and spaces. The operation of a graphic equalizer

is somewhat different from that of a frequency-selective filter used in our earlier

example of the communication system since it amplifies or attenuates selected

frequency components of the input signal. A frequency-selective filter, on the

other hand, attempts to eliminate the frequency components completely within

the stop band of the filter.

This chapter focuses on the design of CT filters. We are particularly interested

in the frequency-selective filters that are categorized in four different categories

(lowpass, highpass, bandpass, and bandstop) in Section 7.1. Practical approxi-

mations to the frequency characteristics of the ideal frequency-selective filters

are presented in Section 7.2, where acceptable levels of distortion is tolerated

320

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321 7 Continuous-time filters

within the pass and stop bands of the ideal filters. Section 7.3 designs three

realizable implementations of an ideal lowpass filter. These implementations

are referred to as the Butterworth, Chebyshev, and elliptic filters. Section 7.4

transforms the frequency characteristics of the highpass, bandpass, and band-

stop filters in terms of the characteristics of the lowpass filters. These transfor-

mations are exploited to design the highpass, bandpass, and bandstop filters.

Finally, the chapter is concluded with a summary of important concepts in

Section 7.5.

7.1 Filter classification

An ideal frequency-selective filter is a system that passes a prespecified range

of frequency components without any attenuation but completely rejects the

remaining frequency components. As discussed earlier, the range of input fre-

quencies that is left unaffected by the filter is referred to as the pass band of the

filter, while the range of input frequencies that are blocked from the output is

referred to as the stop band of the filter. In terms of the magnitude spectrum, the

absolute value of the transfer function |H (ω)| of the frequency filter, therefore, toggles between the values of A and zero as a function of frequency ω. The gain |H (ω)| is A, typically set to one, within the pass band, while |H (ω)| is zero within the stop band. Depending upon the range of frequencies within the

pass and stop bands, an ideal frequency-selective filter is categorized in four

different categories. These categories are defined in the following discussion.

7.1.1 Lowpass filters

The transfer function Hlp(ω) of an ideal lowpass filter is defined as follows:

Hlp(ω) = {

A |ω| ≤ ωc 0 |ω| > ωc,

(7.1)

where ωc is referred to as the cut-off frequency of the filter. The pass band of

the lowpass filter is given by |ω| ≤ ωc, while the stop band of the lowpass filter is given by ωc < |ω| < ∞. The frequency characteristics of an ideal lowpass filter are plotted in Fig. 7.1(a), where we observe that the magnitude |Hlp(ω)| toggles between the values of A within the pass band and zero within the stop band. The phase <Hlp(ω) of an ideal lowpass filter is zero for all frequencies.

7.1.2 Highpass filters

The transfer function Hhp(ω) of an ideal highpass filter is defined as follows:

Hhp(ω) = {

0 |ω| ≤ ωc A |ω| > ωc,

(7.2)

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w 0

Hlp(w)

wc−wc 0 wc−wc

Hbp(w)

−wc2 −wc1 wc1 wc2 0−wc2 −wc1 wc1 wc2 w

Hbs(w)

Hhp(w)

(a) (b)

Fig. 7.1. Magnitude spectra of

ideal frequency-selective filters.

(a) Lowpass filter; (b) highpass

filter; (c) bandpass filter;

(d) bandstop filter.

where ωc is the cut-off frequency of the filter. In other words, the transfer

function of an ideal highpass filter Hhp(ω) is related to the transfer function of an ideal lowpass filter Hlp(ω) by the following relationship:

Hhp(ω) = A − Hlp(ω). (7.3)

The pass band of the lowpass filter is given by ωc < |ω| < ∞, while the stop band of the lowpass filter is given by |ω| ≤ ωc. The frequency characteristics of an ideal lowpass filter are plotted in Fig. 7.1(b). As was the case for the

ideal lowpass filter, the phase <Hhp(ω) of an ideal highpass filter is zero for all frequencies.

7.1.3 Bandpass filters

The transfer function Hbp(ω) of an ideal bandpass filter is defined as follows:

Hbp(ω) = {

A ωc1 ≤ |ω| ≤ ωc2 0 ωc1 < |ω| and ωc2 < |ω| < ∞,

(7.4)

where ωc1 and ωc2 are collectively referred to as the cut-off frequencies of the

ideal bandpass filter. The lower frequency ωc1 is referred to as the lower cut

off, while the higher frequency ωc2 is referred to as the higher cut off. Unlike

the highpass filter, the bandpass filter has a finite bandwidth as it only allows a

range of frequencies (ωc1 ≤ ω ≤ ωc2) to be passed through the filter.

7.1.4 Bandstop filters

The transfer function Hbs(ω) of an ideal bandstop filter is defined as follows:

Hbp(ω) = {

0 ωc1 ≤ |ω| ≤ ωc2 A ωc1 < |ω| and ωc2 < |ω| < ∞,

(7.5)

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323 7 Continuous-time filters

where ωc1 and ωc2 are, respectively, referred to as the lower cut-off and higher

cut-off frequencies of the ideal bandstop filter. A bandstop filter can be imple-

mented from a bandpass filter using the following relationship:

Hbs(ω) = A − Hbp(ω). (7.6)

The ideal bandstop filter is the converse of the ideal bandpass filter as it elimi-

nates a certain range of frequencies (ωc1 ≤ ω ≤ ωc2) from the input signal.

In the above discussion, we used the transfer function to categorize different

types of frequency selective filters. Example 7.1 derives the impulse response

for ideal lowpass and highpass filters.

Example 7.1

Determine the impulse response of an ideal lowpass filter and an ideal highpass

filter. In each case, assume a gain of A within the pass band and a cut-off frequency of ωc.

Solution

Taking the inverse CTFT of Eq. (7.1), we obtain

hlp(t) = ℑ−1{H (ω)} = 1

2π

ωc∫

−ωc

A · e jωt dω = Ae jωt

j2π t

∣ ∣ ∣ ∣

ωc

= A

j2π t [ejωct − e−jωct ],

which reduces to

hlp(t) = 2jA sin(ωct)

j2π t =

ωc A

π sinc

( ωct

)

. (7.7)

To derive the impulse response hhp(t) of the ideal highpass filter, we take the inverse CTFT of Eq. (7.3). The resulting relationship is given by

hhp(t) = Aδ(t) − hlp(t) = Aδ(t) − ωc A

π sinc

( ωct

)

. (7.8)

( )wct pphlp(t) = sinc

wc −4p

wc −3p

wc − p

wc −2p

4p wc

3p wc

p Awc

Awc

2p 0

( )wct pphhp(t) = A −

wc −4p

wc −3p

wc −

−

p wc

−2p wc

4p wc

3p wc

p Awc

AwcA

(a) (b)

sinc

Fig. 7.2. Impulse responses h(t )

of: (a) ideal lowpass filter and

(b) ideal highpass filter derived

in Example 7.1.

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The impulse responses of ideal lowpass and highpass filters are plotted in

Fig. 7.2. In both cases, we note that the filters have an infinite length in the

time domain. Also, both filters are non-causal since h(t) �= 0 for t < 0.

7.2 Non-ideal filter characteristics

As is true for any ideal system, the ideal frequency-selective filters are not

physically realizable for a variety of reasons. From the frequency characteristics

of the ideal filters, we note that the gain A of the filters is constant within the pass band, while the gain within the stop band is strictly zero. A second issue

with the transfer functions H (ω), specified for ideal filters in Eqs. (7.1)–(7.5), is the sharp transition between the pass and stop bands such that there is a

discontinuity in H (ω) at ω = ωc. In practice, we cannot implement filters with constant gains within the pass and stop bands. Also, abrupt transitions cannot

be designed. This is observed in Example 7.1, where the constant gains and

the sharp transition in the ideal lowpass and highpass filters lead to non-causal

impulse responses which are of infinite length. Clearly, such LTI systems cannot

be implemented in the physical world.

To obtain a physically realizable filter, it is necessary to relax some of the

requirements of the ideal filters. Figure 7.3 shows the frequency characteristics

of physically realizable versions of various ideal filters. The upper and lower

bounds for the gains are indicated by the shaded line, while examples of the

frequency characteristics of physically realizable filters that satisfy the specified

bounds are shown using bold lines. These filters are referred to as non-ideal

or practical filters and are different from the ideal filters in the following two

ways.

(i) The gains of the practical filters within the pass and stop bands are not

constant but vary within the following limits:

pass bands 1 − δp ≤ |H (ω)| ≤ 1 + δp; (7.9) stop bands 0 ≤ |H (ω)| ≤ δs. (7.10)

The oscillations within the pass and stop bands are referred to as ripples. In

Fig. 7.3, the pass band ripples are constrained to a value of δp for lowpass,

highpass, and bandpass filters. In the case of the bandstop filter, the pass

band ripple is limited to δp1 and δp2, corresponding to the two pass bands.

Similarly, the stop band ripples in Fig. 7.3 are constrained to δs for lowpass,

highpass, and bandstop filters. In the case of the bandstop filter, the stop

band ripple is limited to δs1 and δs2 for the two stop bands of the bandstop

filter.

(ii) Transition bands of non-zero bandwidth are included in between the pass

and stop bands of the practical filters. Consequently, the discontinuity at

the cut-off frequency ωc of the ideal filters is eliminated.

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Hlp(w)

Hbp(w)

Hhp(w)

1−dp

1+dp

1+dp 1+dp1

1−dp1

1+dp2 1−dp21−dp

1−dp

1+dp

pass band stop bandtransition

band

w 0

ds2 ds1 ds

wp wp ws wp

stop band pass bandtransition

band

w 0

stop

band I

stop

band II

pass band

w w 0

Hbs(w)

ws1 wp1 ws1 ws2 wp2wp1 wp2 ws2

pass

band I

pass

band II

stop

band

(a)

(b)

Fig. 7.3. Frequency

characteristics of practical filters.

(a) Practical lowpass filter;

(b) practical highpass filter;

(d) practical bandstop filter.

Example 7.2 considers a practical lowpass filter and derives the values for the

pass band and the stop band, and the associated gains of the filter.

Example 7.2 Consider a practical lowpass filter with the following transfer function:

H (s) = 5.018×103s4+2.682×1014s3−1.026×104s+3.196×1024

s5+9.863×104s4+2.107×1010s3+1.376×1015s2+1.026×1020s+3.196×1024 .

Assuming that the ripple δp within the pass band is limited to 1 dB and the

ripple δs within the stop band is limited to 40 dB, determine the pass band,

transition band and stop band of the lowpass filter.

Solution

Recall that the CTFT transfer function H (ω) of the lowpass filter can be obtained by substituting s = jω in the Laplace transfer function. The resulting magnitude spectrum |H (ω)| of the lowpass filter is plotted in Fig. 7.4, where Fig. 7.4(a)

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326 Part II Continuous-time signals and systems

0 p 2p 3p 4p 5p 6p 7p 8p3.4p 0

0.5

1 0.8913

w (×104) 0 p 2p 3p 4p 5p 6p 7p 8p

−80 −60 −40 −20

w (×104)4.12p

(a) (b)

Fig. 7.4. Magnitude spectrum of

the practical lowpass filter in

Example 7.2 using (a) a linear

scale and (b) a decibel scale

along the y-axis.

uses a linear scale for the magnitude. Figure 7.4(b) uses a decibel scale to plot

the magnitude spectrum.

Expressed on a linear scale, the pass-band ripple δp is given by 10 −1/20 or

0.8913. From Fig. 7.4(a), we observe that the pass-band frequency ωp corre-

sponding to |H (ω)| = 0.8913 is given by 3.4π × 104 radians/s. Therefore, the pass band is specified by |ω| ≤ 3.4π × 104 radians/s.

To determine the stop band, we use Fig. 7.4(b), which uses a decibel scale

20 × log10|H (ω)| to plot the magnitude spectrum. Figure 7.4(b) shows that the smallest frequency for which the magnitude spectrum equals a gain of

40 dB is given by 4.12π × 104 radians/s. The stop band is therefore specified by |ω| > 4.12π × 104 radians/s.

Based on the aforementioned results, it is straightforward to derive the tran-

sition band as follows:

3.4π × 104 < |ω| < 4.12π × 104 radians/s.

7.2.1 Cut-off frequency

An important parameter in the design of CT filters is the cut-off frequency

ωc of the filter, which is defined as the frequency at which the gain of the

filter drops to 0.7071 times its maximum value. Assuming a gain of unity

within the pass band, the gain at the cut-off frequency ωc is given by 0.7071 or

−3 dB on a logarithmic scale. Since the cut-off frequency lies typically within the transitional band of the filter, therefore

ωp ≤ ωc ≤ ωs (7.11)

For a lowpass filter. Note that the equality ωp = ωc = ωs implies a transitional band of zero bandwidth and is valid only for ideal filters.

As a side note to our discussion, we observe that in this chapter we only

consider positive values of frequencies ω in plotting the magnitude spectrum.

The majority of our designs are based on real-valued impulse responses, which

lead to frequency spectra that satisfy the Hermitian symmetry. Exploiting the

even symmetry for the magnitude spectrum, it is therefore sufficient to spec-

ify the magnitude spectrum only for positive frequencies in such cases. The

pass-band, stop-band, and cut-off frequencies are also specified by positive

values, though their counter-negative values exist for all three parameters.

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327 7 Continuous-time filters

Example 7.3

Determine the cut-off frequency for the lowpass filter specified in Example 7.2.

Solution

Based on the magnitude spectrum, we note that the maximum gain of the filter

is given by 1 or 0 dB. At the cut off frequency ωc,

|H (ωc)| = 0.7071 × 1 = 0.7071,

which implies that ∣ ∣ ∣ ∣

5.018×103( jωc)4+2.682 × 1014( jωc)2−1.026×104( jωc)+3.196 × 1024

( jωc)5+9.863×104( jωc)4+2.107×1010( jωc)3+1.376×1015( jωc)2+1.026×1020( jωc)+3.196×1024

∣ ∣ ∣ ∣

= 0.7071.

The above equality can be solved for ωc using numerical techniques in

M A T L A B . The value of the cut-off frequency is given by ωc = 3.462π × 104 radians/s. Note that the cut-off frequency lies within the transitional

band in between the pass and stop bands of the lowpass filter as derived in

Example 7.2.

7.3 Design of CT lowpass filters

To begin our discussion of the design of CT filters, we consider a prototype or

normalized lowpass filter, defined as a lowpass filter, with a cut-off frequency

of ωc = 1 radians/s. The remaining specifications for the pass and stop bands of the normalized lowpass filter are assumed to be given by

pass band (0 ≤ |ω| ≤ ωp radians/s) 1 − δp ≤ |H (ω)| ≤ 1 + δp; (7.12) stop band (|ω| > ωs radians/s)| H (ω)| ≤ δs, (7.13)

with ωp ≤ ωc ≤ ωs. Using the transfer function of the normalized lowpass filter, it is straightforward to implement any of the more complicated CT filters.

Section 7.4 considers the frequency transformations used to convert a lowpass

filter into another category of frequency-selective filters.

There are several specialized implementations such as Butterworth, Type I

Chebyshev, Type II Chebysev, and elliptic filters, which may be used to design

a normalized lowpass filter. Figure 7.5 shows representative characteristics of

these implementations, where we observe that the Butterworth filter (Fig. 7.5(a))

has a monotonic transfer function such that the gain decreases monotonically

from its maximum value of unity at ω = 0 along the positive frequency axis. The magnitude spectrum of the Butterworth filter has negligible ripples within

the pass and stop bands, but has a relatively lower fall off leading to a wide

transitional band. By allowing some ripples in either the pass or stop band,

the Type I and Type II Chebyshev filters incorporate a sharper fall off. The

Type I Chebyshev filter constitutes ripples within the pass band, while the

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Type II Chebyshev filter allows for the stop-band ripples. Compared with the

Butterworth filter, both Type I and Type II Chebyshev filters have narrower

transitional bands. The elliptic filters allow for the sharpest fall off by incorpo-

rating ripples in both the pass and stop bands of the filter. The elliptic filters

have the narrowest transitional band. To compare the transitional bands, Fig. 7.5

plots the magnitude spectra resulting from the Butterworth, Type I Chebyshev,

Type II Chebysev, and elliptic filters with the same order N . Figure 7.5 confirms our earlier observations that the Butterworth filter

(Fig. 7.5(a)) has the widest transitional band. Both the Type I and Type II

Chebyshev filters (Figs. 7.5(b) and (c)) have roughly equal transitional bands,

which are narrower than the transitional band of the Butterworth filter. The ellip-

tic filter (Fig. 7.5(d)) has the narrowest transitional band but includes ripples in

both the pass and stop bands.

We now consider the design techniques for the four specialized implemen-

tations with a brief explanation of the M A T L A B library functions useful for

computing the transfer functions of the implementations.

7.3.1 Butterworth filters

The frequency characteristics of an N th-order lowpass Butterworth filter are given by

|H (ω)| = 1

√

1 + ( ω

ωc

)2N , (7.14)

where ωc is the cut-off frequency of the filter. Substituting ωc = 1 for the normalized implementation, the transfer function of the normalized lowpass

Butterworth filter of order N is given by

|H (ω)| = 1

√ 1 + ω2N

. (7.15)

To derive the Laplace transfer function H (s) of the normalized Butterworth filter, we use the following relationship:

|H (ω)|2 = H (s)H (−s)|s=jω. (7.16)

Substituting ω = s/j, Eq. (7.16) reduces to

H (s)H (−s) = |H (s/j)|2. (7.17)

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pass band stop band

pass band

stop band

w 0

1+dp

ds ds

wp wp

wpws

ws ws

1−dp 1

1+dp

ds ds

1−dp 1

1+dp

1−dp 1

1+dp

1−dp 1

w 0

(a)

(c)

(b)

(d)

w 0

Fig. 7.5. Frequency

characteristics of standard

implementations of lowpass

filters of order N .

(a) Butterworth filter; (b) Type-I

Chebyshev filter; (c) Type-II

Chebyshev filter; (d) elliptic filter.

Further substituting H (s/j) from Eq. (7.15) leads to the following expression:

H (s)H (−s) = 1

1 + ( s

)2N , (7.18)

where the denominator represents the characteristic function for H (s)H (−s). The poles of H (s)H (−s) occur at

( s

)2N

= −1 = ej(2n−1)π (7.19)

s = j exp [

j (2n − 1)π

]

= exp [

j π

2 + j

(2n − 1)π 2N

]

(7.20)

for 0 ≤ n ≤ 2N−1. It is clear that the 2N poles for H (s)H (−s), specified in Eq. (7.20), are evenly distributed along the unit circle in the complex s-plane.

Of these, N poles would lie in the left half of the s-plane, while the remaining N poles would be in the right half of the s-plane. To ensure a causal and

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Table 7.1. Location of the 2N poles for H(s )H(−s ) in Example 7.4 for N = 7

n 0 1 2 3 4 5 6 7 8 9 10 11 12 13 pn ej3π/7 ej4π/7 ej5π/7 ej6π/7 ejπ e−j6π/7 e−j5π/7 e−j4π/7 e−j3π/7 e−j2π/7 e−jπ/7 1 ejπ/7 e j2π/7

stable implementation, the transfer function H (s) of the normalized lowpass Butterworth filter is determined from the N poles lying in the left half of the s-plane and is given by

H (s) = 1

N∏

n=1 (s − pn)

, (7.21)

wherepn , for 1 ≤ n ≤ N , denotes the location of the poles in the left-half s-plane.

Example 7.4

Determine the Laplace transfer function H (s) for the normalized Butterworth filter with cut-off frequency ωc = 1 and order N = 7.

Solution

Using Eq. (7.20), the poles of H (s)H (−s) are given by

s = exp [

j π

2 + j

(2n − 1)π 14

]

for 0 ≤ n ≤ 13. Substituting different values of n, the locations of the poles are specified by Table 7.1. Figure 7.6 plots the locations of the poles for H (s)H (−s) in the complex s-plane. Allocating the poles located in the left-half s-plane

(1 ≤ n ≤ 7), the Laplace transfer function H (s) of the Butterworth filter is given by

Re{s}

Im{s}

1.0 7π

n = 7

n = 1

n = 2

n = 3

n = 4

n = 5

n = 6

n = 0

n = 13

n = 12

n = 11

n = 10

n = 9

n = 8

Fig. 7.6. Location of the poles

for H(s )H(−s ) in the complex s-plane for N = 7. The poles lying in the left-half s-plane are

allocated to the Butterworth

filter.

H (s) = 1

(s − ej4π/7)(s−ej5π/7)(s−ej6π/7)(s−ejπ )(s−e−j6π/7)(s−e−j5π/7)(s−e−j4π/7) ,

which simplifies to

H (s) = 1

(s+1)[(s−ej4π/7)(s−e−j4π/7)][(s−ej5π/7)(s−e−j5π/7)][(s−ej6π/7)(s−e−j6π/7)]

H (s) = 1

(s + 1)(s2 + 0.4450 s + 1)(s2 + 1.2470 s + 1)(s2 + 1.8019 s + 1) .

In Example 7.4, we observed that the locations of poles for the normalized

Butterworth filter are complex. Since the poles occur in complex-conjugate

pairs, the coefficients of the Laplace transfer function for the normalized

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Table 7.2. Denominator D (s ) for transfer function H (s ) of the Butterworth filter

N D(s)

1 (s + 1) 2 (s2 + 1.414s + 1) 3 (s + 1)(s2 + s + 1) 4 (s2 + 0.7654s + 1)(s2 + 1.8478s + 1) 5 (s + 1)(s2 + 0.6810 s + 1)(s2 + 1.6810 s + 1) 6 (s2 + 0.5176s + 1)(s2 + 1.4142s + 1)(s2 + 1.9319s + 1) 7 (s + 1)(s2 + 0.4450 s + 1)(s2 + 1.2470 s + 1)(s2 + 1.8019 s + 1) 8 (s2 + 0.3902s + 1)(s2 + 1.1111s + 1)(s2 + 1.6629s + 1)(s2 + 1.9616s + 1) 9 (s + 1)(s2 + 0.3473 s + 1)(s2 + s + 1)(s2 + 1.5321 s + 1)(s2 + 1.8794 s + 1)

10 (s2 + 0.3129 s + 1)(s2 + 0.9080s + 1)(s2 + 1.4142s + 1)(s2 + 1.7820s + 1)(s2 + 1.9754 s + 1)

Butterworth filter are all real-valued. In general, Eq. (7.21) can be simplified as

follows:

H (s) = 1

D(s) =

s N + aN−1s N−1 + · · · + a1s + 1 (7.22)

and represents the transfer function of the normalized Butterworth filter of

order N . Repeating Example 7.4 for different orders (1 ≤ N ≤ 10), the transfer func-

tions H (s) of the resulting normalized Butterworth filters can be similarly com- puted. Since the numerator of the transfer function is always unity, Table 7.2

lists the polynomials for the denominator D(s) for 1 ≤ N ≤ 10.

7.3.1.1 Design steps for the lowpass Butterworth filter

In this section, we will design a Butterworth lowpass filter based on the spec-

ifications illustrated in Fig. 7.3(a). Mathematically, the specifications can be

expressed as follows:

pass band (0 ≤ |ω| ≤ ωp radians/s) 1 − δp ≤ |H (ω)| ≤ 1 + δp; (7.23) stop band (|ω| > ωs radians/s) |H (ω)| ≤ δs. (7.24)

At times, Eq. (7.23) is also expressed in terms of the pass-band ripple as

20 log10δp dB. Similarly, Eq. (7.24) is expressed in terms of the stop-band

ripple as 20 log10δs dB. The design of the Butterworth filter consists of the

following steps, which we refer to as Algorithm 7.3.1.1.

Step 1 Determine the order N of the Butterworth filter. To determine the order N of the filter, we calculate the gain of the filter at the corner frequencies ω = ωp

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332 Part II Continuous-time signals and systems

and ω = ωs. Using Eq. (7.15), the two gains are given by

pass-band corner frequency (ω = ωp) |H (ωp)|2 = 1

1 + (ωp/ωc)2N = (1 − δp)2;

(7.25)

stop-band corner frequency (ω = ωs) |H (ωs)|2 = 1

1 + (ωs/ωc)2N = (δs)2.

(7.26)

Equations (7.25) and (7.26) can alternatively be expressed as follows:

(ωp/ωc) 2N =

(1 − δp)2 − 1 (7.27)

and

(ωs/ωc) 2N =

(δs)2 − 1. (7.28)

Dividing Eq. (7.27) by Eq. (7.28) and simplifying in terms of N , we obtain the following expression:

N = 1

2 ×

ln(Gp/Gs)

ln(ωp/ωs) , (7.29)

where the gain terms are given by

Gp = 1

(1 − δp)2 − 1 and Gs =

(δs)2 − 1. (7.30)

Step 2 Using Table 7.2 or otherwise determine the transfer function for the nor- malized Butterworth filter of order N . The transfer function for the normalized Butterworth filter is denoted by H (S) with the Laplace variable S capitalized to indicate the normalized domain.

Step 3 Determine the cut-off frequency ωc of the Butterworth filter using either of the following two relationships:

pass-band constraint ωc = ωp

(Gp)1/2N ; (7.31)

stop-band constraint ωc = ωs

(Gs)1/2N . (7.32)

If Eq. (7.31) is used to compute the cut-off frequency, then the Butterworth filter

will satisfy the pass-band constraint exactly. Similarly, the stop-band constraint

will be satisfied exactly if Eq. (7.32) is used to determine the cut-off frequency.

Step 4 Determine the transfer function H (s) of the required lowpass filter from the transfer function for the normalized Butterworth filter H (S), obtained

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333 7 Continuous-time filters

in Step 2, and the cut-off frequency ωc, using the following transformation:

H (s) = H (S)|S=s/ωc .

Note that the transformation S = s/ωc represents scaling in the Laplace domain. It is therefore clear that the normalized cut-off frequency of 1 radian/s used in

the normalized Butterworth filter is transformed to a value of ωc as required in

Step 3.

Step 5 Sketch the magnitude spectrum from the transfer function H (s) deter- mined in Step 4. Confirm that the transfer function satisfies the initial design

specifications.

Examples 7.5 and 7.6 illustrate the application of the design algorithm.

Example 7.5

Design a Butterworth lowpass filter with the following specifications:

pass band (0 ≤ |ω| ≤ 5 radians/s) 0.8 ≤ |H (ω)| ≤ 1; stop band (|ω| > 20 radians/s) |H (ω)| ≤ 0.20.

Solution

Using Step 1 of Algorithm 7.3.1.1, the gain terms Gp and Gs are given by

Gp = 1

(1 − δp)2 − 1 =

0.82 − 1 = 0.5625

and

Gs = 1

(δs)2 − 1 =

0.22 − 1 = 24.

Using Eq. (7.29), the order of the Butterworth filter is given by

N = 1

2 ×

ln(Gp/Gs)

ln(ωp/ωs) =

2 ×

ln(0.5625/24)

ln(5/20) = 1.3538.

We round off the order of the filter to the higher integer value as N = 2. Using Step 2 of Algorithm 7.3.1.1, the transfer function H (S) of the normal-

ized Butterworth filter with a cut-off frequency of 1 radian/s is given by

H (S) = 1

S2 + 1.414S + 1 .

Using the pass-band constraint, Eq. (7.31), in Step 3 of Algorithm 7.3.1.1, the

cut-off frequency of the required Butterworth filter is given by

ωc = ωp

(Gp)1/2N =

(0.5625)1/4 = 5.7735 radians/s.

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0 5 10 15 20 25 30 35 40 45 50 0

0.2

0.4

0.6

0.8

0 5 10 15 20 25 30 35 40 45 50 0

0.2

0.4

0.6

0.8

(a) (b)

Fig. 7.7. Magnitude spectra of

the Butterworth lowpass filters,

designed in Example 7.5, as a

function of ω. Part (a) satisfies

the constraint at the pass-band

corner frequency, while part (b)

satisfies the magnitude

constraint at the stop-band

corner frequency.

Using Step 4 of Algorithm 7.3.1.1, the transfer function H (s) of the required Butterworth filter is obtained by the following transformation:

H (s) = H (S)|S=s/ωc = 1

S2 + 1.414S + 1

∣ ∣ ∣ ∣

S=s/5.7735 ,

which simplifies to

H (s) = 1

(s/5.7735)2 + 1.414s/5.7735 + 1 =

33.3333

s2 + 8.1637s + 33.3333 .

Step 5 plots the magnitude spectrum of the Butterworth filter. The CTFT transfer

function of the Butterworth filter is given by

H (ω) = H (s)|s=jω = 33.3333

(jω)2 + 8.1637(jω) + 33.3333 .

The magnitude spectrum |H (ω)| is plotted in Fig. 7.7(a) with the specifications shown by the shaded lines. We observe that the design specifications are indeed

satisfied by the magnitude spectrum.

Alternative implementation An alternative implementation of the aforemen- tioned Butterworth filter can be obtained by using the stop-band constraint,

Eq. (7.32), in Step 3 of Algorithm 7.3.1.1. The cut-off frequency of the alter-

native implementation of the Butterworth filter is given by

ωc = ωs

(Gs)1/2N =

(24)1/4 = 9.0360 radians/s.

Using Step 4 of Algorithm 7.3.1.1, the transfer function H (s) of the alternative implementation is obtained by the following transformation:

H (s) = H (S)|S=s/ωc = 1

S2 + 1.414S + 1

∣ ∣ ∣ ∣

S=s/9.0360 ,

which simplifies to

H (s) = 1

(s/9.0360)2 + 1.414s/9.0360 + 1 =

81.6497

s2 + 12.7769s + 81.6497 .

Step 5 plots the magnitude spectrum of the alternative implementation of the

Butterworth filter in Fig. 7.7(b), which satisfies the initial design specifications.

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Example 7.6

Design a lowpass Butterworth filter with the following specifications:

pass band (0 ≤ |ω| ≤ 50 radians/s) −1 dB ≤ 20 log10|H (ω)| ≤ 0; stop band (|ω| > 100 radians/s) 20 log10|H (ω)| ≤ −15 dB.

Solution

Expressed on a linear scale, the pass-band gain is given by (1 − δp) = 10−1/20 = 0.8913. Similarly, the stop-band gain is given by δs = 10−15/20 = 0.1778.

Using Step 1 of Algorithm 7.3.1.1, the gain terms Gp and Gs are given by

Gp = 1

(1 − δp)2 − 1 =

0.89132 − 1 = 0.2588

and

Gs = 1

(δs)2 − 1 =

0.17782 − 1 = 30.6327.

The order N of the Butterworth filter is obtained using Eq. (7.29) as follows:

N = 1

2 ×

ln(Gp/Gs)

ln(ωp/ωs) =

2 ×

ln(0.2588/30.6327)

ln(50/100) = 3.4435.

We round off the order of the filter to the higher integer value as N = 4. Using Step 2 of Algorithm 7.3.1.1, the transfer function H (S) of the normal-

ized Butterworth filter with a cut-off frequency of 1 radian/s is given by

H (S) = 1

(S2 + 0.7654S + 1)(S2 + 1.8478S + 1) .

Using the pass-band constraint, Eq. (7.31), in Step 3 of Algorithm 7.3.1.1, the

cut-off frequency of the required Butterworth filter is given by

ωc = ωp

(Gp)1/2N =

(0.2588)1/8 = 59.2038 radians/s.

Using Step 4 of Algorithm 7.3.1.1, the transfer function H (s) of the required Butterworth filter is obtained by the following transformation:

H (s) = H (S)|S=s/ωc = 1

(S2 + 0.7654S + 1)(S2 + 1.8478S + 1)

∣ ∣ ∣ ∣

S=s/59.2038 ,

which simplifies to

H (s) = (3.5051 × 103)2

(s2 + 45.3146s + 3.5051 × 103)(s2 + 109.396s + 3.5051 × 103) or

H (s) = 1.2286 × 107

s4 + 154.7106 s3 + 1.1976 × 104s2 + 5.4228 × 105s + 1.2286 × 107 .

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0 50 100 150 200 250 0

0.1778

0.4

0.6

0.8913 1

0 50 100 150 200 250 0

0.1778

0.4

0.6

0.8913 1

(a) (b)

Fig. 7.8. Magnitude spectra of

the Butterworth lowpass filters,

designed in Example 7.6, as a

function of ω. Part (a) satisfies

the constraint at the pass-band

corner frequency, while part (b)

satisfies the magnitude

constraint at the stop-band

corner frequency.

Step 5 plots the magnitude spectrum of the Butterworth filter. The CTFT transfer function of the Butterworth filter is given by

H (ω) = H (s)|s=jω

= 1.2286×107

( jω)4+154.7106 ( jω)3+1.1976×104( jω)2+5.4228×105( jω)+1.2286×107 .

The magnitude spectrum |H (ω)| is plotted in Fig. 7.8(a), where the labels on the y-axis are chosen to correspond to the specified gains for the filter. We observe that the design specifications are satisfied by the magnitude spectrum.

Alternative implementation An alternative implementation of the aforemen- tioned Butterworth filter can be obtained by using the stop-band constraint,

Eq. (7.32), in Step 3 of Algorithm 7.3.1.1. The cut-off frequency of the alter-

native implementation of the Butterworth filter is given by

ωc = ωs

(Gs)1/2N =

100

(30.6327)1/4 = 65.1969 radians/s.

Using Step 4 of Algorithm 7.3.1.1, the transfer function H (s) of the alternative implementation is obtained by the following transformation:

H (s) = H (S)|S=s/ωc = 1

(S2 + 0.7654S + 1)(S2 + 1.8478S + 1)

∣ ∣ ∣ ∣

S=s/65.1969 ,

which simplifies to

H (s) = (4.2506 × 103)2

(s2 + 49.9017s + 4.2506 × 103)(s2 + 120.4708s + 4.2506 × 103) or

H (s) = 1.8068 × 107

s4 + 170.3725 s3 + 1.4513 × 104s2 + 7.2419 × 105s + 1.8068 × 107 .

Step 5 plots the magnitude spectrum of the alternative implementation of the

Butterworth filter in Fig. 7.8(b), which satisfies the initial design specifications.

7.3.1.2 Butterworth filter design using M ATL AB

M A T L A B incorporates a number of functions to implement the design algo-

rithm for the Butterworth filter specified in Section 7.3.1.1. The order N and the

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cut-off wc frequency for the filter in Step 1 of Algorithm 7.3.1.1 can be deter-

mined using the library function buttord, which has the following calling

syntax:

>> [N,wc] = buttord(wp,ws,Rp,Rs,‘s’);

where wp is the corner frequency of the pass band, ws is the corner frequency

of the stop band, Rp is the permissible ripple in the pass band in decibels,

and Rs is the permissible attenuation in the stop band in decibels. The last

argument ‘s’ specifies that a CT filter in the Laplace domain is to be designed.

In determining the cut-off frequency, M A T L A B uses the stop-band constraint,

Eq. (7.32).

Having determined the order and the cut-off frequency, the coefficients

of the numerator and denominator polynomials of the Butterworth filter can

be determined using the library function butter with the following calling

syntax:

>> [num,den] = butter(N,wc,‘s’);

where num is a vector containing the coefficients of the numerator and den is

a vector containing the coefficients of the denominator in decreasing powers

of s. Finally, the transfer function H (s) can be determined using the library func-

tion tf as follows:

>> H = tf(num,den).

For Example 7.5, the M A T L A B commands for designing the Butterworth filter

are given by

>> wp=5; ws=20; Rp=1.9382; Rs=13.9794; % specify design parameters

% Rp = -20*log10(0.8) % = 1.9382dB

% Rs = -20*log10(0.2) % = 13.9794dB

>> [N,wc]=buttord (wp,ws,Rp,Rs,‘s’); % determine order and

% cut-off freq

>> [num,den]=butter (N,wc,‘s’); % determine num and denom

% coeff.

>> Ht = tf(num,den); % determine transfer % function

>> [H,w] = freqs(num,den); % determine magnitude % spectrum

>> plot(w,abs(H)); % plot magnitude spectrum

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Stepwise implementation of the above code returns the following values for

different variables:

Instruction II: N = 2; wc = 9.0360; Instruction III: num = [0 0 81.6497]; den = [1.0000

12.7789 81.6497];

Instruction IV: Ht = 1/(s 2̂ + 12.78s + 81.65);

The magnitude spectrum is the same as that given in Fig. 7.7(b).

7.3.2 Type I Chebyshev filters

Butterworth filters have a relatively low roll off in the transitional band, which

leads to a large transitional bandwidth. Type I Chebyshev filters reduce the

bandwidth of the transitional band by using an approximating function, referred

to as the Type I Chebyshev polynomial, with a magnitude response that has

ripples within the pass band. We start with the definition of the Chebyshev

polynomial.

7.3.2.1 Type I Chebyshev polynomial

The N th-order Type I Chebyshev polynomial is defined as

TN (ω) = {

cos(N cos−1(ω)) |ω| ≤ 1 cosh(N cosh−1(ω)) |ω| > 1, (7.33)

where cosh(x) denotes the hyperbolic cosine function, which is given by

cosh(x) = cos( jx) = ex + e−x

2 . (7.34)

Starting from the initial values of T0(ω) = 1 and T1(ω) = ω, the higher orders of the Type I Chebyshev polynomial can be recursively generated using the

following expression:

Tn(ω) = 2ωTn−1(ω) − Tn−2(ω). (7.35)

Table 7.3 lists the Chebyshev polynomial for different values of n within the range 0 ≤ n ≤ 10.

Using Eq. (7.33), the roots of the Type I Chebyshev polynomial TN (ω) can be derived as follows:

ωn = cos [

(2n + 1)π 2N

]

, (7.36)

for 0 ≤ n ≤ N − 1.

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Table 7.3. Chebyshev polynomial T N (ω) for different

values of N

N TN (ω)

0 1

1 ω

2 2ω2 − 1 3 4ω3 − 3ω 4 8ω4 − 8ω2 + 1 5 16ω5 − 20ω3 + 5ω 6 32ω6 − 48ω4 + 18ω2 − 1 7 64ω7 − 112ω5 + 56ω3 − 7ω 8 128ω8 − 256ω6 + 160ω4 − 32ω2 + 1 9 256ω9 − 576ω7 + 432ω5 − 120ω3 + 9ω

10 512ω10 − 1280ω8 + 1120ω6 − 400ω4 + 50ω2 − 1

7.3.2.2 Type I Chebyshev filter

The frequency characteristics of the Type I Chebyshev filter of order N are defined as follows:

|H (ω)| = 1

√

1 + ε2T 2N (ω/ωp) , (7.37)

where ωp is the pass-band corner frequency and ε is the ripple control parameter

that adjusts the magnitude of the ripple within the pass band. Substituting

ωp = 1, the frequency characteristics of the normalized Type I Chebyshev filter of order N are expressed in terms of the Chebyshev polynomial as follows:

|H (ω)| = 1

√

1 + ε2T 2N (ω) . (7.38)

Based on Eqs. (7.35) and (7.38), we make the following observations for the

frequency characteristics of the normalized Type I Chebyshev filter.

(1) For ω = 0, the Chebyshev polynomial TN (ω) has a value of ±1 or 0. This can be shown by substituting ω = 0 in Eq. (7.33), which yields

TN (0) = cos(N cos−1(0)) = cos (

N (2n + 1)π 2

)

= {

±1 N is even 0 N is odd.

(7.39)

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Equation (7.37) implies that the dc component |H (0)| of the Type I Chebyshev filter is given by

|H (0)| =







1 √

1 + ε2 N is even

1 N is odd. (7.40)

(2) For ω = 1 radian/s, the value of the Chebyshev polynomial TN (ω) is given by

TN (1) = cos(N cos−1(1)) = cos(2nNπ ) = 1. (7.41)

Therefore, the magnitude |H (ω)| of the normalized Type I Chebyshev filter at ω = 1 radian/s is given by

|H (1)| = 1

√ 1 + ε2

, (7.42)

irrespective of the order N of the normalized Chebyshev filter. (3) For large values of ω within the stop band, the magnitude response of the

normalized Type I Chebyshev filter can be approximated by

|H (ω)| ≈ 1

εTN (ω) , (7.43)

since εTN (ω) ≫ 1. If N ≫ 1, then a second approximation can be made by ignoring the lower degree terms in TN (ω) and using the approximation TN (ω) ≈ 2N−1ωN . Equation (7.43) is therefore simplified as follows:

|H (ω)| ≈ 1

ε ×

2N−1ωN . (7.44)

(4) Since

H (s)H (−s)|s=jω = |H (ω)|2,

H (s)H (−s) can be derived from Eq. (7.38) as follows:

H (s)H (−s) = 1

1 + ε2T 2N (s/j) . (7.45)

The 2N poles of H (s)H (−s) are obtained by solving the characteristic equation,

1 + ε2T 2N (s/j) = 0, (7.46)

and are given by

sn = sin (

2n − 1 2N

)

sinh

( 1

N sinh−1

( 1

))

+ j cos (

2n − 1 2N

)

cosh

( 1

N sinh−1

( 1

))

(7.47)

for 1 ≤ n ≤ 2N−1. To derive a stable implementation of the normalized Type I Chebyshev filter, the N poles in the left-hand s-plane are included

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in the Laplace transfer function of H (s). From Eq. (7.45), it is clear that there are no zeros for the normalized Type I Chebyshev filter.

Properties (1)–(4) are used to derive the design algorithm for the Type I Cheby-

shev filter, which is explained in the following.

7.3.2.3 Design steps for the lowpass filter

In this section, we will design a lowpass Type I Chebyshev filter based on the

following specifications:

pass band (0 ≤ |ω| ≤ ωp radians/s) 1 − δp ≤ |H (ω)| ≤ 1 + δp; stop band (|ω| > ωs radians/s) |H (ω)| ≤ δs.

Since the Type I Chebyshev filter is designed in terms of its normalized version,

Eq. (7.37), we normalize the aforementioned specifications by the pass-band

corner frequency ωp. The normalized specifications are as follows:

pass band (0 ≤ |ω| ≤ 1) 1 − δp ≤ |H (ω)| ≤ 1 + δp; stop band (|ω| > ωs/ωp) |H (ω)| ≤ δs.

Step 1 Determine the value of the ripple control factor ε. Equation (7.42) computes the value of the ripple control factor ε:

ε = √

Gp with Gp = 1

(1 − δp)2 − 1. (7.48)

Step 2 Calculate the order N of the Chebyshev polynomial. The gain at the normalized stop-band corner frequency ωs/ωp is obtained from Eq. (7.37) as

|H (ωs/ωp)|2 = 1

1 + ε2T 2N (ωs/ωp) = (δs)2. (7.49)

Substituting the value of the Chebyshev polynomial TN (ω) from Eq. (7.33) and simplifying the resulting equation, we obtain

N = cosh−1[(Gs/Gp)0.5]

cosh−1[ωs/ωp] , (7.50)

where the gain terms Gp and Gs are given by

Gp = 1

(1 − δp)2 − 1 with Gs =

(δs)2 − 1. (7.51)

Step 3 Determine the location of the 2N poles of H (S)H (−S) using Eq. (7.47). To derive a stable implementation for the normalized Type I Chebyshev filter

H (S), the N poles lying in the left-half s-plane are selected to derive the transfer function H (S). If required, a constant gain term K is also multiplied with H (S)

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342 Part II Continuous-time signals and systems

such that the gain |H (0)| of the normalized Type I Chebyshev filter is unity at ω = 0.

Step 4 Derive the transfer function H (s) of the required lowpass filter from the transfer function H (S) of the normalized Type I Chebyshev filter, obtained in Step 3, using the following transformation:

H (s) = H (S)|S=s/ωp . (7.52)

Step 5 Sketch the magnitude spectrum from the transfer function H (s) deter- mined in Step 4. Confirm that the transfer function satisfies the initial design

specifications.

Example 7.7

Repeat Example 7.6 using the Type I Chebyshev filter.

Solution

For the given specifications, Example 7.6 calculates the pass-band and stop-

band gain on a linear scale as (1 − δp) = 0.8913 and δs = 10−15/20 = 0.1778 with the gain terms given by Gp = 0.2588 and Gs = 30.6327.

Step 1 determines the value of the ripple control factor ε:

ε = √

Gp = √

0.2588 = 0.5087.

Step 2 determines the order N of the Chebyshev polynomial:

N = cosh−1[(30.6327/0.2588)0.5]

cosh−1 [100/50] = 2.3371.

We round off N to the closest higher integer, N = 3. Step 3 determines the location of the six poles of H (S)H (−S):

[−0.2471 + j0.9660, −0.2471 − j0.9660, 0.2471 + j0.9660, 0.2471 − j0.9660, 0.4943, −0.4943].

The three poles lying in the left-half s-plane are included in the transfer

function H (S) of the normalized Type I Chebyshev filter. These poles are located at

[−0.2471 + j0.9660, −0.2471 − j0.9660, −0.4943] .

The transfer function for the normalized Type I Chebyshev filter is therefore

given by

H (S) = K

(S + 0.2472 + j0.9660)(S + 0.2472 − j0.9660)(S + 0.4943) ,

which simplifies to

H (S) = K

S3 + 0.9885S2 + 1.2386S + 0.4914 .

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0 50 100 150 200 250 0

0.1778

0.4

0.6

0.8913 1Fig. 7.9. Magnitude spectrum of

the Type I Chebyshev lowpass

filter designed in Example 7.7.

Since |H (ω)| at ω = 0 is K /0.4914, K is set to 0.4914 to make the dc gain equal to unity. The new transfer function with unity gain at ω = 0 is given by

H (S) = 0.4914

S3 + 0.9885S2 + 1.2386S + 0.4914 .

Step 4 transforms the normalized Type I Chebyshev filter using the following

relationship:

H (s) = H (S)|S=s/50 = 0.4914

(s/50)3 + 0.9885(s/50)2 + 1.2386(s/50) + 0.4914 or

H (s) = 6.1425 × 104

s3 + 49.425s2 + 3.0965 × 103s + 6.1425 × 104 ,

which is the transfer function of the required lowpass filter.

The magnitude spectrum of the Type I Chebyshev filter is plotted in Fig. 7.9.

It is observed that Fig. 7.9 satisfies the initial design specifications.

Examples 7.6 and 7.7 used the Butterworth and Type I Chebyshev implemen-

tations to design a lowpass filter based on the same specifications. Comparing

the magnitude spectra (Figs. 7.8 and 7.9) for the resulting filters, we note that

the Butterworth filter has a monotonic gain with negligible ripples in the pass

and stop bands. By introducing pass-band ripples, the Type I Chebyshev imple-

mentation is able to satisfy the design specifications with a lower order N for the lowpass filter, thus reducing the complexity of the filter. However, savings

in the complexity are achieved at the expense of ripples, which are added to the

the pass band of the frequency characteristics of the Type I Chebyshev filter.

7.3.2.4 Type I Chebyshev filter design using M ATL AB

M A T L A B uses the cheb1ord and cheby1 functions to implement the

Type I Chebyshev filter. The cheb1ord function determines the order N of

the Type I Chebyshev filter from the pass-band corner frequency wp, stop-band

corner frequency ws, pass-band attenuation rp, and the stop-band attenuation

rs. In terms of the filter specifications, Eqs. (7.23) and (7.24), the values of the

pass-band attenuation rp and the stop-band attenuation rs are given by

rp = 20 × log10(δp) and rs = 20 × log10(δs).

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The cheb1ord also returns wn, another design parameter referred to as the

Chebyshev natural frequency to use with cheby1 to achieve the design spec-

ifications. The syntax for cheb1ord is given by

>> [N,wn] = cheb1ord(wp,ws,rp,rs,‘s’);

To determine the coefficients of the numerator and denominator of the

Type I Chebyshev filter, M A T L A B uses thecheb1 function with the following

syntax:

>> [num,den] = cheby1(N,rp,wn,‘s’);

The transfer function H (s) can be determined using the library function tf as follows:

>> H = tf(num,den);

For Example 7.7, the M A T L A B commands for designing the Butterworth filter

are given by

>> wp=50; ws=100; rp=1; rs=15; % specify design parameters

>> [N,wn] = cheb1ord (wp,ws,rp,rs,‘s’); % determine order and

% natural freq

>> [num,den] = cheby1 (N,rp,wn,‘s’); % determine num and denom

% coeff.

>> Ht = tf(num,den); % determine transfer % function

>> [H,w] = freqs(num,den); % determine magnitude % spectrum

>> plot(w,abs(H)); % plot magnitude spectrum

Stepwise implementation of the above code returns the following values for

different variables:

Instruction II: N = 3; wn = 50; Instruction III: num = [0 0 0 61413.3]; den =

[1.0000 49.417 3096 61413.3];

Instruction IV: Ht = 61413.3/ (s 3̂ + 49.417s 2̂ + 3096s + 61413.3);

The magnitude spectrum is the same as that given in Fig. 7.9.

7.3.3 Type II Chebyshev filters

The Type II Chebyshev filters, or the inverse Chebyshev filters, are monotonic

within the pass band and introduce ripples in the stop band. Such an imple-

mentation is preferred over the Type I Chebyshev filter in applications where a

constant gain is desired within the pass band.

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The frequency characteristics of the Type II Chebyshev filter are given by

|H (ω)| = 1

√

1 + [

ε2T 2N (ωs/ω) ]−1

√

ε2T 2N (ωs/ω)

1 + ε2T 2N (ωs/ω) , (7.53)

whereωs is the lower corner frequency of the stop band. To derive the normalized

version of the Type II Chebyshev filter, we set ωs = 1 in Eq. (7.53) leading to the following expression for the frequency characteristics of the normalized

Type II Chebyshev filter:

|H (ω)| = 1

√

1 + [

ε2T 2N (1/ω) ]−1

√

ε2T 2N (1/ω)

1 + ε2T 2N (1/ω) . (7.54)

In the following section, we list the steps involved in the design of the Type II

Chebyshev filter.

7.3.3.1 Design steps for the lowpass filter

The design of the lowpass Type II Chebyshev filter is based on the following

specifications:

pass band (0 ≤ |ω| ≤ ωp radians/s) 1 − δp ≤ |H (ω)| ≤ 1 + δp; stop band (|ω| > ωs radians/s) |H (ω)| ≤ δs.

Normalizing the specifications with the stop-band corner frequency ωs, we

obtain

pass band (0 ≤ |ω| ≤ ωp/ωs) 1 − δp ≤ |H (ω)| ≤ 1 + δp; stop band (|ω| > 1) |H (ω)| ≤ δs.

Step 1 Compute the value of the ripple factor by setting the normalized fre- quency ω = 1 in Eq. (7.54). Since the Type II Chebyshev filter is normalized with respect to ωs, the normalized frequency ω = 1 corresponds to ωs and the filter gain H (1) = δs. Substituting H (1) = δs in Eq. (7.54), we obtain

|H (1)| =

√

ε2

1 + ε2 = δs,

which simplifies to

ε = 1

√ Gs

, (7.55)

with the gain term specified in Eq. (7.51).

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Step 2 Compute the order N of the Type II Chebyshev filter. To derive an expression for the order N , we compute the gain |H (ω)| at the normalized pass- band corner frequency ωp/ωs. Substituting |H (ω)| = (1 − δp) at ω = ωp/ωs, we obtain

ε2T 2N (ωs/ωp)

1 + ε2T 2N (ωs/ωp) = (1 − δp)2.

Substituting the value of the Chebyshev polynomial from Eq. (7.33) and sim-

plifying the resulting expression with respect to N yields

N = cosh−1[(Gs/Gp)0.5]

cosh−1[ωs/ωp] , (7.56)

where the gain terms Gp and Gs are defined in Eq. (7.51). Note that the expres- sion for the order of the filter for the Type II Chebyshev filter is the same as the

corresponding expression, Eq. (7.50), for the Type I Chebyshev filter.

Step 3 Determine the location of the poles and zeros of the transfer function H (S) of the normalized Type II Chebyshev filter. Substituting

H (s)H (−s)|s=jω = |H (ω)|2,

the Laplace transfer function for the normalized Type II Chebyshev filter is

given by

H (s)H (−s) = ε2T 2N ( j/s)

1 + ε2T 2N ( j/s) . (7.57)

The poles of H (s)H (−s) are obtained by solving for the roots of the character- istic equation,

1 + ε2T 2N ( j/s) = 0. (7.58)

Comparing with the characteristic equation for H (s)H (−s) of the Type I Cheby- shev filter, Eq. (7.46), we note that (s/j) in the Chebyshev polynomial of Eq. (7.46) is replaced by (j/s) in Eq. (7.58). This implies that the poles of the normalized Type II Chebyshev filter are simply the inverse of the poles of

the Type I Chebyshev filter. Hence, the location of the poles for the normalized

Type II Chebyshev filter can be computed by determining the locations of the

poles for the normalized Type I Chebyshev filter and then taking the inverse.

The zeros of H (s)H (−s) are obtaining by solving

T 2N ( j/s) = 0. (7.59)

The zeros of H (s)H (−s) are therefore the inverse of the roots of the Chebyshev polynomial TN (ω) = TN (s/j), which are given by

ω = cos [

(2n + 1)π 2N

]

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The zeros of H (s) are therefore given by

s = j

cos

( 2n + 1π

) (7.60)

for 0 ≤ n ≤ N− 1. The poles and zeros are used to evaluate the transfer function H (S) for the normalized Type II Chebyshev filter. If required, a constant gain term K is also multiplied by H (S) such that the gain |H (0)| of the normalized Type II Chebyshev filter is unity at ω = 0.

Step 4 Derive the transfer function H (s) of the required lowpass filter from the transfer function H (S) of the normalized Type II Chebyshev filter, obtained in Step 3, using the following transformation:

H (s) = H (S)|S=s/ωs . (7.61)

Step 5 Sketch the magnitude spectrum from the transfer function H (s) deter- mined in Step 4. Confirm that the transfer function satisfies the initial design

specifications.

Example 7.8

Repeat Example 7.6 using the Type II Chebyshev filter.

Solution

As calculated in Example 7.6, the pass-band and stop-band gain are (1 −δp) = 0.8913 and δs = 10−15/20 = 0.1778. The gain terms are also calculated as Gp = 0.2588 and Gs = 30.6327.

Step 1 determines the value of the ripple control factor ε:

ε = 1

√ Gs

= 1

√ 30.6327

= 0.1807.

Step 2 determines the order N of the Chebyshev polynomial:

N = cosh−1[(30.6327/0.2588)0.5]

cosh−1[100/50] = 2.3371.

We round off N to the closest higher integer, N = 3. Step 3 determines the location of the poles and zeros of H (S)H (−S). We

first determine the location of poles for the Type I Chebyshev filter with ε = 0.1807 and N = 3. Using Eq. (7.47), the location of poles for H (s)H (−s) of the Type I Chebyshev filter is given by

[−0.4468 + j1.1614, −0.4468 − j1.1614, 0.4468 + j1.1614, 0.4468 −j1.1614, 0.8935, −0.8935].

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Selecting the poles located in the left-half s-plane, we obtain

[−0.4468 + j1.1614, −0.4468 + j1.1614, −0.8935] .

The poles of the normalized Type II Chebyshev filter are located at the inverse

of the above locations and are given by

[−0.2885 − j0.7501, −0.2885 + j0.7501, −1.1192] .

The zeros of the normalized Chebyshev Type II filter are computed using

Eq. (7.60) and are given by

[−j1.1547, +j1.1547, ∞] .

The zero at s = ∞ is neglected. The transfer function for the normalized Type II Chebyshev filter is given by

H (S) = K (S + j1.1547)(S − j1.1547)

(S + 0.2885 + j0.7501)(S + 0.2885 − j0.7501)(S + 1.1192) ,

which simplifies to

H (S) = K (S2 + 1.3333)

S3 + 1.6962S2 + 1.2917S + 0.7229 .

Since |H (ω)| at ω = 0 is 1.3333/0.7229 = 1.8444, K is set to 1/1.8444 = 0.5422 to make the dc gain equal to unity. The new transfer function with unity

gain at ω = 0 is given by

H (S) = 0.5422(S2 + 1.3333)

S3 + 1.6962S2 + 1.2917S + 0.7229 .

Step 4 normalizes H (S) based on the following transformation:

H (s) = H (S)|S=s/100 = 0.5422((s/100)2 + 1.3333)

(s/100)3 + 1.6962(s/100)2 + 1.2917(s/100) + 0.7229 ,

which simplifies to

H (s) = 54.22(s2 + 1.3333 × 104)

s3 + 1.6962 × 102s2 + 1.2917 × 104s + 0.7229 × 106 .

Step 5 plots the magnitude spectrum, which is shown in Fig. 7.10. As expected,

the frequency characteristics in Fig. 7.10 have a monotonic gain within the

pass band and ripples within the stop band. Also, it is noted that the magnitude

spectrum |H (ω)|= 0 between the frequencies of ω = 100 and ω = 150 radians/s. This zero value corresponds to the location of the complex zeros in H (s). Setting

0 50 100 150 200 250 0

0.1778

0.4

0.6

0.8913 1

Fig. 7.10. Magnitude spectrum

of the Type II Chebyshev lowpass

filter designed in Example 7.8.

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349 7 Continuous-time filters

the numerator of H (s) equal to zero, we get two zeros at s = ±j115.4686, which lead to a zero magnitude at a frequency of ω = 115.4686.

7.3.3.2 Type II Chebyshev filter design using M ATL AB

M A T L A B provides the cheb2ord and cheby2 functions to implement the

Type II Chebyshev filter. The usage of these functions is the same as the

cheb1ord and cheby1 functions for the Type I Chebyshev filter except for

the cheby2 function, for which the stop-band constraints (stop-band ripple rs

and stop-band corner frequency ws) are specified. The code for Example 7.8 is

as follows:

>> wp=50; ws=100; rp=1; rs=15; % specify design parameters

>> [N,wn] = cheb2ord (wp,ws,rp,rs,‘s’); % determine order and

% natural freq

>> [num,den] = cheby2(N,rs,ws,‘s’); % determine num and denom

% coeff.

>> Ht = tf(num,den); % determine transfer % function

>> [H,w] = freqs(num,den); % determine magnitude % spectrum

>> plot(w,abs(H)); % plot magnitude spectrum

Stepwise implementation of the above code returns the following values for

different variables:

Instruction II: N = 3; wn = 78.6980; Instruction III: num = [0 54.212 0 722835];

den = [1.0000 169.63 12917 722835]; Instruction IV: Ht = (54.21sˆ2 + 722800) /(sˆ3 + 169.6sˆ2

+ 12920s + 722800);

The magnitude spectrum is the same as that given in Fig. 7.10.

7.3.4 Elliptic filters

Elliptic filters, also referred to as Cauer filters, include both pass-band and stop-

band ripples. Consequently, elliptic filters can achieve a very narrow bandwidth

for the transition band. The frequency characteristics of the elliptic filter are

given by

|H (ω)| = 1

√

1 + ε2U 2N (ω/ωp) , (7.62)

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where UN (ω) is an N th-order Jacobian elliptic function. By setting ωp = 1, we obtain the frequency characteristics of the normalized elliptic filter as follows:

|H (ω)| = 1

√

1 + ε2U 2N (ω) . (7.63)

The design procedure for elliptic filters is similar to that for Type I and

Type II Chebyshev filters. Since UN (1) = 1, for all N , it is straightforward to derive the value of the ripple control factor as

ε = √

Gp, (7.64)

where Gp is the pass-band gain term defined in Eq. (7.51). The order N of the elliptic filter is calculated using the following expression:

N = ψ[(ωp/ωs)

2]ψ⌊ √

1 − Gp/Gs⌋ ψ[Gp/Gs]ψ[

√

1 − (ωp/ωs)2] , (7.65)

where ψ[x] is referred to as the complete elliptic integral of the first kind and is given by

ψ[x] = π/2∫

dφ √

1 − x2 sin φ . (7.66)

M A T L A B provides the ellipke function to compute Eq. (7.66) such that

ψ[x] = ellipke(xˆ2). Finding the transfer function H (s) for the elliptic filters of order N and

ripple control factor ε requires the computation of its poles and zeros from

non-linear simultaneous integral equations, which is beyond the scope of the

text. In Section 7.3.4.1, which follows Example 7.9, we provide a list of library

functions in M A T L A B that may be used to design the elliptic filters.

Example 7.9

Calculate the ripple control factor and order of the elliptic filter that satisfies

the filter specifications listed in Example 7.6.

Solution

Example 7.6 computes the gain terms as Gp = 0.2588 and Gs = 30.6327. The pass-band and stop-band corner frequencies are specified as ωp = 50 radians/s and ωs = 100 radians/s. Using Eq. (7.65), the ripple control factor is given by

ε = √

Gp = √

0.2588 = 0.5087.

Using Eq. (7.65) with ωp/ωs = 0.5 and Gp/Gs = 0.0085, the order N of the elliptic filter is given by

N = ψ[(ωp/ωs)

2] ψ⌊ √

1 − Gp/Gs⌋ ψ[Gp/Gs] ψ[

√

1 − (ωp/ωs)2] =

ψ[0.25] ψ[0.9958]

ψ[0.0085] ψ[0.8660] .

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Table 7.4. Comparison of the different implementations of a lowpass filter

Type of filter Order Pass band Transition band Stop band

Butterworth highest order (4) monotonic gain widest width monotonic gain

either pass or stop bands specs are met; the other is overdesigned

Type I Chebyshev moderate order (3) ripples are present;

exact specs are met

narrow width monotonic gain;

overdesigned specs

Type II Chebyshev moderate order (3);

same as Type I

montonic gain;

overdesigned specs

narrow width; similar

to Type I

ripples are present;

exact specs are met

Elliptic lowest order (2) ripples are present;

exact specs are met

narrowest width ripples are present;

exact specs are met

Using M A T L A B , ψ[0.25] = ellipke(0.25ˆ2)= 1.5962, ψ[0.9958] = ellipke(0.9968ˆ2)= 3.9175, ψ[0.0085] = ellipke(0.0085ˆ2) = 1.5708, and ψ[0.8660] = ellipke(0.8660ˆ2) = 2.1564. The value of N is given by

N = 1.5962 × 3.9715 1.5708 × 2.1564

= 1.8715.

Rounding off to the nearest higher integer, the order N of the filter equals 2.

Examples 7.6 to 7.9 designed a lowpass filter for the same specifications based

on four different implementations derived from the Butterworth, Type I Cheby-

shev, Type II Chebyshev, and elliptic filters. Table 7.4 compares the properties

of these four implementations with respect to the frequency responses within

the pass, transition, and stop bands.

In terms of the complexity of the implementations, the elliptic filters provide

the lowest order at the expense of equiripple gains in both the pass and stop

bands. The Chebyshev filters provide monotonic gain in either the pass or stop

band, but increase the order of the implementation. The Butterworth filters

provide monotonic gains of maximally flat nature in both the pass and stop

bands. However, the Butterworth filters are of the highest order and have the

widest transition bandwidth.

Another factor considered in choice of implementation is the phase response

of the filter. Generally, ripples add non-linearity to the phase responses. There-

fore, the elliptic filter may not be the best choice in applications where a linear

phase is important.

7.3.4.1 Elliptic filter design using M ATL AB

M A T L A B provides the ellipord and ellip functions to implement the

elliptic filters. The usage of these functions is similar to the cheb1ord and

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0 50 100 150 200 250 0

0.1778

0.4

0.6

0.8913 1Fig. 7.11. Magnitude spectrum

of the elliptic lowpass filter

designed in Example 7.9.

cheby1 functions used to design Type I Chebyshev filters. The code to imple-

ment an elliptic filter for Example 7.9 is as follows:

>> wp=50; ws=100; rp=1; rs=15; % specify design parameters

>> [N,wn] = ellipord (wp,ws,rp,rs,‘s’); % determine order and

% natural freq

>> [num,den] = ellip(N,rp,rs,wn,‘s’); % determine num and denom

% coeff.

>> Ht = tf(num,den); % determine transfer % function

>> [H,w] = freqs(num,den); % determine magnitude % spectrum

>> plot(w,abs(H)); % plot magnitude spectrum

Stepwise implementation of the above code returns the following values for

different variables:

Instruction II: N = 2; wn = 50; Instruction III: num = [0.1778 0 2369.66];

den = [1.0000 48.384 2961.75]; Instruction IV: Ht = (0.1778sˆ2 + 2640)/(sˆ2 + 48.38s

+ 2962);

The magnitude spectrum is plotted in Fig. 7.11.

7.4 Frequency transformations

In Section 7.3, we designed a collection of specialized CT lowpass filters. In

this section, we consider the design techniques for the remaining three cat-

egories (highpass, bandpass, and bandstop filters) of CT filters. A common

approach for designing CT filters is to convert the desired specifications into

the specifications of a normalized or prototype lowpass filter using a frequency

transformation that maps the required frequency-selective filter into a lowpass

filter. Based on the transformed specifications, a normalized lowpass filter is

designed using the techniques covered in Section 7.3. The transfer function

H (S) of the normalized lowpass filter is then transformed back into the original frequency domain. Transformation for converting a lowpass filter to a highpass

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353 7 Continuous-time filters

filter is considered next, followed by the lowpass to bandpass, and lowpass to

bandstop transformations.

7.4.1 Lowpass to highpass filter

The transformation that converts a lowpass filter with the transfer function H (S) into a highpass filter with transfer function H (s) is given by

S = ξp

s , (7.67)

where S = σ + jω represents the lowpass domain and s = γ + jξ represents the highpass domain. The frequency ξ = ξp represents the pass-band corner frequency for the highpass filter. In terms of the CTFT domain, Eq. (7.67) can

be expressed as follows:

ω = − ξp

ξ or ξ = −

ξp

ω . (7.68)

Figure 7.12 shows the effect of applying the frequency transformation in

Eq. (7.68) to the specifications of a highpass filter. Equation (7.68) maps the

highpass specifications in the range −∞ < ξ ≤ 0 to the specifications of a lowpass filter in the range 0 ≤ ω < ∞. Similarly, the highpass specifications for the positive range of frequencies (0 < ξ ≤ ∞) are mapped to the lowpass specifications within the range −∞ ≤ ω < 0. Since the magnitude spectra are symmetrical about the y-axis, the change from positive ξ frequencies to negative ω frequencies does not affect the nature of the filter in the entire domain.

Highpass to lowpass transformation

w = −xp/x

pass band

stop band

transition band

Hhp(x)

−xp −xs

1−dp

1+d p

pass band stop bandtransition

band

x 0

w w

|Hlp(w)|

Fig. 7.12. Highpass to lowpass

transformation.

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From Fig. 7.12, it is clear that Eq. (7.68), or alternatively Eq. (7.67), represents

a highpass to lowpass transformation. We now exploit this transformation to

design a highpass filter.

Example 7.10

Design a highpass Butterworth filter with the following specifications:

stop band (0 ≤ |ξ | ≤ 50 radians/s) −1 dB ≤ 20 log10|H (ξ )| ≤ 0; pass band (|ξ | > 100 radians/s) 20 log10|H (ξ )| ≤ −15 dB.

Solution

Using Eq. (7.67) with ξp = 100 radians/s to transform the specifications from the domain s = γ + jξ of the highpass filter to the domain S = σ + jω of the lowpass filter, we obtain

pass band (2 < |ω| ≤ ∞ radians/s) −1 dB ≤ 20 log10|H (ω)| ≤ 0; stop band (|ω| < 1 radian/s) 20 log10|H (ω)| ≤ 15 dB.

The above specifications are used to design a normalized lowpass Butterworth

filter. Expressed on a linear scale, the pass-band and stop-band gains are given

(1 − δp) = 10−1/20 = 0.8913 and δs = 10−15/20 = 0.1778.

The gain terms Gp and Gs are given by

Gp = 1

(1 − δp)2 − 1 =

0.89132 − 1 = 0.2588

and

Gs = 1

(δs)2 − 1 =

0.17782 − 1 = 30.6327.

The order N of the Butterworth filter is obtained using Eq. (7.29) as follows:

N = 1

2 ×

ln(Gp/Gs)

ln(ξp/ξs) =

2 ×

ln(0.2588/30.6327)

ln(1/2) = 3.4435.

We round off the order of the filter to the higher integer value as N = 4. Using the pass-band constraint, Eq. (7.31), the cut-off frequency of the

required Butterworth filter is given by

ωc = ωs

(Gs)1/2N =

(30.6327)1/8 = 1.3039 radians/s.

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0 50 100 150 200 250 0

0.1778

0.4

0.6

0.8913 1Fig. 7.13. Magnitude spectrum

of the Butterworth highpass

filter designed in Example 7.10.

The poles of the lowpass filter are located at

S = ωc exp [

j π

2 + j

(2n − 1)π 8

]

for 1 ≤ n ≤ 4. Substituting different values of n yields

S = [−0.4990 + j1.2047 −1.2047 + j0.4990 −1.2047 −j0.4990 −0.4990 − j1.2047].

The transfer function of the lowpass filter is given by

H (S) = K

(S+0.4490−j1.2047)(S+0.4490+j1.2047)(S+1.2047−j0.4990)(S+1.2047+j0.4990)

H (S) = K

S4 + 3.4074S3 + 5.8050S2 + 5.7934S + 2.8909 .

To ensure a dc gain of unity for the lowpass filter, we set K = 2.8909. The transfer function of a unity gain lowpass filter is given by

H (S) = 2.8909

S4 + 3.4074S3 + 5.8050S2 + 5.7934S + 2.8909 .

To derive the transfer function of the required highpass filter, we use Eq. (7.67)

with ξp = 100 radians/s. The transfer function of the highpass filter is given by

H (s) = H (S)|S=100/s

= 2.8909

(100/s)4 + 3.4074(100/s)3 + 5.8050(100/s)2 + 5.7934(100/s) + 2.8909 or

H (s)= s4

s4 + 2.004 × 102s3 + 2.008 × 104s2 + 1.179 × 106s + 3.459 × 107 .

The magnitude spectrum of the highpass filter is given in Fig. 7.13, which

confirms that the given specifications are satisfied.

7.4.1.1 M ATL AB code for designing highpass filters

The M A T L A B code for the design of the highpass filter required in

Example 7.10 using the Butterworth, Type I Chebyshev, Type II Chebyshev,

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356 Part II Continuous-time signals and systems

and elliptic implementations is included below. In each case, M A T L A B auto-

matically designs the highpass filter. No explicit transformations are needed.

>> % Matlab code for designing highpass filter

>> wp=100; ws=50; Rp=1; Rs=15; % design

% specifications

>> % for Butterworth

% filter

>> [N, wc] = buttord(wp,ws,Rp,Rs,‘s’); % determine order

% and cut off

>> [num1,den1] = butter(N,wc,‘high’,‘s’); % determine

% transfer

% function

>> H1 = tf(num1,den1); >> %%%%% % Type I Chebyshev

% filter

>> [N, wn] = cheb1ord(wp,ws,Rp,Rs,‘s’); >> [num2,den2] = cheby1(N,Rp,wn,‘high’,‘s’); >> H2 = tf(num2,den2); >> %%%%% % Type II Chebyshev

% filter

>> [N,wn] = cheb2ord(wp,ws,Rp,Rs,‘s’) ; >> [num3,den3] = cheby2(N,Rs,wn,‘high’,‘s’) ; >> H3 = tf(num3,den3); >> %%%%% % Elliptic filter

>> [N,wn] = ellipord(wp,ws,Rp,Rs,‘s’) ; >> [num4,den4] = ellip(N,Rp,Rs,wn,‘high’,‘s’) ; >> H4 = tf(num4,den4);

In the above code, note thatwp > ws. Also, an additional argument of‘high’

is included in the design statements for different filters, which is used to specify

a highpass filter. The aforementioned M A T L A B code results in the following

transfer functions for the different implementations:

Butterworth

H (s) = s4

s4 + 2.004 × 102s3 + 2.008 × 104s2 + 1.179 × 106s + 3.459 × 107 ;

Type I Chebyshev H (s) = s3

s3 + 252.1s2 + 2.012 × 104s + 2.035 × 106 ;

Type II Chebyshev H (s) = s3 + 3.027 × 103s

s3 + 113.5s2 + 9.473 × 103s + 3.548 × 105 ;

elliptic H (s) = 0.8903s2 + 1501

s2 + 81.68s + 8441 .

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The transfer function for the Butterworth filter is the same as that derived by

hand in Example 7.9.

7.4.2 Lowpass to bandpass filter

The transformation that converts a lowpass filter with the transfer function H (S) into a bandpass filter with transfer function H (s) is given by

S = s2 + ξp1ξp2 s(ξp2 − ξp1)

, (7.69)

where S = S = σ + jω represents the lowpass domain and s = γ + jξ repre- sents the bandpass domain. The frequency ξ = ξp1 and ξp2 represents the two pass-band corner frequencies for the bandpass filter with ξp2 > ξp1. In terms of

the CTFT variables ω and ξ , Eq. (7.69) can be expressed as follows:

ωs1 = ξp1ξp2 − ξ 2

ξ (ξp2 − ξp1) . (7.70)

From Eq. (7.70), it can be shown that the pass-band corner frequencies ξp1 and

−ξp2 of the bandpass filter are both mapped in the lowpass domain to ω = 1, whereas the pass-band corner frequencies −ξp1 and ξp2 are mapped to ω = −1. Also, the pass band ξp1 ≤ |ξ | = ξp2 of the bandpass filter is mapped to the pass band −1 ≤ |ξ | ≤ 1 of the lowpass filter. These results can be confirmed by substituting different values for the bandpass domain frequencies ξ and

evaluating the corresponding lowpass domain frequencies.

Considering the stop-band corner frequencies of the bandpass filter,

Eq. (7.70) can be used to show that the stop-band corner frequency ±ξs1 is mapped to

ωs1 = ∣ ∣ ∣ ∣

ξp1ξp2 − ξ 2s1 ξs1(ξp2 − ξp1)

∣ ∣ ∣ ∣ , (7.71)

and that the stop-band corner frequency ±ξs2 is mapped to

ωs2 = ∣ ∣ ∣ ∣

ξp1ξp2 − ξ 2s2 ξs2(ξp2 − ξp1)

∣ ∣ ∣ ∣ . (7.72)

As a lower value for the stop-band frequency for the lowpass filter leads to

more stringent requirements, the stop-band corner frequency for the lowpass

filter is selected from the minimum of the two values computed in Eqs. (7.71)

and (7.72). Mathematically, this implies that

ωs = min(ωs1, ωs2). (7.73)

Example 7.11 designs a bandpass filter.

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Example 7.11

Design a bandpass Butterworth filter with the following specifications:

stop band I (0 ≤ |ξ | ≤ 50 radians/s) 20 log10|H (ξ )| ≤ −20 dB; pass band (100 ≤ |ξ | ≤ 200 radians/s) −2 dB ≤ 20 log10|H (ξ )| ≤ 0; stop band II (|ξ | ≥ 380 radians/s) 20 log10|H (ξ )| ≤ −20 dB.

Solution

For ξp1 = 100 radians/s and ξp2 = 200 radians/s, Eq. (7.70) becomes

ω = 2 × 104 − ξ 2

100ξ ,

to transform the specifications from the domain s = γ + jξ of the bandpass filter to the domain S = σ + jω of the lowpass filter. The specifications for the normalized lowpass filter are given by

pass band (0 ≤ |ω| < 1 radian/s) −2 ≤ 20 log10|H (ω)| ≤ 0; stop band (|ω| ≥ min(3.2737, 3.5) radians/s 20 log10|H (ω)| ≤ −20.

The above specifications are used to design a normalized lowpass Butterworth

filter. Expressed on a linear scale, the pass-band and stop-band gains are given

(1 − δp) = 10−2/20 = 0.7943 and δs = 10−20/20 = 0.1.

The gain terms Gp and Gs are given by

Gp = 1

(1 − δp)2 − 1 =

0.79432 − 1 = 0.5850

and

Gs = 1

(δs)2 − 1 =

0.17782 − 1 = 99.

The order N of the Butterworth filter is obtained using Eq. (7.29) as follows:

N = 1

2 ×

ln(Gp/Gs)

ln(ξp/ξs) =

2 ×

ln(0.5850/99)

ln(1/3.2737) = 2.1232.

We round off the order of the filter to the higher integer value as N = 3. Using the stop-band constraint, Eq. (7.31), the cut-off frequency of the low-

pass Butterworth filter is given by

ωc = ωs

(Gs)1/2N =

3.2737

(99)1/6 = 1.5221 radians/s.

The poles of the lowpass filter are located at

S = ωc exp [

j π

2 + j

(2n − 1)π 6

]

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0 50 100 150 200 250 300 350 400 450 0

0.1

0.4

0.6 0.7943

1 Fig. 7.14. Magnitude spectrum

of the Butterworth bandpass

filter designed in Example 7.11.

for 1 ≤ n ≤ 3. Substituting different values of n yields

S = [−0.7610 + j1.3182 −0.7610 − j1.3182 −1.5221].

The transfer function of the lowpass filter is given by

H (S) = K

(S + 0.7610 + j1.3182)(S + 0.7610 + j1.3182)(S + 1.5221)

H (S) = K

S3 + 3.0442S2 + 4.6336S + 3.5264 .

To ensure a dc gain of unity for the lowpass filter, we set K = 3.5364. The transfer function of the unity gain lowpass filter is given by

H (S) = 3.5264

S3 + 3.0442S2 + 4.6336S + 3.5264 .

To derive the transfer function of the required bandpass filter, we use Eq. (7.69)

with ξp1 = 100 radians/s and ξp2 = 200 radians/s. The transformation is given by

S = s2 + 2 × 104

100 s ,

from which the transfer function of the bandpass filter is calculated as follows:

H (s) = H (S)| S= s2+2×104

100 s

= 3.5264

[ s2 + 2 × 104

100 s

+ 3.0442 [

s2 + 2 × 104

100 s

+ 4.6336 [

s2 + 2 × 104

100 s

]

+ 3.5264 ,

which reduces to

H (s) = 3.5264 × 106s3

s6 + 3.0442 × 102s5 + 1.0633 × 105s4 + 1.5703 × 107s3 + 2.1267 × 109s2 + 1.2177 × 1011s + 8 × 1012 .

The magnitude spectrum of the bandpass filter is given in Fig. 7.14, which

confirms that the given specifications for the bandpass filter are satisfied.

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7.4.2.1 M ATL AB code for designing bandpass filters

The M A T L A B code for the design of the bandpass filter required in

Example 7.11 using the Butterworth, Type I Chebyshev, Type II Chebyshev,

and elliptic implementations is as follows:

>> % MATLAB code for designing bandpass filter

>> wp=[100 200]; ws=[50 380]; Rp=2; Rs=20;

% Specifications

>> % Butterworth filter

>> [N, wc] = buttord(wp,ws,Rp,Rs,‘s’); >> [num1,den1] = butter(N,wc,‘s’); >> H1 = tf(num1,den1); >> % Type I Chebyshev filter

>> [N, wn] = cheb1ord(wp,ws,Rp,Rs,‘s’); >> [num2,den2] = cheby1(N,Rp,wn,‘s’); >> H2 = tf(num2,den2); >> % Type II Chebyshev filter

>> [N,wn] = cheb2ord(wp,ws,Rp,Rs,‘s’); >> [num3,den3] = cheby2(N,Rs,wn,‘s’); >> H3 = tf(num3,den3); >> % Elliptic filter

>> [N,wn] = ellipord(wp,ws,Rp,Rs,‘s’); >> [num4,den4] = ellip(N,Rp,Rs,wn,‘s’); >> H4 = tf(num4,den4);

The type of filter is specified by the dimensions of the pass-band and stop-band

frequency vectors. Since wp and ws are both vectors, M A T L A B knows that

either a bandpass or bandstop filter is being designed. From the range of the

values within wp and ws, M A T L A B is also able to differentiate whether a

bandpass or a bandstop filter is being specified. In the above example, since

the range (50–380 Hz) of frequencies specified within the stop-band frequency

vector ws exceeds the range (100–200 Hz) specified within the pass-band fre-

quency vector wp, M A T L A B is able to make the final determination that a

bandpass filter is being designed. For a bandstop filter, the converse would hold

true. The aforementioned M A T L A B code produces bandpass filters with the fol-

lowing transfer functions:

Butterworth H (s) = 3.526×106s4

s6+304.4s5+1.063×105s4+1.57×107s3+2.127×109s2+1.218×1011s+8×1012 ;

Type I Chebyshev H (s) = 6.538 × 103s2

s4 + 80.38s3 + 4.8231 × 104s2 + 1.608 × 106s + 4 × 108 ;

Type II Chebyshev H (s) = 0.1s4 + 1.801 × 104s2 + 4 × 107

s4 + 1.588 × 102s3 + 5.4010 × 104s2 + 3.176 × 106s + 4 × 108 ;

elliptic H (s) = 0.1s4 + 1.101 × 104s2 + 4 × 107

s4 + 74.67s3 + 4.8819 × 104s2 + 1.493 × 106s + 4 × 108 .

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Note that the transfer function for the bandpass Butterworth filter is the same

as that derived by hand in Example 7.9.

7.4.3 Lowpass to bandstop filter

The transformation to convert a lowpass filter with the transfer function H (S) into a bandstop filter with transfer function H (s) is given by the following expression:

S = s(ξp2 − ξp1) s2 + ξp1ξp2

, (7.74)

where S = σ + jω represents the lowpass domain and s = γ + jξ represents the bandstop domain. The frequency ξ = ξp1 and ξp2 represents the two pass- band corner frequencies for the bandpass filter with ξp2 > ξp1. Note that the

transformation in Eq. (7.74) is the inverse of the lowpass to bandpass transfor-

mation specified in Eq. (7.69).

In terms of the CTFT domain, Eq. (7.74) can be expressed as follows:

ω = ξ (ξp2 − ξp1) ξp1ξp2 − ξ 2

, (7.75)

which can be used to confirm that Eq. (7.74) is indeed a lowpass to bandstop

transformation.

As for the bandpass filter, Eq. (7.75) leads to two different values of the

stop-band frequencies,

ωs1 = ∣ ∣ ∣ ∣

ξs1(ξp2 − ξp1) ξp1ξp2 − ξ 2s1

∣ ∣ ∣ ∣

and ωs2 = ∣ ∣ ∣ ∣

ξs2(ξp2 − ξp1) ξp1ξp2 − ξ 2s2

∣ ∣ ∣ ∣

(7.76)

for the lowpass filter. The smaller of the two values is selected as the stop-band

corner frequency for the normalized lowpass filter. Example 7.12 designs a

bandstop filter.

Example 7.12

Design a bandstop Butterworth filter with the following specifications:

pass band I (0 ≤ |ξ | ≤ 100 radians/s) −2 dB ≤ 20 log10|H (ξ )| ≤ 0; stop band (150 ≤ |ξ | ≤ 250 radians/s) 20 log10|H (ξ )| ≤ −20 dB; pass band II (|ξ | ≥ 370 radians/s) −2 dB ≤ 20 log10|H (ξ )| ≤ 0.

Solution

For ξp1 = 100 radians/s and ξp2 = 370 radians/s, Eq. (7.70) becomes

ω = 270ξ

3.7 × 104 − ξ 2 ,

to transform the specifications from the domain s = γ + jξ of the bandstop filter to the domain S = σ + jω of the lowpass filter. The specifications for the

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normalized lowpass filter are given by

pass band (0 ≤ |ω| < 1 radian/s) − 2 ≤ 20 log10 |H (ω)| ≤ 0; stop band (|ω| ≥ min(2.7931, 2.6471) radians/s 20 log10 |H (ω)| ≤ −20.

The above specifications are used to design a normalized lowpass Butterworth

filter. Since the pass-band and stop-band gains of the transformed lowpass filter

are the same as the ones used in Example 7.11, i.e.

(1 − δp) = 10−2/20 = 0.7943 and δs = 10−20/20 = 0.1,

with gain terms

Gp = 0.5850 and Gs = 99,

the order N = 3 of the Butterworth filter is the same as in Example 7.11. Using the stop-band constraint, Eq. (7.31), the cut-off frequency of the low-

pass Butterworth filter is given by

ωc = ωs

(Gs)1/2N =

2.6471

(99)1/6 = 1.2307 radians/s.

The poles of the lowpass filter are located at

S = ωc exp [

j π

2 + j

(2n − 1)π 6

]

for 1 ≤ n ≤ 3. Substituting different values of n yields

S = [−0.6153 + j0.06581 −0.6153 − j0.06581 −1.2307].

The transfer function of the lowpass filter is given by

H (S) = K

(S + 0.6153 − j0.06581)(S + 0.6153 + j0.06581)(S + 1.2307) or

H (S) = K

S3 + 2.4614S2 + 3.0292S + 1.8640 .

To ensure a dc gain of unity for the lowpass filter, we set K = 1.8640. The transfer function of the unity gain lowpass filter is given by

H (S) = 1.8640

S3 + 2.4614S2 + 3.0292S + 1.8640 .

To derive the transfer function of the required bandstop filter, we use Eq. (7.74)

with ξp1 = 100 radians/s and ξp2 = 370 radians/s. The transformation is given by

S = 270s

s2 + 3.7 × 104 ,

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363 7 Continuous-time filters

0 50 100 150 200 250 300 350 400 450 500 0

0.1

0.4

0.6 0.7943

1 Fig. 7.15. Magnitude spectrum

of the Butterworth bandstop

filter designed in Example 7.12.

with the transfer function of the bandstop filter is given by

H (s) = H (S)|S=270s/s2+3.7×104

= 1.8640

[ 270s

s2+3.7×104

+ 2.4641 [

270s

s2+3.7×104

+ 3.0292 [

270s

s2+3.7×104

]

+1.8640 ,

which reduces to

H (s) = s6+1.11×105s4+4.107×109s2+5.065×1013

s6+4.388×102s5+2.0737×105s4+4.302×107s3+7.673×109s2+6 × 1011s+5.065×1013 .

The magnitude spectrum of the bandstop filter is included in Fig. 7.15, which

confirms that the given specifications are satisfied.

7.4.3.1 M ATL AB code for designing bandstop filters

The M A T L A B code for the design of the bandstop filter required in

Example 7.12 using the Butterworth, Type I Chebyshev, Type II Chebyshev,

and elliptic implementations is as follows:

% MATLAB code for designing bandstop filter

>> wp=[100 370]; ws=[150 250]; Rp=2; Rs=20;

% Specifications

>> % Butterworth Filter

>> [N, wn] = buttord(wp,ws,Rp,Rs,‘s’);

>> [num1,den1] = butter(N,wn, ‘stop’,‘s’);

>> H1 = tf(num1,den1);

>> % Type I Chebyshev filter

>> [N, wn] = cheb1ord(wp,ws,Rp,Rs,‘s’);

>> [num2,den2] = cheby1(N,Rp,wn, ‘stop’,‘s’);

>> H2 = tf(num2,den2);

>> % Type II Chebyshev filter

>> [N,wn] = cheb2ord(wp,ws,Rp,Rs,‘s’);

>> [num3,den3] = cheby2(N,Rs,wn, ‘stop’,‘s’);

>> H3 = tf(num3,den3);

>> % Elliptic filter

>> [N,wn] = ellipord(wp,ws,Rp,Rs,‘s’);

>> [num4,den4] = ellip(N,Rp,Rs,wn, ‘stop’,‘s’);

>> H4 = tf(num4,den4);

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364 Part II Continuous-time signals and systems

The aforementioned M A T L A B code produces the following transfer functions for the four filters:

Butterworth H (s) = s6 +1.125×105s4 +4.219×109s2 +5.273×1013

s6 +430.2s5 +2.05×105s4 +4.221×107s3 +7.688×109s2 +6.049×1011s+5.273×1013 ;

Type I Chebyshev H (s) = 0.7943s4 + 5.957 × 104s2 + 1.117 × 109

s4 + 262.4s3 + 1.627 × 105s2 + 9.839 × 106s + 1.406 × 109 ;

Type II Chebyshev H (s) = 0.7943s4 + 8.015 × 104s2 + 1.406 × 109

s4 + 304.5s3 + 1.265 × 105s2 + 1.142 × 106s + 1.406 × 109 ;

elliptic H (s) = 0.7943s4 + 6.776 × 104s2 + 1.117 × 109

s4 + 227.5s3 + 1.568 × 105s2 + 8.53 × 106s + 1.406 × 109 .

7.5 Summary

Chapter 7 defines the CT filters as LTI systems used to transform the frequency

characteristics of the CT signals in a predefined manner. Based on the magnitude

spectrum |H (ω)|, Section 7.1 classifies the frequency-selective filters into four different categories.

(1) An ideal lowpass filter removes frequency components above the cut-off

frequency ωc from the input signal, while retaining the lower frequency

components ω ≤ ωc. (2) An ideal highpass filter is the converse of the lowpass filter since it removes

frequency components below the cut-off frequency ωc from the input signal,

while retaining the higher frequency components ω ≤ ωc. (3) An ideal bandpass filter retains a selected range of frequency components

between the lower cut-off frequency ωc1 and the upper cutoff frequency

ωc2 of the filter. All other frequency components are eliminated from the

input signal.

(4) A bandstop filter is the converse of the bandpass filter, which rejects all

frequency components between the lower cut-off frequency ωc1 and the

upper cut-off frequency ωc2 of the filter. All other frequency components

are retained at the output of the bandstop filter.

The ideal frequency filters are not physically realizable. Section 7.2 introduces

practical implementations of the ideal filters obtained by introducing ripples

in the pass and stop bands. A transition band is also included to eliminate the

sharp transition between the pass and stop bands.

In Section 7.3, we considered the design of practical lowpass filters. We pre-

sented four implementations of practical filters: Butterworth, Type I Chebyshev,

Type II Chebyshev, and elliptic filters, for which the design algorithms were

covered. The Butterworth filters provide a maximally flat gain within the pass

band but have a higher-order N than the Chebyshev and elliptic filters designed with the same specifications. By introducing ripples within the pass band,

Type I Chebyshev filters reduce the required order N of the designed filter. Alternatively, Type II Chebyshev filters introduce ripples within the stop band

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365 7 Continuous-time filters

to reduce the order N of the filter. The elliptic filters allow ripples in both the pass and stop bands to derive a filter with the lowest order N among the four implementations. M A T L A B instructions to design the four implementations

are also presented in Section 7.3.

In Section 7.4, we covered three transformations for converting a highpass

filter to a lowpass filter, a lowpass to a bandpass filter, and a bandstop to a lowpass

filter. Using these transformations, we were able to map the specifications of

any type of the frequency-selective filters in terms of a normalized lowpass

filter. After designing the normalized lowpass filter using the design algorithms

covered in Section 7.3, the transfer function of the lowpass filter is transformed

back into the original domain of the frequency-selective filter.

Problems

7.1 Determine the impulse response of an ideal bandpass filter and an ideal bandstop filter. In each case, assume a gain of A within the pass bands and cut off frequencies of ωc1 and ωc2.

7.2 Derive and sketch the location of the poles for the Butterworth filters of orders N = 12 and 13 in the complex s-plane.

7.3 Show that a lowpass Butterworth filter with an odd value of order N will always have at least one pole on the real axis in the complex s-plane.

7.4 Show that all complex poles of the lowpass Butterworth filter occur in conjugate pairs.

7.5 Show that the N th -order Type I Chebyshev polynomial TN (ω) has N simple roots in the interval [−1, 1], which are given by

ωn = cos [

(2n + 1)π 2N

]

0 ≤ n ≤ N − 1.

7.6 Show that the roots of the characteristic equation

1 + ε2T 2N ( j/s) = 0

for the Type II Chebyshev filter are the inverse of the roots of the charac-

teristic equation

1 + ε2T 2N (s/j) = 0

for the Type I Chebyshev filter.

7.7 Design a Butterworth lowpass filter for the following specifications:

pass band (0 ≤ |ω| ≤ 10 radians/s) 0.9 ≤ |H (ω)| ≤ 1; stop band (|ω| > 20 radians/s) |H (ω)| ≤ 0.10,

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366 Part II Continuous-time signals and systems

by enforcing the pass-band requirements. Repeat for the stop-band

requirements. Sketch the magnitude spectrum and confirm that the mag-

nitude spectrum satisfies the design specifications.

7.8 Repeat Problem 7.7 for the following specifications:

pass band (0 ≤ |ω| ≤ 50 radians/s) −1 ≤ 20 log10|H (ω)| ≤ 0; stop band (|ω| > 65 radians/s) 20 log10|H (ω)| ≤ −25.

7.9 Repeat (a) Problem 7.7 and (b) Problem 7.8 for the Type I Chebyshev filter.

7.10 Repeat (a) Problem 7.7 and (b) Problem 7.8 for the Type II Chebyshev filter.

7.11 Determine the order of the elliptic filters for the specifications included in (a) Problem 7.7 and (b) Problem 7.8.

7.12 Using the results in Problems 7.7–7.11, compare the implementation complexity of the Butterworth, Type I Chebyshev, Type II Chebyshev,

and elliptic filters for the specifications included in (a) Problem 7.7 and

(b) Problem 7.8.

7.13 By selecting the corner frequencies of the pass and stop bands, show that the transformation

S = s(ξp2 − ξp1) s2 + ξp1ξp2

maps a normalized lowpass filter into a bandstop filter.

7.14 Design a Butterworth highpass filter for the following specifications:

stop band (0 ≤ |ω| ≤ 15 radians/s) |H (ω)| ≤ 0.15; pass band (|ω| > 30 radians/s) 0.85 ≤ |H (ω)| ≤ 1.

Sketch the magnitude spectrum and confirm that it satisfies the design

specifications.

7.15 Repeat Problem 7.14 for the Type I Chebyshev filter.

7.16 Repeat Problem 7.14 for the Type II Chebyshev filter.

7.17 Design a Butterworth bandpass filter for the following specifications:

stop band I (0 ≤ |ξ | ≤ 100 radians/s) 20 log10|H (ω)| ≤ −15; pass band (100 ≤ |ξ | ≤ 150 radians/s) −1 ≤ 20 log10|H (ω)| ≤ 0; stop band II (|ξ | ≥ 175 radians/s) 20 log10|H (ω)| ≤ −15.

Sketch the magnitude spectrum and confirm that it satisfies the design

specifications.

7.18 Repeat Problem 7.17 for the Type I Chebyshev filter.

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367 7 Continuous-time filters

7.19 Repeat Problem 7.17 for the Type II Chebyshev filter.

7.20 Design a Butterworth bandstop filter for the following specifications:

pass band I (0 ≤ |ξ | ≤ 25 radians/s) −4 ≤ 20 log10|H (ω)| ≤ 0; stop band (100 ≤ |ξ | ≤ 250 radians/s) 20 log10|H (ω)| ≤ −20; pass band II (|ξ | ≥ 325 radians/s) −4 ≤ 20 log10|H (ω)| ≤ 0.

Sketch the magnitude spectrum and confirm that it satisfies the design

specifications.

7.21 Repeat Problem 7.20 for the Type I Chebyshev filter.

7.22 Repeat Problem 7.20 for the Type II Chebyshev filter.

7.23 Determine the transfer function of the four implementations: (a) Butter- worth, (b) Type I Chebyshev, (c) Type II Chebyshev, and (d) elliptic, of

the lowpass filter specified in Problem 7.7 using M A T L A B . Plot the fre-

quency characteristics and confirm that the specifications are satisfied by

the designed implementations.

7.24 Determine the transfer function of the four implementations: (a) Butter- worth, (b) Type I Chebyshev, (c) Type II Chebyshev, and (d) elliptic, of

the lowpass filter specified in Problem 7.8 using M A T L A B . Plot the fre-

quency characteristics and confirm that the specifications are satisfied by

the designed implementations.

7.25 Determine the transfer function of the four implementations: (a) Butter- worth, (b) Type I Chebyshev, (c) Type II Chebyshev, and (d) elliptic, of

the highpass filter specified in Problem 7.14 using M A T L A B . Plot the

frequency characteristics and confirm that the specifications are satisfied

by the designed implementations.

7.26 Determine the transfer function of the four implementations (a) Butter- worth, (b) Type I Chebyshev, (c) Type II Chebyshev, and (d) elliptic, of

the bandpass filter specified in Problem 7.17 using M A T L A B . Plot the

frequency characteristics and confirm that the specifications are satisfied

by the designed implementations.

7.27 Determine the transfer function of the four implementations: (a) Butter- worth, (b) Type I Chebyshev, (c) Type II Chebyshev, and (d) elliptic, of

the bandstop filter specified in Problem 7.20 using M A T L A B . Plot the

frequency characteristics and confirm that the specifications are satisfied

by the designed implementations.

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C H A P T E R

8 Case studies for CT systems

Several aspects of continuous-time (CT) systems were covered in Chapters

3–7. Among the concepts covered, we used the convolution integral in

Chapter 3 to determine the output y(t) of a linear time-invariant, continuous- time (LTIC) system from the input signal x(t) and the impulse response h(t). Chapters 4 and 5 defined the frequency representations, namely the CT Fourier

transform (CTFT) and the CT Fourier series (CTFS) and evaluated the out-

put signal y(t) of the LTIC system in the frequency domain. The CTFT was also used to estimate the frequency characteristics of the LTIC system by plot-

ting the magnitude and phase spectra. Chapter 6 introduced the Laplace trans-

form widely used as an alternative for the CTFT in control systems, where

the analysis of the transient response is of paramount importance. Chapter 7

presented techniques for designing LTIC systems based on the specified fre-

quency characteristics. When an LTIC system is described in terms of its fre-

quency characteristics, it is referred to as a frequency-selective filter. Design

techniques for four types of analog filters, lowpass filters, highpass filters, band-

pass filters, and bandstop filters, were also covered in Chapter 7. In this chap-

ter, we provide applications for the LTIC systems. Our goal is to illustrate

how the tools developed in the earlier chapters can be utilized in real-world

applications.

The organization of this chapter is as follows. Section 8.1 considers analog

communication systems. In particular, we illustrate the use of amplitude modu-

lation in communication systems for transmitting information to the receivers.

Based on the CTFT, spectral analysis of the process of modulation provides

insight into the performance of the communication systems. Section 8.2 intro-

duces a spring damper system and shows how the Laplace transform is useful

in analyzing the stability of the system. Section 8.3 analyzes the armature-

controlled, direct current (dc) motor by deriving its impulse response and trans-

fer function. The immune system of humans is considered in Section 8.4. Ana-

lytical models for the immune system are considered and later analyzed using

the simulink toolbox available in M A T L A B .

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369 8 Case studies for CT systems

s(t)

modulated

signal

modulator

offset for

modulation Acos(2pfct)

(1+km(t)) attenuator

km(t) m(t)

modulating

signal

+ Fig. 8.1. Schematic diagram

modeling the process of

amplitude modulation.

8.1 Amplitude modulation of baseband signals

Section 2.1.3 introduced modulation as a frequency-shifting operation where

the frequency contents of the information-bearing signal are moved to a higher

frequency range. Modulation leads to two main advantages in communications.

First, since the length of the antenna is inversely proportional to the frequency

of the information signal, transmitting information bearing low-frequency

baseband signals directly leads to antennas with impractical lengths. By shift-

ing the frequency content of the information signal to a higher frequency range,

the length of the antenna is considerably reduced. Secondly, modulation leads

to frequency division multiplexing (FDM), where multiple signals are coupled

together by shifting them to a range of different frequencies and are then trans-

mitted simultaneously. This provides considerable savings in the transmission

time and the power consumed by the communication systems. In this section,

we consider a common form of modulation, referred to as amplitude modulation

(AM), used frequently in radio communications.

Amplitude modulation is a popular technique used for broadcasting radio

stations within a local community. In North America, a frequency range of 520

to 1710 kHz is assigned to the AM stations. Typically, each station occupies a

bandwidth of 10 kHz. To limit the range of transmission to a few kilometers,

the transmitted power for a station ranges from 0.1 to 50 kW, such that the same

AM band can be reused by another community without interference. In this

section, we use the CTFT to analyze AM-based communication systems.

A schematic diagram of an AM system is shown in Fig. 8.1, where m(t) represents a baseband signal with non-zero frequency components within the

range –ωmax ≤ ω ≤ ωmax. The output of the AM system is given by

s(t) = A[1 + km(t)] cos(ωct + φc). (8.1)

The multiplication term A cos(ωct + φc) represents the sinusoidal carrier, whose amplitude is denoted by A, and the radian frequency is given by ωc = 2π fc.† The constant phase term φc is referred to as the epoch of the carrier, while the factor k is referred to as the modulation index, which is adjusted such that the intermediate signal (1 + km(t)) is always positive for all t ≥ 0.

† Note that ωc represents the fundamental frequency of the sinusoidal carrier signal c(t), and should not be confused with ωc , used to denote the cut-off frequency of the CT filter.

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370 Part II Continuous-time signals and systems

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

−1

t (ms)

m(t)

(a) (b)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

−1

t (ms)

s(t)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

−1

t (ms)

m(t)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

−1

t (ms)

s(t)

Fig. 8.2. Amplitude modulation

in time domain for two different

modulating signals: (a) pure

sinusoidal signal with

fundamental frequency of 2 kHz;

(b) synthetic audio signal. The

modulated signal for the pure

sinusoidal signal is shown in (c)

and that for real speech is

shown in (d).

Figure 8.2 shows the results of amplitude modulation for two different modu-

lating signals: a pure sinusoidal signal with the fundamental frequency of 2 kHz

is plotted in Fig. 8.2(a) and a synthetic audio signal is plotted in Fig. 8.2(b). Both

signals are amplitude modulated with the carrier signal cos(ωct + φc) having a fundamental frequency of fc = 40 kHz and an epoch of φc = 0 radians. In the case of the pure sinusoidal signal, the modulation index k is selected to have a value of 0.2, while for the real audio signal the modulation index is set

to 0.7. The results of amplitude modulation are shown in Fig. 8.2(c) for the

pure sinusoidal signal and in Fig 8.2(d) for the audio signal. In both cases, we

observe that the amplitude of the carrier signal is adjusted according to the

magnitude of the modulating signal. In other words, the modulating signal acts

as an envelope and controls the amplitude of the carrier.

To illustrate the effect of modulation on the frequency content of the modu-

lating signal, we use the CTFT. Equation (8.1) is expressed as follows:

s(t) = A cos(ωct + φc) + Akm(t) cos(ωct + φc). (8.2)

Without loss of generality, we set A = 1 and φc = 0. Using the multiplication property for the CTFT, we obtain

S(ω) = π [δ(ω − ωc) + δ(ω + ωc)] + 1

2 k[M(ω − ωc) + M(ω + ωc)]. (8.3)

Equation (8.3) proves that the spectrum of the modulated signal s(t) is the sum of three components: the scaled spectrum of the carrier signal, the scaled

replica of the modulating signal m(t) shifted to +ωc, and the scaled replica of the modulating signal m(t) shifted to −ωc. This result is illustrated in Fig. 8.3(c) for the baseband signal m(t), which is band-limited to ωmax and has the spectrum shown in Fig. 8.3(a). The two replicas of the CTFT M(ω) of the modulating signal in Fig. 8.3(c) are referred to as the side bands of the AM signal.

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371 8 Case studies for CT systems

w −wmax wmax

M(w)

C(w)

wc−wc

S(w)

p 0.5k

− w

c − w

m ax

− w

c + w

m ax

w c −

w m

w c +

w m

(a) (b)

(c)

Fig. 8.3. Amplitude modulation

in the frequency domain.

(a) Spectrum of the baseband

information signal.

(b) Spectrum of the carrier

signal. (c) Spectrum of the

modulated signal.

We now consider the extraction of the information signal x(t) from the mod- ulated signal s(t). This procedure is referred to as demodulation, which is explained in Sections 8.1.1 and 8.1.2.

8.1.1 Synchronous demodulation

The objective of demodulation is to reconstruct m(t) from s(t). Analyzing the spectrum S(ω) of the modulated signal s(t), the following method extracts the information-bearing signal m(t) from s(t).

(1) Frequency shift the modulated signal s(t) by ωc (or −ωc). If the modulated signal is frequency-shifted by ωc, one of the side bands is shifted to zero

frequency, while the second side band is shifted to 2ωc. Conversely, if

the modulated signal is frequency-shifted by −ωc, the two side bands are shifted to zero and −2ωc.

(2) In order to remove the side band shifted to the non-zero frequency, the

result obtained in Step (1) is passed through a lowpass filter having a pass

band of (−ωmax ≤ ω ≤ ωmax). The output of the lowpass filter consists of a scaled version of the modulating signal and an impulse at ω = 0. The impulse represents the dc component and is removed by subtracting a

constant value in the time domain as shown in Step (3).

(3) A constant voltage equal to the dc component is subtracted from the output

of the lowpass signal.

Step (1) can be performed by multiplying the AM signal s(t) by the demodulat- ing carrier cos(ωct) having the same fundamental frequency and phase as the

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372 Part II Continuous-time signals and systems

C(w)

wc−wc

wc−wc w

M(w)

p 0.5k

− w

c − w

m ax

− w

c + w

m ax

w c −

w m

w c +

w m

(a) (b)

(c)

D(w)

p 5.0k

− w

m ax

w m

2 w

c + w

m ax

2 w

c − w

m ax

2 w

− 2 w

c + w

m ax

− 2 w

c − w

m ax

− 2 w

Fig. 8.4. Demodulation in the

frequency domain. (a) Spectrum

of the modulated signal.

(b) Spectrum of the carrier

signal. (c) Spectrum of the

demodulated signal.

modulating carrier. In the time domain, the result of the multiplication is given

d(t) = s(t)c(t) = [cos(ωct) + km(t) cos(ωct)] cos(ωct), (8.4)

which is expressed as

d(t) = s(t)c(t) = 1

2 [1 + km(t)]

︸︷︷︸

dlow(t)

+ 1

2 [1 + km(t)] cos(2ωct)

︸︷︷︸

dhigh(t)

. (8.5)

Equation (8.5) shows that the demodulated signal d(t) has two components. The first component dlow(t) is the low-frequency component, which consists of a constant factor of 1/2 and a scaled replica of the modulated signal. The second

component dhigh(t) is the higher-frequency component and can be filtered out, as explained next. Taking the CTFT of Eq. (8.5) yields

D(ω) = 1

2π [S(ω) ∗ C(ω)] =

[

πδ(ω) + k

2 M(ω)

]

︸︷︷︸

dlow(t)

+ [π

2 δ(ω − 2ωc) +

4 M(ω − 2ωc)

]

+ [π

2 δ(ω + 2ωc) +

4 M(ω + 2ωc)

]

︸︷︷︸

dhigh(t)

(8.6)

which is plotted in Fig. 8.4(c). Recall that Fig. 8.4(a) represents the spectrum

of the modulated signal m(t) and that Fig. 8.4(b) represents the spectrum of the carrier signal c(t). By filtering d(t) with a lowpass filter having a pass band of

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−ωc ≤ ω ≤ ωc, the lowpass component dlow(t) is extracted. The information signal m(t) is then obtained from dlow(t) using the following relationship:

m(t) = 2[dlow(t) − 1]. (8.7)

8.1.2 Synchronous demodulation with non-zero epochs

In synchronous demodulation, the epoch φc of the modulating carrier is assumed

to be identical to the epoch of the demodulating carrier. In practice, perfect syn-

chronization between the carriers is not possible, which leads to distortion in

the signal reconstructed from demodulation. To illustrate the effect of distor-

tion introduced by unsynchronized carriers, consider the following modulated

signal:

s(t) = A cos(ωct + φc) + Akm(t) cos(ωct + φc), (8.8)

as derived in Eq. (8.2). Assume that the demodulator carrier is given by

c2(t) = A cos(ωct + θc(t)), (8.9)

which has a time-varying epoch θc(t) �= φc. Using c2(t), the demodulated signal is given by

d(t) = s(t)c2(t) = [A cos(ωct+φc) + Akm(t) cos(ωct + φc)] cos(ωct+θc(t)), (8.10)

which simplifies to

d(t) = A

2 [1 + km(t)] cos(φc − θc(t))

︸︷︷︸

dlow(t)

+ A

2 [1 + km(t)] cos(2ωct + φc + θc(t))

︸︷︷︸

dhigh(t)

(8.11)

Equation (8.11) illustrates that the demodulated signal contains a low-frequency

component dlow(t) and a higher-frequency component dhigh(t). By passing the demodulated signal through a lowpass filter, the higher-frequency component

is removed. The output of the lowpass filter is given by

y(t) = A

2 [1 + km(t)] cos(φc − θc(t)). (8.12)

Even after eliminating the dc component, the reconstructed signal has the fol-

lowing form:

y(t) = A

2 km(t) cos(φc − θc(t)), (8.13)

where distortion is caused by the factor of cos(φc − θc(t)). Since the epoch θc(t) is time-varying, it is difficult to eliminate the distortion. To reconstruct x(t) precisely, the phase difference between the carrier signals used at the mod- ulator and demodulator must be kept equal to zero over time. In other words,

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374 Part II Continuous-time signals and systems

s(t) d(t)R C

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

−1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 t (ms)

d(t)

(a) (b)

Fig. 8.5. Asynchronous AM

demodulation. (a) RC parallel

circuit coupled with a diode to

implement the envelope

detector. (b) Reconstructed

signal d(t ) is shown as a solid

line. For comparison, the

information component

[1 + km(t )] is shown as the envelope of the AM signal

(dashed line).

perfect synchronization between the modulator and demodulator is essential to

retrieve the information signal m(t). For this reason, the aforementioned mod- ulation scheme based on multiplying the modulated signal by the carrier signal

and lowpass filtering is referred to as synchronous demodulation. Although syn-

chronous demodulation is an elegant way of retrieving the information signal

m(t), the demodulator has a high implementation cost due to the synchroniza- tion required between the two carriers. An alternative scheme, which does not

require synchronization of the modulating and demodulating carriers, is referred

to as asynchronous demodulation, which is considered in the following section.

8.1.3 Asynchronous demodulation

In amplitude modulation, the information-bearing signal m(t) modulates the magnitude of the carrier signal c(t). This is illustrated in Figs. 8.2(c) and (d), where the envelope of the amplitude modulated signal follows the informa-

tion component [1 + km(t)]. In asynchronous demodulation, we reconstruct the information signal m(t) by tracking the envelope of the modulated signal.

Figure 8.5(a) shows a parallel RC circuit used to reconstruct the information-

bearing signal m(t) from the amplitude modulated signal s(t) applied at the input of the RC circuit. The diode acts as a half-wave rectifier removing the negative

values from the modulated signal, while the capacitor C tracks the envelope of the AM signal by charging to the peak of the sinusoidal carrier during the

positive transition of the signal. During the negative transitions of the carrier,

the capacitor discharges slightly, but is again recharged by the next positive

transition. The process is illustrated in Fig. 8.5(b), where the demodulated signal

is represented by a solid line. We observe that the demodulated signal closely

follows the envelope of the modulated signal and is a good approximation of

the information-bearing signal.

8.2 Mechanical spring damper system

The spring damping system, considered in Section 2.1.5, is a classic example

of a second-order system; the schematic diagram for such a system is shown in

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375 8 Case studies for CT systems

Fig. 8.6. The input–output relationship of the spring damping system is given

by Eq. (2.16), which for convenience of reference is repeated below:

M d2 y

dt + r

dt + ky(t) = x(t). (8.14)

In Eq. (8.14), M is the mass attached to the spring, r is the frictional coefficient, k is the spring constant, x(t) is the force applied to pull the mass, and y(t) is the displacement of the mass caused by the force. In this section, we analyze

the spring damping system using the methods discussed in Chapters 5 and 6.

Using the Laplace transform, the transfer function determines the stability of

the system.

8.2.1 Transfer function

Taking the Laplace transform of both sides of Eq. (8.14) and assuming zero

initial conditions, we obtain

(Ms2 + rs + k)Y (s) = X (s), (8.15)

which results in the following transfer function:

x(t)

kyby

y(t)

x(t)

r y(t)wall friction

kspring

constant

(a)

(b)

Fig. 8.6. Mechanical spring

damping system.

H (s) = Y (s)

X (s) =

(Ms2 + rs + k) . (8.16)

Alternatively, Eq. (8.16) can be expressed as follows:

H (s) = 1/M

s2 + (r/M)s + k/M =

1/M

s2 + 2ξnωns + ω2n , (8.17)

where

ωn = √

M and ξn =

2 √

k M .

The characteristic equation of the mechanical spring damping system is given

s2 + 2ξnωns + ω2n = 0, (8.18)

which has two poles at

p1 = −ξnωn + ωn √

ξ 2n − 1 and p2 = −ξnωn − ωn √

ξ 2n − 1 . (8.19)

Depending on the value of ξn , the poles p1 and p2 may lie in different locations within the s-plane. If ξn = 1, poles p1 and p2 are real-valued and identical. If ξn > 1, poles p1 and p2 are real-valued but not equal. Finally, if ξn < 1, poles p1 and p2 are complex conjugates of each other. In the following, we calculate the impulse response h(t) of the spring damping system for three sets of values of ξn and show that the characteristics of the system depend on the value of ξn .

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376 Part II Continuous-time signals and systems

Case 1 (ξn = 1) For ξn = 1, Eq. (8.17) reduces to

H (s) = 1/M

s2 + 2ωns + ω2n =

1/M

(s + ωn)2 , (8.20)

with repeated roots at s = −ωn , −ωn .Taking the inverse transform, the impulse response is given by

h(t) = 1

M te−ωn t u(t). (8.21)

Case 2 (ξn > 1) For ξn > 1, the poles p1 and p2 of the spring damping system are real-valued and given by

p1 = −ξnωn + ωn √

ξ 2n − 1 and p2 = −ξnωn − ωn √

ξ 2n − 1 . (8.22)

The transfer function of the spring damping system can be expressed as follows:

H (s) = 1/M

s2 + 2ξnωns + ω2n =

1/M

(s − p1)(s − p2) , (8.23)

which leads to the impulse response

h(t) = 1

(p1 − p2) [ep1t − ep2t ] u(t) =

2ωn M √

ξ 2n − 1 e−ξ nωn t

× [

eωn √

ξ 2n −1 t − e−ωn √

ξ 2n −1 t ]

u(t). (8.24)

Case 3 (ξn < 1) For ξn > 1, the poles p1 and p2 of the spring damping system are complex and are given by

p1 = −ξnωn + jωn √

1 − ξ 2n and p2 = −ξnωn − jωn √

1 − ξ 2n . (8.25)

By repeating the procedure for Case 2, the impulse response of the spring

damping system is given by

H (s) = 1/M

s2 + 2ξnωns + ω2n =

1/M

(s − p1)(s − p2) , (8.26)

which leads to the impulse response

h(t) = 1

ωn M √

1 − ξ 2n e−ξnωn t sin

[

ωn

√

1 − ξ 2n t ]

u(t). (8.27)

Figure 8.7 shows the impulse response of the spring damping system for the

three cases considered earlier. We set M = 10 and ωn = 0.3 radians/s in each case. For Case 1 with ξn = 1, the impulse response decreases monotonically approaching the steady state value of zero. Such systems are referred to as

critically damped systems.

For Case 2 with ξn = 4, the impulse response of the spring damping system again approaches the steady state value of zero. Initially, the deviation from the

steady state value is smaller than that of the critically damped system, but the

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377 8 Case studies for CT systems

0 5 10 15 20 25 30 35 40 45 50

−0.1

0.1

0.2 h(t)

t (s)

x = 2.0 x = 1 x = 4

Fig. 8.7. Impulse responses of

the spring damping system for

M = 10 and ωn = 0.3.

overall duration over which the steady state value is achieved is much longer.

Such systems are referred to as overdamped systems.

For Case 3 with ξn = 0.2, the spring acts as a flexible system. The impulse response approaches the steady state value of zero after several oscillations.

Such systems are referred to as underdamped systems. Since the fundamental

frequency ωn is 0.3 radians/s, the period of oscillation is given by

T = 2π

ωn =

2π

0.3 = 21.95 seconds. (8.28)

Based on Eq. (8.28), parameter ωn is referred to as the fundamental frequency of

the spring damping system. Since parameter ξn determines the level of damping,

it is referred to as the damping constant.

8.3 Armature-controlled dc motor

Electrical motors form an integral component of most electrical and mechan-

ical devices such as automobiles, ac generators, and power supplies. Broadly

speaking, electrical motors can be classified into two categories: direct current

(dc) motors and alternating current (ac) motors. Within each category, there are

additional subclassifications covering different applications. In this section, we

analyze the armature-controlled dc motor by deriving its transfer function and

impulse response.

Figure 8.8(a) shows an armature-controlled dc motor, in which an armature,

consisting of several copper conducting coils, is placed within a magnetic field

generated by a permanent or an electrical magnet. A voltage applied across the

armature results in a flow of current through the armature circuit. Interaction

between the electrical and magnetic fields causes the armature to rotate, the

direction of rotation being determined by the following empirical rule, derived

by Faraday.

Extend the thumb, index finger, and middle finger of the right hand such that

the three are mutually orthogonal to each other. If the index finger points in the

direction of the current and the middle finger in the direction of the magnetic

field, then the thumb points in the direction of motion of the armature.

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378 Part II Continuous-time signals and systems

(a)

(b)

Load

motor

torque

angular

velocity

w(t)

kfw(t) viscous

friction

Ra La

+ Vemf

− ia(t)

va(t)

−

Fig. 8.8. Armature-controlled dc

motor. (a) Cross-section; (b)

schematic representation.

8.3.1 Mathematical model

The linear model of the armature-controlled dc motor is shown in Fig. 8.8(b),

where a load J is coupled to the armature through a shaft. Rotation of the arma- ture of the dc motor causes the desired motion in the attached load J . Moving a conductor within a magnetic field also generates a back electromagnetic field

(emf) to be induced in the dc motor. The back emf results in an opposing emf

voltage, which is denoted by Vemf in Fig. 8.8(b). In the following analysis, we decompose the motors into three components: armature, motor, and load. The

equations for the three components are presented below.

Armature circuit Applying Kirchhoff’s voltage law to the armature circuit, we obtain

La dia dt

+ Raia + kfω(t) ︸︷︷︸

Vemf(t)

= Va(t), (8.29)

where Va(t) denotes the armature voltage and ia(t) denotes the armature current. The electrical components of the armature circuit are given by La and Ra, where La denotes the self inductance of the armature and Ra denotes the self resistance

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379 8 Case studies for CT systems

of the armature. The emf voltage Vemf(t) is approximated by the product of the feedback factor kf and the angular velocity ω(t).

Motor circuit The torque Tm, induced by the applied voltage across the arma- ture, is given by

Tm = kmia(t), (8.30)

where km is referred to as the motor or armature constant and ia(t) is the armature current. The armature constant km depends on the physical properties of the dc motor such as the strength of the magnetic field and the density of the armature

coil.

Load The load component of the dc motor is obtained by applying Newton’s third law of motion, which states that the sum of the applied and reactive forces

is zero. The applied forces are the torques around the motor shaft. The reactive

force causes acceleration of the armature and equals the product of the inertial

load J and the derivative of the angular rate ω(t). In other words,

∑

Tp = J dω

dt , (8.31)

where J denotes the inertia of the rotor. There are three different torques, i.e. p = 3, observed at the shaft: (i) motor torque Tm represented by Eq. (8.30); (ii) frictional torque Tf given by rω(t), r being the frictional constant; and (iii) load disturbance torque Td. In other words, Eq. (8.31) can be expressed as follows:

J dω

dt = Tm − rω(t) − Td. (8.32)

Since the angular velocity ω(t) is related to the shaft position θ (t) by the fol- lowing expression:

ω(t) = dθ

dt , (8.33)

Eq. (8.32) can be expressed as follows:

J d2θ

dt2 + r

dθ

dt = Tm − Td = TL, (8.34)

where TL denotes the difference between the motor torque Tm and the load disturbance torque Td.

8.3.2 Transfer function

The dc motor shown in Fig. 8.8 is modeled as a linear time-invariant (LTI)

system with the armature voltage va(t) considered as the input signal and the shaft position θ (t) as the output signal. We now derive the transfer function of the linearized model.

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380 Part II Continuous-time signals and systems

Taking the Laplace transform of Eq. (8.29) yields

[sLa + Ra] Ia(s) + kfΩ(s) = Va(s), (8.35)

where Va(s), Ia(s), andΩ(s) are the Laplace transforms of va(t), ia(t), and ω(t), respectively. Substituting the value of Ia(s) from Eq. (8.30), Eq. (8.35) can be expressed as follows:

km [sLa + Ra]Tm(s) + kfΩ(s) = Va(s). (8.36)

We also take the Laplace transform of Eq. (8.34) to derive

[Js2 + rs]θ (s) = Tm(s), (8.37)

where we have ignored the disturbance torque Td(s), which will later be approx- imated as noise to the system’s input. Substituting θ (s) = Ω(s)/s, Eq. (8.37) is expressed as follows:

Tm(s) = [Js + r ]Ω(s). (8.38)

Finally, substituting the value of Tm(s) from Eq. (8.38) into Eq. (8.35) yields

[sLa + Ra] [Js + r ]Ω(s) + kmkfΩ(s) = kmVa(s)

Ω(s)

Va(s) =

km La Js2 + [Ra J + Lar ]s + [Rar + kmkf]

. (8.39)

The transfer function H (s) can therefore be expressed as follows:

H (s) = θ (s)

Va(s) =

Ω(s)

sVa(s) =

km/J La

s3 + [

Ra J + Lar La J

]

s2 + [

Rar + kmkf La J

]

H (s) = k ′m

s3 + 2ξnωns2 + ω2ns , (8.40)

where

k ′m = km J La

, ξn = 1

2ωn

[ Ra La

+ r

]

and ωn = √

Rar + kmkf La J

From Eq. (8.40), we note that the system transfer function H (s) has a third- order characteristic equation with one pole at the origin (s = 0) of the s-plane. The remaining two poles are located at

p1 = −ξnωn + ωn √

ξ 2n − 1 and p2 = −ξnωn − ωn √

ξ 2n − 1. (8.41)

As ξn and ωn are positive, the two non-zero poles lie in the left half of the

complex s-plane. Due to the zero pole, however, the system is a marginally

stable system.

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381 8 Case studies for CT systems

Impulse response Since

H (s) = 1

s ×

k ′m s2 + 2ξnωns + ω2n ︸︷︷︸

H ′(s)

the impulse response of the dc motor equals the integral of the impulse response

h′(t), the inverse Laplace transform of H′(s). Since the form of H′(s) is similar to the transfer function of the spring damping system, Eqs. (8.20)–(8.27) are

used to derive the impulse response h(t) of the dc motor. Depending upon the value of ξn , we consider three different cases.

Case 1 (ξn = 1) As derived in Eq. (8.21), the inverse Laplace transform of H′(s) for ξn = 1 is given by

k ′mte −ωn t u(t)

L←→ k ′m

s2 + 2ξnωns + ω2n .

Taking the integral of h′(t) yields

h(t) = ∫

k ′m te −ωn t dt = −k ′m

[ t

ωn +

ω2n

]

e−ωn t + C for t ≥ 0. (8.42)

Case 2 (ξn > 1) Equation (8.24) derives the inverse Laplace transform of H′(s) for ξn > 1 as follows:

k ′m 2ωn

√

ξ 2n − 1 e−ξnωn t

[

eωn √

ξ 2n−1 t − e−ωn √

ξ 2n−1 t ]

u(t) L←→

k ′m s2 + 2ξnωns + ω2n

The impulse response of the dc motor is given by

h(t) = k ′m

2ωn √

ξ 2n − 1

∫

e−ξnωn t [

eωn √

ξ 2n−1 t − e−ωn √

ξ 2n−1 t ]

= k ′me

−ξnωn t

2ωn √

ξ 2n − 1

[ e−ωn

√ ξ 2n−1 t

ξnωn + ωn √

ξ 2n − 1 −

eωn √

ξ 2n−1 t

ξnωn − ωn √

ξ 2n − 1

]

+ C for t ≥ 0. (8.43)

Case 3 (ξn < 1) Equation (8.27) derives the inverse Laplace transform of H′(s) for ξn < 1 as follows:

k ′m ωn

√

1 − ξ 2n e−ξnωn t sin

[

ωn

√

1 − ξ 2n t ]

u(t) L←→

k ′m s2 + 2ξnωns + ω2n

The impulse response of the dc motor is given by

h(t) = k ′m

ωn √

1 − ξ 2n

∫

e−ξnωn t sin [

ωn

√

1 − ξ 2n t ]

= − k ′me

−ξnωn t

ω2n

√

1 − ξ 2n

[√

1 − ξ 2n cos [

ωn

√

1 − ξ 2n t ]

+ ξn sin [

ωn

√

1 − ξ 2n t ]]

+ C for t ≥ 0. (8.44)

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382 Part II Continuous-time signals and systems

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0

0.2

0.4

0.6

0.8 h(t)

t (s)

x = 0.2 x = 1

x = 4

Fig. 8.9. Impulse response of

the armature-controlled dc

motor for k ′m = 1000 and ωn = 50.

In Eqs. (8.42)–(8.44), C is the integration constant, which can be computed from the initial conditions.

Figure 8.9 plots the impulse response of the armature-controlled dc motor for

k ′

m = 1000 and ωn = 50. Three different values of ξn are chosen and the value of the integration constant C is set to 0.4. As is the case for the spring damping system, the impulse response is critically damped for ξn = 1, underdamped for ξn = 0.2, and overdamped for ξn = 4. In the case of the underdamped system, the frequency of oscillations is given by ωn = 50 radians/s, with the fundamental period given by 2π/50, or 0.126 seconds.

Block diagram To derive the feedback representation of the armature- controlled dc motor, Eq. (8.35) is expressed as follows:

[sLa + Ra] Ia(s) = Va(s) − kfΩ(s). (8.45)

Substituting Ia(s) = Tm(s)/km, Eq. (8.35) is given by

[sLa + Ra]Tm(s) = km[Va(s) − kfΩ(s)]. (8.46)

Taking the Laplace transform of Eq. (8.32), we obtain

Tm(s) = [s J + r ]Ω(s) + Td. (8.47)

Substituting Eq. (8.47) into Eq. (8.46), the relationship between the input volt-

age Va(s) and the angular velocity Ω(s) is given by

Ω(s) = 1

[Js + r ]

[ km

[Las + Ra] [Va(s) − kfΩ(s)] − Td

]

. (8.48)

Equation (8.48) is used to develop the block diagram representation for the

transfer function:

H ′(s) = Ω(s)

Va(s) =

k ′m s2 + 2ξnωns + ω2n

which is shown in Fig. 8.10. In this case, the system has two poles, both are

in the left-half of the s-plane. Therefore, the system is a stable system. The

block diagram representation of the transfer function H (s) can be obtained by integrating ω(t), which in the Laplace domain is equivalent to multiplyingΩ(s) by a factor of 1/s. A block with the transfer function 1/s can, therefore, be cascaded at the end of Fig. 8.10 to derive the feedback configuration for H (s).

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383 8 Case studies for CT systems

Va(t)

applied

voltage armature load

w(t) output

angular velocity

sLa+ Ra

sJ + r 1

+ ++ −

−

Vemf (t)

Fig. 8.10. Schematic models for

the dc motor with transfer

function H ′(s ).

8.4 Immune system in humans

We now apply the Laplace transform to model a more natural system, such as

the human immune system. The human immune system is non-linear but, with

some assumptions, it can be modeled as a linear time-invariant (LTI) system.

Below we provide the biological working of the human immune system, which

is followed by an explanation of its linearized model.

Human blood consists of a suspension of specialized cells in a liquid, referred

to as plasma. In addition to the commonly known erythrocytes (red blood

cells) and leukocytes (white blood cells), blood contains a variety of other

cells, including lymphocytes. The lymphocytes are the main constituents of

the immune system, which provides a natural defense against the attack of

pathogenic microorganisms such as viruses, bacteria, fungi, and protista. These

pathogenic microorganisms are referred to as antigens. When the lymphocytes

come into contact with the foreign antigens, they yield antibodies and arrange

the antibodies on their membrane. The antibody is a molecule that binds itself

to antigens and destroys them in the process. When sufficient numbers of anti-

bodies are produced, the destruction of the antigens occurs at a higher rate than

their creation, resulting in the suppression of the disease or infection. Based on

this simplified explanation of the human immune system, we now develop the

system equations.

8.4.1 Mathematical model

The following notation is used to develop a mathematical model for the human

immune system:

g(t) = number of antigens entering the human body; a(t) = number of antigens already existing within the human body; l(t) = number of active lymphocytes; p(t) = number of plasma cells; b(t) = number of antibodies.

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384 Part II Continuous-time signals and systems

Number of antigens At any given time, the total number of antigens present in the human body depends on three factors: (i) external antigens entering

the human body from outside; (ii) reproduced antigens produced within the

human body by the already existing antigens; and (iii) destroyed antigens that

are eradicated by the antibodies. The net change in the number of antigens is

modeled by the following equation:

dt = αa(t) − ηb(t) + g(t), (8.49)

where α denotes the reproduction rate at which the antigens are multiplying

within the human body and η is the destruction rate at which the antigens are

being destroyed by the antibodies.

Number of lymphocytes Assuming that the number of lymphocytes is pro- portional to the number of antigens, the number of lymphocytes present within

the human body is given by

l(t) = βa(t), (8.50)

where β is the proportionality constant relating the number of lymphocytes to

the number of antigens. The value of β generally depends on many factors,

including the health of the patient and external stimuli. In general, β varies

with time in a non-linear fashion. For simplicity, however, we can assume that

β is a constant.

Number of plasma cells The change in the number of plasma cells is pro- portional to the number of lymphocytes l(t). Typically, there is a delay of τ seconds between the instant that the antigens are detected and the instant that

the plasma cells are generated. Therefore, the number of plasma cells depends

on l(t − τ ), where the proportionality constant is assumed to be unity. Also, a large portion of plasma cells die due to aging. The number of plasma cells at

any time t can therefore be expressed as follows:

dt = l(t − τ ) − γ p(t), (8.51)

where γ denotes the rate at which the plasma cells die due to aging.

Number of antibodies The number of antibodies depends on three factors: (i) new antibodies being generated by the human body (the rate of generation µ

of the new antibodies is proportional to the number of plasma cells in the human

body); (ii) destroyed antibodies lost to the antigens (the rate of destruction σ

of such antibodies is proportional to the number of existing antigens); and

(iii) dead antibodies lost to aging. We assume that the antibodies die at the rate

of λ because of aging. Combining the three factors, the number of antibodies

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385 8 Case studies for CT systems

at any time t is given by

dt = µp(t) − σa(t) − λb(t). (8.52)

8.4.2 Transfer function

Equations (8.49)–(8.52) describe the linearized model used to analyze the

human immune system. To develop the transfer function, we take the Laplace

transform of Eqs. (8.49)–(8.52). The resulting expressions can be expressed as

follows:

number of antigens A(s) = 1

(s − α) [G(s) − ηB(s)]; (8.53)

number of lymphocytes L(s) = β A(s); (8.54)

number of antigens P(s) = e−τ s

(s + γ ) L(s); (8.55)

number of antibodies B(s) = 1

(s + λ) [µP(s) − σ A(s)]. (8.56)

In Eqs. (8.53)–(8.56), variables A(s), G(s), L(s), P(s), and B(s) are, respec- tively, the Laplace transforms of the number of antigens a(t) present within the human body, the number of antigens g(t) entering the human body, the number of lymphocytes l(t) within the blood, the total number of antigens p(t) within the human body, and the number of antibodies b(t) in the blood. Assuming the number of antigens g(t) entering the human body to be the input and the number of antibodies b(t) produced to be the output, the human immune system can be modeled by the schematic diagram shown in Fig. 8.11(a). Figure 8.11(b) is the

simplified version of Fig. 8.11(a), which yields the following transfer function

for the human immune system:

T (s) = M(s)

(1 + η M(s)) =

µβe−τ s − σ (s + γ ) (s − α)(s + λ)(s + γ ) + η[µβe−τ s − σ (s + γ )]

(8.57)

8.4.3 System simulations

The simplified model of the human immune system is still a fairly complex

system to be analyzed analytically. The characteristic equation of the human

immune system is not a polynomial of s, therefore evaluation of its poles is difficult. In this section, we simulate the human immune system using the

simulink toolbox available in M A T L A B .

8.4.3.1 Simulation 1

In simulink, a system is simulated using a block diagram where the subblocks

represent different subsystems. Figure 8.12 shows the simulink representation of

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386 Part II Continuous-time signals and systems

s−a 1 bexp(−ts)

m s +l

1 ++

−

+ −

g(t)

number of

antigens

b(t)

number of

antibodies

(s−a)(s+l)(s+g) mbexp(−ts) − s(s+g)

M(s) =+ +

−

g(t)

number of

antigens

b(t)

number of

antibodies

(a)

(b)

s +g

Fig. 8.11. Schematic models for

the immune response system.

(a) Detailed model;

(b) simplified model.

the human immune system shown in Fig. 8.11. We have assumed a hypothetical

case with the values of the proportionality constants given by

α = 0.1, β = 0.5, γ = 0.1, µ = 0.5, τ = 0.2, λ = 0.1, σ = 0.1, and η = 0.5.

The proportionality constants α, γ , σ , and λ related to the antigens are deliber-

ately kept smaller than the proportionality constants β, η, and µ related to the

antibodies for quick recovery from the infection. The input signal g(t) modeling the number of antigens entering the human body is approximated by a pulse

and is shown in Fig. 8.13(a). The duration of the pulse is 0.5 s, implying that

the antigens keep entering the human body at a constant level for the first 0.5 s.

The outputs a(t), p(t), and b(t) are monitored by the simulated scope available in simulink. The output of the scope is shown in Fig. 8.13(b), where we observe

+germs

input germs antigen

generation

antigen

generation

scope

lymphocyte delay

_ +_ error

a(t)

a(t) p(t)

a(t)

b(t)

b(t)1

s−0.1

0.5 0.5

0.1

0.5

gain2

gain3

gain1 s+ 0.1

s+ 0.1

Fig. 8.12. Simulink model for

Simulation 1 modeling the

immune response system of

humans.

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387 8 Case studies for CT systems

0 2 4 6 8 10 12 0

800

1200

1600

2000

400 a(t)

p(t)

b(t)

t (s)

−1000

1000

2000

3000

4000

a(t)

p(t)

b(t)

0 2 4 6 8 10 12

0 0.5 2 4 6 8 10 12

1000

2000

3000

g(t)

t (s)

(a) (b)

(c)

Fig. 8.13. Results of Simulations

1 and 2. (a) Number of antigens

g(t ) entering the human body.

(b) Time evolution of the

number of antigens a(t ), plasma

cells p(t ), and antibodies b(t ) in

Simulation 1. (c) Same as (b) for

Simulation 2.

that the number of antigens increases linearly for the initial duration of 0.5 s.

Since the human body generates lymphocytes with a delay τ , which is 0.2 s in

our simulation, the number of plasma cells p(t) starts rising with a delay of 0.2 s. After 0.5 s, no external antigens enter the human body. However, new antigens

are being reproduced by the already existing antigens present inside the human

body. As a result, the number of antigens a(t) keeps rising, even after 0.5 s. After roughly 3 s, the respective strengths of lymphocytes and plasma cells is

high enough to impact the overall population of the antigens. The number of

antigens a(t) starts decreasing after 3 s. After 5.3 s, all antigens in the body are destroyed. At this time, the body stops producing any further plasma cells.

After this stage, the number of plasma cells p(t) starts decreasing, as some of these cells die naturally due to aging. As the number of plasma cells decreases,

the number of antibodies b(t) also decreases such that after 10 s no antibodies are present in the simulation.

8.4.3.2 Simulation 2

Simulation 1 models successful eradication of the antigens. Let us now consider

the other extreme, where the antigens are lethal such that the human immune

system is unable to terminate the infection. The proportionality constants β and

µ related to the antibodies have lower values than those specified in Simulation

1. Also, the delay τ between the instant when the antigens are detected to the

instant when antibodies are produced is increased to 1 s. The simulated values

of the constants are given by

α = 0.1, β = 0.1, γ = 0.1, µ = 0.3, τ = 1, λ = 0.1, σ = 0.1, and η = 0.5.

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388 Part II Continuous-time signals and systems

As in Simulation 1, the input signal g(t) representing the number of antigens entering the human body is assumed to be a pulse of duration 0.5 s. The numbers

of antigens a(t), plasma cells p(t), and antibodies b(t) are monitored with the simulated scope available in simulink and are plotted in Fig. 8.13(c). We observe

that the number of antigens a(t) increases at an exponential rate. Although the number of plasma cells p(t), and consequently the number of antibodies b(t), also increases, it does so at a slower pace due to the small value of β and large

delay τ . Since the number of antigens exceeds the number of plasma cells, the

antibodies are destroyed by the antigens. This is shown by negative values for

the number of antibodies b(t). In reality, the minimum number of antibodies is zero. The negative values are observed because of the unconstrained analytical

model. We can make Simulation 2 more realistic by constraining the number

of antigens, plasma cells, and antibodies to be greater than zero.

In summary, Simulation 1 presents a scenario where the patient will survive,

whereas Simulation 2 presents a scenario where the patient will die. Although

this model presents a very simplistic view of a highly complex system, it is pos-

sible to improve the model by using more accurate model parameters. Similar

mathematical models can be used in several applications, such as population

prediction, ecosystem analysis, and weather forecasting.

8.5 Summary

We have presented applications of signal processing in analog communica-

tions, mechanical systems, electrical machines, and human immune systems.

In particular, the CTFT and Laplace transform were used to analyze these

systems. Section 8.1 introduced amplitude modulation (AM) and used the

CTFT to analyze the frequency characteristics of AM-based communication

systems. Both synchronous and asynchronous detection schemes for recon-

structing the information-bearing signals were developed. Sections 8.2 and 8.3

used the Laplace transform to analyze the spring damping system and armature-

controlled dc motor. For the two applications, the transfer function and impulse

response of the overall systems were derived. Section 8.4 used the Laplace

transform to model the human immune system. An analytical model for the

human immune system was developed and later analyzed using the simulink

toolbox available in M A T L A B .

Problems

8.1 The information signal given by

x(t) = 3 sin(2π f1t) + 2 cos(2π f2t)

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389 8 Case studies for CT systems

modulates the carrier signal c(t) = cos(2π fct) with the AM signal s(t) given by Eq. (8.1).

(a) Determine the value of the modulation index k to ensure |s(t)| ≥ 0 for all t .

(b) Determine the ratio of the power lost because of the transmission of

the carrier in s(t) versus the total power of s(t). (c) Sketch the spectrum of x(t) and s(t) for f1 = 10 kHz, f2 = 20 kHz,

and fc = 50 kHz. (d) Show how synchronous demodulation can be used to reconstruct x(t)

from s(t).

8.2 Repeat Problem 8.1 for the information signal

x(t) = sinc(5 × 103t)

if the fundamental frequency of the carrier is given by fc = 20 kHz.

8.3 An AM station uses a modulation index k of 0.75. What fraction of the total power resides in the information signal? By repetition for different

values of k within the range 0 ≤ k ≤ 1, deduce whether low or high values of modulation index are better for improved efficiency.

8.4 Synchronous demodulation requires both phase and frequency coherence for perfect reconstruction of the information signal. Assume that the infor-

mation signal

x(t) = 2 sin(2π f1t)

is used to modulate the carrier c(t) = cos(2π fct). However, the demod- ulating carrier has a frequency offset given by c(t) = cos[2π fc + � f )t]. Determine the spectrum of the demodulated signal. Can the information

signal be reconstructed in such situations?

8.5 A special case of amplitude modulation, referred to as the quadrature ampli- tude modulation (QAM), modulates two information-bearing signals x1(t) and x2(t) simultaneously using two different carriers c1(t) = A1 cos(2π fct) and c2(t) = A2 sin(2π fct). The QAM signal is given by

s(t) = A1[1 + k1x1(t)] cos(2π fct) + A2[1 + k2x2(t)] sin(2π fct),

where k1 and k2 are the two modulation indexes used for modulating x1(t) and x2(t). Draw the block diagram of the demodulator that reconstructs x1(t) and x2(t) from the modulated signal.

8.6 Assume the frictional coefficient r of the spring damping system, shown in Fig. 8.6, to equal zero. Determine the transfer function H (s) and impulse response h(t) for the modified model. Based on the location of the poles, comment on the stability of the spring damping system.

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390 Part II Continuous-time signals and systems

f1(t) input

phase loop filter

v(t)

output

voltage

∫ dt −∞

+ +

−

integrator

G(s)

q(t)

× ×

Fig. P8.11. Block diagram

representation of a phase-locked

loop. 8.7 By integrating the impulse response h(t) of the armature-controlled dc

motor, derive Eq. (8.42) for ξn = 1.

8.8 Assume that the inductance La of the induction motor, shown in Fig. 8.8(b), is zero. Determine the transfer function H (s) and impulse response h(t) for the modified model. Based on the location of the poles, comment on

the stability of the induction motor.

8.9 Repeat Problem 8.7 for Eq. (8.43) with ξn > 1 and Eq. (8.44) with ξn < 1.

8.10 Based on Eqs. (8.53)–(8.56), derive the expression for the transfer function H (s) of the human immune system shown in Eq. (8.57).

8.11 In order to achieve synchronization between the modulating and demod- ulating carriers, a special circuit referred to as a phase-locked loop (PLL)

is commonly used in communications. The block diagram representing

the PLL is shown in Fig. P8.11.

Show that the transfer function of the PLL is given by

V (s)

φ(s) = K1 K2

sG(s)

s + K1G(s) ,

where K1 and K2 are gain constants and G(s) is the transfer function of a loop filter. Specify the condition under which the PLL acts as an ideal

differentiator. In other words, derive the expression for G(s) when the transfer function of the PLL equals Ks, with K being a constant.

8.12 Repeat the simulink simulation for the human immune system for the following values of the proportionality constants:

α = 0.3, β = 0.1, γ = 0.25, µ = 0.6, τ = 1, λ = 0.1, σ = 0.4, and η = 0.2

Sketch the time evolution of the antigens, plasma cells, and antibodies.

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P A R T I I I

Discrete-time signals and systems

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C H A P T E R

9 Sampling and quantization

Part II of the book covered techniques for the analysis of continuous-time (CT)

signals and systems. In Part III, we consider the corresponding analysis tech-

niques for discrete-time (DT) sequences and systems. A DT sequence may

occur naturally. Examples are the one-dimensional (1D) hourly measurements

x[k] made with an electronic thermometer, or the two-dimensional (2D) image

x[m, n] recorded with a digital camera, as illustrated earlier in Fig. 1.1. Alter-

natively, a DT sequence may be derived from a CT signal by a process known as

sampling. A widely used procedure for processing CT signals consists of trans-

forming these signals into DT sequences by sampling, processing the resulting

DT sequences with DT systems, and converting the DT outputs back into the

CT domain. This concept of DT processing of CT signals is illustrated by the

schematic diagram shown in Fig. 9.1. Here, the input CT signal x(t) is con-

verted to a DT sequence x[k] by the sampling module, also referred to as the

A/D converter. The DT sequence is then processed by the DT system module.

Finally, the output y[k] of the DT module is converted back into the CT domain

by the reconstruction module. The reconstruction module is also referred to as

the D/A converter. Although the intermediate waveforms, x[k] and y[k], are

DT sequences, the overall shaded block may be considered as a CT system

since it accepts a CT signal x(t) at its input and produces a CT output y(t). If

the internal working of the shaded block is hidden, one would interpret that the

overall operation of Fig. 9.1 results from a CT system.

In practice, a CT signal can either be processed by using a full CT setup,

in which the individual modules are themselves CT systems (as explained in

Chapters 3–8), or by using a CT–DT hybrid setup (as shown in Fig. 9.1). Both

approaches have advantages and disadvantages. The primary advantage of CT

signal processing is its higher speed as DT systems are not as fast as their

counterparts in the CT domain due to limits on the sampling rate of the A/D

converter and the clock rate of the processor used to implement the DT systems.

In spite of its limitation in speed, there are important advantages with DT sig-

nal processing, such as improved flexibility, self-calibration, and data-logging.

Whereas CT systems have a limited performance range, DT systems are more

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394 Part III Discrete-time signals and systems

samplingx(t) DT

system reconst. y(t)

xx[k] y[k]

Fig. 9.1. Processing CT signals

using DT systems.

flexible and can be reprogrammed such that the same hardware can be used in a

variety of different applications. In addition, the characteristics of CT systems

tend to vary with changes in the operating conditions and with age. The DT

systems have no such problems as the digital hardware used to implement these

systems does not drift with age or with changes in the operating conditions and,

therefore, can be self-calibrated easily. Digital signals, obtained by quantizing

DT sequences, are less sensitive to noise and interference than analog signals

and are widely used in communication systems. Finally, the data available from

the DT systems can be stored in a digital server so that the performance of the

system can be monitored over a long period of time. In summary, the advan-

tages of the DT system outweigh their limitations in most applications. Until

the late 1980s, most signal processing applications were implemented with CT

systems constructed with analog components such as resistors, capacitors, and

operational amplifiers. With the recent availability of cheap digital hardware,

it is a common practice now to perform signal processing in the DT domain

based on the hybrid setup shown in Fig. 9.1.

Although, a CT–DT hybrid setup similar to Fig. 9.1 is advantageous in many

applications, care should be taken during the design stage. For example, during

the sampling process some loss of information is generally inevitable. Conse-

quently, if the system is not designed properly, the performance of a CT–DT

hybrid setup may degrade significantly as compared with a CT setup. In this

chapter, we focus on the analysis of the sampling process and the converse

step of reconstructing a CT signal from its DT version. In addition, we also

analyze the process of quantization for converting an analog signal to a digi-

tal signal. Both time-domain and frequency-domain analyses are used where

appropriate.

The organization of Chapter 9 is as follows. Section 9.1 introduces the

impulse-train sampling process and derives a necessary condition, referred to as

the sampling theorem, under which a CT signal can be perfectly reconstructed

from its sampled DT version. We observe that violating the sampling theorem

leads to distortion or aliasing in the frequency domain. Section 9.2 introduces

the practical implementations for impulse-train sampling. These implementa-

tions are referred to as pulse-train sampling and zero-order hold.

In Section 9.3, we introduce another discretization process called quantiza-

tion, which, in conjunction with sampling, converts a CT signal into a digital

signal. In Section 9.4, we present an application of sampling and quantization

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395 9 Sampling and quantization

used in recording music on a compact disc (CD). Finally, Section 9.5 con-

cludes our discussion with a summary of the key concepts introduced in the

chapter.

9.1 Ideal impulse-train sampling

In this section, we consider sampling of a CT signal x(t) with a bounded CTFT

X (ω) such that

X (ω) = 0 for |ω| > 2πβ. (9.1)

A CT signal x(t) satisfying Eq. (9.1) is referred to as a baseband signal, which

is band-limited to 2πβ radians/s or β Hz. In the following discussion, we prove

that a baseband signal x(t) can be transformed into a DT sequence x[k] with

no loss of information if the sampling interval Ts satisfies the criterion that

Ts ≤ 1/2β. To derive the DT version of the baseband signal x(t), we multiply x(t) by an

impulse train:

s(t) = ∞∑

k=−∞ δ(t − kTs), (9.2)

where Ts denotes the separation between two consecutive impulses and is called

the sampling interval. Another related parameter is the sampling rate ωs, with

units of radians/s, which is defined as follows:

ωs = 2π

Ts . (9.3)

Mathematically, the resulting sampled signal, xs(t) = x(t) · s(t), is given by

xs(t) = x(t) ∞∑

k=−∞ δ(t − kTs) =

∞∑

k=−∞ x(kTs)δ(t − kTs). (9.4)

Figure 9.2 illustrates the time-domain representation of the process of the

impulse-train sampling. Figure 9.2(a) shows the time-varying waveform repre-

senting the baseband signal x(t). In Figs. 9.2(b) and (c), we plot the sampled

signal xs(t) for two different values of the sampling interval. In Fig. 9.2(b), the

sampling interval Ts = T and the sampled signal xs(t) provides a fairly good approximation of x(t). In Fig. 9.2(c), the sampling interval Ts is increased to

2T . With Ts set to a larger value, the separation between the adjacent samples

in xs(t) increases. Compared to Fig. 9.2(b), the sampled signal in Fig. 9.2(c)

provides a coarser representation of x(t). The choice of Ts therefore determines

how accurately the sampled signal xs(t) represents the original CT signal x(t).

To determine the optimal value of Ts, we consider the effect of sampling in the

frequency domain.

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396 Part III Discrete-time signals and systems

x(t)

t 0 2T 4T 6T−2T−4T−6T

xs(t) with Ts = T

xs(t) with Ts = 2T

t 0 2T 4T 6T−2T−4T−6T

(a) (b)

(c)

Fig. 9.2. Time-domain

illustration of sampling as a

product of the band-limited

signal and an impulse train.

(a) Original signal x(t );

(b) sampled signal xs(t ) with

sampling interval T s = T ; (c) sampled signal xs(t ) with

sampling interval T s = 2T .

Calculating the CTFT of Eq. (9.4), the CTFT Xs(ω) of the sampled signal

xs(t) is given by

Xs(ω) = ℑ

{

x(t) ∞∑

k=−∞ δ(t − kTs)

}

= 1

2π F{x(t)} ∗ ℑ

{ ∞∑

k=−∞

δ(t − kTs)

}

= 1

2π

[

X (ω) ∗ 2π

∞∑

m=−∞

(

ω − 2mπ

) ]

= 1

∞∑

m=−∞

(

ω − 2mπ

)

(9.5)

where ∗ denotes the CT convolution operator. In deriving Eq. (9.5), we used

the following CTFT pair:

∞∑

k=−∞

δ(t − kTs) CTFT ←→

2π

∞∑

m=−∞

(

ω − 2mπ

)

based on entry (19) of Table 5.2. Equation (9.5) implies that the spectrum Xs(ω)

of the sampled signal xs(t) is a periodic extension, consisting of the shifted

replicas of the spectrum X (ω) of the original baseband signal x(t). Figure 9.3

illustrates the frequency-domain interpretation of Eq. (9.5). The spectrum of the

original signal x(t) is assumed to be an arbitrary trapezoidal waveform and is

shown in Fig. 9.3(a). The spectrum Xs(ω) of the sampled signal xs(t) is plotted

X(w)

2pb−2pb

Xs(w) with ws ≥ 4pb Xs(w) with ws < 4pb

2pb−2pb ws−ws

1/Ts

2pb ws−ws−2ws 2ws

(ws − 2pb)

1/Ts

(a) (b) (c)

w ww

Fig. 9.3. Frequency-domain

illustration of the impulse-train

sampling. (a) Spectrum X(ω) of

the original signal x(t );

(b) spectrum Xs(ω) of the

sampled signal xs(t ) with

sampling rate ωs ≥ 4πβ ; (c)

spectrum Xs(ω) of the sampled

signal xs(t ) with sampling rate

ωs < 4πβ .

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397 9 Sampling and quantization

in Figs. 9.3(c) and (d) for the following two cases:

case I ωs ≥ 4πβ;

case II ωs < 4πβ.

When the sampling rate ωs ≥ 4πβ, no overlap exists between consecutive repli-

cas in Xs(ω). However, as the sampling rate ωs is decreased such that ωs < 4πβ,

adjacent replicas overlap with each other. The overlapping of replicas is referred

to as aliasing, which distorts the spectrum of the original baseband signal x(t)

such that x(t) cannot be reconstructed from its samples. To prevent aliasing, the

sampling rate ωs ≥ 4πβ. This condition is referred to as the sampling theorem

and is stated in the following.†

Sampling theorem A baseband signal x(t), band-limited to 2πβ radians/s, can

be reconstructed accurately from its samples x(kT) if the sampling rate ωs, in

radians/s, satisfies the following condition:

ωs ≥ 4πβ. (9.6a)

Alternatively, the sampling theorem may be expressed in terms of the sampling

rate fs = ωs/2π in samples/s, or the sampling interval Ts. To prevent aliasing,

sampling rate (samples/s) fs ≥ 2β; (9.6b)

sampling interval Ts ≤ 1/2β. (9.6c)

The minimum sampling rate fs (Hz) required for perfect reconstruction of the

original band-limited signal is referred to as the Nyquist rate.

The sampling theorem is applicable for baseband signals, where the sig-

nal contains low-frequency components within the range 0 − β Hz. In some

applications, such as communications, we come across bandpass signals that

also contain a band of frequencies, but the occupied frequency range lies

within the band β2 − β1 Hz with β1 = 0. In these cases, although the max-

imum frequency of β2 Hz implies the Nyquist sampling rate of 2β2 Hz it

is possible to achieve perfect reconstruction with a lower sampling rate (see

Problem 9.8).

† The sampling theorem was known in various forms in the mathematics literature before its

application in signal processing, which started much later, in the 1950s. Several people

developed independently or contributed towards its development. Notable contributions,

however, were made by E. T. Whittaker (1873–1956), Harry Nyquist (1889–1976), Karl

Küpfmüller (1897–1977), V. A. Kotelnikov (1908–2005), Claude Shannon (1916–2001), and

I. Someya.

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398 Part III Discrete-time signals and systems

9.1.1 Reconstruction of a band-limited signal from its samples

Figure 9.3(b) illustrates that the CTFT Xs(ω) of the sampled signal xs(t) is a

periodic extension of the CTFT of the original signal x(t). By eliminating the

replicas in Xs(ω), we should be able to reconstruct x(t). This is accomplished

by applying the sampled signal xs(t) to the input of an ideal lowpass filter (LPF)

with the following transfer function:

H (ω) = {

Ts |ω| ≤ ωs/2 0 elsewhere.

(9.7)

The CTFT Y (ω) of the output y(t) of the LPF is given by Y (ω) = Xs(ω)H (ω),

and therefore all shifted replicas at frequencies ω > ωs/2 are eliminated. All

frequency components within the pass band ω ≤ ωs/2 of the LPF are amplified

by a factor of Ts to compensate for the attenuation of 1/Ts introduced during

sampling. The process of reconstructing x(t) from its samples in the frequency

domain is illustrated in Fig. 9.4. We now proceed to analyze the reconstruction

process in the time domain.

According to the convolution property, multiplication in the frequency

domain transforms to convolution in the time domain. The output y(t) of

the lowpass filter is therefore the convolution of its impulse response h(t)

with the sampled signal xs(t). Based on entry (17) of Table 5.2, the impulse

response of an ideal lowpass filter with the transfer function given in Eq. (9.7) is

given by

h(t) = sinc

( ωst

2π

)

. (9.8)

Convolving the impulse response h(t) with the sampled signal, xs(t) = ∞∑

k=−∞

x(kTs)δ(t − kTs) yields

y(t) = sinc

( ωst

2π

)

∗

∞∑

k=−∞

x(kTs)δ(t − kTs), (9.9)

which reduces to

y(t) = ∞∑

k=−∞

x(kTs)

[

sinc

( ωst

2π

)

∗ δ(t − kTs)

]

(9.10)

w 0

Y(w)

2pb−2pb

w 0

Xs(w) H(w)

2pb−2pb ws−ws

1/Ts

w 0−ws/2 ws/2

Fig. 9.4. Reconstruction of the

original baseband signal x(t ) by

ideal lowpass filtering.

(a) Spectrum of the sampled

signal xs(t ); (b) transfer function

H(ω) of the lowpass filter;

reconstructed signal x(t ).

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399 9 Sampling and quantization

0 Ts 2Ts 3Ts 4Ts3Ts2TsTs−Ts −Ts−2Ts −2Ts−3Ts −3Ts−4Ts

xs(t) ( )wst2ph(t) = sinc

(a)

(c)

(b)

t t

4Ts3Ts2TsTs−Ts−2Ts−3Ts−4Ts 0

y(t)

Fig. 9.5. Reconstruction of the

band-limited signal in the time

domain. (a) Sampled signal

xs(t ); (b) impulse response h(t )

of the lowpass filter;

obtained by convolving xs(t )

with h(t ).

y(t) = ∞∑

k=−∞

x(kTs)

[

sinc

( ωs(t − kTs)

2π

)]

. (9.11)

Equation (9.11) implies that the original signal x(t) is reconstructed by adding

a series of time-shifted sinc functions, whose amplitudes are scaled according

to the values of the samples at the center location of the sinc functions. The

sinc functions in Eq. (9.11) are called the interpolating functions and the over-

all process is referred to as the band-limited interpolation. The time-domain

interpretation of the reconstruction of the original band-limited signal x(t) is

illustrated in Fig. 9.5. At t = kTs, only the kth sinc function, with amplitude

x(kTs), is non-zero. The remaining sinc functions are all zero. The value of the

reconstructed signal at t = kTs is therefore given by x(kTs). In other words,

the values of the reconstructed signal at the sampling instants are given by the

respective samples. The values in between two samples are interpolated using

a linear combination of the time-shifted sinc functions.

Example 9.1

Consider the following sinusoidal signal with the fundamental frequency f0 of

4 kHz:

g(t) = 5 cos(2π f0t) = 5 cos(8000π t).

(i) The sinusoidal signal is sampled at a sampling rate fs of 6000 samples/s

and reconstructed with an ideal LPF with the following transfer function:

H1(ω) =

{

1/6000 |ω| ≤ 6000π

0 elsewhere.

Determine the reconstructed signal.

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400 Part III Discrete-time signals and systems

(ii) Repeat (i) for a sampling rate fs of 12 000 samples/s and an ideal LPF with

the following transfer function:

H2(ω) = {

1/12 000 |ω| ≤ 12 000π 0 elsewhere.

Solution

(i) The CTFT G(ω) of the sinusoidal signal g(t) is given by

G(ω) = 5π [δ(ω − 8000π ) + δ(ω + 8000π )].

Using Eq. (9.4), the CTFT Gs(ω) of the sampled signal with a sampling rate

ωs = 2π (6000) radians/s (Ts = 1/6000 s) is expressed as follows:

Gs(ω) = 6000 ∞∑

m=−∞

G(ω − 2πm(6000)) = 6000 ∞∑

m=−∞

G(ω − 12 000mπ ).

Substituting the value of G(ω) in the above expression yields

Gs(ω) = 6000 ∞∑

m=−∞

5π [δ(ω − 8000π − 12 000 mπ )

+ δ(ω + 8000π − 12 000 mπ )]

= 6000(5π )



· · · + δ(ω + 16 000π ) + δ(ω + 32 000π ) ︸︷︷︸

m=−2

+ δ(ω + 4000π ) + δ(ω + 20 000π ) ︸︷︷︸

m=−1

+ δ(ω − 8000π ) + δ(ω + 8000π ) ︸︷︷︸

m=0

+ δ(ω − 20 000π ) + δ(ω − 4000π ) ︸︷︷︸

m=1

+ δ(ω − 32 000π ) + δ(ω − 16 000π ) ︸︷︷︸

m=2

+ · · ·



 .

When the sampled signal is passed through the ideal LPF with transfer func-

tion H1(ω), all frequency components |ω| > 6000π radians/s) are eliminated

from the output. The CTFT Y (ω) of the output y(t) of the LPF is given

Y (ω) = H1(ω)Gs(ω) = 1

6000 · 6000(5π )[δ(ω + 4000π ) + δ(ω − 4000π )].

Calculating the inverse CTFT, the reconstructed signal is given by y(t) =

5 cos(4000π t).

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401 9 Sampling and quantization

G(w)

(× 1000p)8 16 24−8−16−24 0

S(w)

2p(6000)

…

(× 1000p)8 16 24−8−16−24 0

5p(6000) Gs(w)

(× 1000p)8 16 24−8−16−24 0

Y(w)

5p H1(w)

(× 1000p)8 16 24−8−16−24 0

…

(a)

(c)

(b)

(d)

w w

Fig. 9.6. Sampling and

reconstruction of a sinusoidal

signal g(t ) = 5 cos(8000πt ) at a sampling rate of

6000 samples/s. CTFTs of:

(a) the sinusoidal signal g(t );

(b) the impulse train s(t ); (c) the

sampled signal gs (t );

and (d) the signal reconstructed

with an ideal LPF H 1(ω) with a

cut-off frequency of

6000π radians/s.

The graphical representation of the sampling and reconstruction of the sinu-

soidal signal in the frequency domain is illustrated in Fig. 9.6. The CTFTs of

the sinusoidal signal g(t) and the impulse train s(t) are plotted, respectively,

in Fig. 9.6(a) and Fig. 9.6(b). Since the CTFT S(ω) of s(t) consists of several

impulses, the CTFT Gs(ω) of the sampled signal gs(t) is obtained by convolving

the CTFT G(ω) of the sinusoidal signal g(t) separately with each impulse in

Gs(ω) and then applying the principle of superposition. To emphasize the results

of individual convolutions, a different pattern is used in Fig. 9.6(b) for each

impulse in S(ω). For example, the impulse δ(ω) located at origin in S(ω) is

shown in Fig. 9.6(b) by a solid line. Convolving G(ω) with δ(ω) results in two

impulses located at ω = ±8000π , which are shown in Fig. 9.6(c) by solid lines. Similarly for the other impulses in S(ω).

The output y(t) is obtained by applying Gs(ω) to the input of an ideal LPF

with a cut-off frequency of 6000π radians/s. Clearly, only the two impulses at

ω = ± 4000π , corresponding to the sinusoidal signal cos(4000π t), lie within the pass band of the lowpass filter. The remaining impulses are eliminated from

the output. This results in an output, y(t) = cos(4000π t), which is different from the original signal.

(ii) The CTFT Gs(ω) of the sampled signal with ωs = 2π (12 000) radians/s (Ts = 1/12 000 s) is given by

Gs(ω) = 12 000 ∞∑

m=−∞

G(ω − 2πm(12 000))

= 12 000

∞∑

m=−∞

G(ω − 24 000mπ ).

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402 Part III Discrete-time signals and systems

Substituting the value of the CTFT G(ω) = 5π [δ(ω − 8000π ) + δ(ω + 8000π )] in the above equation, we obtain

Gs(ω) = 12 000 ∞∑

m=−∞

5π [δ(ω − 8000π − 24 000mπ )

+ δ(ω + 8000π − 24 000mπ )]

= 12 000(5π )



· · · + δ(ω + 40 000π ) + δ(ω + 56 000π ) ︸︷︷︸

m=−2

+ δ(ω + 16 000π ) + δ(ω + 32 000π ) ︸︷︷︸

m=−1

+ δ(ω − 8000π ) + δ(ω + 8000π ) ︸︷︷︸

m=0

+ δ(ω − 32 000π ) + δ(ω − 16 000π ) ︸︷︷︸

m=1

+ δ(ω − 56 000π ) + δ(ω − 40 000π ) ︸︷︷︸

m=2

+ · · ·



 .

To reconstruct the original sinusoidal signal, the sampled signal is passed

through an ideal LPF H2(ω). The frequency components outside the pass-band

range |ω| ≤ 12 000π radians/s are eliminated from the ouput. The CTFT Y (ω)

of the output y(t) of the LPF is therefore given by

Y (ω) = 5π [δ(ω + 8000π ) + δ(ω − 8000π )],

which results in the reconstructed signal

y(t) = 5 cos(8000π t).

The graphical interpretation of the aforementioned sampling and reconstruction

process is illustrated in Fig. 9.7.

As the signal g(t) is a sinusoidal signal with frequency 4 kHz, the Nyquist

sampling rate is 8 kHz. In part (i), the sampling rate (6 kHz) is lower than the

Nyquist rate, and consequently the reconstructed signal is different from the

original signal due to the aliasing effect. In part (ii), the sampling rate is higher

than the Nyquist rate, and as a result the original sinusoidal signal is accurately

reconstructed.

9.1.2 Aliasing in sampled sinusoidal signals

As demonstrated in Example 9.1, undersampling of a baseband signal at a

sampling rate less than the Nyquist rate leads to aliasing. Under such conditions,

perfect reconstruction of the baseband signal is not possible from its samples.

In this section, we consider undersampling of a sinusoidal signal

x(t) = cos(2π f0t)

with a fundamental frequency of f0 Hz. The sampling rate fs, in samples/s,

is assumed to be less than the Nyquist rate of 2 f0, i.e. fs < 2 f0. We show

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403 9 Sampling and quantization

Y(w)

5p H2(ω)

w (×1000p)8 16 24−8−16−24 0 32−32

G(w)

(× 1000p)8 16 24−8−16−24 0 32−32

S(w) 12 000

…

8 16 24−8−16−24 0 32−32

5π(12 000)

Gs(w)

(× 1000p)8 16 24−8−16−24 0 32−32

w (× 1000p)

(a) (b)

Fig. 9.7. Sampling and

reconstruction of a sinusoidal

signal g(t ) = 5 cos(8000π t ) at a sampling rate of

12 000 samples/s. CTFTs of:

(a) the sinusoidal signal g(t );

(b) the impulse train s(t );

(d) the signal reconstructed with

an ideal LPF H2(ω) with a cut-off

frequency of 12 000π radians/s.

that the reconstructed signal is sinusoidal but with a different fundamental

frequency.

Using Eq. (9.4), the CTFT Xs(ω) of the sampled sinusoidal signal xs(t) is

given by

Xs(ω) = fs ∞∑

m=−∞

X (ω − 2mπ fs). (9.12)

In Eq. (9.12), we substitute the CTFT, X (ω) = π [δ(ω – 2π f0) + δ(ω + 2π f0)],

of the sinusoidal signal x(t). The resulting expression is as follows:

Xs(ω) = π fs

∞∑

m=−∞

δ(ω + 2π ( f0 − m fs)) + π fs

∞∑

k=−∞

δ(ω − 2π ( f0 + k fs)).

(9.13)

To reconstruct x(t), the sampled signal xs(t) is filtered with an ideal LPF with

transfer function

H (ω) =

{

Ts |ω| ≤ π fs 0 elsewhere.

(9.14)

Within the pass band |ω| ≤ π fs of the LPF, the input frequency components are

amplified by a factor of Ts or 1/ fs. All frequency components within the stop

band |ω| > π fs are eliminated from the reconstructed signal y(t). In addition,

the CT FT of the reconstructed signal y(t) satisfies the following properties.

(1) The CTFT Y (ω) consists of impulses located at frequencies ω = −2π ( f0 −

m fs) and ω = 2π ( f0 + k fs), where m and k are integers such that |( f0 −

m fs)| ≤ fs/2 and |( f0 + k fs)| ≤ fs/2. Since the two conditions are satisfied

only for m = −k, the locations of the impulses are given by ω = ±2π ( f0 −

m fs).

1) fs)| > fs/2. Combined with (1), this implies that only two impulses at

ω = ±2π ( f0 − m fs) will be present in Y (ω).

(3) Each impulse in Y (ω) will have a magnitude (enclosed area) of π .

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404 Part III Discrete-time signals and systems

Based on properties (1)–(3) listed above, the spectrum of the reconstructed

signal is given by

Y (ω) = π [δ(ω + 2π ( f0 − m fs)) + δ(ω − 2π ( f0 − m fs))]. (9.15)

Calculating the inverse CTFT of Eq. (9.15) leads to the following sinusoidal

signal:

y(t) = cos(2π ( f0 − m fs)t), (9.16)

where m is an integer such that |( f0 − m fs)| ≤ fs/2.

Lemma 9.1 If a sinusoidal signal x(t) = cos(2π f0t) is undersampled such that

the sampling rate fs < 2 f0, then the signal reconstructed with an ideal LPF,

with pass band |ω| ≤ π fs, is another sinusoidal signal

y(t) = cos(2π ( f0 − m fs)t),

where m is a positive integer satisfying the condition |( f0 − m fs)| < fs/2.

In Example 9.1(i), for example, the fundamental frequency f0 = 4000 Hz and

the sampling rate fs = 6000 samples/s is less than the Nyquist rate. Selecting

m = 1, the reconstructed signal y(t) is given by

y(t) = cos(2π ( f0 − m fs)t) = cos(2π (4000 − 6000)t) = cos(4000π t).

The result obtained from Lemma 9.1 is in agreement with the expression derived

in Example 9.1(i).

Example 9.2

A signal generator produces a sinusoidal tone x(t) = cos(2π f0t) with funda-

mental frequency f0 between 1 Hz and 1000 kHz. The signal is sampled with a

sampling rate fs = 6000 samples/s and is reconstructed using an ideal LPF with

a cut-off frequency ωc = π fs = 6000π radians/s. Determine the reconstructed

signal for f0 = 500 Hz, 2.5 kHz, 2.8 kHz, 3.2 kHz, 3.5 kHz, 7 kHz, 10 kHz,

20 kHz, and 1000 kHz.

Solution

Table 9.1 lists the reconstructed signals obtained by applying Lemma 9.1. The

sampling frequency fs in the top three entries of Table 9.1 satisfies the sampling

theorem. Therefore, the original signal is reconstructed without any distortion.

In the remaining entries, the sampling theorem is violated. Lemma 9.1 is used

to determine the fundamental frequency of the reconstructed sinusoidal sig-

nal, which is different from that of the original signal due to aliasing. The

reconstructed signals are tabulated in entries (4)–(9) of Table 9.1. An inter-

esting observation is that the reconstructed signals for the sinusoidal wave-

forms x(t) = cos(5600π t) and x(t) = cos(6400π t), listed in entries (3)–(4)

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405 9 Sampling and quantization

Table 9.1. Signals reconstructed from samples of a sinusoidal tone x (t ) = cos(2π f0t ) for different values of the fundamental frequency f0; the sampling frequency fs is kept constant at 6000 samples/s

Funadmental Original Reconstructed

frequency ( f0) signal |( f0 − m fs)| < fs/2 signal Comments

(1) 500 Hz cos(1000π t) fs > 2 f0 cos(1000π t) no aliasing

(2) 2.5 kHz cos(5000π t) fs > 2 f0 cos(5000π t) no aliasing

(3) 2.8 kHz cos(5600π t) fs > 2 f0 cos(5600π t) no aliasing

(4) 3.2 kHz cos(6400π t) |3200 − 1 × 6000| cos(5600π t) aliasing (5) 3.5 kHz cos(7000π t) |3500 − 1 × 6000| cos(5000π t) aliasing (6) 7 kHz cos(14000π t) |7000 − 1 × 6000| cos(2000π t) aliasing (7) 10 kHz cos(20000π t) |10000 − 2 × 6000| cos(4000π t) aliasing (8) 20 kHz cos(40000π t) |20000 − 3 × 6000| cos(4000π t) aliasing (9) 1000 kHz cos(2 × 106π t) |106 − 167 × 6000| cos(4000π t) aliasing

of Table 9.1, are identical. Similarly, the reconstructed signals for the sinu-

soidal waveforms x(t) = cos(5000π t) and x(t) = cos(7000π t), listed in entries (2) and (5) of Table 9.1, are also identical. Finally, the reconstructed signals

for the sinusoidal waveforms x(t) = cos(20 000π t), x(t) = cos(40 000π t), and x(t) = cos(2 × 106π t), listed in entries (7)–(9) of Table 9.1, are the same. The identical waveforms are the consequences of aliasing.

9.2 Practical approaches to sampling

Section 9.1 introduced the impulse-train sampling used to derive the DT version

of a band-limited CT signal. In practice, impulses are difficult to generate and

are often approximated by narrow rectangular pulses. The resulting approach

is referred to as pulse-train sampling, which is discussed in Section 9.2.1. A

second practical implementation, referred to as the zero-order hold, is discussed

in Section 9.2.2.

9.2.1 Pulse-train sampling

In pulse-train sampling, the impulse train s(t) is approximated by a rectangular

pulse train of the form

r (t) = ∞∑

k=−∞

p1(t − kTs) =

[

p1(t) ∗ ∞∑

k=−∞

δ(t − kTs)

]

, (9.17)

where p1(t) represents a rectangular pulse of duration τ ≪ Ts, which is given

p1(t) = rect

( t

)

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x(t) ∑ ∞

k = −∞

p(t−kTs)r(t) =

0 Ts 2Ts 3Ts−Ts−2Ts−3Ts

xs(t)

(a) (b)

(c)

Fig. 9.8. Time-domain

illustration of the pulse-train

sampling of a CT signal.

(a) Original signal x(t ); (b) pulse

train r(t ); (c) sampled signal

xs(t ) = x(t )r(t ).

As in impulse-train sampling, the sampled signal xs(t) is obtained by multiply-

ing the reference signal x(t) by r (t) such that

xs(t) = x(t)r (t) = x(t)

[

p1(t) ∗ ∞∑

k=−∞

δ(t − kTs)

]

. (9.18)

Based on Eq. (9.18), the time-domain representation of the process of pulse-

train sampling is shown in Fig. 9.8. The sampled signal, shown in Fig. 9.8(c),

consists of several pulses of duration τ . The magnitude of the rectangular pulses

in xs(t) follows the reference signal x(t) within the duration of the pulses.

To analyze the process in the frequency domain, we consider the CTFS

expansion of the periodic pulse train. The exponential CTFS representation of

r (t) is given by(see Example 9.14)

r (t) = ∞∑

n=−∞

Dne jnωst with Dn =

ωsτ

2π sinc

(nωsτ

2π

)

, (9.19)

where ωs is the sampling rate in radians/s and is given by ωs = 2π fs = 2π /Ts.

the CTFT of r (t) is given by

R(ω) = 2π ∞∑

n=−∞

Dnδ(ω − nωs) with Dn = ωsτ

2π sinc

( nωsτ

2π

)

. (9.20)

Based on Eq. (9.18), the CTFT Xs(ω) of sampled signal xs(t) is given by

Xs(ω) = 1

2π X (ω) ∗ R(ω). (9.21a)

Substituting the value of R(ω) from Eq. (9.20) yields

Xs(ω) = 1

2π X (ω) ∗ R(ω) =

∞∑

n=−∞

Dn X (ω − nωs). (9.21b)

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407 9 Sampling and quantization

R(w)

0 2pb−2pb ws−ws

2ws 3ws−3ws −2ws

Xs(w) with ws > 2pb

0 2pb−2pb ws−ws

2ws 3ws−3ws −2ws

(a)

(b)

X(w)

2pb−2pb

(c)

Fig. 9.9. Frequency-domain

illustration of the pulse-train

sampling of a CT signal.

Spectrum of (a) the original

signal x(t ); (b) the pulse train

r(t ); (c) the sampled signal

xs(t ) = x(t )r(t ).

Based on Eq. (9.21b), Fig. 9.9 illustrates the frequency-domain interpretation

of the pulse-train sampling. The spectrum X (ω) of the original signal x(t) is

shown in Fig. 9.9(a), while the spectrum R(ω) of the pulse train r (t) is shown

in Fig. 9.9(b). The spectrum Xs(ω) of the sampled signal xs(t) is obtained

by convolving X (ω) with R(ω). As shown in Fig. 9.9(c), Xs(ω) consists of

several shifted replicas of X (ω) attenuated with a factor of Dn . Compared to

the impulse-train sampling, the spectra of the two sampled signals are identical

except for a varying attenuation factor of Dn introduced by the pulse-train

sampling.

Reconstruction of the original signal x(t) from the pulse-train sampled signal

xs(t) is achieved by filtering xs(t) with an ideal LPF having a cut-off frequency

ωc = ωs/2 and a gain of 1/D0 in the pass band. The LPF eliminates all shifted replicas present at frequencies |ω| > ωs/2. This leaves a single replica at ω = 0, which is the same as the CTFT of the original signal. For perfect reconstruc-

tion, pulse-train sampling should not introduce any aliasing. To prevent alias-

ing between different replicas, the sampling rate fs must satisfy the sampling

theorem, i.e. ωs = 2π fs ≥ 4πβ.

9.2.2 Zero-order hold

A second practical implementation of sampling is achieved by the sample-and-

hold circuit, which samples the band-limited input signal x(t) at discrete time

(t = kTs) and maintains the sampled value for the next Ts seconds. To pre-

vent aliasing, the sampling interval Ts must satisfy the sampling theorem. This

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408 Part III Discrete-time signals and systems

x(t) xs(t)

0−3Ts 3Ts−2Ts 2Ts−Ts Ts

(a) (b)

Fig. 9.10. Time-domain

illustration of the zero-order

hold operation for a CT signal.

(a) Original signal x(t );

(b) zero-order hold output xs(t ).

zero-order hold operation is illustrated in Fig. 9.10. Unlike the pulse-train sam-

pling, the amplitude of the sampled signal is maintained constant for Ts seconds

until the next sample is taken.

For mathematical analysis, the zero-order hold operation can be modeled by

the following expression:

xs(t) = ∞∑

k=−∞

x(kTs)p2(t − kTs) (9.22a)

xs(t) = p2(t) ∗ ∞∑

k=−∞

x(kTs)δ(t − kTs) = p2(t) ∗

[

x(t) ∞∑

k=−∞

δ(t − kTs)

]

(9.22b)

where p2(t) represents a rectangular pulse given by

p2(t) = rect

( t − 0.5 Ts

)

. (9.23)

Equation (9.22b) models the zero-hold operation and is different from Eq. (9.18)

in two ways. First, the duration of the pulse p2(t) in Eq. (9.22b) is the same as

the sampling interval Ts, whereas the duration of the pulse p1(t) is much smaller

than Ts in pulse-train sampling. Secondly, the order of operation in the sampled

signal xs(t) is different from that used in the corresponding sampled signal in

pulse-train sampling. In Eq. (9.22b), the sampled signal xs(t) is obtained by

convolving p2(t) with a periodic impulse train, which is scaled by the values

of the reference signal at the location of the impulse functions. In Eq. (9.18),

on the other hand, xs(t) is obtained by multiplying the original signal directly

by the periodic pulse train r (t).

The CTFT of Eq. (9.22b) is given by

Xs(ω) = P2(ω) · 1

2π

[

X (ω) ∗ 2π

∞∑

k=−∞

(

ω − 2kπ

) ]

, (9.24)

where P2(ω) denotes the CTFT of the rectangular pulse p2(t). Based on entry

(16) of Table 5.2, the CTFT of p2(t) is given by the following transform pair:

rect

( t − 0.5 Ts

)

CTFT ←→ Ts sinc

( ωTs

2π

)

e−j 0.5 ωTs .

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409 9 Sampling and quantization

Xs(w) with ws > 2pb

∑ ∞

−∞=k Ts

X (w − )2kp

0 2pb−2pb−ws ws 2ws−2ws

0 2pb−2pb ws 2ws−2ws

X(w)

0 2pb−2pb w

(a)

(b)

(c)

Fig. 9.11. Frequency-domain

illustration of the zero-order

hold operation for a CT signal.

CTFTs of the: (a) original signal

x(t ); (b) periodic replicas; and

Substituting the value of P2(ω), Eq. (9.23) can be expressed as follows:

Xs(ω) = e−j 0.5 ωTs sinc (

ωTs

2π

)

· ∞∑

k=−∞

(

ω − 2kπ

)

. (9.25)

Based on Eq. (9.25), Fig. 9.11 illustrates the frequency-domain interpretation

of the zero-hold operation. The spectrum Xs(ω) of the sampled signal is shown

in Fig. 9.11(c), which contains scaled replicas of the CTFT of the original base-

band signal. Unlike the pulse-train sampling, some distortion in the amplitude

is introduced in the central replica located at ω = 0. This distortion can be

minimized by increasing the width of the main lobe of the sinc function in Eq.

(9.25). Since the width of the main lobe is given by 2π /Ts, it is equivalent to

reducing the sampling interval Ts.

To recover the original CT signal, the sampled signal is filtered with an

LPF having a cut-off frequency ωc = ωs/2. Due to the amplitude distortion

introduced in the central replica, ideal lowpass filtering recovers an approximate

version of the original CT signal. For perfect reconstruction, the filter with the

transfer function given by

H (ω) =







sinc(ωTs/2π ) |ω| ≤ ωs/2

0 elsewhere

(9.26)

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410 Part III Discrete-time signals and systems

output

rk+4 rk+4

dk+4 dk+4

rL−1

rL−2 rL−2

dL−1 dL−1

input

dL dL

output

input

r2 r2

d0 d1 d2 dk d0 d1 d2

(a) (b)

Fig. 9.12. Input–output

relationship of an L-level

quantizer used to discretize the

sample values x[kTs] of a DT

sequence x[k ]. (a) Uniform

quantizer; (b) non-uniform

quantizer.

is used. The above filter is referred to as the compensation, or anti-imaging,

filter. Filtering Xs(ω) with the anti-imaging filter introduces a linear phase −ωTs corresponding to the exponential term exp(−jωTs). Inclusion of a linear phase in the frequency domain is equivalent to a delay in the time domain and is

therefore harmless and not considered as a distortion.

9.3 Quantization

The process of sampling, discussed in Sections 9.1 and 9.2, converts a CT signal

x(t) into a DT sequence x[k], with each sample representing the amplitude of

the CT signal x(t) at a particular instant t = kTs. The amplitude x[kTs] of a sample in x[k] can still have an infinite number of possible values. To produce

a true digital sequence, each sample in x[k] is approximated to a finite set

of values. The last step is referred to as quantization and is the focus of our

discussion in this section.

9.3.1 Uniform and non-uniform quantization

Figure 9.12(a) illustrates the input–output relationship for an L-level uniform

quantizer. The peak-to-peak range of the input sequence x[k] is divided uni-

formly into (L + 1) quantization levels {d0, d1, . . . , dL} such that the sepa- ration � = (dm+1 – dm) is the same between any two consecutive levels. The separation � between two quantization levels is referred to as the quantile inter-

val or quantization step size. For a given input, the output of the quantizer is

calculated from the following relationship:

y[k] = rm = 1

2 [dm + dm+1] for dm ≤ x[k] < dm+1 and 0 ≤ m < L .

(9.27)

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411 9 Sampling and quantization

In other words, the quantized value of the input lying within the levels dm and

dm+1 is given by rm , which equals 0.5(dm + dm+1). The quantization levels {d0, d1, . . . , dL} are referred to as the decision levels, while the output levels

{r0, r1, . . . , rL−1} are referred to as the reconstruction levels.

Equation (9.27) approximates the analog sample values by using a finite

number of quantization levels. The approximation introduces a distortion, which

is referred to as the quantization error. The peak value of the quantization error

is one-half of the quantile interval in the positive or negative direction.

The quantizer illustrated in Fig. 9.12(a) is called a uniform quantizer because

the quantization levels are uniformly distributed between the minimum and

maximum ranges of the input sequence. In most practical applications, the

distribution of the amplitude of the input sequence is skewed towards low

values. In speech communication, for example, low speech volumes dominate

the sequence most of the time. Large-amplitude values are extremely rare and

typically occupy only 15% to 25% of the communication time. A uniform

quantizer will be wasteful, with most of the quantization levels rarely used.

In such applications, we use non-uniform quantization, which provides fine

quantization at frequently occurring lower volumes and coarse quantization at

higher volumes. The input–output relationship of a non-uniform quantizer is

shown in Fig. 9.12(b). The quantile interval is small at low values of the input

sequence and large at high values of the sequence.

Example 9.3

Consider an audio recording system where the microphone generates a CT

voltage signal within the range [−1, 1] volts. Calculate the decision and recon- struction levels for an eight-level uniform quantizer.

Solution

For an L = 8 level quantizer with peak-to-peak range of [−1, 1] volts, the quantile interval � is given by

� = 1 − (−1)

8 = 0.25 V.

Starting with the minimum voltage of −1 V, the decision levels dm are uniformly distributed between −1 V and 1 V. In other words,

dm = −1 + m� for 0 ≤ m ≤ L.

Substituting different values of m, we obtain

dm = −1 V, −0.75 V, −0.5 V, −0.25 V, 0 V, 0.25 V,

0.50 V, 0.75 V, 1 V.

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412 Part III Discrete-time signals and systems

x(t)

−1.00

−0.75

−0.50

−0.25

0.25

0.50

0.75

1.00

−0.875

−0.625

−0.375

−0.125

0.125

0.375

0.625

0.875

decision

levels

reconstruction

levels

3-bit

codewords

000

001

010

011

100

101

110

111

PCM output 011 010 000 001 111 111001 110 101 100

Fig. 9.13. Derivation of a PCM

sequence from a CT signal x(t ).

The original CT signal x(t ) is

shown by the dotted line, while

the PCM sequence is shown as a

stem plot.

Using Eq. (9.27), the reconstruction levels rm are given by

rm = −0.875 V, −0.625 V, −0.375 V, −0.125 V, 0.125 V, 0.375 V, 0.625 V, 0.875 V.

The maximum quantization error is one-half of the quantile interval � and is

given by 0.125 V.

9.3.1.1 Pulse code modulation

Pulse code modulation (PCM) is the analog-to-digital conversion of a CT signal,

where the quantized samples of the CT signal are represented by finite-length

digital words. The essential features of PCM are illustrated in Fig. 9.13, where

a CT signal, with a peak-to-peak range of ±1 V, is sampled and quantized by an eight-level uniform quantizer. As derived in Example 9.3, the decision levels

dm are located at [−1 V, −0.75 V, −0.5 V, −0.25 V, 0 V, 0.25 V, 0.50 V, 0.75 V, 1 V], while the corresponding reconstruction levels rm are located at [−0.875 V, −0.625 V, −0.375 V, −0.125 V, 0.125 V, 0.375 V, 0.625 V, 0.875 V]. Since there are eight reconstruction levels, each quantized sample can be encoded by a

minimum of ℓ = log2(L) = log2(8) = 3-bit word. We assign the 3-bit word 000 to the reconstruction level r0 = −0.875 V, 001 to the reconstruction level r1 = −0.625, and so on for the remaining reconstruction levels as shown in Fig. 9.13. The PCM representation of the waveform x(t) shown in Fig. 9.13 is

therefore given by the following bits:

[011 010 001 000 001 111 111 110 101 100],

where the final output is parsed in terms of 3-bit codewords.

9.3.2 Fidelity of quantized signal

In Table 9.2, we list the sampling frequency, the total number of quantization

levels, and the resulting raw (uncompressed) data rate for a number of commer-

cial audio applications. Low-fidelity applications, for example the telephone

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413 9 Sampling and quantization

Table 9.2. Raw data rates for digital audio used in commercial applications

Bandwidth Sampling rate Quantization Raw data rate

Applications (Hz) (samples/s) levels (L) (bytes/s)

Telephone 200–3400 8000 28 8000

AM radio 11 025 28 11 025

FM radio (stereo) 22 050 216 88 200

CD (stereo) 20–20 000 44 100 216 176 400

Digital audio tape (stereo) 20–20 000 48 000 216 192 000

and the AM radio, are sampled at a relatively low sampling rate followed by

a coarse quantizer to generate the PCM sequence. The quality of the recon-

structed audio is moderate in such applications. In high-fidelity applications,

for example the FM radio, compact disc (CD), and digital audio tape (DAT),

the sampling rate is much higher to ensure accurate reconstruction of the high-

frequency components. The number of levels in the quantizer is also increased

to 216 to reduce the effect of the quantization error. Two channels, one for

the right speaker and the other for the left speaker, are transmitted for high-

fidelity applications. Compared to a single channel, the data rate is effectively

doubled with the transmission of two channels. The CD and DAT provide

excellent audio quality and are generally recognized as world standards for

achieving fidelity of audio reproduction that surpasses any other existing tech-

nique. In the following section, we discuss the CD digital audio system in more

detail.

9.4 Compact discs

The compact disc (CD) digital audio system was defined jointly in 1979 by the

Sony Corporation of Japan and the Philips Corporation of the Netherlands. The

most important component of the CD digital audio system is an optical disc

about 120 mm in diameter, which is used as the storage medium for recording

data. The optical disc is referred to as the compact disc (CD) and stores about

1010 bits of data in the form of minute pits. To read data, the CD is optically

scanned with a laser.

Before music can be recorded on a CD, it is preprocessed and converted

into PCM data. The schematic diagram of the preprocessing stage for a single

music channel is illustrated in Fig. 9.14(a). Each channel of the music signal is

amplified and applied at the input of a lowpass filter (LPF), referred to as the anti-

aliasing filter. Since the human ear is only sensitive to frequency components

within the range 20 Hz–20 kHz, the anti-aliasing filter limits the bandwidth of

the input channel to 20 kHz. Following the anti-aliasing filter is the PCM system,

which converts the CT music channel into binary data. The sampling rate used in

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414 Part III Discrete-time signals and systems

pulse code modulation (PCM)

amplifier anti-aliasing

filter

sample and

hold

analog to digital

conversion channel Cm of

CT music signal

digital data

for channel Cm

error

protection

PCM

multiplexer

channel C1

channel C2

channel CN

(a)

(b)

Fig. 9.14. Storing digital music

on a compact disk.

(a) Preprocessing stage to

convert CT music channels into

PCM data. (b) Multiplexing stage

to interleave data from multiple

channels.

the sample-and-hold circuit is 44.1 ksamples/s, which exceeds the Nyquist rate

by a margin of 4.1 ksamples/s. The additional margin reduces the complexity

of the anti-aliasing filter by allowing a fair transition bandwidth between the

pass and stop bands of the filter. The audio samples obtained from the sample-

and-hold circuit are quantized using 216-level uniform quantization. Finally,

each quantized sample is encoded with a 16-bit codeword, which results in a

raw data rate of (44 100 samples/s × 16 bits/sample) = 705.6 kbits per second (kbps) or 705.6/8 = 88.2 kBytes per second (kBps).

For high-fidelity performance, several channels of the music signal are

recorded on a CD. For commonly used stereo systems, only two channels

corresponding to the left and right speakers are recorded. Many home theatre

systems now record a much higher number of channels to simulate surround

sound and other audio effects. Each channel of the music signal is prepro-

cessed by the system illustrated in Fig. 9.14(a) and converted into raw PCM

data. Figure 9.14(b) illustrates the multiplexing stage, where data streams from

different channels are interleaved together into a single continuous bit stream.

The final step in the multiplexing stage is an error control scheme, which adds

an additional layer of protection to the music data. Any scanning errors that

were introduced whilst data were being read out from the CD are concealed

by the error control scheme. The output of the error control circuit is stored on

the CD. To record more music on a single CD, PCM data may be compressed

using an audio compression standard such as MP3.

A CD player reverses each step illustrated in Fig. 9.14. Data read from the

CD is checked for possible scanning errors. After correcting or concealing

the detected errors, the data streams for the individual channels are derived from

the interleaved bit stream. By following the reconstruction procedure outlined

in Section 9.1.1, each data stream is used to reconstruct the corresponding music

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channel. The reconstructed channels are played simultaneously to simulate the

effect of real audio.

Example 9.4

Consider a digital monochrome CCD camera that records an image x[m, n]

at a resolution of 800 × 1200 picture elements (pixels). In other words, each image consists of 800 × 1200 = 0.96 × 106 pixels. Assuming that the human visual system cannot distinguish between more than 200 different shades of

gray, determine how many bytes are required to store a single image. If the

CCD camera has 32 million bytes of memory space to store images, how many

images can be saved simultaneously in the camera?

Solution

An image pixel can have 200 different shades of gray. The number of bits

required to represent the intensity value of each pixel is given by ⌈log2(200)⌉

or ⌈7.64⌉ or 8 bits;

space required to save one image

= 0.96 × 106 pixels × 8 bits/pixel

= 7.68 × 106 bits or 0.96 × 106 bytes.

Since the disc space for storing images is 32 × 106 bytes,

number of images that can be stored simultaneously

= 32 × 106 bytes/0.96 × 106 bytes

= 33.

9.5 Summary

In this chapter, we introduced the principle of sampling that is used to transform

a CT baseband signal into an equivalent DT sequence. Section 9.1 discussed

the ideal impulse-train sampling, where a periodic impulse train is multiplied

by a CT baseband signal, resulting in a sequence of equally spaced samples at

the location of the impulses (t = kTs). In the frequency domain, the spectrum

of the sampled signal consists of several shifted replicas of the spectrum of the

original signal. We observe that the original CT signal is recoverable from its

DT version by ideal lowpass filtering if the sampling rate fs = 1/Ts is greater

than twice the highest frequency present in the baseband signal. This condition

is referred to as the sampling theorem. Violating the sampling theorem distorts

the spectrum of the original baseband signal; a phenomenon known as aliasing.

In practice, impulses are difficult to generate and are often approximated by

narrow rectangular pulses. This leads to a more practical approach to sampling,

covered in Section 9.2, in which a periodic rectangular pulse train is multiplied

by the CT baseband signal to produce the sampled signal. Compared with the

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416 Part III Discrete-time signals and systems

ideal impulse train sampling, the spectra of the two sampled signals are identical,

except that the shifted replicas in the spectrum of the pulse-train are attenuated

by a sinc function. Reconstruction of the original signal in rectangular pulse-

train sampling is also achieved by lowpass filtering the sampled signal. A second

practical implementation of sampling uses a zero-order hold circuit to sample

the CT signal; this is covered in Section 9.2.2.

To encode a CT signal into a digital waveform, Section 9.3 introduces the

process of quantization, in which the values of the samples are approximated to a

finite set of levels. This involves replacing the exact sample value with the closest

level defined by the L-level quantizer. In uniform quantization, the quantization

levels are distributed uniformly between the maximum and minimum ranges

of the input sequence. A uniform quantizer results in high quantization error

in most practical applications, where the distribution of the sample values is

skewed towards low amplitudes. In such cases, most of the quantization levels

in the uniform quantizer are rarely used. A non-uniform quantizer reduces the

overall quantization error by providing finer quantization at frequently occurring

lower amplitudes and coarser quantization at less frequent higher amplitudes.

Sampling is used in a number of important applications. Section 9.4 intro-

duces the compact disc (CD) and illustrates how sampling and quantization are

used to convert an analog music signal into binary data, which can be stored on

a CD. Since digital signals are less sensitive to distortion and interference than

analog signals, the audio CD provides excellent audio quality that surpasses

most analog storage mechanisms.

Problems

9.1 For the following CT signals, calculate the maximum sampling period Ts that produces no aliasing:

(a) x1(t) = 5 sinc(200t); (b) x2(t) = 5 sinc(200t) + 8 sin(100π t); (c) x3(t) = 5 sinc(200t) sin(100π t); (d) x4(t) = 5 sinc(200t) ∗ sin(100π t), where ∗ denotes the CT convolu-

tion operation.

9.2 A famous theorem known as the uncertainty principle states that a baseband signal cannot be time-limited. By calculating the inverse CTFT of the

following baseband signals, show that the uncertainty principle is indeed

satisfied by the following signals (assume that ω0 and W are real, positive

constants):

(a) X1(ω) = rect ( ω

)

e−j2ω;

(b) X2(ω) =

{

1 |ω| ≤ W

0 elsewhere;

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417 9 Sampling and quantization

ω − ω0 2W

)

+ rect (

ω + ω0 2W

)

;

(d) X4(ω) = u(ω − ω0) − u(ω − 2ω0).

9.3 The converse of the uncertainty principle, explained in Problem 9.2, is also true. In other words, a time-limited signal cannot be band-limited. By

calculating the CTFT of the following time-limited signals, show that the

converse of the uncertainty principle is indeed true (assume that τ, T , and

α are real, positive constants):

(a) x1(t) = cos(ω0t)[u(t + T ) − u(t − T )];

(b) x2(t) = rect (

)

∗ rect

( t

)

(∗ denotes the CT convolution operator);

( t

)

;

(d) x4(t) = δ(t − 5) + δ(t + 5).

9.4 The CT signal x(t) = v1(t) v2(t) is sampled with an ideal impulse train:

s(t) = ∞∑

k=−∞

δ(t − kTs).

(a) Assuming that v1(t) and v2(t) are two baseband signals band-limited

to 200 Hz and 450 Hz, respectively, compute the minimum value of

the sampling rate fs that does not introduce any aliasing.

(b) Repeat part (a) if the waveforms for v1(t) and v2(t) are given by the

following expression:

v1(t) = sinc(600t) and v2(t) = sinc(1000t).

x(t) = v1(t)v2(t) specified in part (b), sketch the spectrum of the sam-

pled signal. Can x(t) be accurately recovered from the sampled signal?

(d) Repeat part (c) for a sampling interval of Ts = 0.1 ms.

9.5 The CT signal x(t) = sin(400π t) + 2 cos(150π t) is sampled with an ideal impulse train. Sketch the CTFT of the sampled signal for the following

values of the sampling rate:

(a) fs = 100 samples/s;

(b) fs = 200 samples/s;

(d) fs = 500 samples/s.

In each case, calculate the reconstructed signal using an ideal LPF with the

transfer function given in Eq. (9.7) and a cut-off frequency of ωs/2 = π fs.

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418 Part III Discrete-time signals and systems

9.6 Consider the following CT signal:

x(t) = {

0.25(3 − |t |) 0 ≤ |t | ≤ 3 0 otherwise.

(a) Calculate the CTFT X (ω). Determine the bandwidth of the signal and

the ideal Nyquist sampling rate.

(b) If the bandwidth is infinite, approximate the bandwidth as β Hz, such

that

|X (ω)| < 0.01 max|X (ω)| for |ω| > 2πβ

and recalculate a practical Nyquist sampling rate.

DT sequence x[k] corresponding to the duration −5 ≤ t ≤ 5.

(d) Quantize the signal x[k] obtained in (c) with the uniform quantizer

derived in Example 9.3. Plot the quantization error with respect to k.

What is the maximum value of the quantization error?

(e) Repeat (d) using a uniform quantizer with L = 16 reconstruction levels

defined within the dynamic range [−1, 1]. Plot the quantization error

with respect to k. What is the maximum value of the quantization error?

Compare the plot with your answer obtained in (d).

9.7 Show that the CTFS representation of the rectangular pulse train r (t) as defined in Eq. (9.17) is given by Eq. (9.19).

9.8 The spectrum of a CT signal x(t) satisfies the following conditions:

X (ω) = 0 for |ω| < ω1 or |ω| > ω2 with ω2 > ω1 > 0.

In other words, the CTFT X (ω) of x(t) is non-zero only within the range

of frequencies ω1 ≤ |ω| ≤ ω2. Such a signal is referred to as a bandpass

signal.

(a) Show that the bandpass signal x(t) can be sampled with an ideal

impulse train at a rate less than the Nyquist rate of 2(ω2/2π )

samples/s and can be perfectly reconstructed with a bandpass filter

with the following transfer function:

Hbp(ω) =

{

p ωℓ ≤ |ω| ≤ ωu 0 elsewhere.

(b) Determine the minimum sampling rate for which perfect reconstruction

is possible.

transfer function of the bandpass filter.

9.9 An alternative to the bandpass sampling procedure introduced in Problem 9.8 is the system illustrated in Fig. P9.9. For a real-valued

bandpass signal x(t) with the spectrum shown in Fig. P9.9(a), the

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419 9 Sampling and quantization

X(w)

−w2 −w1 w1 w2

x(t) w

wc−wc

××

j(w 1 +w

2 )t

q(t) = e 2 1−

∑ ∞

d(t − kTs) k=−∞

s(t) =

xs(t)

Hlp(w)

(a) (b)

Fig. P. 9.9. (a) Spectrum of a

bandpass signal x(t ); (b) ideal

sampling of a bandpass

baseband signal.

cut-off frequency of the ideal LPF Hlp(ω) in Fig. P9.9(b) is given by

ωc = 0.5(ω2 − ω1). (a) Sketch the spectrum of the sampled signal xs(t).

(b) Determine the maximum value of the sampling interval Ts that intro-

duces no aliasing. Compare this sampling interval with that obtained

from the Nyquist rate.

signal xs(t).

9.10 An alternative to ideal impulse train sampling is sawtooth wave sampling. Here, a CT signal x(t) is multiplied with a periodic sawtooth wave s(t)

(shown in Fig. P9.10). Denote the resulting signal by z(t) = x(t) ∗ s(t). s(t)

−2Ts −Ts 0 Ts 2Ts t

Fig. P. 9.10. Sawtooth function

used in sawtooth wave

sampling.

(a) Derive an expression for the CTFT Z (ω) of the signal z(t) in terms

of the CTFT of the original signal x(t).

(b) Assuming that the CTFT of the original signal x(t) is shown in

Fig. 9.3(a), sketch the spectrum of the CTFT of the signal z(t).

If yes, state the conditions under which x(t) may be reconstructed.

Sketch the block diagram of the reconstruction system including the

specifications of any filters used.

(d) By comparing the CTFTs, state how z(t) relates to the sampled signal

xs(t) obtained by ideal impulse train sampling.

9.11 Repeat Problem 9.10 with an alternating sign impulse train,

s(t) = ∞∑

k=−∞

(−1)kδ(t − kTs),

as the sampling signal.

9.12 Repeat Problem 9.10 with the periodic signal,

s(t) = ∞∑

k=−∞

[δ(t − kTs) + δ(t − � − kTs)],

as the sampling signal.

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420 Part III Discrete-time signals and systems

9.13 A CT band-limited signal x(t) is sampled at its Nyquist rate fs and trans- mitted over a band-limited channel modeled with the transfer function

Hch(ω) = {

1 4π fs ≤ |ω| ≤ 8π fs 0 otherwise.

Let the signal received at the end of the channel be xch(t). Determine the

reconstruction system that recovers the CT signal x(t) from xch(t).

9.14 If the quantization noise needs to be limited to ±p% of the peak-to-peak value of the input signal, show that the number of bits in each PCM word

must satisfy the following inequality:

n ≥ 3.32 log10

( 50

)

9.15 A voice signal with a bandwidth of 4 kHz and an amplitude range of ±20 mV is converted to digital data using a PCM system.

(a) Determine the maximum sampling interval Ts that can be used to

sample the voice signal.

(b) If the PCM system has an accuracy of ±5% during the quantization

step, determine the length of the codewords in bits.

9.16 A baseband signal with a bandwidth of 100 kHz and an amplitude range of ±1 V is to be transmitted through a channel which is constrained to

a maximum transmission speed of 2 Mbps. Your task is to design a uni-

form quantizer that introduces minimum quantization error. Determine

the maximum number of levels L in the uniform quantizer. What is the

maximum distortion introduced by the uniform quantizer? Assume the

Nyquist rate for sampling.

9.17 Consider the input–output relationship of an ideal sampling system given by

xs(t) = x(t) ∞∑

k=−∞

δ(t − kTs) = ∞∑

k=−∞

x(kTs)δ(t − kTs).

Determine if the ideal sampling system is (i) linear, (ii) time-invariant,

(iii) memoryless, (iv) causal, (v) stable, and (vi) invertible.

9.18 Consider the input–output relationship of a DT quantizer with L decision levels, given by

y[k] = Q{x[k]} = 1

2 [dm + dm+1] for dm ≤ x[k] < dm+1 and

0 ≤ m < L .

Determine if the DT quantizer is (i) linear, (ii) time-invariant, (iii) mem-

oryless, (iv) causal, (v) stable, and (vi) invertible.

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421 9 Sampling and quantization

9.19 Consider a digital mp3 player that has 1024 × 106 bytes of memory. Assume that the audio clips stored in the player have an average duration

of five minutes.

(a) Assuming a sampling rate of 44 100 samples/s and 16 bits/sample/

channel quantization, determine the average storage space required

(without any form of compression) to store a stereo (i.e. two-channel)

audio clip.

(b) Assume that the audio clips are stored in the mp3 format, which

reduces the audio file size to roughly one-eighth of its original size.

Calculate the storage space required to store an mp3-compressed

audio clip.

player?

9.20 Consider a digital color camera with a resolution of 2560 × 1920 pixels. (a) Calculate the storage space required to store an image in the camera

without any compression. Assume three color channels and quanti-

zation of 8 bit/pixel/channel.

(b) Assume that the images are stored in the camera in the JPEG format,

which reduces an image to roughly one-tenth of its original size.

Calculate the storage space required to store a JPEG-compressed

image.

(c) If the camera has 512 × 106 bytes of memory, determine the number of JPEG-compressed images that can be stored in the camera.

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C H A P T E R

10 Time-domain analysis of discrete-time systems

An important subset of discrete-time (DT) systems satisfies the linearity and

time-invariance properties, discussed in Chapter 2. Such DT systems are

referred to as linear, time-invariant, discrete-time (LTID) systems. In this chap-

ter, we will develop techniques for analyzing LTID systems. As was the case

for the LTIC systems discussed in Part II, we are primarily interested in cal-

culating the output response y[k] of an LTID system to a DT sequence x[k]

applied at the input of the system.

In the time domain, an LTID system is modeled either with a linear, constant-

coefficient difference equation or with its impulse response h[k]. Section 10.1

covers linear, constant-coefficient difference equations and develops numer-

ical techniques for solving such equations. Section 10.2 defines the impulse

response h[k] as the output of an LTID system to an unit impulse function δ[k]

applied at the input of the system and shows how the impulse response can

be derived from a linear, constant-coefficient difference equation. Section 10.3

proves that any arbitrary DT sequence can be represented as a linear combina-

tion of time-shifted DT impulse functions. This development leads to a second

approach for calculating the output y[k] based on convolving the applied input

sequence x[k] with the impulse response h[k] in the DT domain. The resulting

operation is referred to as the convolution sum and is defined in Section 10.4.

Section 10.5 introduces two graphical methods for calculating the convolution

sum, and Section 10.6 lists several important properties of the convolution sum.

A special case of convolution sum, referred to as the periodic or circular con-

volution, occurs when the two operands are periodic sequences. Section 10.7

develops techniques for computing the periodic convolution and shows how

it may be used to compute the linear convolution. In Section 10.8, we revisit

the causality, stability, and invertibility properties of LTID systems and express

these properties in terms of the impulse response h[k]. M A T L A B instructions

for computing the convolution sum are listed in Section 10.9. The chapter is

concluded in Section 10.10 with a summary of the important concepts covered

in the chapter.

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423 10 Time-domain analysis of DT systems

10.1 Finite-difference equation representation of LTID systems

As discussed in Section 3.1, an LTIC system can be modeled using a linear,

constant-coefficient differential equation. Likewise, the input–output relation-

ship of a linear DT system can be described using a difference equation, which

takes the following form:

y[k + n] + an−1 y[k + n − 1] + · · · + a0 y[k] = bm x[k + m] + bm−1x[k + m − 1] + · · · + b0x[k], (10.1)

where x[k] denotes the input sequence and y[k] denotes the resulting out-

put sequence, and coefficients ar (for 0 ≤ r ≤ n − 1), and br (for 0 ≤ r ≤ m)

are parameters that characterize the DT system. The coefficients ar and br

are constants if the DT system is also time-invariant. For causal signals and

systems analysis, the following n initial (or ancillary) conditions must be spec-

ified in order to obtain the solution of the nth-order difference equation in

Eq. (10.1):

y[−1], y[−2], . . . , y[−n].

We now consider an iterative procedure for solving linear, constant-coefficient

difference equations.

Example 10.1

The DT sequence x[k] = 2ku[k] is applied at the input of a DT system described

by the following difference equation:

y[k + 1] − 0.4y[k] = x[k].

By iterating the difference equation from the ancillary condition y[−1] = 4,

compute the output response y[k] of the DT system for 0 ≤ k ≤ 5.

Solution

Express y[k + 1] − 0.4y[k] = x[k] as follows:

y[k] = 0.4y[k − 1] + x[k − 1]

= 0.4y[k − 1] + 2(k − 1) u(k − 1) { . .. x[k] = 2k u[k]} ,

which can alternatively be expressed as

y[k] =

{

0.4y[k − 1] k = 0

0.4y[k − 1] + 2(k − 1) k ≥ 1.

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424 Part III Discrete-time signals and systems

y[k]

0 1 2 3 4 5−5 −4 −3 −2 −1

2.3 4.9

7.9

11.2

0.64 1.6 k

y[k]

0 1 2 3 4 5−5 −4 −3 −2 −1

8 6

4 2

(a) (b)

Fig. 10.1. Input and output

sequences for Example 10.1.

(a) Input sequence x[k ];

(b) output sequence y [k ].

By iterating from k = 0, the output response is computed as follows: y[0] = 0.4y[−1] = 1.6, y[1] = 0.4y[0] + 2 × 0 = 0.64, y[2] = 0.4y[1] + 2 × 1 = 2.256, y[3] = 0.4y[2] + 2 × 2 = 4.902, y[4] = 0.4y[3] + 2 × 3 = 7.961, y[5] = 0.4y[4] + 2 × 4 = 11.184.

Additional values of the output sequence for k > 5 can be similarly evaluated

from further iterations with respect to k. The input and output sequences are

plotted in Fig. 10.1 for 0 ≤ k ≤ 5.

In Chapter 3, we showed that the output response of a CT system, represented by

the differential equation in Eq. (3.1), can be decomposed into two components:

the zero-state response and the zero-input response. This is also valid for the

DT systems represented by the difference equation in Eq. (10.1). The output

response y[k] can be expressed as

y[k] = yzi[k] ︸︷︷︸

zero-input response

+ yzs[k], ︸︷︷︸

zero-state response

(10.2)

where yzi[k] denotes the zero-input response (or the natural response) of the

system and yzs[k] denotes the zero-state response (or the forced response) of

the DT system.

The zero-input component yzi[k] for a DT system is the response produced by

the system because of the initial conditions, and is not due to any external input.

To calculate the zero-input component yzi[k], we assume that the applied input

sequence x[k] = 0. On the other hand, the zero-state response yzs[k] arises due

to the input sequence and does not depend on the initial conditions of the system.

To calculate the zero-state response yzs[k], the initial conditions are assumed

to be zero. Based on Eq. (10.2), a DT system represented by Eq. (10.1) can

be considered as an incrementally linear system (see Section 2.2.1) where the

additive offset is caused by the initial conditions (see Fig. 2.10). If the initial

conditions are zero, the DT system becomes on LTID system. We now solve

Example 10.1 in terms of the zero-input and zero-state components of the output.

Example 10.2

Repeat Example 10.1 to calculate (i) the zero-input response yzi[k], (ii) the zero-

state response yzs[k], and (iii) the overall output response y[k] for 0 ≤ k ≤ 5.

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425 10 Time-domain analysis of DT systems

Solution

(i) The zero-input response of the system is obtained by solving the following

difference equation:

y[k + 1] − 0.4y[k] = x[k],

with input x[k] = 0 and ancillary condition y[−1] = 4. The difference equation reduces to

yzi[k] = 0.4yzi[k − 1],

with ancillary condition yzi[−1] = 4. Iterating for k = 0, 1, 2, 3, 4, and 5 yields

yzi[0] = 0.4yzi[−1] = 1.6, yzi[1] = 0.4yzi[0] = 0.64, yzi[2] = 0.4yzi[1] = 0.256, yzi[3] = 0.4yzi[2] = 0.1024, yzi[4] = 0.4yzi[3] = 0.0410, yzi[5] = 0.4yzi[4] = 0.0164.

(ii) The zero-state response of the system is calculated by solving the fol-

lowing difference equation:

yzs[k] = 0.4yzs[k − 1] + 2(k − 1)u[k − 1],

with ancillary condition yzs[−1] = 0. Iterating the difference equation for k = 0, 1, 2, 3, 4, and 5 yields

yzs[0] = 0.4yzs[−1] + 2 × (−1) × 0 = 0, yzs[1] = 0.4yzs[0] + 2 × 0 × 1 = 0, yzs[2] = 0.4yzs[1] + 2 × 1 × 1 = 2, yzs[3] = 0.4yzs[2] + 2 × 2 × 1 = 4.8, yzs[4] = 0.4yzs[3] + 2 × 3 × 1 = 7.92, yzs[5] = 0.4yzs[4] + 2 × 4 × 1 = 11.168.

(iii) Adding the zero-input and zero-state components obtained in parts

(i) and (ii), yields

y[0] = yzi[0] + yzs[0] = 1.6, y[1] = yzi[1] + yzs[1] = 0.64, y[2] = yzi[2] + yzs[2] = 2.256, y[3] = yzi[3] + yzs[3] = 4.902, y[4] = yzi[4] + yzs[4] = 7.961, y[5] = yzi[5] + yzs[5] = 11.184.

Note that the overall output response y[k] is identical to the output response

obtained in Example 10.1. By iterating with respect to k, additional values for

the output response y[k] for k > 5 can be computed.

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426 Part III Discrete-time signals and systems

In Section 10.1, we used a linear, constant-coefficient difference equation to

model an LTID system. A second model is based on the impulse response h[k]

of a system. This alternative representation leads to a different approach for

analyzing LTID systems. Section 10.2 presents this alternative approach.

10.2 Representation of sequences using Dirac delta functions

In this section, we show that any arbitrary sequence x[k] may be represented as

a linear combination of time-shifted, DT impulse functions. Recall that a DT

impulse function is defined in Eq. (1.51) as follows:

δ[k] = {

1 k = 0 0 k �= 0.

(10.3)

We are interested in representing any DT sequence x[k] as a linear combina-

tion of shifted impulse functions, δ[k − m], for −∞ < m < ∞. We illustrate

the procedure using the arbitrary function x[k] shown in Fig. 10.2(a). Figures

10.2(b)–(f) represent x[k] as a linear combination of a series of simple functions

xm[k], for−∞ < m < ∞. Since xm[k] is non-zero only at one location (k = m),

it represents a scaled and time-shifted impulse function. In other words,

xm[k] = x[m]δ[k − m]. (10.4)

In terms of xm[k], the DT sequence x[k] is, therefore, represented by

x[k] = · · · + x−2[k] + x−1[k] + x0[k] + x1[k] + x2[k] + · · ·

= · · · + x[−2]δ[k + 2] + x[−1]δ[k + 1] + x[0]δ[k]

+ x[1]δ[k − 1] + x[2]δ[k − 2] + · · · ,

x[k]

0 32

4 5−5 −4 −3 −2 −1 0 321 4 5−5 −4 −3 −2 −1

x−2[k] = x[−2]d[k+2]

0 321 4 5−5 −4 −3 −2 −1 k

−

x−1[k] = x[−1]d[k+1]

0 321 4 5−5 −4 −3 −2 −1 k

x0[k] = x[0]d[k]

0 32 4 5−5 −4 −3 −2 −1 k

x1[k] = x[1]d[k−1]

0 321 4 5−5 −4 −3 −2 −1 k

x2[k] = x[2]d[k−2]

(a) (b) (c)

(d) (e) (f)

Fig. 10.2. Representation of a

DT sequence as a linear

combination of time-shifted

impulse functions. (a) Arbitrary

sequence x[k ]; (b)–(f) its

decomposition using DT impulse

functions.

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427 10 Time-domain analysis of DT systems

which reduces to

x[k] = ∞∑

m=−∞

x[m]δ[k − m]. (10.5)

Equation (10.5) provides an alternative representation of an arbitrary DT func-

tion using a linear combination of time-shifted DT impulses. In Eq. (10.5),

variable m denotes the dummy variable for the summation that disappears as

the summation is computed. Recall that a similar representation exists for the

CT functions and is given by Eq. (3.24).

10.3 Impulse response of a system

In Section 10.1, a constant-coefficient difference equation is used to specify the

input–output characteristics of an LTID system. An alternative representation of

an LTID system is obtained by specifying its impulse response. In this section,

we will formally define the impulse response and illustrate how the impulse

response of an LTID system can be derived directly from the difference equation

modeling the LTID system.

Definition 10.1 The impulse response h[k] of an LTID system is the output of

the system when a unit impulse δ[k] is applied at the input of the LTID system.

Following the notation introduced in Eq. (2.1b), the impulse response can be

expressed as follows:

δ[k] → h[k], (10.6)

with zero ancillary conditions.

Note that an LTID system satisfies the linearity and the time-shifting properties.

Therefore, if the input is a scaled and time-shifted impulse function aδ[k − k0],

the output, Eq. (10.6), of the DT system is also scaled by a factor of a and

time-shifted by k0, i.e.

aδ[k − k0] → ah[k − k0], (10.7)

for any arbitrary constants a and k0. Section 10.4 illustrates how Eq. (10.7) can

be generalized to calculate the output of LTID systems for any arbitrary input.

Example 10.3

Consider the LTID systems with the following input–output relationships:

(i) y[k] = x[k − 1] + 2x[k − 3]; (10.8)

(ii) y[k + 1] − 0.4y[k] = x[k]. (10.9)

Calculate the impulse responses for the two LTID systems. Also, determine

the output responses of the LTID systems when the input is given by x[k] =

2δ[k] + 3δ[k − 1].

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428 Part III Discrete-time signals and systems

Solution

(i) The impulse response of a system is the output of the system when the input

sequence x[k] = δ[k]. Therefore, the impulse response h[k] of system (i) can be obtained by substituting y[k] by h[k] and x[k] by δ[k] in Eq. (10.8). In other

words, the impulse response for system (i) is given by

h[k] = δ[k − 1] + 2δ[k − 3].

To evaluate the output response resulting from the input sequence x[k] = 2δ[k] + 3δ[k − 1], we use the linearity and time-invariance properties of the system. The outputs resulting from the two terms 2δ[k] and 3δ[k − 1] in the input sequence are as follows:

2δ[k] → 2h[k] = 2δ[k − 1] + 4δ[k − 3]

and

3δ[k − 1] → 3h[k − 1] = 3δ[k − 2] + 6δ[k − 4].

Applying the superposition principle, the output y[k] to input x[k] = 2δ[k] +

3δ[k − 1] is given by

2δ[k] + 3δ[k − 1] → 2h[k] + 3h[k − 1]

y[k] = (2δ[k − 1] + 4δ[k − 3]) + (3δ[k − 2] + 6δ[k − 4])

= 2δ[k − 1] + 3δ[k − 2] + 4δ[k − 3] + 6δ[k − 4]).

(ii) On substituting y[k] by h[k] and x[k] by δ[k] in Eq. (10.9), the impulse

response of the LTID system (ii) is represented by the following recursive

equation:

h[k + 1] − 0.4h[k] = δ[k]. (10.10a)

Equation (10.10a) is a difference equation, which can be solved by substituting

k = m − 1. The resulting equation is given by

h[m] = δ[m − 1] + 0.4h[m − 1]. (10.10b)

To solve for the delayed response h[m − 1], we substitute k = m − 2 in

Eq. (10.10a). The resulting expression is given by

h[m − 1] = δ[m − 2] + 0.4h[m − 2]. (10.10c)

Substituting the above value of h[m− 1] from Eq. (10.10c) in Eq. (10.10a)

yields

h[m] = δ[m − 1] + 0.4δ[m − 2] + 0.42h[m − 3].

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429 10 Time-domain analysis of DT systems

The aforementioned procedure can be repeated for the delayed impulse response

h[m − 3] on the right-hand side of the equation, then for the resulting h[m − 4], and so on. The final result is as follows:

h[m] = δ[m − 1] + 0.4δ[m − 2] + 0.42δ[m − 3] + 0.43δ[m − 4] + · · ·

h[m] = ∞∑

ℓ=1

0.4ℓ−1δ[m − ℓ] = 0.4m−1u[m − 1]

h[k] = 0.4k−1u[k − 1],

which is the required expression for the impulse response of the system.

Next, we proceed to calculate the output of the LTID system for the

input sequence x[k] = 2δ[k] + 3δ[k − 1]. Because the system is linear and

time-invariant, the output sequence y[k] resulting from input x[k] = 2δ[k] +

3δ[k − 1] is given by

2δ[k] + 3δ[k − 1] → 2h[k] + 3h[k − 1]

y[k] = 2 × 0.4k−1u[k − 1] + 3 × 0.4k−2u[k − 2]

= 2 × 0.40δ[k − 1] + (2 × 0.4k−1u[k − 2] + 3 × 0.4k−2u[k − 2])

= 2δ[k − 1] + 3.8 × 0.4k−2u[k − 2].

Example 10.4

The impulse response of an LTID system is given by h[k] = 0.5ku[k]. Deter-

mine the output of the system for the input sequence x[k] = δ[k − 1] + 3δ[k −

2] + 2δ[k − 6].

Solution

Because the system is LTID, it satisfies the linearity and time-shifting properties.

The individual responses to the three terms δ[k − 1], 3δ[k − 2], and 2δ[k − 6]

in the input sequence x[k] are given by

δ[k − 1] → h[k − 1] = 0.5k−1u[k − 1],

3δ[k − 2] → 3h[k − 2] = 3 × 0.5k−2u[k − 2],

and

2δ[k − 6] → 2h[k − 6] = 2 × 0.5k−6u[k − 6].

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430 Part III Discrete-time signals and systems

y[k]

3.5

1.75

0.88

0.44

2.22

1.11

0.55

k 3 41 2 5 6 7 8−1 0−2 3 41 2 5 6 7 8−1 0−2

h[k]1

0.5 0.52

0.530.54

(a) (b)

Fig. 10.3. (a) Impulse response

h[k ] of the LTID system specified

in Example 10.4. (b) Output y [k ]

of the LTID system for input

x[k ] = δ[k − 1] + 3δ[k − 2] + 2δ[k − 6].

Applying the principle of superposition, the overall response to the input

sequence x[k] is given by

y[k] = h[k − 1] + 3h[k − 2] + 2h[k − 6].

Substituting the value of h[k] = 0.5ku[k] results in the output response:

y[k] = 0.5k−1u[k − 1] + 3 × 0.5k−2u[k − 2] + 2 × 0.5k−6u[k − 6] .

The impulse response h[k] and the resulting output sequence are plotted in

Figs 10.3(a) and (b).

10.4 Convolution sum

Examples 10.3 and 10.4 compute the output of an LTID system for relatively

elementary input sequences x[k] consisting of a few scaled and time-shifted

impulses. In this section, we extend the approach to more complex input

sequences.

It was shown in Eq. (10.5), which is reproduced below for clarity, that any

arbitrary input sequence can be represented as a linear combination of time-

shifted impulse functions as follows:

x[k] = ∞∑

m=−∞

x[m]δ[k − m]. (10.11)

Note that in Eq. (10.11), x[m] is a scalar representing the magnitude of the

impulse δ[k − m] located at k = m. In terms of the impulse response h[k], the

output resulting from a single impulse x[m]δ[k − m] is given by

x[m]δ[k − m] −→ x[m]h[k − m]. (10.12)

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431 10 Time-domain analysis of DT systems

DT system

h[k] ∞

m=−∞

∞

m=−∞ x[k] = ∑ x[m]d[k−m] y[k]= ∑x[m]h[k −m] = x[m]∗h [m]

Fig. 10.4. Output response of a

system to an arbitrary input

sequence x[k ].

Applying the principle of superposition, the overall output y[k] resulting from

the input sequence x[k], represented by Eq. (10.11), is given by

∞∑

m=−∞

x[m]δ[k − m]

︸︷︷︸

x[k]

−→

∞∑

m=−∞

x[m]h[k − m]

︸︷︷︸

y[k]

, (10.13)

where the summation on the right-hand side, used to compute the output

response y[k], is referred to as the convolution sum. Equation (10.13) pro-

vides us with a second approach for calculating the output y[k]. It states that

the output y[k] can be calculated by convolving the input sequence x[k] with

the impulse response h[k] of the LTID system. Mathematically, Eq. (10.13) is

expressed as follows:

y[k] = x[k] ∗ h[k] =

∞∑

m=−∞

x[m]h[k − m], (10.14)

where ∗ denotes the convolution sum. Figure 10.4 illustrates the process of

convolution. The convolution operation defined in Eq. (10.14) is commonly

referred to as the linear convolution, in contrast to a special type of convolution

known as periodic convolution, which is discussed in Section 10.6.

We now consider several examples to illustrate the steps involved in com-

puting the convolution sum.

Example 10.5

Assuming that the impulse response of an LTID system is given by h[k] =

0.5ku[k], determine the output response y[k] to the input sequence x[k] =

0.8ku[k].

Solution

Using Eq. (10.14), the output response y[k] of the LTID system is given by

y[k] =

∞∑

m=−∞

x[m]h[k − m] =

∞∑

m=−∞

0.8mu[m]0.5k−mu[k − m].

Using the values of the unit step function u[m], the above summation simplifies

as follows:

y[k] =

∞∑

m=0

0.8m0.5k−mu[k − m].

Depending on the value of k, the output response y[k] of the system may take

two different forms for k ≥ 0 or k < 0. We consider the two cases separately.

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432 Part III Discrete-time signals and systems

10 2 3 4 5−3 −2 −1 6−4

1.31.291.16 0.99

0.82 0.67

3 [0.8k+1 − 0.5k+1]u[k]=

Fig. 10.5. Output of an LTID

system, with impulse response

h[k ] = 0.2ku[k ], to the input sequence x[k ] = 0.5ku[k ] as calculated in Example 10.5.

Case 1 (k < 0) When k < 0, the unit step function u[k − m] = 0 within the limits of summation (0 ≤ m ≤ ∞). Therefore, the output sequence y[k] = 0

for k < 0.

Case II (k ≥ 0) When k ≥ 0, the unit step function u[k − m] has the following values:

u[k − m] =

{

1 m ≤ k

0 m > k.

The output sequence y[k] is therefore given by

y[k] =

k∑

m=0

0.8m0.5k−m = 0.5k k∑

m=0

( 0.8

0.5

for k ≥ 0. The above summation represents a geometric progression (GP) series.

Using the GP series sum formula provided in Appendix A, Section A.3, the

output response y[k] is calculated as follows:

y[k] = 0.5k [

1 − (0.8/0.5)k+1

1 − (0.8/0.5)

]

= 10

3 [0.8k+1 − 0.5k+1].

Combining the two cases (k < 0 and k ≥ 0), the output response y[k] is given

y[k] =







0 k < 0 10

3 [0.8k+1 − 0.5k+1] k ≥ 0

= 10

3 [0.8k+1 − 0.5k+1]u[k].

The output response of the system is plotted in Fig. 10.5.

Example 10.5 shows how to calculate the convolution sum analytically. In

many situations, it is more convenient to use a graphical approach to evaluate

the convolution sum. Section 10.5 describes the graphical approach.

10.5 Graphical method for evaluating the convolution sum

The graphical approach for calculating the convolution sum is similar to the

graphical procedure for calculating the convolution integral for the LTIC system,

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433 10 Time-domain analysis of DT systems

discussed in Chapter 3. In the following, we highlight the main steps in calcu-

lating the convolution sum between two sequences x[k] and h[k].

Algorithm 10.1 Graphical procedure for computing the linear convolution

(1) Sketch the waveform for input x[m] by changing the independent variable

of x[k] from k to m and keep the waveform for x[m] fixed during steps

(2)–(7).

(2) Sketch the waveform for the impulse response h[m] by changing the inde-

pendent variable from k to m.

(3) Reflect h[m] about the vertical axis to obtain the time-inverted impulse

response h[−m]. (4) Shift the sequence h[−m] by a selected value of k. The resulting function

represents h[k − m]. (5) Multiply the input sequence x[m] by h[k − m] and plot the product function

x[m]h[k − m]. (6) Calculate the summation

∑∞

m=−∞ x[m]h[k − m].

(7) Repeat steps (4)–(6) for −∞ ≤ k ≤ ∞ to obtain the output response y[k]

over all time k.

The graphical approach for calculating the output response is illustrated through

a series of examples.

Example 10.6

Repeat Example 10.5 with input x[k] = 0.8ku[k] and impulse response h[k] =

0.5ku[k] to determine the output of the LTID system using the graphical con-

volution approach.

Solution

Following steps (1)–(3) of Algorithm 10.1, the DT sequences x[m] = 0.8mu[m],

h[m] = 0.5mu[m] and its time reflection h[−m] = 0.5−mu[−m] are plotted in

Fig. 10.6. Based on step (4), the sequence h[k − m] = h[−(m − k)] is obtained

by shifting h[−m] by k samples. To compute the output sequence, we consider

two cases based on the values of k.

Case 1 For k < 0, the waveform h[k − m] is on the left-hand side of the vertical axis. As is apparent in Fig. 10.6, step (5a), waveforms for h[k − m] and x[m] do

not overlap. In other words, the product x[m]h[k − m] = 0, for −∞ ≤ m ≤ ∞,

as long as k < 0. The output sequence y[k] is therefore zero for k < 0.

Case 2 For k ≥ 0, we see from Fig. 10.6, step (5b), that the non-zero parts of h[k − m] and x[m] overlap over the range m = [0, k]. Therefore,

y[k] =

k∑

m=0

x[m]h[k − m] =

k∑

m=0

0.8m0.5k−m .

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434 Part III Discrete-time signals and systems

0 1 2 3 4−4 −3 −2 −1 5−5 0 1 2 3 4−4 −3 −2 −1 5−5

h[m] = 0.5mu[m]

h[−m] = 0.5−mu[−m]

h[k−m] = 0.5k−mu[k−m]

kk −

k −

k +

x[m]h[k− m]

mkk −

k −

43210

x[m]h[k− m]

mkk −

k −

43210

k −

5 k

− 6

k −

x[m] = 0.8mu[m]

m 0 1 2 3 4−4 −3 −2 −1 5−5

step (1)

step (4)

step (6)

step (5a) step (5b)

step (2) step (3)

k 10 2 3 4 5−3 −2 −1 6−4

1.3 1.29 1.16

0.99 0.82

0.67

y[k]

Fig. 10.6. Convolution of the

input sequence x[k ] with the

impulse response h[k ] in

Example 10.6.

As shown in Example 10.5, the above summation simplifies to

y[k] = 10

3 [0.8k+1 − 0.5k+1] for k ≥ 0.

Combining Cases 1 and 2, the overall output sequence is given by

y[k] = 10

3 [0.8k+1 − 0.5k+1]u[k].

The final output response is plotted in Fig. 10.6, step (6).

Example 10.7

For the following DT sequences:

x[k] =

{

2 0 ≤ k ≤ 2

0 otherwise and h[k] =

{

k + 1 0 ≤ k ≤ 4

0 otherwise,

calculate the convolution sum y[k] = x[k] ∗ h[k] using the graphical approach.

Solution

Following steps (1)–(3) of Algorithm 10.1, the sequences x[m], h[m], and its

reflection h[−m] are plotted as a function of the independent variable m in

Fig. 10.7, steps (1)–(3). The DT sequence h[k − m] = h[−(m − k)] is obtained

by shifting the time-reflected function h[−m] by k. Depending on the value of

k, five special cases arise. We consider these cases separately.

Case 1 For k < 0, we see from Fig. 10.7, step (5a), that the non-zero parts of h[k − m] and x[m] do not overlap. In other words, output y[k] = 0 for k < 0.

Case 2 For 0 ≤ k ≤ 2, we see from Fig. 10.7, step (5b), that the non-zero parts of h[k − m] and x[m] overlap over the duration m = [0, k]. Therefore, the

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435 10 Time-domain analysis of DT systems

step (1) step (2) step (3)

−3−2 −1 0 1 32 4 65 7 8−4 m

−5

x[m]

−3−2 −1 0 1 32 4 65 7 8−4 m

−5

h[m]

−3−2 −1 0 1 32 4 65 7 8−4 m

−5

h[−m]

step (4) step (5a) step (5b)

step (6)

k +

k k +

3 k

+ 4

k +

5 k

+ 6

k +

k −

5 k

− 4

k −

3 k

− 2

k −

h[k−m]

kk −

4 k

− 3

k −

h[k−m]x[m]

0 1 2 3 4 5 6 7 8

kk −

4 k

− 3

k −

h[k − m]x[m]

0 1−1−2−3−4−5 2 3 4 5 6 7 8

18 10 2418

62 12

y[k]

0 1−1−2−3−4−5 2 3 4 5 6 7 8

≈≈ ≈≈ ≈≈ ≈≈ ≈ ≈≈ ≈

step (5c) step (5d) step (5e)

k −

3 k

− 2

k −

1 k

−3−2 −1 0 1 32 4 65 7 8−4 m

−5

h[k−m]x[m]

k −

3 k

− 2

k −

1 k

−3−2 −1 0 1 32 4 65 7 8−4 m

−5

h[k−m]x[m]

k −

3 k

− 2

k −

1 k

−3−2 −1 0 1 32 4 65 7 8−4 m

−5

h[k−m]x[m]

Fig. 10.7. Convolution of the

input sequence x [k ] with the

impulse response h[k ] in

Example 10.7.

output response for 0 ≤ k ≤ 2 is given by

y[k] =

k∑

m=0

x[m]h[k − m] =

k∑

m=0

2 × (k − m + 1) = 2(k + 1)

k∑

m=0

1 − 2

k∑

m=0

= 2(k + 1)2 − 2

k∑

m=1

The summation ∑k

m=1 m is an arithmetic progression (AP) series. Using the

AP series summation formula provided in Appendix A, Section A.3, the output

response y[k] for 0 ≤ k ≤ 2 is calculated as follows:

y[k] = 2(k + 1)2 − k(k + 1) = k2 + 3k + 2.

Case 3 For 2 ≤ k ≤ 4, we see from Fig. 10.7, step (5c), that the non-zero part of h[k − m] completely overlaps x[m] over the region m = [0, 2]. The output

response y[k] for 2 ≤ k ≤ 4 is given by

y[k] =

2∑

m=0

x[m]h[k − m] =

2∑

m=0

2 × (k − m + 1)

= 2(k + 1)

2∑

m=0

1 − 2

2∑

m=0

m = 6(k + 1) − 6 = 6k.

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436 Part III Discrete-time signals and systems

Case 4 For 4 ≤ k ≤ 6, we see from Fig. 10.7, step (5d), that the non-zero part of h[k − m] partially overlaps x[m] over the region m = [k − 4, 2]. The output

y[k] for 5 ≤ k ≤ 6 is given by

y[k] =

2∑

m=k−4

x[m]h[k − m] =

2∑

m=k−4

2 × (k − m + 1)

= 2(k + 1)

2∑

m=k−4

1 − 2

2∑

m=k−4

= 2(k + 1)(7 − k) − (7 − k)(k − 2) = −k2 + 3k + 8.

Case 5 For k > 6, we see from Fig. 10.7, step (5e), that the non-zero parts of h[k − m] and x[m] do not overlap. Therefore, the product x[m]h[k − m] = 0

for all values of m. The value of the output sequence y[k] = 0 for k > 6.

Combining the above five cases, we obtain

y[k] =



  

  

0 k < 0, k > 6

k2 + 3k + 2 0 ≤ k ≤ 2

6k 2 ≤ k ≤ 4

−k2 + 3k + 8 4 ≤ k ≤ 6,

which is plotted in Fig. 10.7, step (6).

10.5.1 Sliding tape method

The graphical convolution approach, illustrated in Examples 10.6 and 10.7,

for LTID systems is similar to the graphical convolution procedure for LTIC

systems. However, sketching the figures for the time-reversed and time-shifted

impulse functions may prove to be difficult in certain cases. There is a variant

of the graphical method for DT convolution, known as the sliding tape method,

which is convenient in cases where the convolved sequences are relatively

short in length. Instead of drawing the figures in such cases, we compute the

convolution sum using a table whose entries are the values of the DT sequences

at different instances. We illustrate the sliding tape method in Examples 10.8

and 10.9.

Example 10.8

For the two sequences x[k] and h[k] defined in Example 10.7, calculate the

convolution y[k] = x[k] ∗ h[k] using the sliding tape method.

Solution

The convolution of x[k] and h[k] using the sliding tape method is illustrated

in Table 10.1. The first row represents the m-axis; the second row represents

the input sequence x[m]; and the third row represents the impulse response

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437 10 Time-domain analysis of DT systems

Table 10.1. Convolution of x [k ] and h[k ] using the sliding tape method for Example 10.8

m . . . −5 −4 −3 −2 −1 0 1 2 3 4 5 6 7 . . . k y[k]

x[m] 2 2 2

h[m] 1 2 3 4 5

h[−m] 5 4 3 2 1 h[−1 − m] 5 4 3 2 1 −1 0 h[0 − m] 5 4 3 2 1 0 2 h[1 − m] 5 4 3 2 1 1 6 h[2 − m] 5 4 3 2 1 2 12 h[3 − m] 5 4 3 2 1 3 18 h[4 − m] 5 4 3 2 1 4 24 h[5 − m] 5 4 3 2 1 5 18 h[6 − m] 5 4 3 2 1 6 10 h[7 − m] 5 4 3 2 1 7 0

h[m] for different values of m. Following the steps involved in convolution,

we generate the values for the sequence h[k − m] and store the value in a row. To generate the values of h[k − m], we first form the function h[−m], which is obtained by time-inverting h[m]. The result is illustrated in the fourth row

of Table 10.1. The time-reversed function h[−m] is used to generate h[k − m] by right-shifting h[−m] by k time units. For example, the fifth row contains the values of the function h[−1 − m] = h[−(m + 1)]. Similarly, rows (6)–(13) contain the values of the function h[k − m] = h[−(m − k)] for the range 0 ≤ k ≤ 7. In order to calculate y[k] for a fixed value of k, we multiply the entries

in the row containing x[m] by the corresponding entries contained in the row

for h[k − m] and then evaluate the summation:

y[k] =

∞∑

m=−∞

x[m]h[k − m].

For k = −1, we note that the non-zero entries of x[m] and h[k − m] do not

overlap. Therefore, y[k] = 0 for k = −1. Since there is also no overlap for

k < −1, the output y[k] = 0 for k ≤ −1.

The aforementioned multiplication process is repeated for different values

of k. For k = 0, we note that the only overlap between the non-zero values of

x[m] and h[−m] occurs for m = 0. The output response is therefore given by

y[0] = 2 · 1 = 2.

These values of time instant k = 0 and the output response y[0] = 2 are stored

in the last two columns of row (6), corresponding to the entries of h[0 − m]

in Table 10.1. Similarly, for k = 1, we observe that the overlap between the

non-zero values of x[m] and h[1 − m] occurs for m = 0 and 1. The output

response is given by

y[0] = 2 · 2 + 2 · 1 = 6

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438 Part III Discrete-time signals and systems

Table 10.2. Convolution of x [k ] and h[k ] using the sliding tape method for Example 10.9

m . . . −5 −4 −3 −2 −1 0 1 2 3 4 5 6 . . . k y[k]

h[m] 3 1 −2 3 −2 x[m] −1 1 2 x[−m] 2 1 −1 x[−3 − m] 2 1 −1 −3 0 x[−2 − m] 2 1 −1 −2 −3 x[−1 − m] 2 1 −1 −1 2 x[0 − m] 2 1 −1 0 9 x[1 − m] 2 1 −1 1 −3 x[2 − m] 2 1 −1 2 1 x[3 − m] 2 1 −1 3 4 x[4 − m] 2 1 −1 4 −4 x[5 − m] 2 1 −1 5 0

and is stored in the last column of Table 10.1. We repeat the process for increas-

ing values of k until the overlap between x[m] and h[k − m] is eliminated. In Table 10.1, this occurs for k > 7, beyond which the output response y[k] is

zero.

By comparison with the result obtained in Example 10.7, we note that the

output response y[k] obtained using the sliding tape method is identical to the

one obtained using the graphical approach.

Example 10.9

For the following pair of the input sequence x[k] and impulse response h[k]:

x[k] =



  

  

−1 k = −1 1 k = 0 2 k = 1 0 otherwise

and h[k] =



  

  

3 k = −1, 2 1 k = 0

−2 k = 1, 3 0 otherwise,

calculate the output response using the sliding tape method.

Solution

The output y[k] can be calculated by convolving the input sequence x[k] with

the impulse response h[k]. Since convolution satisfies the distributive property,

i.e.

y[k] = x[k] ∗ h[k] = h[k] ∗ x[k],

Table 10.2 reverses the role of the input sequence x[k] with that of the impulse

response h[k] and computes the following summation:

y[k]

k 0 2 3

−2

−1

1 4

5−4 −3−5

−3 −4

−3

4≈≈

Fig. 10.8. Output response

calculated using the sliding tape

method in Example 10.9.

y[k] =

∞∑

m=−∞

h[m]x[k − m],

implying that the input sequence is time-reversed and time-shifted, while the

impulse response is kept fixed. The results of Table 10.2 are plotted in Fig. 10.8.

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439 10 Time-domain analysis of DT systems

10.6 Periodic convolution

Linear convolution is used to convolve aperiodic sequences. If the convolv-

ing sequences are periodic, the result of linear convolution is unbounded. In

such cases, a second type of convolution, referred to as periodic or circular

convolution, is generally used.

Consider two periodic sequences xp[k] and hp[k], with identical fundamental

period K0. The subscript p denotes periodicity. The relationship for the periodic

convolution between two periodic sequences is defined as follows:

yp[k] = xp[k] ⊗ hp[k] = ∑

m=〈K0〉

xp[m]hp[k − m], (10.15)

where the summation on the right-hand side of Eq. (10.15) is defined over

one complete period K0. In calculating the summation, we can, therefore, start

from any arbitrary position (say m = m0) as long as one complete period of

the sequences is covered by the summation. For the lower limit m = m0, the

upper limit is given by m = m0 + K0− 1. In the text, the periodic convolu-

tion is denoted by the operator ⊗, whereas the linear convolution is denoted

by ∗.

The steps involved in calculating the periodic convolution are given in the

following algorithm.

Algorithm 10.2 Graphical procedure for computing the periodic convolution

(1) Sketch the waveform for input xp[m] by changing the independent vari-

able of xp[k] from k to m and keep the waveform for xp[m] fixed during

steps (2)–(7).

(2) Sketch the waveform for the impulse response hp[m] by changing the inde-

pendent variable from k to m.

(3) Reflect hp[m] about the vertical axis to obtain the time-inverted impulse

response hp[−m]. Set the time index k = 0.

(4) Shift the function hp[−m] by a selected value of k. The resulting sequence

represents hp[k − m].

(5) Multiply input sequence xp[m] by hp[k − m] and plot the product function

xp[m]hp[k − m].

(6) Calculate the summation ∑

m=〈K0〉 xp[m]hp[k − m] for m = [m0, m0 +

K0 − 1] to determine yp[k] for the value of k selected in step (4).

(7) Increment k by one and repeat steps (4)–(6) till all values of k in the specified

range (0 ≤ k ≤ K0 − 1) are exhausted.

(8) Since yp[k] is periodic with period K0, the values of yp[k] outside the range

0 ≤ k ≤ K0 − 1 are determined from the values obtained in steps (6) and

(7).

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440 Part III Discrete-time signals and systems

step (1)

step (4a) step (4b) step (4c)

step (2) step (3)

step (4d) step (8)step (7)

1 2

1 2 3 4 5 6 70−1−2−3−4

2 3

xp[m]

1 2 3 4 5

6 70−1−2−3−4

hp[m]

1 2 3 4 5 6 70−1−2−3−4

hp[− m]

1 2 3 4 5 6 70−1−2−3−4

xp[− m]hp[0−m]

1 2 3 4 5 6 70−1−2−3−4

xp[−m]hp[1−m]

1 2 3 4 5 6 70−1−2−3−4

xp[− m]hp[2−m]

1 2 3 4 5 6 70−1−2−3−4

xp[−m]hp[3−m]

1 2 3 4 5 6 70−1−2−3−4

yp[k] yp[k]

5 5

15 15 15 15

25 25

20 1 3

Fig. 10.9. Periodic convolution

of the periodic sequences x[k ]

and h[k ] in Example 10.10.

By comparing the aforementioned procedure for computing the periodic con-

volution with the procedure specified for evaluating the linear convolution in

Section 10.5, we observe that steps (4), (6), and (7) are different in the two

algorithms. In the linear convolution, the summation

∞∑

m=−∞

x[m]h[k − m]

is computed within the limits m = [−∞, ∞] for different values of k in the

range −∞ ≤ k ≤ ∞. In the periodic convolution, however, the summation is

computed over one complete period, say m = [m0, m0 + K0 − 1] for a reduced

range (0 ≤ k ≤ K0 − 1).

Example 10.10

Determine the periodic convolution between the following periodic sequences:

xp[k] = k, for 0 ≤ k ≤ 3 and hp[k] =

{

5 k = 0, 1

0 k = 2, 3,

with the fundamental period K0 = 4.

Solution

Following steps (1)–(3), the periodic sequences xp[m], hp[m], and its reflected

version hp[−m] are plotted in Fig. 10.9, steps (1)–(3). Since the fundamental

period K0 = 4, we compute the result of the periodic convolution as follows:

yp[k] = xp[k] ⊗ hp[k] =

3∑

m=0

xp[m]hp[k − m] (10.16)

for 0 ≤ k ≤ 3. The DT periodic sequences hp[k − m] and xp[m] for k = 0, 1,

2, and 3 are plotted, respectively, in Fig. 10.9, steps 4(a)–(d). The convolution

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441 10 Time-domain analysis of DT systems

summation, Eq. (10.16), has the following values:

(k = 0) yp[0] = xp[0]hp[0] + xp[1]hp[−1] + xp[2]hp[−2] + xp[3]hp[−3]

= 0 × 5 + 1 × 0 + 2 × 0 + 3 × 5 = 15;

(k = 1) yp[1] = xp[0]hp[1] + xp[1]hp[0] + xp[2]hp[−1] + xp[3]hp[−2]

= 0 × 0 + 1 × 5 + 2 × 0 + 3 × 0 = 5;

(k = 2) yp[2] = xp[0]hp[2] + xp[1]hp[1] + xp[2]hp[0] + xp[3]hp[−1]

= 0 × 0 + 1 × 5 + 2 × 5 + 3 × 0 = 15;

(k = 3) yp[3] = xp[0]hp[3] + xp[1]hp[2] + xp[2]hp[1] + xp[3]hp[0]

= 0 × 0 + 1 × 0 + 2 × 5 + 3 × 5 = 25.

The remaining values of yp[k] are easily determined by exploiting the period-

icity property of yp[k]. The output yp[k] is plotted in Fig. 10.9, step (8).

An alternative procedure for computing the periodic convolution can be

obtained by calculating the limits of Eq. (10.15) for m = 0 to m = K0 − 1. The resulting expression is given by

yp[k] = K0−1∑

m=0 xp[m]hp[k − m]

yp[k] = xp[0]hp[k] + xp[1]hp[k − 1] + xp[2]hp[k − 2] + · · · + xp[K0 − 1]hp[k − (K0 − 1)],

for 0 ≤ k ≤ K0 − 1. Expanding the above equation in terms of the time index

k yields

yp[0] = xp[0]hp[0] + xp[1]hp[−1] + xp[2]hp[−2] + · · ·

+ xp[K0 − 1]hp[−(K0 − 1)],

yp[1] = xp[0]hp[1] + xp[1]hp[0] + xp[2]hp[−1]

+ · · · + xp[K0 − 1]hp[−(K0 − 2)],

yp[2] = xp[0]hp[2] + xp[1]hp[1] + xp[2]hp[0] + · · ·

+ xp[K0 − 1]hp[−(K0 − 3)],

...

yp[K0 − 1] = xp[0]hp[K0 − 1] + xp[1]hp[K0 − 2]

+ xp[2]hp[K0 − 3] + · · · + xp[K0 − 1]hp[0].



                 

                 

(10.17)

Since hp[k] is periodic,

hp[k] = hp[k + K0] or hp[−k] = hp[K0 − k]. (10.18)

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442 Part III Discrete-time signals and systems

Equation (10.18) is referred to as periodic or circular reflection. Before pro-

ceeding with the alternative algorithm for periodic convolution, we explain

circular reflection in more detail.

2 30 k

hp[k]

2 30 k

hp[k]

2 30 k

hp[k]

hp[−k]

2 30 1

hp[k−1]

(a)

(c)

(b)

(e)

(d)

Fig. 10.10. Circular reflection

and shifting for a periodic

sequence. (a) Original periodic

sequence hp[k ]. (b) Procedure

to determine circularly reflected

sequence hp[−k ] from hp[k ]. (c) Circularly reflected sequence

hp[−k ]. (d) Procedure to determine circularly shifted

sequence hp[k − 1] from hp[k ]. (e) Circularly shifted sequence

hp[k − 1].

Example 10.11

For the periodic sequence

hp[k] = {

5 k = 0, 1 0 k = 2, 3,

with fundamental period K0 = 4, determine the circularly reflected sequence hp[−k] and the circular shifted sequence hp[k−1].

Solution

Let vp[k] denote the circular reflected sequence hp[−k]. Using vp[k] = hp[−k] = hp[K0 − k], the values of the circularly reflected signals are given by

k = 0 vp[0] = hp[K0] = hp[0] = 5; k = 1 vp[1] = hp[K0 − 1] = hp[3] = 0; k = 2 vp[2] = hp[K0 − 2] = hp[2] = 0; k = 3 vp[3] = hp[K0 − 3] = hp[1] = 5.

The original sequence hp[k] is plotted in Fig. 10.10(a), and the circularly

reflected sequence hp[−k] is plotted in Fig. 10.10(c). Note that the circu- larly reflected signal hp[−k] can be obtained directly from hp[k] by keep- ing the value of hp[0] fixed and then reflecting the remaining values of

hp[k] for 1 ≤ k ≤ K0 − 1 about k = K0/2. This procedure is illustrated in

Fig. 10.10(b).

Substituting 0 ≤ k ≤ K0 − 1, the values for the circularly shifted signal wp[k]

= hp[k−1] are obtained as follows:

k = 0 wp[0] = hp[−1] = hp[K0 − 1] = 0;

k = 1 wp[1] = hp[0] = 5;

k = 2 vp[2] = hp[1] = 5;

k = 3 vp[3] = hp[2] = 0.

The circularly shifted sequence hp[k − 1] is plotted in Fig. 10.10(e). The circu-

larly shifted signal hp[k − 1] can also be obtained directly from hp[k] by shift-

ing hp[k] towards the left by one time unit and moving the overflow value of

hp[K0 − 1] back into the sequence. This procedure is illustrated in Fig. 10.10(d).

To derive the alternative algorithm for periodic convolution, we substitute

different values of k within the range 1 ≤ k ≤ K0 − 1 in Eq. (10.18). The

resulting equations are given by

hp[−1] = hp[K0 − 1]; hp[−2] = hp[K0 − 2]; . . . ; hp[−(K0 − 1)] = hp[1],

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443 10 Time-domain analysis of DT systems

which are substituted in Eq. (10.18) to obtain

yp[0] = xp[0]hp[0] + xp[1]hp[K0 − 1] + xp[2]hp[K0 − 2]

+ · · · xp[K0 − 1]hp[1],

yp[1] = xp[0]hp[1] + xp[1]hp[0] + xp[2]hp[K0 − 1]

+ · · · xp[K0 − 1]hp[2],

yp[2] = xp[0]hp[2] + xp[1]hp[1] + xp[2]hp[0] + · · · xp[K0 − 1]hp[3],

...

yp[K0 − 1] = xp[0]hp[K0 − 1] + xp[1]hp[K0 − 2]

+ xp[2]hp[K0 − 3] + · · · xp[K0 − 1]hp[0].



                  

                  

(10.19)

These expressions require values from only one period (0 ≤ k ≤ K0 − 1) of

the input sequence xp[k] and the impulse response hp[k]. Therefore, we can

implement the periodic convolution from a single period of the convolving

functions. The main steps involved in such an implementation are listed in the

following algorithm.

Algorithm 10.3 Alternative procedure for computing the periodic convolution

(1) Sketch one period of the waveform for input xp[m] by changing the inde-

pendent variable of xp[k] from k to m within the range 0 ≤ k ≤ K0 − 1.

(2) Sketch one period of the waveform for the impulse response hp[m] by

changing the independent variable from k to m within the range 0 ≤ k ≤

K0 − 1.

(3) Reflect hp[m] such that hp[−m] = hp[K0 − m] as defined by the circular

reflection. Set k = 0.

(4) Using the circularly reflected function hp[−m], determine the waveform

for hp[k − m] = hp[−(m − k)].

(5) Multiply the function xp[m] by hp[k − m] for 0 ≤ m ≤ K0 − 1 and plot

the product function xp[m]hp[k − m].

(6) Calculate the summation ∑K0−1

m=0 xp[m]hp[k − m] to determine yp[k] for

the value of k selected in step (4).

(7) Increment k by one and repeat steps (4)–(6) till all values of k within the

range 0 ≤ k ≤ K0 − 1 are exhausted.

(8) Since yp[k] is periodic with period K0, the values of yp[k] outside the range

0 ≤ k ≤ K0 − 1 are determined from the values obtained in steps (7).

We illustrate the alternative implementation by repeating Example 10.12 and

using the modified algorithm.

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444 Part III Discrete-time signals and systems

2 30 1

1 2

3 xp[m]

2 30 1

hp[m]55

2 30 1

hp[−m]

2 30 1

hp[−m]

2 30 1

xp[m]hp[0 − m]

2 30 1

hp[1−m]

2 30 1

xp[m]hp[1− m]

2 30 1

hp[2−m]

2 30 1

xp[m]hp[2 − m]

2 30 1

hp[3 −m]

2 30 1

xp[m]hp[3−m]

step (1)

step (4a) step (4b)

step (2) step (3)

step (4c) step (4d)

steps (5)–(8)

4 7−2 −1−3−4 5 6

yp[k]

1515

20 21 30

15 15 15 15

25 25

5 5 Fig. 10.11. Periodic convolution

using circular shifting in Example

10.12.

Example 10.12

Using Algorithm 10.3, determine the periodic convolution of the periodic

sequences

xp[k] = k (0 ≤ k ≤ 3) and hp[k] = {

5 k = 0, 1

0 k = 2, 3,

with fundamental period K0 = 4.

Solution

Following steps (1) and (2), the applied input and the impulse response are

plotted as a function of m in Fig. 10.11, steps (1) and (2).

Following step (3), the circularly reflected impulse response vp[m] =

hp[−m] = hp[K0 − m] for 0 ≤ m ≤ 3 is calculated as follows:

vp[0] = hp[0] = 1; vp[1] = hp[−1] = hp[3] = 0; vp[2] = hp[−2]

= hp[2] = 0; and vp[3] = hp[−3] = hp[1] = 3.

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445 10 Time-domain analysis of DT systems

For k = 0, the DT sequence hp[k − m] = hp[−m]. The value of the output response at k = 0 is given by

yp[0] = K0−1∑

m=0 xp[m]hp[−m] = 0(5) + 1(0) + 2(0) + 3(5) = 15.

For k = 1, the DT sequence hp[k − m] = hp[1 − m]. The new sequence hp[1 − m] = hp[−(m − 1)] is obtained by circularly shifting hp[−m] towards the right by one sample, with the last sample at m = 3 taking the place of the first sample at m = 0. The sequence hp[1 − m] is plotted in Fig. 10.11, step (4b). Multiplying by hp[m], the value of the output response at k = 1 is given by

yp[1] = K0−1∑

m=0 xp[m]hp[1 − m] = 0(5) + 1(5) + 2(0) + 3(0) = 5.

For k = 2, the DT sequence hp[k − m] = hp[2 − m]. The new sequence hp[2 − m] is obtained by circularly shifting hp[1 − m] towards the right by one sample, with the last sample at m = 3 taking the place of the first sample at m = 0. The sequence hp[2 − m] is plotted in Fig. 10.11, step (4c). Multiplying by hp[m], the value of the output response at k = 2 is given by

yp[2] = K0−1∑

m=0 xp[m]hp[2 − m] = 0(0) + 1(5) + 2(5) + 3(0) = 15.

For k = 3, the DT sequence hp[k − m] = hp[3 − m]. The new sequence hp[3 − m] is obtained by circularly shifting hp[2 − m] towards the right by one sample, with the last sample at m = 3 taking the place of the first sample at m = 0. The sequence hp[3 − m] is plotted in Fig. 10.11, step (4d). Multiplying by hp[m], the value of the output response at k = 3 is given by

yp[3] = K0−1∑

m=0 xp[m]hp[3 − m] = 0(0) + 1(0) + 2(5) + 3(5) = 25.

The final output yp[k], obtained from steps (5)–(8) of Algorithm 10.3, is

plotted in Fig. 10.11, Steps (5)–(8). Observe that the result is identical to that in

Fig. 10.9, which was obtained using the full periodic convolution.

10.6.1 Linear convolution through periodic convolution

In this chapter, we have introduced two types of DT convolution. The linear

convolution, defined in Eq. (10.14), is used to convolve aperiodic sequences,

while the periodic convolution, defined in Eq. (10.15), is used for convolving

periodic sequences. Definition 10.3 states a condition under which the results

of the periodic and linear convolution are the same.

Definition 10.3 Assume that x[k] and h[k] are two aperiodic DT sequences of

finite length such that the following are true.

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446 Part III Discrete-time signals and systems

(i) The DT sequence x[k] = 0 outside the range kℓ1 ≤ k ≤ ku1. Note that it is possible for x[k] to have some zero values within the range kℓ1 ≤ k ≤ ku1.

The length Kx of x[k] is given by Kx = (ku1 − kℓ1 + 1) samples.

(ii) The DT sequence h[k] = 0 outside the range kℓ2 ≤ k ≤ ku2. As for x[k], it

is possible for h[k] to have intermittent zero values within the range kℓ2 ≤

k ≤ ku2. The length Kh of h[k] is given by Kh = ku2 − kℓ2 + 1 samples.

Add the appropriate number of zeros to the two sequences x[k] and h[k] so

that they have the same length K0 ≥ (Kx + Kh − 1). The procedure of adding

zeros to a sequence is referred to as zero padding. The periodic extensions of

zero-padded x[k] and h[k] are denoted by xp[k] and hp[k], which have the

same fundamental period of K0 ≥ (Kx + Kh − 1). Mathematically, the single

periods of xp[k] and hp[k] are defined as follows:

xp[k] =

{

x[k] kℓ1 ≤ k ≤ ku1

0 ku1 < k ≤ K0 + kℓ1 − 1 (10.20a)

and

hp[k] =

{

h[k] kℓ2 ≤ k ≤ ku2

0 ku2 < k ≤ K0 + kℓ2 − 1. (10.20b)

It can be shown that the linear convolution between x[k] and h[k] can be

obtained from the periodic convolution between xp[k] and hp[k] using the fol-

lowing relationship:

x[k] ∗ h[k] = xp[k] ⊗ hp[k],

for (kℓ1 + kℓ2) ≤ k ≤ (ku1 + ku2).

Definition 10.3 provides us with an alternative algorithm for implementing

the linear convolution through the periodic convolution. The advantage of the

above approach lies in computationally efficient implementations of the peri-

odic convolution, which are much faster than the implementations of the linear

convolution. Chapter 12 presents one such approach using the discrete Fourier

transform (DFT) to compute the periodic convolution.

Algorithm 10.4 Computing linear convolution from periodic convolution

(1) Consider two time-limited DT sequences x[k] and h[k]. The DT sequence

x[k] = 0 outside the range kℓ1 ≤ k ≤ ku1 of length Kx = ku1 − kℓ1 + 1

samples. Similarly, the DT sequence h[k] = 0 outside the range kℓ2 ≤ k ≤

ku2 of length Kh = ku2 − kℓ2 + 1 samples.

(2) Select an arbitrary integer K0 ≥ Kx + Kh− 1.

(3) Compute the periodic extension xp[k] of x[k] using Eq. (10.20a).

(4) Compute the periodic extension hp[k] of h[k] using Eq. 10.20b).

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447 10 Time-domain analysis of DT systems

x[k]

k 1−1

−1 −1

k 1−1

−1 −1 2 3 4 5 6

h[k] xp[k]

k 10−1

−2 2

−1−1

3 2

hp[k]

k 0

−2

3 4 5

−1−1

3 2

−1

step (1)

step (5)

step (3) step (4)

yp[k]

k −8

1−1

0 2

3 4 5 97

11 12 13

−7−9 −5 −4 −3

−2 6 10−6 1

−2

−5

−2

−5

1 1

−5

−2

−5

1 1 1

−5

Fig. 10.12. Periodic convolution

using circular shifting in Example

10.13.

(5) Calculate the periodic convolution yp[k] = xp [k] ⊗ hp[k]. The result of the linear convolution is obtained by selecting the range kℓ1 + kℓ2 ≤ k ≤

ku1 + ku2 of yp[k].

Example 10.13 illustrates the aforementioned procedure.

Example 10.13

Compute the linear convolution of the following DT sequences:

x[k] =







2 k = 0

−1 |k| = 1

0 otherwise

and h[k] =



  

  

2 k = 0

3 |k| = 1

−1 |k| = 2

0 otherwise,

using the periodic convolution method outlined in Algorithm 10.4.

Solution

The DT sequences x[k] and h[k] are plotted in Fig. 10.12, step (1). We observe

that the length Kx of x[k] is 3, while the length Kh of h[k] is 5.

Based on step (2), the value of K0 ≥ 3 + 5 − 1 or 7. We select K0 = 8.

Following step (3), we form xp[k] by padding x[k] with K0 − Kx or five

zeros. The resulting sequence xp[k] is shown in Fig. 10.12, step (3).

Following step (4), we form hp[k] by padding h[k] with K0 − Kh , or three

zeros. The resulting sequence hp[k] is shown in Fig. 10.12, step (4).

Following step (5), the periodic convolution of the DT sequences xp[k] and

hp[k] is performed using the sliding tape method. The final result is shown

in Table 10.3, where only one period (K0 = 8) of each sequence within the

duration k = [−3, 4] is considered.

The sliding tape approach illustrated in Table 10.3 is slightly different from

that of Table 10.2. The reflection and shifting operations in Table 10.3 are

based on circular reflection and circular shifting since periodic sequences are

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448 Part III Discrete-time signals and systems

Table 10.3. Periodic convolution of xp[k ] and hp[k ] in Example 10.13

m −3 −2 −1 0 1 2 3 4 k yp[k]

hp[k] 0 −1 3 2 3 −1 0 0 xp[k] 0 0 −1 2 −1 0 0 0 xp[−k] 0 0 −1 2 −1 0 0 0 xp[−4 − k] −1 0 0 0 0 0 −1 2 −4 0 xp[−3 − k] 2 −1 0 0 0 0 0 −1 −3 1 xp[−2 − k] −1 2 −1 0 0 0 0 0 −2 −5 xp[−1 − k] 0 −1 2 −1 0 0 0 0 −1 5 xp[0 − k] 0 0 −1 2 −1 0 0 0 0 −2 xp[1 − k] 0 0 0 −1 2 −1 0 0 1 5 xp[2 − k] 0 0 0 0 −1 2 −1 0 2 −5 xp[3 − k] 0 0 0 0 0 −1 2 −1 3 1 xp[4 − k] −1 0 0 0 0 0 −1 2 4 0

being convolved. The values of the output sequence yp[k] over one period

(−3 ≤ k ≤ 4) are listed in the right-hand column of Table 10.3. The plot of the periodic output yp[k] is sketched in Fig. 10.12, step (5).

The result of the linear convolution y[k] = x[k] ∗ h[k] is obtained by selecting

one period of the periodic output yp[k] within the duration kℓ1 + kℓ2 ≤ k ≤

ku1 + ku2, which equals −3 ≤ k ≤ 3.

10.7 Properties of the convolution sum

The properties of the DT linear convolution sum are similar to the proper-

ties of the CT convolution integral presented in Chapter 3. In the following,

we list the properties of linear convolution for DT sequences followed by the

corresponding properties for the periodic convolution.

Commutative property

x1[k] ∗ x2[k] = x2[k] ∗ x1[k]. (10.21)

The commutative property states that the order of the convolution operands

does not affect the result of the convolution. In the context of LTID systems, the

commutative property implies that the input sequence and the impulse response

of the DT system may be interchanged without affecting the output response.

The periodic convolution also satisfies the commutative property provided that

the two sequences have the same fundamental period K0.

Distributive property

x1[k] ∗ {x2[k] + x3[k]} = x1[k] ∗ x2[k] + x1[k] ∗ x3[k]. (10.22)

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449 10 Time-domain analysis of DT systems

The distributive property states that convolution is a linear operation with respect

to addition. The periodic convolution also satisfies the distributive property

provided that the three sequences have the same fundamental period K0.

Associative property

x1[k] ∗ {x2[k] ∗ x3[k]} = {x1[k] ∗ x2[k]} ∗ x3[k]. (10.23)

This property states that changing the order of the linear convolution operands

does not affect the result of the linear convolution. The periodic convolution

also satisfies the associative property provided that the three sequences have

the same fundamental period K0.

Shift property If x1[k] ∗ x2[k] = g[k], then

x1[k − k1] ∗ x2[k − k2] = g[k − k1 − k2] (10.24)

for any arbitrary integer constants k1 and k2. In other words, if the two operands

of the linear convolution sum are shifted then the result of the convolution sum

is shifted in time by a duration that is the sum of the individual time shifts

introduced in the operands. The periodic convolution satisfies the shift property

with respect to the circular shift operation.

Length of convolution Let the non-zero lengths of the convolution operands x1[k] and x2[k] be denoted by K1 and K2 time units, respectively. It can be shown

that the non-zero length of the linear convolution (x1[k] ∗ x2[k]) is K1 + K2 − 1

time units. The periodic convolution does not satisfy the length property. The

circular convolution of two periodic sequences with fundamental period K0 is

also of length K0.

Convolution with impulse function

x1[k] ∗ δ[k − k0] = x1[k − k0]. (10.25)

In other words, convolving a DT sequence with a unit impulse function whose

origin is located at k = k0 shifts the DT sequence by k0 time units. Since periodic

convolution is defined in terms of periodic sequences and the impulse function

is not a periodic sequence, Eq. (10.25) is not valid for the periodic convolution.

Convolution with unit step function

x1[k] ∗ u[k] =

∞∑

m=−∞

x[m]u[k − m] =

k∑

m=−∞

x[m]. (10.26)

Equation (10.26) states that convolving a DT sequence x[k] with a unit step

function produces the running sum of the original sequence x[k] as a function

of time k. Since periodic convolution is defined in terms of periodic sequences

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450 Part III Discrete-time signals and systems

and the unit step function is not periodic, Eq. (10.26) is not valid for the periodic

convolution.

Causal functions If one of the sequences is causal, the expression for linear convolution, Eq. (10.14), can be written in a simpler form. For example, if

h[k] = 0 for k < 0, the convolution sum y[k] in Eq. (10.14) is expressed as follows:

y[k] = x[k] ∗ h[k] = ∞∑

m=−∞

h[m]x[k − m]

∞∑

m=0

h[m]x[k − m]. (10.27a)

However, if h[k] is both causal and time-limited, i.e. if h[k] = 0 for k < 0 and

k > K , then the convolution sum is expressed as follows:

y[k] =

K∑

m=0

h[m]x[k − m]. (10.27b)

Since periodic convolution is defined in terms of periodic sequences, which are

not causal, Eqs. (10.27a) and (10.27b) are not valid for the periodic convolution.

Example 10.14

Simplify the following expressions using the properties of the discrete-time

convolution:

(i) (x[k] + 2δ[k − 1]) ∗ δ[k − 2],

(ii) (x[k − 1] − 3δ[k + 1]) ∗ (δ[k − 2] + u[k − 1]),

where x[k] is an arbitrary function and δ[k] is the unit impulse function.

Solution

(i) Applying the distributive property,

(x[k] + 2δ[k − 1]) ∗ δ[k − 2] = x[k] ∗ δ[k − 2] ︸︷︷︸

term I

+ 2δ[k − 1] ∗ δ[k − 2] ︸︷︷︸

term II

In both terms I and II, convolution with an impulse function is involved.

Equation (10.25) yields

term I = x[k] ∗ δ[k − 2] = x[k − 2]

and

term II = 2δ[k − 1] ∗ δ[k − 2] = 2δ[k − 3].

The simplified expression for (i) is as follows:

(x[k] + 2δ[k − 1]) ∗ δ[k − 2] = x[k − 2] + 2δ[k − 3].

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(ii) Applying the distributive property,

(x[k − 1] − 3δ[k + 1]) ∗ (δ[k − 2] + u[k − 1])

= x[k − 1] ∗ δ[k − 2] ︸︷︷︸

term I

− 3δ[k + 1] ∗ δ[k − 2] ︸︷︷︸

term II

+ x[k − 1] ∗ u[k − 1] ︸︷︷︸

term III

− 3δ[k + 1] ∗ u[k − 1] ︸︷︷︸

term IV

Terms I, II, and IV involve convolution with an impulse function. Equation

(10.24) yields

term I = x[k − 1] ∗ δ[k − 2] = x[k − 3],

term II = 3δ[k + 1] ∗ δ[k − 2] = 3δ[k − 1],

and

term IV = 3δ[k + 1] ∗ u[k − 1] = 3u[k].

Term III involves convolution with a unit step function. We express term III as

follows:

term III = x[k − 1] ∗ u[k − 1] = (δ[k − 1] ∗ x[k]) ∗ (u[k] ∗ δ[k − 1])

= (x[k] ∗ u[k]) ∗ (δ[k − 1] ∗ δ[k − 1]) = (x[k] ∗ u[k]) ∗ δ[k − 2].

Using Eq. (10.26) we can further simplify term III to obtain

term III = (x[k] ∗ u[k]) ∗ δ[k − 2] =

( k∑

m=−∞

x[m]

)

∗ δ[k − 2]

k−2∑

m=−∞

x[m].

The simplified expression for (ii) is given by

(x[k − 1] − 3δ[k + 1]) ∗ (δ[k − 2] + u[k − 1])

= x[k − 3] − 3δ[k − 1] + 3u[k] +

k−2∑

m=−∞

x[m].

10.8 Impulse response of LTID systems

In Section 2.2, we considered several properties of DT systems. Since the char-

acteristics of an LTID system is completely specified by its impulse response, it

is logical to assume that its properties can also be completely determined from

its impulse response. In this section, we express some of the basic properties of

the LTID systems defined in Section 2.2 in terms of the impulse response of the

LTID systems. We consider the memory, causality, stability, and invertibility

properties for the LTID systems.

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10.8.1 Memoryless LTID systems

A DT system is said to be memoryless if its output y[k] at time instant k = k0 depends only on the value of the applied input sequence x[k] at the same time

instant k = k0. In other words, a memoryless LTID system typically has the input–output relationship of the following form:

y[k] = ax[k],

where a is a constant. By substituting x[k] = δ[k], the impulse response h[k] of a memoryless system can be expressed as

h[k] = aδ[k]. (10.28)

An LTID system will be memoryless if and only if its impulse response

h[k] = aδ[k]. Equivalently, an LTID system is memoryless if and only if h[k] = 0 for k �= 0.

10.8.2 Causal LTID systems

A DT system is said to be causal if the output at time instant k = k0 depends

only on the value of the applied input sequence x[k] at and before the time

instant k = k0. Using the reasoning similar to that given in Section 3.7.2 for the

CT system, the following can be stated.

An LTID system will be causal if and only if its impulse response h[k] = 0

for k < 0.

10.8.3 Stable LTID systems

A DT system is BIBO stable if an arbitrary bounded input sequence always

produces a bounded output sequence. Consider a bounded sequence x[k] with

|x[k]| < Bx , for all k, applied as the input to an LTID system with impulse

response h[k]. The magnitude of the output y[k] is given by

|y[k]| =

∣ ∣ ∣ ∣ ∣

∞∑

m=−∞

h[m]x[k − m]

∣ ∣ ∣ ∣ ∣ .

Using the traingle inequality, we can say that the output is bounded by the

following limit:

|y[k]| ≤

∞∑

m=−∞

|h[m]| x[k − m]|.

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Since |x[k]| < Bx , the above inequality reduces to

|y[k]| ≤ Bx ∞∑

m=−∞

|h[m]|.

It is clear from the above expression that for the output y[k] to be bounded

(i.e. |y[k]| < ∞), the summation ∑∞

m=−∞ |h[m]| needs to be bounded. The

stability condition can therefore be stated as follows.

If the impulse response h[k] of an LTID system satisfies the following

condition:

∞∑

k=−∞

|h[k]| < ∞, (10.29)

the LTID system is BIBO stable.

Example 10.15

Determine which of the LTID systems with impulse responses, shown in

Figs 10.13(a)–(c), are memoryless, causal, and stable.

h1[k]

k −1−2−3−4−5 210 3 4 5

2222

−1−2−3−4−5 210 3 4 5

h2[k]

3 2

−1−2−3−4−5 210 3 4 5

h3[k]

(a)

(b)

(c)

Fig. 10.13. Impulse responses

for systems considered in

Example 10.15.

Solution

(a) Memoryless: since h1[k] �= 0 for k �= 0, the DT system in Fig. 10.13(a) is

not memoryless. In fact, the impulse response h1[k] extends to −∞, therefore

this system has an infinite memory.

Causality: since h1[k] �= 0 for all k < 0, the system is not causal.

Stability: using Eq. (10.29),

∞∑

k=−∞

|h1[k]| =

2∑

k=−∞

|h1[k]| =

2∑

k=−∞

k is even

2 = ∞.

Therefore, the system is not stable.

(b) Memoryless: since h2[k] �= 0 for k �= 0, the DT system in Fig. 10.13(b)

is not memoryless. The impulse response h2[k] has a finite memory of two time

units.

Causality: since h2[k] = 0 for all k < 0, the system is causal.

Stability: using Eq. (10.29),

∞∑

k=−∞

|h2[k]| =

2∑

k=0

|h2[k]| = 3 + 2 + 1 = 6.

Therefore, the system is BIBO stable.

is memoryless.

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Causality: since h3[k] = 0 for all k < 0, the system is causal. Also note that all memoryless systems are causal.

Stability using Eq. (10.29),

∞∑

k=−∞

|h3[k]| = |h3[0]| = 5.

Therefore, the system is BIBO stable.

10.8.4 Invertible LTID systems

Consider an LTID system with impulse response h[k]. The output y1[k] of the

system for an input sequence x[k] is given by y1[k] = x[k] ∗ h[k]. To check its

invertibility property, we cascade a second LTID system with impulse response

hi[k] in series with the original system. The output of the second system is

given by

y2[k] = y1[k] ∗ hi[k] = (x[k] ∗ h[k]) ∗ hi[k]

= x[k] ∗ (h[k] ∗ hi[k]),

based on the associative property.

For the second system to be an inverse of the original system, the final output

y2[k] should be the same as x[k], the input to the first LTID system. This is

possible only if

h[k] ∗ hi[k] = δ[k]. (10.30)

The existence of hi[k] proves that an LTID system is invertible. At times, it is

difficult to determine the inverse system hi[k] in the time domain. In Chapter 11,

when we introduce the discrete Fourier transform, we will revisit the topic and

illustrate how the impulse response of the inverse system can be evaluated with

relative ease in the frequency domain.

Example 10.16

Determine which of the following systems is invertible:

(i) h[k] = δ[k − 3];

(ii) h[k] = δ[k] + δ[k − 1].

Solution

(i) Because δ[k − 3] ∗ δ[k + 3] = δ[k], system (i) is invertible. The impulse

response hi[k] of the inverse of system (i) is given by

hi[k] = δ[k + 3].

(ii) It is difficult to calculate the impulse response of the inverse system in

the time domain. Using the DTFT introduced in Chapter 11, we can show that

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the impulse response of the inverse of system (ii) is given by

hi[k] = ∞∑

m=0

(−1)mδ[k − m] = δ[k] − δ[k − 1] + δ[k − 2] − δ[k − 3] ± · · ·

We can show indirectly that hi[k] is indeed the impulse response of the inverse

of system (ii) by proving that h[k] ∗ hi[k] = δ[k]:

h[k] ∗ hi[k] = (δ[k] + δ[k − 1]) ∗ hi[k] = hi[k] + hi[k − 1]

= (δ[k] − δ[k − 1] + δ[k − 2] − δ[k − 3] ± · · ·) + (δ[k − 1]

− δ[k − 2] + δ[k − 3] − δ[k − 4] ± · · ·)

= δ[k].

10.9 Experiments with M A T L A B

M A T L A B provides several functions (also referred to as M-files) for processing

DT signals and LTID systems. In this section, we will focus on the M A T L A B

implementations of the difference equations with known ancillary conditions,

convolution of two DT signals, and deconvolution.

10.9.1 Difference equations

Consider the following linear, constant-coefficient difference equation:

y[k + n] + an−1 y[k + n − 1] + · · · + a0 y[k]

= bm x[k + m] + bm−1x[k + m − 1] + · · · + b0x[k], (10.31)

which models the relationship between the input sequence x[k] and the output

response y[k] of an LTID system. The ancillary conditions y[−1], y[−2], . . . ,

y[−n] are also specified.

To solve the difference equation, M A T L A B provides a built-in function

filter with the syntax

>> [y] = filter(B,A,X,Zi);

In terms of the difference equation, Eq. (10.31), the input variables B and A are

defined as follows:

A = [1, an−1, . . . , a0] and B = [bm, bm−1, . . . , b0],

while X is the vector containing the values of the input sequence and Zi denotes

the initial conditions of the delays used to implement the difference equation.

The initial conditions used by the filter function are not the past values of

the output y[k] but a modified version of these values. The initial conditions

used by M A T L A B can be obtained by using another built-in function,filtic.

The calling syntax for the filtic function is as follows:

>> [Zi] = filtic(B,A,yinitial);

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For an n-order difference equation, the input variable yinitial is set to

yinitial = [y[−1], y[−2], . . . , y[−n]].

To illustrate the usage of the built-in function filter, let us repeat

Example 10.1 using M A T L A B .

Example 10.17

The DT sequence x[k] = 2ku[k] is applied at the input of an LTID system described by the following difference equation:

y[k + 1] − 0.4 y[k] = x[k],

with the ancillary condition y[−1] = 4. Compute the output response y[k] of the LTID system for 0 ≤ k ≤ 50 using M A T L A B.

Solution

The M A T L A B code used to solve the difference equation is listed below. The

explanation follows each instruction in the form of comments.

>> k = [0:50]; % time index k = [-1, 0, 1,

% ...50]

>> X = 2*k.*(k>=1); % Input signal

>> A = [1 -0.4]; % Coefficients with y[k]

>> B = [0 1]; % Coefficients with x[k]

>> Zi = filtic(B,A,4); % Initial condition

>> Y = filter(B,A,X,Zi); % Calculate output

The output response is stored in the vector Y. Printing the first six values of the

output response yields

Y = [1.6 0.6400 2.2560 4.9024 7.9610 11.1844],

which corresponds to the values of the output response y[k] for the duration

0 ≤ k ≤ 5. Comparing with the numerical solution obtained in Example 10.1,

we observe that the two results are identical.

Next we proceed with a second-order difference equation.

Example 10.18

The DT sequence x[k] = 0.5ku[k] is applied at the input of an LTID system

described by the following second-order difference equation:

y[k + 2] + y[k + 1] + 0.25y[k] = x[k + 2],

with ancillary conditions y[−1] = 1 and y[−2] = −2. Compute the output

response y[k] of the LTID system for 0 ≤ k ≤ 50 using M A T L A B.

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Solution

The M A T L A B code used to solve the difference equation is listed below. The

explanation follows each instruction in the form of comments.

>> k = [0:50]; % time index k = [-1, 0, 1,

% ...50]

>> X = 0.5.ˆk.*(k>=0); % Input signal

>> A = [1 1 0.25]; % Coefficients with y[k]

>> B = [1 0 0]; % Coefficients with x[k]

>> Zi = filtic(B,A,[1 -2]); % Initial condition

>> Y = filter(B,A,X,Zi); % Calculate output

The output response is stored in the vector Y. Printing the first five values of

the output response yields

Y = [0.5000 -0.2500 0.3750 -0.1875 0.1563

-0.0781 0.0547].

To confirm if the M A T L A B code is correct, we also compute the values of

the output response in the range 0 ≤ k ≤ 5. We express y[k + 2] + y[k + 1] +

0.25y[k]=x[k + 2] as follows:

y[k] = −y[k − 1] − 0.25y[k − 2] + x[k],

with ancillary conditions y[−1] = 1 and y[−2] = −2. Solving the difference

equation iteratively yields

y[0] = −y[−1] − 0.25y[−2] + x[0] = −1 − 0.25(−2) + 1 = 0.5,

y[1] = −y[0] − 0.25y[−1] + x[1] = −0.5 − 0.25(1) + 0.5 = −0.25,

y[2] = −y[1] − 0.25y[0] + x[2] = −(−0.25) − 0.25(0.5) + 0.25 = 0.375,

y[3] = −y[2] − 0.25y[1] + x[3] = −0.375 − 0.25(−0.25) + 0.125

= −0.1875,

y[4] = −y[3] − 0.25y[2] + x[4] = −(−0.1875) − 0.25(0.375) + 0.0625

= 0.1563,

and

y[5] = −y[4] − 0.25y[3] + x[2] = −0.1563 − 0.25(−0.1875) + 0.031 25

= −0.0782,

which are the same as the values computed using M A T L A B .

The expressions for the initial conditions for the higher-order difference equa-

tions are more complex. Fortunately, most systems are causal with zero ancillary

conditions. The initial conditions Zi are zero in such cases.

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10.9.2 Convolution

Consider two time-limited DT sequences x1[k] and x2[k], where x1[k] �= 0

within the range kℓ1 ≤ k ≤ ku1 and x2[k] �= 0 within the range kℓ2 ≤ k ≤ ku2.

The length K1 of the DT sequence x1[k] is given by K1 = ku1 − kℓ1 + 1 samples,

while the length K2 of the DT sequence x2[k] is K2 = ku2 − kℓ2 + 1 samples.

In M A T L A B , two vectors are required to represent each DT signal. The first

vector contains the sample values, while the second vector stores the time

indices corresponding to the sample values. For example, the following DT

sequence:

x[k] =



  

  

−1 k = −1

1 k = 0

2 k = 1

0 otherwise

has the following M A T L A B representation:

>> kx = [-1 0 1]; % time indices where x is nonzero

>> x = [-1 1 2]; % Sample values for DT sequence x

To perform DT convolution, M A T L A B provides a built-in function conv. We

illustrate its usage by repeating Example 10.9 with M A T L A B .

Example 10.19

Consider the following two DT sequences x[k] and h[k] specified in

Example 10.9:

x[k] =



  

  

−1 k = −1

1 k = 0

2 k = 1

0 otherwise

and h[k] =



  

  

3 k = −1, 2

1 k = 0

−2 k = 1, 3

0 otherwise.

Compute the convolution y[k] = x[k] ∗ h[k] using M A T L A B.

Solution

The M A T L A B code used to convolve the two functions is given below. As

before, the explanation follows each instruction in the form of comments.

>> kx = [-1 0 1]; % time indices where x is nonzero

>> x = [-1 1 2]; % Sample values for DT sequence x

>> kh = [-1 0 1 2 3]; % time indices where y is nonzero

>> h = [3 1 -2 3 -2]; % Sample values for DT sequence y

>> y = conv(x,h); % Convolve x with h

>> ky = kx(1)+kh(1):kx(length(kx))+kh(length(kh));

% ky= time indices for y

In the above instructions, note that M A T L A B does not calculate the indices of

the result of convolution. These indices have to be calculated separately based

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on the observation that we made on the starting and last indices of the convolved

result.

The computed values of y are given by

y = [-3 2 9 -3 1 4 -4],

with the computed indices

ky = [-2 -1 0 1 2 3 4].

Note that the above result is the same as the one obtained in Example 10.9.

The function deconv performs the inverse of the convolution sum. Given a DT

input sequence x and the output sequence y, for example, the impulse response

h can be determined using the following instructions:

>> h2 = deconv(y,x); % Deconvolve x out of y

>> kh2 = ky(1)-kx(1):ky(length(ky)) -kx(length(kx));

% kh2 = indices for h2

Note that h2 has the same sample values and indices kh2 as those of h.

10.10 Summary

In this chapter, we developed analytical techniques for LTID systems. We saw

that the output sequence y[k] of an LTID system can be calculated analytically

in the time domain using two different methods. In Section 10.1, we determined

the output of a DT system by solving a linear, constant-coefficient difference

equation. The solution of such a difference equation can be expressed as a sum

of two components: the zero-input response and the zero-state response. The

zero-input response is the output produced by the DT system because of the

initial conditions. For most DT systems, the zero-input response decays to zero

with increasing time. The zero-state response results from the input sequence.

The overall output of a DT system is the sum of the zero-input response and

the zero-state response. A DT system, of the form shown in Eq. (10.1), will be

an LTID system if all initial conditions are zero. In other words, the zero-input

response of an LTID system is always zero.

An alternative representation for determining the output of an LTID system

is based on the impulse response of the system. In Section 10.3, we defined the

impulse response h[k] as the output of an LTID system when a unit impulse δ[k]

is applied at the input of the system. In Section 10.4, we proved that the output

y[k] of an LTID system could be obtained by convolving the input sequence

x[k] with its impulse response h[k]. The resulting convolution sum can either

be solved analytically or by using a graphical approach. The graphical approach

was illustrated through several examples in Section 10.5. In discrete time, the

convolution of two periodic functions is also defined and is known as periodic

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460 Part III Discrete-time signals and systems

or circular convolution. The periodic convolution is discussed in Section 10.6,

where we mentioned that the linear convolution may be efficiently calculated

through periodic convolution. The convolution sum satisfies the commutative,

distributive, associative, and time-shifting properties.

(1) The commutative property states that the order of the convolution operands

does not affect the result of the convolution.

(2) The distributive property states that convolution is a linear operation with

respect to addition.

(3) The associative property is an extension of the commutative property to

more than two convolution operands. It states that changing the order of

the convolution operands does not affect the result of the convolution sum.

(4) The time-shifting property states that if the two operands of the convolution

sum are shifted in time then the result of the convolution sum is shifted

by a duration that is the sum of the individual time shifts introduced in the

convolution operands.

(5) If the lengths of the two functions are K1 and K2 samples, the convolution

sum of these two functions will have a length of K1 + K2 − 1 samples. (6) Convolving a sequence with a unit DT impulse function with the origin at

k = k0 shifts the sequence by k0 time units. (7) Convolving a sequence with a unit DT step function produces the running

sum of the original sequence as a function of time k.

Finally, in Section 10.8, we expressed the memoryless, causality, stability, and

invertibility properties of an LTID system in terms of its impulse response.

(1) An LTID system will be memoryless if and only if its impulse response

h[k] = 0 for k �= 0. (2) An LTID system will be causal if and only if its impulse response h[k] = 0

for k < 0.

(3) The impulse response h[k] of a (BIBO) stable LTID system is absolutely

summable, i.e.

∞∑

k=−∞

|h[k]| < ∞.

(4) An LTID system will be invertible if there exists another LTID system with

impulse response hi[k] such that h[k] ∗ hi[k] = δ[k]. The system with the

impulse response hi[k] is the inverse system.

In the next chapter, we consider the frequency representations of DT sequences

and systems.

Problems

10.1 Consider the input sequence x[k] = 2u[k] applied to a DT system modeled with the following input–output relationship:

y[k + 1] − 2y[k] = x[k],

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and ancillary condition y[−1] = 2. (a) Determine the response y[k] by iterating the difference equation for

0 ≤ k ≤ 5.

(b) Determine the zero-state response yzi[k] for 0 ≤ k ≤ 5.

(d) Verify that y[k] = yzi[k] + yzs[k].

10.2 Repeat Problem 10.1 for the applied input x[k] = 0.5ku[k] and the input– output relationship

y[k + 2] − y[k + 1] + 0.5y[k] = x[k],

with ancillary conditions y[−1] = 0 and y[−2] = 1.

10.3 Repeat Problem 10.1 for the applied input x[k] = (−1)ku[k] and the input–output relationship

y[k + 2] − 0.75y[k + 1] + 0.125y[k] = x[k],

with ancillary conditions y[−1] = 1 and y[−2] = −1.

10.4 Show that the convolution of two sequences aku[k] and bku[k] is given by

(aku[k]) ∗ (bku[k]) =







(k + 1)aku[k] a = b 1

a − b (ak+1 − bk+1)u[k] a �= b.

10.5 Calculate the convolution (x1[k] ∗ x2[k]) for the following pairs of sequences:

(a) x1[k] = u[k + 2] − u[k − 3], x2[k] = u[k + 4] − u[k − 5];

(b) x1[k] = 0.5 ku[k], x2[k] = 0.8

ku[k − 5];

ku[k − 4];

(d) x1[k] = 0.6 ku[k], x2[k] = sin(πk/2)u[−k];

(e) x1[k] = 0.5 |k|, x2[k] = 0.8

|k|.

10.6 For the following pairs of sequences:

(a) x[k] =

{

k 0 ≤ k ≤ 3

0 otherwise and h[k] =

{

2 −1 ≤ k ≤ 2

0 otherwise;

(b) x[k] =

{

|k| |k| ≤ 2

0 otherwise and h[k] =

{

2−k 0 ≤ k ≤ 3

0 otherwise,

calculate the DT convolution y[k] = x[k] ∗ h[k] using (i) the graphical

approach and (ii) the sliding tape method.

10.7 Using the sliding tape method and the following equation:

y[k] =

∞∑

m=−∞

h[m]x[k − m],

calculate the convolution of the sequences in Example 10.8 and show that

the convolution output is identical to that obtained in Example 10.8.

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462 Part III Discrete-time signals and systems

10.8 Using the sliding tape method and the following equation:

y[k] = ∞∑

m=−∞

x[m]h[k − m],

calculate the convolution of the sequences in Example 10.9 and show

that the convolution output is identical to that obtained in Example 10.9.

10.9 The linear convolution between two sequences x[k] and h[k] of lengths K1 and K2, respectively, can be performed using periodic convolution by

considering periodic extensions of the two zero-padded sequences. Cal-

culate the linear convolution of the sequences defined in Example 10.8

using the periodic convolution approach with the fundamental period

K0 set to 10. Repeat for K0 set to 13.

10.10 Repeat Example 10.7 using the periodic convolution approach with K set to 10.

10.11 Repeat Example 10.7 using the periodic convolution approach with K set to 15.

10.12 Repeat Example 10.12 with K set to 8.

10.13 Calculate the unit step response of the DT systems with the following impulse responses:

(a) h[k] = u[k + 7] − u[k − 8];

(b) h[k] = 0.4ku[k];

(d) h[k] = 0.6|k|;

(e) h[k] =

∞∑

m=−∞

(−1)mδ(k − 2m).

10.14 Simplify the following expressions using the properties of discrete-time convolution:

(a) (x[k] + 2δ[k − 1]) ∗ δ[k − 2];

(b) (x[k] + 2δ[k − 1]) ∗ (δ[k + 1] + δ[k − 2]);

(d) (x[k] − x[k − 1]) ∗ u[k],

where x[k] is an arbitrary function, δ[k] is the unit impulse function, and

u[k] is the unit step function.

10.15 Prove Definition 10.3 by expanding the right-hand side of the periodic convolution and showing it to be equal to the left-hand side.

10.16 Prove the time-shifting property stated in Eq. (10.24).

10.17 Show that the linear convolution y[k] of a time-limited DT sequence x1[k] that is non-zero only within the range kℓ1 ≤ k ≤ ku1 with another

time-limited DT sequence x2[k] that is non-zero only within the range

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463 10 Time-domain analysis of DT systems

kℓ2 ≤ k ≤ ku2 is time-limited, and is non-zero only within the range

kℓ1 + kℓ2 ≤ k ≤ ku1 + ku2.

10.18 For each of the following impulse responses, determine if the DT system is (i) memoryless; (ii) causal; and (iii) stable:

(a) h[k] = u[k + 7] − u[k − 8];

(b) h[k] = sin (

πk 8

)

u[k];

(d) h[k] = 0.9|k|;

(e) h[k] =

∞∑

m=−∞

(−1)mδ(k − 2m).

10.19 Determine which of the following pair of impulse responses correspond to inverse systems:

(a) h1[k] = u[−k − 1], h2[k] = δ[k − 1] − δ[k];

(b) h1[k] = 0.5 ku[k], h2[k] = δ[k] − 0.5δ[k − 1];

+ 1.25δ[k + 1];

(d) h1[k] = ku[k], h2[k] = δ[k + 1] − 2δ[k] + δ[k − 1];

(e) h1[k] = (k + 1)0.8 ku[k], h2[k] = δ[k] − 1.6δ[k − 1]

+ 0.64δ[k − 2].

10.20 Repeat Problems 10.1–10.3 to compute the first 50 samples of the out- put response using the filter and filtic functions available in

M A T L A B.

10.21 Repeat Problem 10.5 using the conv function available in M A T L A B . For a sequence with infinite length, you may truncate the sequence when

the value of the sequence is less than 0.1% of its maximum value.

10.22 The M A T L A B function impz can be used to determine the impulse response of an LTID system from its difference equation representation.

Determine the first 50 samples of the impulse response of the LTID

systems with the difference equations specified in Problems 10.1–10.3.

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C H A P T E R

11 Discrete-time Fourier series and transform

In Chapter 10, we developed analysis techniques for LTID systems based on the

convolution sum by representing the input sequence x[k] as a linear combination

of time-shifted unit impulse functions. In this chapter, we introduce frequency-

domain representations for DT sequences and LTID systems based on weighted

superpositions of complex exponential functions. For periodic sequences, the

resulting representation is referred to as the discrete-time Fourier series (DTFS),

while for aperiodic sequences the representation is called the discrete-time

Fourier transform (DTFT). We exploit the properties of the discrete-time Fourier

series and Fourier transform to develop alternative techniques for analyzing DT

sequences. The derivations of these results closely parallel the development of

the CT Fourier series (CTFS) and CT Fourier transform (CTFT) as presented

in Chapters 4 and 5.

The organization of this chapter is as follows. In Section 11.1, we intro-

duce the exponential form of the DTFS and illustrate the procedure used to

calculate the DTFS coefficients through a series of examples. The DTFT pro-

vides frequency representations for aperiodic sequences and is presented in

Section 11.2. Section 11.3 defines the condition for the existence of the DTFT,

and Section 11.4 extends the scope of the DTFT to represent periodic sequences.

Section 11.5 lists the properties of the DTFT and DTFS, including the time-

convolution property, which states that the convolution of two DT sequences

in the time domain is equivalent to the multiplication of the DTFTs of the

two sequences in the frequency domain. The convolution property provides

us with an alternative technique to compute the output response of the LTID

system. The DTFT of the impulse response is referred to as the transfer func-

tion, which is covered in Section 11.6. Section 11.7 defines the magnitude and

phase spectra for LTID systems, and Section 11.8 relates the CTFT and DTFT

of periodic and aperiodic waveforms to each other. Finally, the chapter is con-

cluded in Section 11.9 with a summary of important concepts covered in the

chapter.

464

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465 11 Discrete-time Fourier series and transform

11.1 Discrete-time Fourier series

In Example 4.4, we proved that the set of complex exponential functions

exp(jnω0t), n ∈ Z , defines an orthonormal set of functions over the interval

t = (t0, t0 + T0) with duration T0 = 2π/ω0. This orthonormal set of exponen-

tials was used to derive the CT Fourier series. In the same spirit, we now show

that the discrete-time (DT) complex exponential sequences form an orthonor-

mal set in the DT domain and are used to derive the DTFS. We start with the

definition of the orthonormal sequences.

Definition 11.1 Two sequences p[k] and q[k] are said to be orthogonal over

interval k = [k1, k2] if

orthogonality property

k2∑

k=k1

p[k]q∗[k] = k2∑

k=k1

p∗[k]q[k] = 0, p[k] �= q[k],

(11.1)

where the superscript ∗ denotes complex conjugation. In addition to Eq. (11.1), both signals p[k] and q[k] must also satisfy the following unit magnitude prop-

erty to satisfy the orthonormality condition:

unit magnitude property

k2∑

k=k1

p[k]p∗[k] = k2∑

k=k1

q∗[k]q[k] = 1. (11.2)

Definition 11.2 A set comprising an arbitrary number of N functions, say{p1[k],

p2[k], . . . , pN [k]}, is mutually orthogonal over interval k = [k1, k2] if

k2∑

k=k1

pm[k]p ∗ n[k] =

{

En �= 0 m = n 0 m �= n, (11.3)

for 1 ≤ m, n ≤ N. In addition, if En = 1 for all n, the orthogonal set is referred to as an orthonormal set.

Based on Definitions 11.1 and 11.2, we show that the DT complex sequences

form an orthogonal set.

Proposition 11.1 The set of discrete-time complex exponential sequences

{exp(jnΩ0k), n ∈ Z}, is orthogonal over the interval [r , r + K0 − 1], where the duration K0 = 2π/Ω0 and r is an arbitrary integer.

Proof

Consider the following summation:

r+K0−1∑

k=r e jmΩ0ke−jnΩ0k =

r+K0−1∑

k=r e j(m−n)Ω0k .

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466 Part III Discrete-time signals and systems

Substituting p = k − r to make the lower limit of the summation equal to zero yields

r+K0−1∑

k=r e j(m−n)Ω0k =

K0−1∑

p=0 e j(m−n)Ω0(p+r ) = e j(m−n)Ω0r

K0−1∑

p=0 e j(m−n)Ω0 p.

The above summation is solved for two different cases, m = n and m �= n.

Case I For m = n, the summation reduces to

e j(m−n)Ω0r K0−1∑

p=0

e j(m−n)Ω0 p = 1 ·

K0−1∑

p=0

1 = K0.

Case II For m �= n, the summation forms a GP series and is simplified as follows:

e j(m−n)Ω0r K0−1∑

p=0

e j(m−n)Ω0 p = e j(m−n)Ω0r [

1 − e j(m−n)Ω0 K0

1 − e j(m−n)Ω0

]

Because Ω0 K0 = 2π and indices m and n are integers, the exponential term in

the numerator is given by

e j(m−n)Ω0 K0 = e j(m−n)2π = 1.

Therefore, for m �= n the summation reduces to

e j(m−n)Ω0r K0−1∑

p=0

e j(m−n)Ω0 p = e j(m−n)Ω0r [

1 − 1

1 − e j(m−n)Ω0

]

= 0.

Combining the results of cases I and II, we obtain

r+K0−1∑

k=r

e jmΩ0ke−jnΩ0k =

{

K0 if m = n

0 if m �= n.

In other words, the set of DT complex exponential sequences {exp(jnΩ0k),

n ∈ Z} is orthogonal over the specified interval [r, r + K0 − 1].

An important difference between the DT and CT complex exponential functions

lies in the frequency–periodicity property of the DT exponential sequences.

Since

e jn(Ω0+2π )k = e jnΩ0ke jn2πk = e jnΩ0k,

the exponential sequence exp(jnΩ0k) is identical to exp(jn(Ω0 + 2π )k). This is in contrast to the CT exponentials, where exp(jnω0t) is different from

exp(jn(ω0 + 2π )t). The following example illustrates the frequency period- icity for the DT sinusoidal signals. Using the Euler property, a DT complex

exponential exp(jnΩ0k) can be expressed as follows:

e jnΩ0k = cos(nΩ0k) + j sin(nΩ0k).

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467 11 Discrete-time Fourier series and transform

Example 11.1 shows that both the real and imaginary components of the com-

plex exponential satisfy the frequency–periodicity property; therefore, the DT

complex exponential should also satisfy the frequency–periodicity property.

Example 11.1

Consider a CT sinusoidal function with a fundamental frequency of 1.4 Hz, i.e.

x(t) = cos(2.8π t + φ),

where φ is the constant phase. Sample the function with a sampling rate of

1 sample/s and determine the fundamental frequency of the resulting DT

sequence.

Solution

In the time domain, the DT sequence is obtained by sampling x(t) at t = kT . Since the sampling interval T = 1 s,

x[k] = x(kT ) = cos(2.8πk + φ),

which is periodic with a periodΩ1 = 2.8π radians/s. Because the CT signal x(t) is a sinusoid with a fundamental frequency of 1.4 Hz, the minimum sampling

rate, required to avoid aliasing, is given by 2.8 samples/s. Since the sampling

rate of 1 samples/s is less than the Nyquist sampling rate, aliasing is introduced

due to sampling. Based on Lemma 9.1, the reconstructed signal is given by

y(t) = cos(2π (1.4 − 1)t) = cos(0.8π t + φ).

Substituting t = kT , the DT representation of the reconstructed signal is given by

y[k] = cos(0.8πk + φ),

which is periodic with a periodΩ2 = 0.8π radians/s. From the above analysis, it is clear that the DT sequences x[k] = cos(2.8πk + φ) and y[k] = cos(0.8πk + φ) are identical. This is because the difference in the fundamental frequenciesΩ1

and Ω2 is 2π .

Proposition 11.2 A discrete-time periodic function x[k] with period K0 can be

expressed as a superposition of DT complex exponentials as follows:

x[k] = ∑

n=<K0> Dne

jnΩ0k, (11.4)

whereΩ0 is the fundamental frequency, given byΩ0 = 2π/K0, and the discrete- time Fourier series (DTFS) coefficients Dn for 1 ≤ n ≤ K0 are given by

Dn = 1

∑

k=〈K0〉 x[k]e−jnΩ0k . (11.5)

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468 Part III Discrete-time signals and systems

In Eq. (11.5), the limit of k = 〈K0〉 implies that the sum can be taken over any K0 consecutive samples of x[k]. Unless otherwise specified, we would consider

the range 0 ≤ k ≤ K0 − 1 in our derivations.

Proof

To verify the DTFS, we expand the right-hand side of Eq. (11.4) by substituting

the value of Dn from Eq. (11.5). With K0 consecutive exponentials in the range

0 ≤ n ≤ k0 − 1, the resulting expression is given by

K0−1∑

n=0 Dne

jnΩ0k = K0−1∑

n=0

[

K0−1∑

m=0 x[m]e−jnΩ0m

]

e jnΩ0k .

Interchanging the order of the summation yields

K0−1∑

n=0 Dne

jnΩ0k = 1

K0−1∑

m=0 x[m]

[ K0−1∑

n=0 e jnΩ0(k−m)

]

. (11.6)

From Proposition 11.1, we have

K0−1∑

n=0 e jnΩ0(k−m) =

{

K0 if k = m 0 if k �= m.

The right-hand side of Eq. (11.6) reduces to

K0−1∑

n=0 Dne

jnΩ0k = 1

K0−1∑

m=0 x[m]K0δ[m − k] =

K0 K0x[k] = x[k],

and therefore proves Proposition 11.2.

Examples 11.2–11.5 calculate the DTFS for selected DT periodic sequences.

Example 11.2

Determine the DTFS coefficients of the following periodic sequence:

h[k] = {

1 |k| ≤ N 0 N + 1 ≤ k ≤ K0 − N − 1,

(11.7)

with a fundamental period K0 > (2N + 1).

Solution

With K0 consecutive samples in the range −N ≤ k ≤ K0 − N − 1, Eq. (11.5) reduces to

Dn = 1

N∑

k=−N 1 · e−jnΩ0k +

K0−N−1∑

k=N+1 0 · e−jnΩ0k =

N∑

k=−N e−jnΩ0k .

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469 11 Discrete-time Fourier series and transform

Table 11.1. Values of Dn for 0 ≤ n ≤ 9 in Example 11.2

n 0 1 2 3 4 5 6 7 8 9

Dn 0.300 0.262 0.162 0.038 −0.062 −0.100 −0.062 0.038 0.162 0.262

h[k]

k 0 2 4 6 8 10 12 14−8 −6 −4 −2−10−12−14

1 11 1 111 11

n 0 2

4 6

8 10 12

−8

−6 −4

−2−10−12

−14

0.16

0.04

0.26

−0.06 −0.1

−0.06

0.04 0.16

0.26 0.3

0.16

0.04

0.26

−0.06 −0.1

0.16 0.04

0.26

−0.1

0.3

−0.06 −0.06

0.04

0.16 0.260.3

−0.06

0.04 0.16

0.26

(a)

(b)

Fig. 11.1. (a) DT periodic

sequence h[k ]; (b) its DTFS

coefficients calculated in

Example 11.2.

The summation represents a GP series and simplifies as follows:

Dn = 1

[

e jnΩ0 N 1 − e−jnΩ0(2N+1)

1 − e−jnΩ0

]

= 1

[

e jnΩ0 N e−jnΩ0(2N+1)/2

e− jnΩ0/2 e jnΩ0(2N+1)/2 − e−jnΩ0(2N+1)/2

e jnΩ0/2 − e−jnΩ0/2

]

= 1



  

sin

( 2N + 1

2 nΩ0

)

sin

( 1

2 nΩ0

)



  

Substituting the value of the fundamental frequency Ω0 = 2π/K0 yields

Dn = 1



  

sin (

2N + 1 K0

nπ )

sin

( 1

K0 nπ

)



  

, (11.8)

which represents a DT sinc function.

As a special case, we plot the values of the coefficients Dn for N = 1 and K0 = 10 in Fig. 11.1. The expression for the DTFS coefficients is given by

Dn = 1

[ sin(0.3nπ )

sin(0.1nπ )

]

with the values for 0 ≤ n ≤ 9 given in Table 11.1.

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470 Part III Discrete-time signals and systems

The value of the DTFS coefficient D0 is calculated using L’Hôpital’s rule as

follows:

D0 = lim n→0

[ sin(0.3nπ )

sin(0.1nπ )

]

= lim n→0

[ (0.3π ) cos(0.3nπ )

(0.1π ) cos(0.1nπ )

]

= 0.3.

In Fig. 11.1(b), we observe that the DTFS coefficients are periodic with a period

of 10, which is the same as the fundamental period of the original sequence

h[k]. One such period is highlighted in Fig. 11.1(b).

11.1.1 Periodicity of DTFS coefficients

In Example 11.2, we noted that the DTFS coefficients Dn of a periodic sequence

are themselves periodic with a period of K0. In Proposition 11.3, we show that

this is true for any DT periodic sequence.

Proposition 11.3 The DTFS coefficients Dn of a periodic sequence x[k], with a

period of K0, are themselves periodic with a period of K0. In other words,

Dn = Dn+mK0 for m ∈ Z . (11.9)

Proof

By definition, the DTFS coefficients are expressed as follows:

Dn+mK0 = 1

∑

k=〈K0〉 x[k]e−j(n+mK0)Ω0k

= 1

∑

k=〈K0〉 x[k]e−jnΩ0ke−jmΩ0 K0k

where the exponential term exp(−jmΩ0 K0k) = exp(−j2mπk) = 1. The above expression reduces to

Dn+mK0 = 1

∑

k=〈K0〉 x[k]e−jnΩ0k,

which, by definition, is Dn .

In the following examples, we calculate the DTFS coefficients Dn over one

period (n = 〈K0〉) and exploit the periodicity property to obtain the DTFS coefficients outside this range.

Example 11.3

Determine the DTFS coefficients of the periodic DT sequence x[k] with one

fundamental period defined as

x[k] = 0.5ku[k], 0 ≤ k ≤ 14. (11.10)

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471 11 Discrete-time Fourier series and transform

x[k]

k 0 2 4 6 8 10 12 14−8 −6 −4 −2−10−12−14−16−18−20 201816

0.5 0.25

0.13

0.5

0.25 0.13

0.5 0.250.13

Fig. 11.2. Periodic DT sequence

defined in Example 11.3. Solution

The DT sequence x[k] is plotted in Fig. 11.2. Since its period K0 = 15, the fundamental frequency Ω0 = 2π/15. The DTFS coefficients Dn are given by

Dn = 1

14∑

k=0 0.5ke−jnΩ0k =

14∑

k=0 (0.5e−jnΩ0 )k,

which is a GP series that simplifies to

Dn = 1

15 ·

1 − (

0.5e−jnΩ0 )15

1 − 0.5e−jnΩ0 =

15 ·

1 − 0.515e−j15nΩ0 1 − 0.5e−jnΩ0

Since Ω0 = 2π/15, the exponential term in the numerator, exp(−j15nΩ0) = exp(−j2nπ ) = 1. Expanding the exponential term in the denominator as exp(−jnΩ0) = cos(nΩ0) − j sin(nΩ0), the DTFS coefficients are given by

Dn = 1

15 ·

1 − 0.515

1 − 0.5 cos(nΩ0) + j0.5 sin(nΩ0)

≈ 1

15 ·

1 − 0.5 cos(nΩ0) + j0.5 sin(nΩ0) . (11.11)

As the DTFS coefficients are complex, we determine the magnitude and phase

of the coefficients as follows:

magnitude |Dn| = 1

15 ·

1 √

(1 − 0.5 cos(nΩ0))2 + (0.5 sin(nΩ0))2

= 1

15 ·

1 √

1.25 − cos(nΩ0) ; (11.12)

phase <Dn = − tan−1 [

0.5 sin(nΩ0)

1 − 0.5 cos(nΩ0)

]

, (11.13)

whereΩ0 = 2π/15. The magnitude and phase spectra of the DTFS coefficients are plotted in Figs. 11.3(a) and (b), in which one period of Dn is highlighted by

a shaded region.

Example 11.4

Determine the DTFS coefficients of the following periodic function:

x[k] = Ae j((2πm/N )k+θ ), (11.14)

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472 Part III Discrete-time signals and systems

n 0 2 4 6 8 10 12 14−8 −6 −4 −2−10−12−14−16−18−20 201816

0.13 0.11

0.090.070.060.050.050.050.050.050.050.06 0.070.09

0.11 0.13

0.11 0.090.070.060.050.050.050.050.050.050.06

0.070.09 0.11

0.050.06 0.070.09

0.11 0.13

0.11 0.090.070.060.05

<Dn

n 0 2 4 6

8 10 12 14

−8

−6 −4 −2

−10−12−14

−16−18−20

2018160.070.21

−0.51 −0.44

−0.33 −0.21

−0.07

0.51 0.44

0.33

0.51

0.36

−0.51

−0.36

0.07 0.21

−0.51 −0.44

−0.33 −0.21

−0.07

0.51 0.44

0.33

0.51

0.36

−0.51

−0.36

−0.51 −0.44

−0.33

−0.51

−0.36

0.51 0.44

0.33

0.51

0.36

(a)

(b)

Fig. 11.3. (a) Magnitude

spectrum and (b) phase

spectrum of the DTFS

coefficients in Example 11.3.

where the greatest common divisor between the fundamental period N and the

integer constant m is one.

Solution

We first show that the DT sequence x[k] is periodic and determine its fundamen-

tal period. It was mentioned in Proposition 1.1 that a DT complex exponential

sequence x[k] = exp(j(Ω0k + θ )) is periodic if 2π/Ω0 is a rational number. In this case, 2π/Ω0 = N/m, which is a rational number as m, K and N are all integers. In other words, the sequence x[k] is periodic. Using Eq. (1.8), the

fundamental period of x[k] is calculated to be

K0 = (2π/Ω0)p = pN/m,

where p is the smallest integer that results in an integer value for K0. Note

that the fraction N/m represents a rational number, which cannot be reduced

further since the greatest common divisor between m and N is given to be one.

Selecting p = m, the fundamental period is obtained as K0 = N . To compute the DTFS coefficients, we express x[k] as follows:

x[k] = Ae jθe j�0mk

and compare this expression with Eq. (11.4). For 0 ≤ n ≤ K0 − 1, we observe that

Dn = {

Ae jθ if n = m 0 if n �= m. (11.15)

As a special case, we consider A = 2, K0 = 6, m = 5, and θ = π/4. The magnitude and phase spectra for the selected values are shown in Figs.

11.4(a) and (b), where we have used the periodicity property of the DTFS

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473 11 Discrete-time Fourier series and transform

0 2 4 6 8 10 12 14−8 −6 −4 −2−10−12−14−16−18−20 201816

2 2 22 2 22

<Dn

p/4 p/4 p/4p/4 p/4 p/4p/4

(a)

(b)

420

Fig. 11.4. (a) Magnitude

spectrum and (b) phase

spectrum of the DTFS

coefficients in Example 11.4.

coefficients to plot the values of the coefficients outside the duration 0 ≤ n ≤ (K0 − 1).

Substituting θ = 0 in Example 11.4 results in Corollary 11.1.

Corollary 11.1 The DTFS coefficients corresponding to the complex exponential

sequence x[k] = A exp(j2πmk/K0) with the fundamental period K0 are given by

Dn = {

A if n = m, m ± K0, m ± 2K0, . . . 0 elsewhere,

(11.16)

provided the greatest common divisor between the m and K0 is one.

Example 11.5

Determine the DTFS coefficients of the following sinusoidal sequence:

y[k] = B sin (

2πm

K0 k + θ

)

, (11.17)

where the greatest common divisor between integers m and N is one. The phase

component θ is constant with respect to time.

Solution

Using Proposition 1.1, it is straightforward to show that the sinusoidal sequence

y[k] is periodic with fundamental period K0. The fundamental frequency is

given by Ω0 = 2π/K0.

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474 Part III Discrete-time signals and systems

Based on Eq. (11.5), and noting thatΩ0 = 2π/K0, the DTFS coefficients are given by

Dn = 1

∑

k=<K0> B sin(mΩ0k + θ ) · e− jnΩ0k

= 1

∑

k=<K0> B

[ e j(mΩ0k+θ ) − e j(mΩ0k+θ )

2 j

]

· e−jnΩ0k

= − j B

2K0 e jθ

∑

k=〈K0〉 e j(m−n)Ω0k

︸︷︷︸

summation I

+ j B

2K0 e−jθ

∑

k=〈K0〉 e−j(m+n)Ω0k

︸︷︷︸

summation II

In proving Proposition 11.2, we used the following summation:

K0−1∑

n=0 e jnΩ0(k−m) =

{

K0 if k = m 0 if k �= m.

Therefore, summations I and II are given by

I = ∑

k=〈K0〉 e j(m−n)Ω0k =

{

K0 if n = m 0 if n �= m;

II = ∑

k=〈K0〉 e−j(m+n)Ω0k =

{

K0 if n = −m 0 if n �= −m,

which results in the following values for the DTFS coefficients:

Dn =



   

   

−j B

2 e jθ for n = m

j B

2 e−jθ for n = −m

0 elsewhere,

(11.18)

within one period (−m ≤ n ≤ (K0 − m − 1)). As a special case, let us consider the DTFS for the following discrete sinu-

soidal sequence:

y[k] = 3 sin (

2π

7 k +

)

which has a fundamental period of K0 = 7. Substituting B = 3, m = 1, and θ = π /4 into Eq. (11.18), we obtain

Dn =



    

    

−j 3

2 e

j π 4 for n = 1

j 3

2 e −j π

4 for n = −1

0 elsewhere,

(11.19)

for −1 ≤ n ≤ 5. The magnitude and phase spectra for the sinusoidal sequence are shown in Figs. 11.5(a) and (b).

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475 11 Discrete-time Fourier series and transform

8 10 12 14−8 −6 −4 −2−10−12−14−16−18−20 201816

1.5 1.51.5 1.5 1.51.5 1.5 1.51.5 1.5 1.51.5

<Dn

0 2 4 6

10 12 14−8

−6

−4 −2−10−12−14−16−18

−20

201816

−p/4 −p/4−p/4 −p/4 −p/4−p/4

p/4 p/4p/4 p/4 p/4p/4

0 2 4 6

(a)

(b)

Fig. 11.5. (a) Magnitude

spectrum and (b) phase

spectrum of the DTFS

coefficients in Example 11.5.

Corollary 11.2 The DTFS coefficients of the sinusoidal sequence x[k] = B sin(2πmk/K0) are given by

Dn =



    

    

−j B

2 for n = m, m ± K0, m ± 2K0, . . .

j B

2 for n = −m, −m ± K0, −m ± 2K0, . . .

0 elsewhere,

(11.20)

provided that the greatest common divisor between integers m and K0 is one.

11.2 Fourier transform for aperiodic functions

In Section 11.1, we used the exponential DTFS to derive the frequency repre-

sentations for periodic sequences. In this section, we consider the frequency

representations for aperiodic sequences. The resulting representation is called

the DT Fourier transform (DTFT).

Figure 11.6(a) shows the waveform of an aperiodic sequence x[k], which

is zero outside the range M1 ≤ k ≤ M2. Such a sequence is referred to as a time-limited sequence having a length of M2 − M1 + 1 samples. As was the case for the CTFT, we consider periodic repetitions of x[k] uniformly spaced

with a duration of K0 between each other; K0 ≥ (M2 − M1 + 1) such that the adjacent replicas of x[k] do not overlap with each other. The resulting sequence

is referred to as the periodic extension of x[k] and is denoted by x̃K0 [k]. If we

increase the value of K0, in the limit, we obtain

lim K0→∞

x̃K0 [k] = x[k]. (11.21)

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476 Part III Discrete-time signals and systems

M1 M20−K0

xK0[k]

M1 M20

x[k]

(a)

(b)

Fig. 11.6. (a) Time-limited

sequence x[k ]; (b) its periodic

extension.

Since x̃K0 [k] is periodic with fundamental period K0 (or fundamental frequency

Ω0), we can express it using the DTFS as follows:

x̃K0 [k] = ∑

n=〈K0〉 Dne

jnΩ0k, (11.22)

where the DTFS coefficients Dn are given by

Dn = 1

∑

k=〈K0〉 x̃K0 [k]e

−jnΩ0k,

for 1 ≤ n ≤ K0. Using Eq. (11.21), the above equation can be expressed as follows:

Dn = lim K0→∞

∞∑

k=−∞ x[k]e−jnΩ0k (11.23)

for 1 ≤ n ≤ K0. Let us now define a new function X (Ω), which is continuous with respect to the independent variable Ω:

X (Ω) = ∞∑

k=−∞ x[k]e−jΩk . (11.24)

In Eq. (11.24), the independent variable Ω is continuous in the range −∞ ≤ Ω ≤ ∞. In terms of X (Ω), Eq. (11.23) can be expressed as follows:

Dn = lim K0→∞

K0 X (nΩ0). (11.25)

The function X (nΩ0) is obtained by sampling X (Ω) at discrete pointsΩ = nΩ0. Given the DTFS coefficients Dn of x̃K0 [k], the aperiodic sequence x[k] can

be obtained by substituting the values of Dn in Eq. (11.22) and solving for

M1 ≤ k ≤ M2. The resulting expression is given by

x[k] = lim K0→∞

x̃K0 [k] = lim K0→∞

∑

n=〈K0〉

K0 X (nΩ0)e

jnΩ0k . (11.26a)

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477 11 Discrete-time Fourier series and transform

In the limit K0 → ∞, the angular frequency Ω0 takes a very small value, say �Ω, with the fundamental period K0 = 2π/�Ω. In the limit K0 → ∞, Eq. (11.26a) can, therefore, be expressed as follows:

x[k] = lim �Ω→0

∑

n=〈K0〉

2π X (n�Ω)e jnk�Ω�Ω. (11.26b)

Substituting Ω = n�Ω and applying the limit �Ω → 0, Eq. (11.26b) reduces to the following integral:

x[k] = 1

2π

∫

〈2π〉

X (Ω)e jkΩdΩ. (11.27)

In Eq. (11.27), the limits of integration are derived by evaluating the duration

n = 〈K0〉 in terms of Ω as follows:

Ω = 〈n�Ω〉|n=〈K0〉 = ⟨

( 2π

)⟩∣ ∣ ∣ ∣ n=〈K0〉

= 〈2π〉,

implying that any frequency range of 2π may be used to solve the integral in

Eq. (11.27). Collectively, Eq. (11.24), in conjunction with Eq. (11.27), is

referred to as the DTFT pair.

Definition 11.3 The DTFT pair for an aperiodic sequence x[k] is given by

DTFT synthesis equation x[k] = 1

2π

∫

〈2π〉

X (Ω)e jkΩdΩ; (11.28a)

DTFT analysis equation X (Ω) = ∞∑

k=−∞ x[k]e−jΩk . (11.28b)

In the subsequent discussion, we will denote the DTFT pair as follows:

x[k] DTFT←−−→ X (Ω). (11.28c)

Example 11.6

Calculate the Fourier transform of the following functions:

(i) unit impulse sequence, x1[k] = δ[k];

(ii) gate sequence, x2[k] = rect (

2N + 1

)

= {

1 |k| ≤ N 0 elsewhere;

(iii) decaying exponential sequence, x3[k] = pku[k] with |p| < 1.

Solution

(i) By definition,

X1(Ω) = ∞∑

k=−∞ δ[k]e−jΩk = e−jΩk |k=0 = 1.

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478 Part III Discrete-time signals and systems

0 2 4 6 8 10−8 −6 −4 −2−10

x2[k]

X2(W)

p 2p 3p0−3p −2p −p W

(a) (b)

Fig. 11.7. (a) Rectangular

sequence x2[k ] with a width of

seven samples. (b) DTFT of the

rectangular sequence derived in

Example 11.6(ii).

(ii) By definition,

X2(Ω) = ∞∑

k=−∞

x2[k]e −jΩk =

N∑

k=−N

1 · e−jΩk .

The summation represents a GP series with exp(−jΩ) as the ratio between two

consecutive terms. The GP series simplifies to

X2(Ω) = (e −jΩ)−N

1 − (e−jΩ)2N+1

1 − e−jΩ

= (e−jΩ)−N (e−jΩ)(2N+1)/2

(e−jΩ)1/2 (e−jΩ)−(2N+1)/2 − (e−jΩ)(2N+1)/2

(e−jΩ)−1/2 − (e−jΩ)1/2

= ej(2N+1)Ω/2 − e−j(2N−1)Ω/2

ejΩ/2 − e−jΩ/2 =

sin

( 2N + 1

2 Ω

)

sin

( 1

2 Ω

) .

As a special case, we assume N = 3 and plot the rectangular sequence x2[k]

and its DTFT X2(Ω) in Fig. 11.7.

(iii) By definition,

X3(Ω) =

∞∑

k=−∞

pku[k]e−jΩk =

∞∑

k=0

(pe−jΩ)k .

The summation represents a GP series, which can be simplified to

X3(Ω) = 1

1 − pe−jΩ =

1 − p cosΩ+ jp sinΩ .

The DTFT X3(Ω) is a complex-valued function of the angular frequency Ω. Its

magnitude and phase spectra are determined below:

magnitude spectrum |X3(Ω)| = |1|

|1 − p cosΩ+ jp sinΩ|

= 1

√

1 − 2p cosΩ+ p2 ;

phase spectrum <X3(Ω) = <1 − <(1 − p cosΩ+ jp sinΩ)

= − tan−1 (

p sinΩ

1 − p cosΩ

)

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479 11 Discrete-time Fourier series and transform

0 2 4 6 8 10−8 −6 −4 −2−10

1 0.6

x3[k] = 0.6 k u[k]

0.36

X3(W)

0 p 2p 3p W

−2p−3p −p

2.5

0.2p

0 p 2p 3p−2p−3p −p

−0.2p

<X3(W)

(a) (b)

(c)

Fig. 11.8. (a) Decaying

exponential sequence x3[k ]

with a decay factor p = 0.6. (b) Magnitude spectrum and

derived in Example 11.6(iii).

As a special case, we plot the DT sequence x3[k] and its magnitude and phase

spectra for p = 0.6 in Figs. 11.8(a)–(c).

In Example 11.6, we calculated the DTFTs for three different sequences and

observed that all three DTFTs are periodic with periodΩ0 = 2π . This property is referred to as the frequency–periodicity property and is satisfied by all DTFTs.

In Section 11.4, we present a mathematical proof verifying the frequency–

periodicity property.

Example 11.7

Calculate the DT sequences for the following DTFTs:

(i) X1(Ω) = 2π ∞∑

m=−∞

δ(Ω− 2mπ );

(ii) X2(Ω) = 2π

∞∑

m=−∞

δ(Ω− Ω0 − 2mπ ).

Solution

(i) Using the synthesis equation, Eq. (11.28a), the inverse DTFT of X1(Ω) is

given by

x1[k] = 1

2π

∫

〈2π〉

X1(Ω)e jkΩdΩ =

2π

π∫

−2π

2π

∞∑

m=−∞ δ(Ω− 2mπ )e jkΩdΩ

= π∫

−π

∞∑

m=−∞ δ(Ω− 2mπ ) ejk2mπ

︸︷︷︸

dΩ [∵ δ(Ω− θ ) f (Ω) = δ(Ω− θ ) f (θ )]

= π∫

−π

∞∑

m=−∞ δ(Ω− 2mπ )dΩ.

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480 Part III Discrete-time signals and systems

The integral on the right-hand side of the above equation includes several

impulse functions located at Ω = 0, ±2π, ±4π, . . . Only the impulse function located at Ω = 0 falls in the frequency range Ω = [−π, π ]. Therefore, x1[k] can be simplified as follows:

x1[k] = π∫

−π

δ(Ω)dΩ = 1.

(ii) Using the synthesis equation, (11.28a), the inverse DTFT of X2(Ω) is

given by

x1[k] = 1

2π

∫

〈2π〉

X2(Ω)e jkΩdΩ =

2π

π∫

−π

2π

∞∑

m=−∞ δ(Ω− Ω0 − 2mπ )e jkΩdΩ.

= π∫

−π

∞∑

m=−∞ δ(Ω− Ω0 − 2mπ )ejk(Ω0−2mπ ) dΩ

= π∫

−π

∞∑

m=−∞ δ(Ω− Ω0 − 2mπ )ejkΩ0 ejk2mπ︸︷︷︸

dΩ

= ejkΩ0 π∫

−π

∞∑

m=−∞ δ(Ω− Ω0 − 2mπ )dΩ.

The integral on the right-hand side of the above equation includes several

impulse functions located at Ω = Ω0 + 2mπ . Only one of these infinite num- ber of impulse functions will be present in the frequency range Ω = [−π, π ]. Therefore, the integral will have a vaue of unity and the function x2[k] can be

simplified as follows:

x2[k] = ejkΩ0 .

Table 11.2 lists the DTFT and DTFS representations for several DT sequences.

In situations where a DT sequence is aperiodic, the DTFS representation is

not possible and therefore not included in the table. The DTFT of the peri-

odic sequences is determined from its DTFS representation and is covered in

Section 11.4.

Table 11.3 plots the DTFT for several DT sequences. In situations where a

DT sequence or its DTFT is complex, we plot both the magnitude and phase

components. The magnitude component is shown using a bold line, and the

phase component is shown using a dashed line.

Example 11.7 illustrates the calculation of a DT function from its DTFT

using Eq. (11.28a). In many cases, it may be easier to calculate a DT

function from its DTFT using the partial fraction expansion and the DTFT

pairs listed in Table 11.2. This procedure is explained in more detail in

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481 11 Discrete-time Fourier series and transform

Table 11.2. DTFTs and DTFSs for elementary DT sequences

Note that the DTFS does not exist for aperiodic sequences

Sequence: x[k] DTFS: Dn = 1

∑

k=〈K0〉 x[k]e−jnΩ0k DTFT: X (Ω) =

∞∑

k=−∞ x[k]e−jΩk

(1) x[k] = 1 Dn = 1 X (Ω) = 2π ∞∑

m=−∞ δ(Ω− 2mπ )

(2) x[k] = δ[k] does not exist X (Ω) = 1 (3) x[k] = δ[k − k0] does not exist X (Ω) = e−jΩk0

(4) x[k] = ∞∑

m=−∞ δ(k − mK0) Dn =

K0 for all n X (Ω) =

2π

∞∑

m=−∞ δ

(

Ω− 2mπ

)

(5) x[k] = u[k] does not exist X (Ω) = π ∞∑

m=−∞ δ(Ω− 2mπ ) +

1 − e−jΩ

(6) x[k] = pku[k] with |p| < 1 does not exist X (Ω) = 1

1 − pe−jΩ

(7) First-order time-rising

decaying exponential

x[k] = (k + 1)pku[k], with |p| < 1.

does not exist X (Ω) = 1

(1 − pe−jΩ)2

(8) Complex exponential

(periodic)

x[k] = e jkΩ0 K0 = 2πp/Ω0

Dn =

{

1 n = p ± r K0 0 elsewhere

for −∞ < r < ∞

X (Ω) = 2π ∞∑

m=−∞ δ(Ω− Ω0 − 2mπ )

(9) Complex exponential

(aperiodic)

x[k] = e jkΩ0 , 2π/Ω0 �= rational

does not exist X (Ω) = 2π ∞∑

m=−∞ δ(Ω− Ω0 − 2mπ )

(10) Cosine (periodic)

x[k] = cos(Ω0k) K0 = 2πp/Ω0

Dn =







2 n = ±p ± r K0

0 elsewhere

for −∞ < r < ∞

X (Ω) = π ∞∑

m=−∞ δ(Ω+ Ω0 − 2mπ )

+ π ∞∑

m=−∞ δ(Ω− Ω0 − 2mπ )

(11) Cosine (aperiodic)

x[k] = cos(Ω0k), 2π/Ω0 �= rational

does not exist X (Ω) = π ∞∑

m=−∞ δ(Ω+ Ω0 − 2mπ )

+ π ∞∑

m=−∞ δ(Ω− Ω0 − 2mπ )

(12) Sine (periodic)

x[k] = sin(Ω0k) K0 = 2πp/Ω0

Dn =







2 j n = ±p ± r K0

0 elsewhere

for −∞ < r < ∞

X (Ω) = jπ ∞∑

m=−∞ δ(Ω+ Ω0 − 2mπ )

− jπ ∞∑

m=−∞ δ(Ω− Ω0 − 2mπ )

(13) Sine (aperiodic)

x[k] = sin(Ω0k), 2π/Ω0 �= rational

does not exist X (Ω) = jπ ∞∑

m=−∞ δ(Ω+ Ω0 − 2mπ )

− jπ ∞∑

m=−∞ δ(Ω− Ω0 − 2mπ )

(cont.)

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482 Part III Discrete-time signals and systems

Table 11.2. (cont.)

Sequence: x[k] DTFS: Dn = 1

∑

k=〈K0〉 x[k]e−jnΩ0k DTFT: X (Ω) =

∞∑

k=−∞ x[k]e−jΩk

(14) Rectangular (periodic)

x[k]= {

1 |k| ≤ N 0 N < |k| ≤ K0/2

x[k]= x[k + K0]

Dn =



  

  

(2N + 1)/K0 k = r K0





sin (

2N+1 K0

nπ )

sin (

K0 nπ

)



 elsewhere

X (Ω) = 2π ∞∑

n=−∞ Dnδ

(

Ω− 2nπ

)

(15) Rectangular (aperiodic)

x[k] = {

1 |k| ≤ N 0 elsewhere

does not exist

X (Ω) = sin

( 2N + 1

2 Ω

)

sin

( 1

2 Ω

)

(16) Sinc

x[k] = W

π sinc

( W k

)

sin(W k)

πk for 0 < W < π

does not exist X (Ω) =

{

1 |Ω| ≤ W 0 W < |Ω| ≤ π

X (Ω) = X (Ω+ 2π )

(17) Arbitrary periodic sequence

with period K0 x[k] =

∑

n=〈K0〉 Dne

jnΩ0k

Dn = 1

∑

k=〈K0〉 x[k]e−jnΩ0k X (Ω) = 2π

∞∑

n=−∞ Dnδ

(

Ω− 2nπ

)

Appendix D, and has been used later in this chapter in solving Examples 11.15

and 11.18.

11.3 Existence of the DTFT

Definition 11.4 The DTFT X (Ω) of a DT sequence x[k] is said to exist if

|X (Ω)| < ∞, for − ∞ < Ω < ∞. (11.29)

The above definition for the existence of the DTFT satisfies our intuition that

a valid function should be finite for all values of the independent variable.

Substituting the value of the DTFT X (Ω) from Eq. (11.28b), Eq. (11.29) can

be expressed as follows: ∣ ∣ ∣ ∣ ∣

∞∑

k=−∞ x[k]e−jΩk

∣ ∣ ∣ ∣ ∣ < ∞,

which is satisfied if

∞∑

k=−∞ |x[k]| · |e−jΩk | < ∞.

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Table 11.3. DTFT spectra for elementary DT sequences

Sequence Time-domain waveform Magnitude and phase spectra

(1) Constant

x[k] = 1 1

x[k] = 1

0 2 4 6−6 −4 −2 k

… … (W−2mp)∑

∞

−∞=

d= m

X(W) ∑ ∞

2π

0 2p 4p 6p−6p −4p −2p

… …

(2) Unit impulse

x[k] = δ[k] 1

x[k] = δ[k]

0 2 4 6−6 −4 −2 k

1=WX 1

( ) 1=X

0 p 2p 3p−3p −2p −p W

(3) Unit step

x[k] = u[k] ][][ kukx =

0 6−−6 −4 2 2 4 k

… ( ) ( )∑

∞

−∞=

+π−Ωδπ=Ω m

mX 2

0 p 2p 3p−3p −2p

( +−Ω 1 − e−jΩ

−p

… …

(4) Decaying exponential

x[k] = pku[k] with |p| < 1

p p

][][ kupkx k=

0 2 4 6 −6 −4 −2

…

( ) =WX 1 − pe−jW

0 p 2p 3p−3p −2p −p W

1− p

= 1

(5) Rectangular

x[k] = {

1 |k| ≤ N 0 elsewhere

≤ =

elsewhere0

1 ][

Nk kx

0 N−N

( ) =WX

0 p 2p 3p− −2p −p

2N+1

( ) = sin((2N + 1)W/2) sin(W/2)

−3p

(6) First-order time-rising

decaying exponential

x[k] = (k + 1)pku[k] with |p| < 1

][)1(][ kupkkx k+=

0 2 4 6−6 −4 −2

2p 3p2

][)1(][ kupkkx k+=

0 2 4 6−6 −4 −2

2p 3p2

0 2 4 6−6 −4 −2

2p 3p2

…

0 p 2p 3p−3p −2p −p

( ) (1− pe−jW)2

=WX

−− W

(1− p)2 1

(7) Sinc

x[k] = W

π sinc

( W k

) =kx sinc][ 1

0 2

k 4 6−6 −4 −2

……

( )pp= WkW 11

( ) p≤W<

≤W =W

W X

= 1

0 2p−2π W

− 2 p

+ W

2 p

+ WW

− 2 p

− W

2 p

− W

(8) Complex exponential

x[k] = e jkΩ0 k

kx ][ W=

… … 1

][kx je 0][ =

0 2

k 4 6−6 −4 −2

][

][kx<

( ) ( )∑ ∞

pm−W−Wd=Ω m=−∞

X 22 0

− 4

p +

W 0

−−= 2p

− 4

0 2p−2p

− 2

p +

W 0

2 p

+ W

W 0

(9) Cosine

x[k] = cos(Ω0k) 1… …

0 2 4 6−6 −4 −2

( )kkx 0cos][ Ω= 11

( ) ( ) ([ )]∑ ∞

−− W Wd+W−WdW 2mp00

0 2p−2p

− 2

p +

W 0

− 4

p +

W 0

2 p

+ W

W 0

− 2

p −

W 0

4 p

− W

2 p

− W

m=−∞

+−+=X 2mpp

2−2

− 2

− 4 2

− 2 42

483

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484 Part III Discrete-time signals and systems

From the Euler’s formula, we know that |exp(−jΩk)| = 1. Therefore, an alter- native expression to verify the existence of the DTFT is given by

∞∑

k=−∞

|x[k]| < ∞.

Condition for the existence of DTFT The DTFT X (Ω) of a DT sequence x[k] exists if

∞∑

k=−∞

|x[k]| < ∞. (11.2830)

Equation (11.2830) is a sufficient condition to verify the existence of the DTFT.

Example 11.8

Determine if the DTFTs exist for the following functions:

(i) causal exponential function, x1[k] = p ku[k].

(ii) cosine waveform, x2[k] = cos(Ω0k).

Solution

(i) Equation (11.2830) yields

∞∑

k=−∞

|x1[k]| =

∞∑

k=−∞

|pku[k]| =

∞∑

k=0

|pk | =







1 − p 0 < |p| > 1

∞ |p| ≥ 1.

Therefore, the DTFT of the exponential sequence x1[k] = p ku[k] exists if 0 <

|p| < 1. Under such a condition, x1[k] is a decaying exponential sequence with

the summation in Eq. (11.2830) having a finite value.

(ii) Equation (11.2830) yields ∞∑

k=−∞

|x2[k]| =

∞∑

k=−∞

|cos(Ω0k)| → ∞.

Therefore, the DTFT does not exist for the cosine waveform. However, this

appears to be in violation of Table 11.2, which lists the following DTFT pair

for the cosine sequence:

cos(Ω0k) DTFT←−−→ π

∞∑

m=−∞ [δ(Ω+ Ω0 − 2πm) + δ(Ω− Ω0 − 2πm)].

Looking closely at the above DTFT pair, we note that the DTFT X (Ω) of the

cosine function consists of continuous impulse functions at discrete frequencies

Ω = (±Ω0 − 2πm), for −∞ < m < ∞. Since the magnitude of a continuous impulse function is infinite, |X (Ω)| is infinite at the location of the impulses. The infinite magnitude of the impulses in the DTFT X (Ω) leads to the violation

of the existence condition stated in Eq. (11.2830).

Example 11.8 has introduced a confusing situation for the cosine sequence.

We proved that the condition of the existence of the DTFT is violated by the

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485 11 Discrete-time Fourier series and transform

cosine waveform, yet its DTFT can be expressed mathematically. A similar

behavior is exhibited by most periodic sequences. So how do we determine

the DTFT for a periodic sequence? We cannot use the definition of the DTFT,

Eq. (11.28b), since the procedure will lead to infinite DTFT values. In such

cases, an alternative procedure based on the DTFS is used; this is explained in

Section 11.4.

11.4 DTFT of periodic functions

Consider a periodic function x[k] with fundamental period K0. The DTFS

representation for x[k] is given by

x[k] = ∑

n=〈K0〉 Dne

jnΩ0k, (11.2831)

where Ω0 = 2π/K0 and the DTFS coefficients are given by

Dn = 1

∑

k=〈K0〉 x[k]e−jnΩ0k . (11.2832)

Calculating the DTFT of both sides of Eq. (11.2831), we obtain

X (Ω) = ℑ

{

∑

n=〈K0〉 Dne

jnΩ0k

}

Since the DTFT satisfies the linearity property, the above equation can be

expressed as follows:

X (Ω) = ∑

n=〈K0〉 Dnℑ{e jnΩ0k}, (11.2833)

where the DTFT of the complex exponential sequence is given by

ℑ{e jnΩ0k} = 2π ∞∑

m=−∞ δ(Ω− nΩ0 − 2πm).

Using the above value for the DTFT of the complex exponential, Eq. (11.2833)

takes the following form:

X (Ω) = ∑

n=〈K0〉 Dn2π

∞∑

m=−∞ δ(Ω− nΩ0 − 2πm).

By changing the order of summation in the above equation and substituting

Ω0 = 2π/K0, we have

X (Ω) = 2π ∞∑

m=−∞

∑

n=〈K0〉 Dnδ

(

Ω− 2nπ

K0 − 2πm

)

Since the DTFT is periodic with a period of 2π , we determine the DTFT in

the range Ω = [0, 2π ] and use the periodicity property to determine the DTFT values outside the specified range. Taking n = 0, 1, 2, . . . , K0 − 1 and m = 0,

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486 Part III Discrete-time signals and systems

the following terms of X (Ω) lie within the range Ω = [0, 2π ]:

X (Ω) = 2π D0δ(Ω) + 2π D1δ (

Ω− 2π

)

+ 2π D2δ (

Ω− 4π

)

+ · · · + 2π DK0−1δ (

Ω− 2(K0 − 1)π

)

(11.34a)

X (Ω) = 2π ∑

n=〈K0〉 Dnδ

(

Ω− 2nπ

)

, (11.34b)

for 0 ≤ Ω ≤ 2π . Since X(Ω) is periodic, Eq. (11.34b) can also be expressed as follows:

X (Ω) = 2π ∞∑

n=−∞ Dnδ

(

Ω− 2nπ

)

, (11.35)

which is the DTFT of the periodic sequence x[k] for the entire Ω-axis. The

values of the DTFS coefficients lying outside the range 0 ≤ n ≤ (K0 − 1) are evaluated from Eq. (11.9) to be

Dn = Dn+mK0 for m ∈ Z .

Definition 11.5 The DTFT X (Ω) of a periodic sequence x[k] is given by

X (Ω) = 2π ∞∑

n=−∞ Dnδ

(

Ω− 2nπ

)

, (11.36a)

where Dn are the DTFS coefficients of x[k]. The DTFS coefficients are given

Dn = 1

∑

k=〈K0〉 x[k]e−jnΩ0k (11.36b)

for 0 ≤ n ≤ K0 − 1 and the values outside the range are evaluated from the following periodicity relationship:

Dn = Dn+mK0 for m ∈ Z . (11.36c)

Example 11.9

Calculate the DTFT of the following periodic sequences:

(i) x1[k] = k for 0 ≤ k ≤ 3, with the fundamental period K0 = 4;

(ii) x2[k] = {

5 k = 0, 1 0 k = 2, 3, with the fundamental period K0 = 4;

(iii) x3[k] = 0.5k for 0 ≤ k ≤ 14, with the fundamental period K0 = 15;

(iv) x4[k] = 3 sin (

2π

7 k +

)

, with the fundamental period K0 = 7.

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487 11 Discrete-time Fourier series and transform

2p p 2p

0 2p−p p−2p 0.5p 1.5p−1.5p −0.5p

X1(W)

<X1(W)

−2p

p 4

3p −

0 2p−p p0.5p

1.5p

−1.5p

−0.5p

(a)

(b)

Fig. 11.9. DTFT of the periodic

sequence x1[k ] = k, 0 ≤ k ≤ 3, with fundamental period K0 = 4. (a) Magnitude spectrum; (b) phase spectrum.

Solution

(i) Using Eq. (11.36a), the DTFT of x1[k] is given by

X1(Ω) = 2π ∞∑

n=−∞ Dnδ

(

Ω− 2nπ

)

= 2π ∞∑

n=−∞ Dnδ

(

Ω− nπ

)

SubstitutingΩ0 = 2π/K0 = π/2 in Eq. (11.36b), the DTFS coefficients Dn for x1[k] are given by

Dn = 1

3∑

k=0 ke−jnπk/2 =

4 [e−jnπ/2 + 2e−jnπ + 3e−j3nπ/2].

For 0 ≤ n ≤ 3, the values of the DTFS coefficients are as follows:

n = 0 D0 = 1

4 [1 + 2 · 1 + 3 · 1] =

2 ;

n = 1 D1 = 1

4 [e−jπ/2 + 2 · e−jπ + 3 · e−j3π/2]

= 1

4 [−j + 2(−1) + 3( j)] = −

2 [1 − j];

n = 2 D2 = 1

4 [e−jπ + 2 · e−j2π + 3 · e−j3π ]

= 1

4 [−1 + 2(1) + 3(−1)] = −

2 ;

n = 3 D3 = 1

4 [e−j3π/2 + 2 · e−j3π + 3 · e−j9π/2]

= 1

4 [ j + 2(−1) + 3(−j)] = −

2 [1 + j].

The values of the DTFS coefficients that lie outside the range 0 ≤ n ≤ 3 can be obtained by using the periodicity property Dn+4 = Dn .

Since X1(Ω) is a complex-valued function, its magnitude and phase spectra

are plotted separately in Figs. 11.9(a) and (b). The area enclosed by the impulse

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488 Part III Discrete-time signals and systems

0 2p−p−2p 0.5p p 1.5p−1.5p −0.5p

X2(W)

5p 5p p

55p

<X2(W)

0.5p−1.5p

p −

p 0 2p−p−2p 1.5p−0.5p

p 2

5 p

(a)

(b)

Fig. 11.10. DTFT of the periodic

sequence x2[k ], with

fundamental period K0 = 4. (a) Magnitude spectrum; (b)

phase spectrum.

functions in the magnitude spectrum is given by 2π Dn and is indicated at the

top of each impulse in Fig. 11.9(a).

(ii) Using Eq. (11.36a), the DTFT of x2[k] is given by

X2(Ω) = 2π ∞∑

n=−∞

Dnδ

(

Ω− 2nπ

)

= 2π

∞∑

n=−∞

Dnδ (

Ω− nπ

)

SubstitutingΩ0 = 2π/K0 = π/2 in Eq. (11.36b), the DTFS coefficients Dn are

as follows:

Dn = 1

1∑

k=0

5e−jnπk/2 = 1

4 [5 + 5e−jπn/2] =

2 e−jπn/4 cos

(πn

)

For 0 ≤ n ≤ 3, the values of the DTFS coefficients are as follows:

n = 0 (Ω = 0) D0 = 5

2 with |D0| =

2 , <D0 = 0;

n = 1 (Ω = 0.5π ) D1 = 5

2 √

2 e−jπ/4 with |D0| =

2 √

2 , <D0 = −

4 ;

n = 2 (Ω = π ) D2 = 0 with |D0| = 0, <D0 = 0;

n = 3 (Ω = 1.5π ) D3 = − 5

2 √

2 e−j3π/4 with |D0| =

2 √

2 ,

<D0 = π − 3π

4 =

4 .

The magnitude and phase spectra are plotted separately in Figs. 11.10(a) and

(b), where the values of the DTFS coefficients lying outside 0 ≤ n ≤ 3 are obtained using the periodicity property Dn+4 = Dn .

(iii) Using Eq. (11.36a), the DTFT of x3[k] is given by

X3(Ω) = 2π ∞∑

n=−∞ Dnδ

(

Ω− 2nπ

)

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489 11 Discrete-time Fourier series and transform

Table 11.4. Values of |D n | and <D n for 0 ≤ n ≤ 14 in Example 11.9(iii) The radian frequency Ω corresponding to each value of n is given in the second row

n 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Ω 0 2π/15 4π /15 6π/15 8π /15 10π/15 12π /15 14π/15 16π/15 18π /15 20π /15 22π/15 24π /15 26π/15 28π/15

|Dn | 0.133 0.115 0.088 0.069 0.057 0.050 0.047 0.045 0.045 0.047 0.050 0.057 0.069 0.088 0.115 <Dn 0 −0.11π −0.16π −0.16π −0.14π −0.11π 0.07π 0.02π 0.02π 0.07π 0.11π 0.14π 0.16π 0.16π 0.11π

The DTFS coefficients of x3[k] are computed in Example 11.3. Substituting

Ω0 = 2π/K0 = 2π/15 in Eqs. (11.11)–(11.13), we obtain

Dn = 1



  

1 − 0.5 cos (

2nπ

)

+ j0.5 sin (

2nπ

)



  

where the magnitude component is given by

|Dn| = 1



    

1 √

1.25 − cos (

2nπ

)



    

and the phase component is given by

<Dn = −tan−1



  

0.5 sin

( 2nπ

)

1 − 0.5 cos (

2nπ

)



  

The magnitude and phase components of the DTFS coefficients Dn for 0 ≤ n ≤ 14 are given in Table 11.4.

The values of the DTFS coefficients, lying outside 0 ≤ n ≤ 14 are obtained using the periodicity property Dn+14 = Dn . The magnitude and phase of X3(Ω) are plotted in Fig. 11.11.

(iv) In Example 11.5, the DTFS coefficients Dn of x4[k] are computed and

are given by Eq. (11.19), which is reproduced here:

Dn =



  

  

−j1.5e jπ/4; for n = 1

j1.5e−jπ/4; for n = −1 for −1 ≤ n ≤ 5

0 elsewhere.

The values of the DTFS coefficients lying outside −1 ≤ n ≤ 5 are obtained using the periodicity property Dn+7 = Dn . Using Eq. (11.36a), the DTFT of

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490 Part III Discrete-time signals and systems

X2(W)

<X3(W)

0.266p 0.266p0.266p

−2p

0.5p−0.5p W

0.163p

−0.163p

0.163p

−0.163p

0 2p−p p p−2p 0.5p 1.5p−1.5p −0.5p

0 2p−p 1.5pp−0.5p

(a)

(b)

Fig. 11.11. DTFT of the periodic

sequence

x3[k ] = 0.5k u[k ], 0 ≤ k ≤ 14, with fundamental period

K0 = 15. (a) Magnitude spectrum; (b) phase spectrum.

X4(W)

<X4(W)

3p 3p3p 3p

0.25p

−0.25p −0.25p

0.25p

0 2p−p p−2p 0.5p 1.5p−1.5p −0.5p

0 2p−p p

p −2p 0.5p 1.5p−1.5p −0.5p

(a)

(b)

Fig. 11.12. DTFT of the periodic

sequence x4[k ], with

fundamental period K0 = 7. (a) Magnitude spectrum;

(b) phase spectrum.

x4[k] is given by

X4(Ω) = 2π ∞∑

n=−∞ Dnδ

(

Ω− 2nπ

)

= 2π ∞∑

n=−∞ n=7m+1

D1δ

(

Ω− 2nπ

)

+ 2π ∞∑

n=−∞ n=7m−1

D−1δ

(

Ω− 2nπ

)

= −j3πe j(π/4) ∞∑

m=−∞ δ

(

Ω− 2π

7 − 2mπ

)

+ j3πe−j(π/4) ∞∑

m=−∞ δ

(

Ω+ 2π

7 − 2mπ

)

The magnitude and phase of X4(Ω) are plotted in Fig. 11.12.

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491 11 Discrete-time Fourier series and transform

11.5 Properties of the DTFT and the DTFS

In this section, we present the properties of the DTFT. These properties are

similar to the properties for the CTFT discussed in Chapter 5. In most cases,

we do not explicitly state the DTFS properties, but a list of the DTFS properties

is included in Table 11.5.

11.5.1 Periodicity

DTFT The DTFT X (Ω) of an arbitrary DT sequence x[k] is periodic with a period Ω0 = 2π . Mathematically,

X (Ω) = X (Ω+ 2π ). (11.37)

DTFS The DTFS coefficients Dn of a periodic sequence x[n] with period K0 are periodic with respect to the coefficient number n and has a period K0. In

other words,

Dn = Dn+mK0 , (11.38)

for 0 ≤ n ≤ K0 − 1 and −∞ < m < ∞. Recall that the coefficient num- ber n = K0 corresponds to the frequency Ωn = 2πn/K0 = 2π . Therefore, the frequency–periodicity property of the DTFS and DTFT are in fact the

same.

11.5.2 Hermitian symmetry

The DTFT X (Ω) of a real-valued sequence x[k] satisfies

X (−Ω) = X∗(Ω), (11.39a)

where X∗(Ω) denotes the complex conjugate of X (Ω). By expressing the DTFT

X (Ω) in terms of its real and imaginary components,

X (Ω) = Re{X (Ω)} + j Im{X (Ω)},

Eq. (11.39a) can be expressed as follows:

Re{X (−Ω)} + j Im{X (−Ω)} = Re{X (Ω)} − j Im{X (Ω)}.

Separating the real and imaginary components yields

Re{X (−Ω)} = Re{X (Ω)} and Im{X (−Ω)} = −Im{X (Ω)}, (11.39b)

which implies that the real component Re{X (Ω)} of the DTFT X (Ω) of a real-

valued sequence x[k] is an even function of frequency Ω and that its imaginary

component Im{X (Ω)} is an odd function of Ω. In terms of the magnitude and

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492 Part III Discrete-time signals and systems

phase spectra of the DTFT X (Ω), the Hermitian symmetry property can be

expressed as follows:

|X (−Ω)| = |X (Ω)| and <X (−Ω) = − <X (Ω), (11.39c)

implying that the magnitude spectrum is even and that the phase spectrum is

odd.

As extensions of the Hermitian symmetry properties, we consider the special

cases when: (a) x[k] is real-valued and even and (b) x[k] is imaginary-valued

and odd.

Case 1 If x[k] is both real-valued and even, then its DTFT X (Ω) is also real- valued and even, with the imaginary component Im{X (Ω)} = 0. In other words,

Re{X (−Ω)} = Re{X (Ω)} and Im{X (−Ω)} = 0. (11.39d)

Case 2 If x[k] is both imaginary-valued and odd, then its DTFT X (Ω) is also imaginary-valued and odd, with the real component Re{X (Ω)} = 0. In other words,

Re{X (−Ω)} = 0 and Im{X (−Ω)} = −Im{X (Ω)}. (11.39e)

11.5.3 Linearity

Like the CTFT, both the DTFT and DTFS satisfy the linearity property.

DTFT If x1[k] and x2[k] are two DT sequences with the following DTFT pairs:

x1[k] DTFT

←−−→ X1(Ω) and x2[k] DTFT←−−→ X2(Ω),

then the linearity property states that

a1x1[k] + a2x2[k] DTFT←−−→ a1 X1(Ω) + a2 X2(Ω), (11.40a)

for any arbitrary constants a1 and a2, which may be complex-valued.

DTFS If x1[k] and x2[k] are two periodic DT sequences with the same funda- mental period K0 and the following DTFS pairs:

x1[k] DTFS←−−→ Dx1n and x2[k]

DTFS←−−→ Dx2n ,

then the DTFS coefficients of the periodic DT sequence x3[k] = a1x1[k] + a2x2[k], which also has a period of K0, are given by

a1x1[k] + a2x2[k] ︸︷︷︸

x3[k]

DTFS←−−→ a1 Dx1n + a2 D x2 n

︸︷︷︸

D n3 n

, (11.40b)

for any arbitrary constants a1 and a2, which may be complex-valued.

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493 11 Discrete-time Fourier series and transform

11.5.4 Time scaling

The time-scaling property of the CTFT, defined in Section 5.4.2, states that if a

CT function x(t) is time-compressed by a factor of a (a �= 0), its CTFT X (ω) is

expanded in the frequency domain by a factor of a, and vice versa. For the DTFT,

the time-scaling property has a limited scope, as illustrated in the following.

Decimation Since decimation is an irreversible nature of the decimation oper- ation, the DTFT of x[k] and the decimated sequence y[k] = x[mk] are not

related to each other.

Interpolation In the DT domain, interpolation is defined in Chapter 1 as follows:

x (m)[k] =



 

 

[ k

]

if k is a multiple of integer m

0 otherwise.

(11.41a)

The interpolated sequence x (m)[k] inserts (m − 1) zeros in between adjacent

samples of the DT sequence x[k]. The time-scaling property for the interpolated

sequence x (m)[k] is given as follows.

x[k] DTFT

←−−→ X (Ω),

then the DTFT X (m)(Ω) of x (m)[k] is given by

X (m)(Ω) = X (mΩ), (11.41b)

for 2 ≤ m < ∞. Equation (11.41b) shows that interpolation in time results in compression in the frequency domain. To demonstrate the application of the

interpolation property, consider the DTFT of a rectangular sequence:

x[k] = rect (

)

DTFT←−−→ sin (3.5Ω)

sin (0.5Ω) .

Using the interpolation property, the DTFT of the interpolated function x (2)[k]

for m = 2 is given by

x (2)[k] DTFT←−−→ X (2Ω) =

sin(3Ω)

sin(Ω) .

The functions x[k] and x (2)[k] and their DTFTs are shown graphically in Fig.

11.13.

11.5.5 Time shifting

The time-shifting operation delays or advances the reference sequence in time.

Given a signal x[k], the time-shifted signal is given by x[k– k0], where k0 is

an integer. If the value of the shift k0 is positive, the reference sequence x[k]

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494 Part III Discrete-time signals and systems

x[k]

0 2 4−6 −4

1 11

6 8−2−8

0 2 4−6 −4

1 11

6 8−2−8

x(2)[k] X (2)(W)

p 2p−2p −p W

0 p 2p−2p p

X [W]

(a)

(b)

Fig. 11.13. Time-scaling

property. (a) DTFT pair for a

rectangular sequence x[k ] with a

length of seven samples. (b)

DTFT pair for x (2)[k ] obtained by

interpolating x[k ] by a factor of

m = 2.

is delayed and shifted towards the right-hand side of the k-axis. On the other

hand, if the value of the shift k0 is negative, sequence x[k] advances and is

shifted towards the left-hand side of the k-axis. The DTFT of the time-shifted

sequence x[k– k0] is related to the DTFT of the reference sequence x[k] using

the following time-shifting property.

x[k] DTFT

←−−→ X (Ω)

then

x[k − k0] DTFT←−−→ e−jk0ΩX (Ω), (11.42)

for integer values of k0.

Example 11.10

Using the time-shifting property, calculate the DTFT of the following sequence:

x[k] =







0.75 (3 ≤ k ≤ 9) 0.5k−12 (12 ≤ k < ∞) 0 elsewhere.

Solution

The DT sequence x[k], plotted in Fig. 11.14, can be expressed as a linear

combination of: (i) a time-shifted gate or rectangular sequence, denoted by

x2[k] in Example 11.6, (ii), as follows:

x2[k] = rect (

2N + 1

)

= {

1 |k| ≤ N 0 elsewhere,

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495 11 Discrete-time Fourier series and transform

x[k]

4 6 10 12 14 16 18−4 −2 0 2 8

0.75 1 0.5

0.25 0.125

Fig. 11.14. DT sequence x[k ]

used in Example 11.10.

with N = 3; and (ii) a time-shifted decaying exponential sequence, denoted by x3[k] in Example 11.6, (iii), as follows:

x3[k] = pku[k],

with decay factor p = 0.5. In terms of x2[k] and x3[k], the expression for x[k] is given by

x[k] = 0.75x2[k − 6] + x3[k − 12].

Using the linearity and time-shifting properties, the DTFT X (Ω) of x[k] is given

X (Ω) = 0.75e−j6ΩX2(Ω) + e−j12ΩX3(Ω).

From the results in Example 11.6, the DTFTs for the sequences x2[k] and x3[k]

are given by

X2(Ω) = sin(3.5Ω)

sin(0.5Ω) and X3(Ω) =

1 − 0.5e−jΩ .

Substituting the values of the DTFTs results in the following:

X (Ω) = 0.75e−j6Ω sin(3.5Ω)

sin(0.5Ω) + e−j12Ω

1 − 0.5e−jΩ .

11.5.6 Frequency shifting

In the time-shifting property, we observed the change in the DTFT when the DT

sequence x[k] is shifted in the time domain. The frequency-shifting property

addresses the converse problem of how shifting the DTFT X (Ω) in the frequency

domain affects the sequence x[k] in the time domain.

x[k] DTFT

←−−→ X (Ω)

then

x[k]e jΩ0k DTFT←−−→ X (Ω− Ω0), (11.43)

for 0 ≤ Ω0 < 2π .

Example 11.11

Using the frequency-shifting property, calculate the DTFT of x[k] = cos(Ω0k) cos(Ω1k) with (Ω0 + Ω1) < π .

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496 Part III Discrete-time signals and systems

Solution

Using Table 11.2, the DTFT of cos(Ω0k) is given by

cos(Ω0k) DTFT

←−−→ π ∞∑

m=−∞ [δ(Ω+ Ω0 − 2mπ ) + δ(Ω− Ω0 − 2mπ )].

Using the frequency-shifting property,

cos (Ω0k)e jΩ1k DTFT←−−→ π

∞∑

m=−∞ [δ(Ω+ Ω0 − Ω1 − 2mπ )

+ δ(Ω− Ω0 − Ω1 − 2mπ )]

and

cos(Ω0k)e −jΩ1k DTFT←−−→ π

∞∑

m=−∞ [δ(Ω+ Ω0 + Ω1 − 2mπ )

+ δ(Ω− Ω0 + Ω1 − 2mπ )].

Adding the two DTFT pairs and noting that [exp(jΩ1k) + exp(−jΩ1k)] = 2 cos(Ω1k), we obtain

cos(Ω0k) cos(Ω1k) DTFT←−−→

∞∑

m=−∞ [δ(Ω+ Ω0 − Ω1 − 2mπ )

+ δ(Ω− Ω0 − Ω1 − 2mπ ) + δ(Ω+ Ω0 + Ω1 − 2mπ ) + δ(Ω− Ω0 + Ω1 − 2mπ )].

The above DTFT can also be obtained by expressing

2 cos(Ω0k) cos(Ω1k) = cos[(Ω0 + Ω1)k] + cos[(Ω0 − Ω1)k]

and calculating the DTFT of the right-hand side of the above expression.

11.5.7 Time differencing

The time differencing in the DT domain is the counterpart of differentiation in

the CT domain. The time-differencing property is stated as follows.

x[k] DTFT←−−→ X (Ω)

then

x[k] − x[k − 1] DTFT←−−→ [1 − e−jΩ]X (Ω). (11.44)

The proof of Eq. (11.44) follows directly from the application of the time-

shifting property.

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Example 11.12

Based on the DTFT of the unit step u[k] and the time-shifting property, calculate

the DTFT of x[k] = δ[k].

Solution

Using Table 11.2, the DTFT of the unit step function is given by

u[k] DTFT

←−−→ π ∞∑

m=−∞ δ(Ω− 2mπ ) +

1 − e−jΩ .

Applying the time-differencing property yields

u[k] − u[k − 1] DTFT←−−→ (1 − e−jΩ)

[

∞∑

m=−∞ δ(Ω− 2mπ ) +

1 − e−jΩ

]

which reduces to

u[k] − u[k − 1] DTFT←−−→ 1 + π ∞∑

m=−∞ δ(Ω− 2mπ )(1 − e−jΩ)|Ω=2mπ .

Since [1 − exp(−j2mπ )] = 0, the above DTFT pair reduces to

δ[k] DTFT←−−→ 1.

11.5.8 Differentiation in frequency

x[k] DTFT←−−→ X (Ω)

then

−jkx [k] DTFT←−−→ dX

dΩ . (11.45)

Example 11.13

Based on the DTFT of the exponential decaying function and the frequency

differentiation property, calculate the DTFT of x[k] = (k + 1)pku[k].

Solution

In Table 11.2, the DTFT of the exponential decaying function is given as

pku[k] DTFT←−−→

1 − pe−jΩ .

Using the frequency-differentiation property, we obtain

(−jk)pku[k] DTFT←−−→ d

dΩ

[ 1

1 − pe−jΩ

]

= −jpe−jΩ

(1 − pe−jΩ)2

kpku[k] DTFT←−−→ j

dΩ

[ 1

1 − pe−jΩ

]

= pe−jΩ

(1 − pe−jΩ)2 .

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Adding the DTFT pairs for pku[k] and kpku[k] yields

(k + 1)pku[k] DTFT←−−→ 1

1 − pe−jΩ +

pe−jΩ

(1 − pe−jΩ)2 =

(1 − pe−jΩ)2 .

11.5.9 Time summation

The time summation in the DT domain is the counterpart of integration in the

CT domain. The time-summation property is defined as follows.

x[k] DTFT←−−→ X (Ω)

then

k∑

n=−∞ x[n]

DTFT←−−→ 1

(1 − e−jΩ) X (Ω) + π X (0)

∞∑

m=−∞ δ(Ω− 2πm). (11.46)

Example 11.14

Based on the DTFT of the unit impulse sequence and the time-summation

property, calculate the DTFT of the unit step sequence.

Solution

Using Table 11.2, the DTFT of the unit impulse sequence is given by

δ[k] DTFT←−−→ 1.

Using the time-summation property, we obtain

k∑

n=−∞ δ[n]

DTFT←−−→ 1

1 − e−jΩ · 1 + π · 1

∞∑

m=−∞ δ(Ω− 2πm),

which yields

u[k] DTFT←−−→

1 − e−jΩ + π

∞∑

m=−∞ δ(Ω− 2πm).

11.5.10 Time convolution

In Section 10.5, we showed that the output response of an LTID system is

obtained by convolving the input sequence with the impulse response of the

system. Sometimes the resulting convolution sum is difficult to solve analyti-

cally in the time domain. The convolution property provides us with an alter-

native approach, based on the DTFT, of calculating the output response. Below

we state the convolution property and explain its application in calculating the

output response of an LTID system using an example.

If x1[k] and x2[k] are two DT sequences with the following DTFT pairs:

x1[k] DTFT←−−→ X1(Ω) and x2[k]

DTFT←−−→ X2(Ω),

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499 11 Discrete-time Fourier series and transform

then the time-convolution property states that

x1[k] ∗ x2[k] DTFT←−−→ X1(Ω)X2(Ω). (11.47)

In other words, the convolution between two DT sequences in the time domain is

equivalent to multiplication of the DTFTs of the two functions in the frequency

domain. Note that the CTFT also has a similar property, as stated in Section

5.4.8.

Equation (11.47) provides us with an alternative technique for calculating the

convolution sum using the DTFT. Expressed in terms of the following DTFT

pairs:

x[k] DTFT←−−→ X (Ω), h[k] DTFT←−−→ H (Ω), and y[k] DTFT←−−→ Y (Ω) ,

the output sequence y[k] can be expressed in terms of the impulse response

h[k] and the input sequence x[k] as follows:

y[k] = x[k] ∗ h[k] DTFT←−−→ Y (Ω) = X (Ω)H (Ω). (11.48)

In other words, the DTFT of the output sequence is obtained by multiplying

the DTFTs of the input sequence and the impulse response. The procedure for

evaluating the output y[k] of an LTID system in the frequency domain therefore

consists of the following four steps.

(1) Calculate the DTFT X (Ω) of the input signal x[k].

(2) Calculate the DTFT H (Ω) of the impulse response h[k] of the LTID system.

The DTFT H (Ω) is referred to as the transfer function of the LTID system

and provides a meaningful insight into the behavior of the system.

(3) Based on the convolution property, the DTFT of the output y[k] is given

by Y (Ω) = H (Ω)X (Ω). (4) The output y[k] in the time domain is obtained by calculating the inverse

DTFT of Y (Ω) obtained in step (3).

Since the DTFTs are periodic with periodΩ = 2π , steps (1)–(4) can be applied only to the frequency range [−π ≤ Ω ≤ π ].

Example 11.15

The exponential decaying sequence x[k] = aku[k], 0 ≤ a ≤ 1, is applied at the input of an LTID system with the impulse response h[k] = bku[k], 0 ≤ b ≤ 1. Using the DTFT approach, calculate the output of the system.

Solution

Based on Table 11.2, the DTFTs for the input sequence and the impulse response

are given by

x[k] DTFT←−−→

1 − ae−jΩ and h[k]

DTFT←−−→ 1

1 − be−jΩ .

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The DTFT Y (Ω) of the output signal is therefore calculated as follows:

y[k] = x[k] ∗ h[k] DTFT←−−→ Y (Ω) = 1

(1 − ae−jΩ)(1 − be−jΩ) .

The inverse of the DTFT Y (Ω) takes two different forms depending on the

values of a and b:

Y (Ω) =



  

  

(1 − ae−jΩ)2 a = b

(1 − ae−jΩ)(1 − be−jΩ) a �= b.

We consider the two cases separately.

Case 1 (a = b) The inverse DTFT follows directly from Table 11.2 as follows:

y[k] = (k + 1)aku[k + 1].

Case 2 (a �= b) Using partial fraction expansion, the DTFT Y (Ω) is expressed as follows:

Y (Ω) = A

1 − ae−jΩ +

1 − be−jΩ , (11.49)

where the partial fraction coefficients are given by

A = 1

1 − be−jΩ

∣ ∣ ∣ ∣ ae−jΩ=1

= a

a − b

and

B = 1

1 − ae−jΩ

∣ ∣ ∣ ∣ be−jΩ=1

= − b

a − b .

Substituting the values of A and B in Eq. (11.49) and calculating the inverse

DTFT yields

y[k] = 1

a − b [ak+1 − bk+1]u[k].

Combining case 1 with case 2, we obtain

y[k] =







(k + 1)aku[k] a = b 1

a − b [ak+1 − bk+1]u[k] a �= b.

(11.50)

11.5.11 Periodic convolution

The time-convolution property, defined by Eq. (11.35), is used to calculate

the output of convolving aperiodic sequences. In Section 10.6, we defined the

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501 11 Discrete-time Fourier series and transform

xp[k]

2 30 1 4 5 6 7−2 −1−3−4

1 2

2 30 1 4 5 6 7−2 −1−3−4

hp[k]

5 5

Fig. 11.15. Periodic sequences

xp[k ] and hp[k ] used in Example

11.16.

periodic, or circular, convolution to convolve periodic sequences. We now show

how the periodic convolution can be calculated using the DTFS.

If x1[k] and x2[k] are two DT periodic sequences with the same fundamental

period K0 and the following DTFS pairs:

x1[k] DTFS←→ Dx1n and x2[k]

DTFS←→ Dx2n ,

then the periodic convolution property states that

x1[k] ⊗ x2[k] DTFS←−−→ K0 Dx1n D

x2 n . (11.51)

We illustrate the application of the periodic convolution property by revisiting

Example 10.10.

Example 11.16

In Example 10.10, we calculated the periodic convolution yp[k] of the two

periodic sequences xp[k] and hp[k], defined over one period (K0 = 4) as xp[k] = k, 0 ≤ k ≤ 3, and hp[k] = 5, 0 ≤ k ≤ 1, in the time domain. Repeat Example 10.10 using the periodic convolution property.

Solution

The periodic sequences xp[k] and hp[k] are shown in Fig. 11.15. In part (i) of

Example 11.9, we calculated the DTFS coefficients of xp[k] as follows:

D xp 0 =

2 , D

xp 1 = −

2 [1 − j], Dxp2 = −

2 , and D

xp 3 = −

2 [1 + j].

Similarly, in part (ii) of Example 11.9 we calculated the DTFS coefficients of

hp[k]:

D hp 0 =

2 , D

hp 1 =

4 [1 − j], Dhp2 = 0, and D

hp 3 =

4 [1 + j].

Using the periodic convolution property, the DTFS coefficients of yp[k] are

D yp 0 = K0 D

xp 0 D

hp 0 = 15;

D yp 1 = K0 D

xp 1 D

hp 1 = j5;

D yp 2 = K0 D

xp 2 D

hp 2 = 0;

D yp 3 = K0 D

xp 3 D

hp 3 = −j5.

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Calculating the inverse DTFS, the DT sequence yp[k] is given by

yp[k] = 3∑

n=0 D

yp n e

−j 2π 4

= [

15 + j5 · e−j π 2

k + 0 · e−jπk − j5 · e−j 3π 2

k ]

Calculating the values of yp[k] within one period (0 ≤ k0 ≤ 3) yields

k = 0 yp[0] = 15 + j5 − j5 = 15;

k = 1 yp[1] = 15 + j5e−j π 2 − j5e−j

3π 2 = 15 − 5 − 5 = 5;

k = 2 yp[2] = 15 + j5e−jπ − j5e−j3π = 15 − j5 + j5 = 15;

k = 3 yp[3] = 15 + j5e−j 3π 2 − j5e−j

9π 2 = 15 + 5 + 5 = 25.

The above result is identical to the result obtained in Example 10.10.

Example 11.16 shows how periodic convolution can be calculated using

the DTFS periodic-convolution property. A more computationally efficient

approach of calculating the periodic convolution is based on the dis-

crete Fourier transform (DFT). The theory of DFT will be presented in

Chapter 12.

11.5.12 Frequency convolution

The time-convolution property (see Section 11.5.10) states that the convolution

between two DT sequences in the time domain is equivalent to the multiplication

of the DTFTs of the two sequences in the frequency domain. The converse of

the time-convolution property is also true, and is referred to as the frequency-

convolution property.

If x1[k] and x2[k] are two DT sequences with the following DTFT pairs:

x1[k] DTFT←−−→ X1(Ω) and x2[k]

DTFT←−−→ X2(Ω),

then the frequency-convolution property states that

x1[k]x2[k] DTFT←−−→

2π

∫

〈2π〉

X1(θ )X2(Ω− θ )dθ. (11.52)

The limits of integration in Eq. (11.52) are given by 〈Ω = 2π〉, which implies that any range of 2π may be chosen during the integration.

The frequency-convolution property is widely used in digital communica-

tions systems, where it is commonly referred to as the modulation property.

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503 11 Discrete-time Fourier series and transform

11.5.13 Parseval’s theorem

If x[k] is an energy signal and x[k] DTFT

←−−→ X (Ω), the energy of the DT signal x[k] is given by

Ex = ∞∑

k=−∞ |x[k]|2 =

2π

∫

〈2π〉

|X (Ω)|2dΩ. (11.53)

Parseval’s theorem states that the DTFT is a lossless transform as there is no

loss of energy if a signal is transformed to the frequency domain.

Example 11.17

Using Parseval’s theorem, evaluate the following integral:

∫

〈2π〉



  

sin

( 2N + 1

2 Ω

)

sin

( 1

2 Ω

)



  

dΩ.

Solution

Since

rect

( k

2N + 1

)

DTFT←−−→ sin

( 2N + 1

2 Ω

)

sin

( 1

2 Ω

) ,

Eq. (11.53) computes the area enclosed by the squared DT sinc function within

one period Ω = 〈2π〉 as follows:

2π

∫

〈2π〉



  

sin

( 2N + 1

2 Ω

)

sin

( 1

2 Ω

)



  

dΩ = ∞∑

k=−∞

∣ ∣ ∣ ∣

rect

[ k

2N + 1

]∣ ∣ ∣ ∣

where

rect

[ k

2N + 1

]

= {

1 |k| ≤ N 0 elsewhere.

Simplifying the summation on the right-hand side of this equation yields

∫

〈2π〉



  

sin

( 2N + 1

2 Ω

)

sin

( 1

2 Ω

)



  

dΩ = 2π (2N + 1).

We have presented several properties in Sections 11.5.1–11.5.12. Table 11.5

lists the properties of the DTFS and Table 11.6 lists the properties of the DTFT.

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Table 11.5. Properties of the DTFS: sequences x [k ], x1[k ], and x2[k ] are periodic with a period of K0

Properties Time domain Frequency domain Comments

x[k] = ∑

n=〈K0〉 Dne

jnΩ0k Dn = 1

∑

k=〈K0〉 x[k]e−jnΩ0k Ω0 = 2π/K0

x1[k] D x1 n Ω0 = 2π/K0

x1[k] D x2 n Ω0 = 2π/K0

Periodicity x[k] Dn = Dn+K0 Linearity a1x1[k] + a2x2[k] a1 Dx1n + a2 Dx2n a1, a2 ∈ C

Scaling x (m)[k]

with period mK0

m Dn m = 1, 2, 3, . . .

Time shifting x[k − k0] exp (

j 2πk0

K0 n

)

Dn k0 ∈ R

Frequency shifting exp

(

j 2πn0

K0 k

)

x[k] Dn−n0 n0 ∈ R

Time differencing x[k] − x[k − 1] [

1 − exp (

j 2π

K0 n

)]

Time summation S = k∑

m=−∞ x[m]

1 − exp (

j 2π

K0 n

) Dn summation S is

finite only if

D0 = 0 Periodic convolution

∑

n=〈K0〉 x1[n]x2[n − k] K0 Dx1n Dx2n convolution over a

period K0

Frequency

convolution

x1[k]x2[k] ∑

m=〈K0〉 Dx1m D

x2 m−n multiplication in

time domain

Parseval’s relationship 1

∑

k=〈K0〉 |x[k]|2 =

∑

n=〈K0〉 |Dn|2 power of a periodic

sequence

Symmetry properties

DTFS: D−n = D∗n Comments

real and imaginary

components: {

Re{D−n} = Re{Dn} Im{D−n} = −Im{Dn}

real component is

even; imaginary

component is odd

Hermitian property x[k] is a real-valued sequence

magnitude and phase spectra: {

|D−n| = |Dn| <D−n = − <Dn

magnitude spectrum

is even; phase

spectrum is odd

Real-valued and even

function

x[k] is an even and real-valued

sequence

{

Re{D−n} = Re{Dn} Im{D−n} = 0

DTFS is real-valued

and even

Real-valued and odd

function

x[k] is an odd and real-valued

sequence

{

Re{D−n} = 0 Im{D−n} = −Im{Dn}

DTFS is imaginary

and odd

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505 11 Discrete-time Fourier series and transform

Table 11.6. Properties of the discrete-time Fourier transform (DTFT)

Transformation

properties Time domain Frequency domain Comments

x[k] = 1

2π

∫

〈2π〉

X (Ω)e jkΩdΩ X (Ω) = ∞∑

k=−∞ x[k]e−jΩk

x1[k] X1(�)

x2[k] X2(Ω)

Periodicity x[k] X (Ω) = X (Ω+ 2π ) Linearity a1x1[k] + a2x2[k] a1 X1(Ω) + a2 X2(Ω) a1, a2 ∈ C Scaling x (m)[k] X (mΩ) m = 1, 2, 3, . . . Time shifting x[k − k0] exp(− jk0Ω)X (Ω) k0 ∈ R Frequency shifting exp( jkΩ0)x[k] X (Ω− Ω0) Ω0 ∈ R Time differencing x[k] − x[k − 1] [1 − exp( jΩ)]X (Ω)

Time summation S = k∑

m=−∞ x[m]

1 − exp( jΩ) X (Ω) +

π X (0)

∞∑

m=−∞ δ(Ω− 2πm)

provided summation

S is finite

Time convolution x1[k] ∗ x2[k] X1(Ω)X2(Ω) Periodic convolution x1[k] ⊗ x2[k] X1(Ω)X2(Ω) over period K0

Frequency

convolution

x1[k]x2[k] 1

2π

∫

〈2π〉

X1(θ )X2(Ω− θ )dθ multiplication in time domain

Parseval’s

relationship

Ex = ∞∑

k=−∞ |x[k]|2=

2π

∫

〈2π〉

|X (Ω)|2 dΩ energy in a signal

Symmetry properties

DTFT: X (−Ω) = X∗(Ω)

real and imaginary component: {

Re{X (−Ω)} = Re{X (Ω)} Im{X (−Ω)} = −Im{X (Ω)}

real component is

even: imaginary

component is odd

Hermitian property x[k] is a real-valued function

magnitude and phase spectra: {

|X (−Ω)| = |X (Ω)| <X (−Ω) = − <X (Ω)

magnitude spectrum

is even; phase

spectrum is odd

Real-valued and

even function

x[k] is even and real-valued

{

Re{X (−Ω)} = Re{X (Ω)} Im{X (Ω)} = 0 DTFT is real-valued

and even

Real-valued and odd

function

x[k] is odd and real-valued

{

Re{X (−Ω)} = 0 Im{X (−Ω)} = −Im{X (Ω)} DTFT is imaginary

and odd

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506 Part III Discrete-time signals and systems

11.6 Frequency response of LTID systems

In Chapter 10, we presented two different representations to specify the input–

output relationship of an LTID system. Section 10.1 used a linear, constant-

coefficient difference equation, while Section 10.3 used the impulse response

h[k] to model an LTID system. A third representation for an LTID system is

obtained by calculating the DTFT of the impulse response,

h[k] DTFT

←−−→ H (Ω). The DTFT H (Ω) is referred to as the Fourier transfer function of the LTID

system. In conjunction with the linear convolution property, the transfer function

H (Ω) can be used to determine the output response y[k] of the LTID system

due to the input sequence x[k]. In the time domain, the output response y[k] is

given by

y[k] = x[k] ∗ h[k].

Calculating the DTFT of both sides of the equation, we obtain

Y (Ω) = X (Ω)H (Ω) (11.54)

H (Ω) = Y (Ω)

X (Ω) , (11.55)

where Y (Ω) and X (Ω) are, respectively, the DTFTs of the output response y[k]

and the input signal x[k]. Equation (11.55) provides an alternative definition

for the transfer function as the ratio of the DTFT of the output signal and the

DTFT of the input signal.

Given one representation for an LTID system, it is straightforward to derive

the remaining two representations based on the DTFT and its properties. In the

following, we derive a formula to calculate the transfer function of an LTID

system from its difference equation representation.

Consider an LTID system whose input–output relationship is given by the

following difference equation:

y[k + n] + an−1 y[k + n − 1] + · · · + a0 y[k] = bm x[k + m] + bm−1x[k + m − 1] + · · · + b0x[k]. (11.56)

Calculating the DTFT of both sides of the above equation, we obtain

{

e jnΩ + an−1e j(n−1)Ω + · · · + a0 }

Y (Ω) = {

bme jmΩ + bm−1e j(m−1)Ω

+ · · · + b0 }

X (Ω),

which reduces to the following transfer function:

H (Ω) = Y (Ω)

X (Ω) =

bme jmΩ + bm−1e j(m−1)Ω + · · · + b0

e jnΩ + an−1e j(n−1)Ω + · · · + a0 . (11.57)

The impulse response h[k] of the LTID system can be obtained by calculating

the inverse DTFT of the transfer function H (Ω).

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507 11 Discrete-time Fourier series and transform

h[k]

4 6 8−2 0

10 122−4

2.5 2.13

1.66 1.260.95

0.710.54 0.4 0.30.230.170.13

Fig. 11.16. Impulse response

h[k ] of the LTID system derived

in Example 11.18.

Example 11.18

The input–output relationship of an LTID system is given by the following

difference equation:

y[k + 2] − 3

4 y[k + 1] +

8 y[k] = 2x[k + 2]. (11.58)

Determine the transfer function and the impulse response of the system.

Solution

Calculating the DTFT of Eq. (11.58) yields {

e j2Ω − 3

4 e jΩ +

}

Y (Ω) = 2e j2ΩX (Ω),

which results in the following transfer function:

H (Ω) = 2e j2Ω

e j2Ω − 3

4 e jΩ +

= 2

1 − 3

4 e−jΩ +

8 e−j2Ω

= 2

(

1 − 1

2 e−jΩ

) (

1 − 1

4 e−jΩ

) .

To calculate the impulse response of the LTID system, we calculate the partial

fraction of H (Ω) as follows:

H (Ω) = 4

1 − 1

2 e−jΩ

− 2

1 − 1

4 e−jΩ

By calculating the inverse DTFT of both sides, the impulse response h[k] is

given by

h[k] = 4 (

u[k] − 2 (

u[k],

which is plotted in Fig. 11.16.

11.7 Magnitude and phase spectra

The Fourier transfer function H (Ω) provides a complete description of an LTID

system. In most cases, H (Ω) is a complex function of the angular frequencyΩ.

Therefore, it is difficult to analyze the frequency characteristics of the transfer

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508 Part III Discrete-time signals and systems

LTID system

H(W)x[k] = cos(W0k)

H(W0) y[k] = cos(W0k + <H(W0))

Fig. 11.17. Gain and phase

responses of an LTID system.

The LTID system provides a gain

of |H (Ω)| to the magnitude and a phase change of <H (Ω) to the

phase of the sinusoidal input. function directly from the mathematical expression. By expressing the transfer

function H (Ω) as

H (Ω) = |H (Ω)|e j<H (Ω), (11.59)

the LTID system is analyzed by plotting the magnitude |H (Ω)| and phase <H (Ω) as functions of frequency Ω. The plot of the magnitude |H (Ω)| with respect to frequency Ω is referred to as the magnitude spectrum, while the plot

of the phase <H (Ω) with respect to frequency Ω is referred to as the phase

spectrum. Collectively, magnitude and phase spectra are used to analyze the

LTID system.

A second interpretation of the magnitude and phase spectra is obtained by

considering a sinusoidal sequence x[k] = cos(Ω0k) applied at the input of an LTID system with transfer function H (Ω). The DTFT of the output of the LTID

system is given by

Y (Ω) = ℑ{x[k]}H (Ω)

= π [δ(Ω− Ω0) + δ(Ω+ Ω0)]|H (Ω)|e j<H (Ω)

= π [

δ(Ω− Ω0)|H (Ω0)|e j<H (Ω0) + δ(Ω+ Ω0)|H (−Ω0)|e

j<H (−Ω0) ]

Assuming that the impulse response h[k] of the LTID system is real-valued and

then applying the Hermitian symmetry property, we observe that the magnitude

response |H (Ω)| is an even function of Ω while the phase response <H (Ω) is

an odd function of Ω. Mathematically,

|H (−Ω)| = |H (Ω)| and <H (−Ω) = − < H (Ω).

The DTFT Y (Ω) of the output of the LTID system is therefore given by

Y (Ω) = π |H (Ω0)| [

δ(Ω− Ω0)e j<H (Ω0) + δ(Ω+ Ω0)e

−j<H (Ω0) ]

Calculating the inverse DTFT, the output of the LTID system is given by

y[k] = 1

2 |H (Ω0)|

[

e j(Ω0k+<H (Ω0)) + e−j(Ω0k+<H (Ω0)) ]

(11.60a)

y[k] = |H (Ω0)| cos(Ω0k+ <H (Ω0)). (11.60b)

Figure 11.17 is a schematic diagram of the gain and phase changes in the sinu-

soidal input caused by an LTID system. Computed at the fundamental frequency

Ω = Ω0 of the sinusoidal input, the magnitude |H (Ω)| of the transfer function

determines the gain introduced by the LTID system, while the phase <H (Ω) at

Ω = Ω0 determines the phase change in the applied sinusoidal sequence. The

magnitude |H (Ω)| and phase <H (Ω) are therefore also referred to as the gain

and phase responses of the LTID system.

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509 11 Discrete-time Fourier series and transform

Example 11.19

Plot the magnitude and phase spectra of the LTID system specified in

Example 11.18.

Solution

From Example 11.18, the transfer function of the LTID system is given by

H (Ω) = 2

1 − 3

4 e−jΩ +

8 e−j2Ω

Using Euler’s formula exp(jΩ) = cosΩ+ j sinΩ and similarly for exp(j2Ω), yields

H (Ω) = 2

1 − 3

4 cosΩ+

8 cos(2Ω) + j

[ 3

4 sinΩ−

8 sin(2Ω)

] ,

which leads to the following expressions for the magnitude and phase responses:

|H (Ω)| = 2

√ [

1 − 3

4 cosΩ+

8 cos(2Ω)

+ [

4 sinΩ−

8 sin(2Ω)

= 2

√

101

64 −

16 cosΩ+

4 cos(2Ω)

;

<H (Ω) = < 2− < {

1 − 3

4 cosΩ+

8 cos(2Ω) + j

[ 3

4 sinΩ−

8 sin(2Ω)

]}

= − tan−1



 

 

4 sinΩ−

8 sin(2Ω)

1 − 3

4 cosΩ+

8 cos(2Ω)



 

 

Figures 11.18(a) and (b) plot the magnitude and phase spectra in the frequency

range Ω = [−π, π ]. Because the DTFT is periodic with period Ω0 = 2π , the magnitude and phase spectra at other frequencies can be calculated using the

periodicity property. It is observed that the gain |H (Ω)| of the LTID system has the maximum value of 16/3 at frequency Ω = 0. The gain |H (Ω)| at Ω = 0 is also referred to as the dc component of the impulse response h[k], and is

the sum ∑

h[k] over the duration of the impulse response. As the frequency

increases to π (or decreases to −π ), the gain decreases monotonically and has a minimum value of 16/15 at Ω = ±π radians/s. For LTID systems, the fre- quency Ω = ±π radians/s corresponds to the maximum frequency. The trans- fer function H (Ω) represents a non-uniform amplifier as the lower-frequency

components are amplified at a relatively higher scale than the high-frequency

components.

The phase response <H (Ω) of the LTID system has a value of zero atΩ = 0. As the frequency increases from zero, the phase decreases to its minimum

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510 Part III Discrete-time signals and systems

0 p/2 p−p −p/2 p/2 p−p −p/2 0 W

16 H(W) <H(W)

(a) (b)

.0.245p

−0.245p

Fig. 11.18. (a) Magnitude

spectrum and (b) phase

spectrum of the LTID system

considered in Example 11.19. The

responses are shown in the

frequency rangeΩ = [−π, π ].

value of –0.245π radians atΩ = 0.37π radians/s. FromΩ = 0.37π radians/s to Ω = π radians/s, the phase increases and approaches zero at Ω = π radians/s. For negative frequencies, the phase increases to its maximum value of

0.245 π radians at � = −0.37π radians/s, after which the phase decreases and approaches zero at Ω = −π radians/s.

It is also observed that the transfer function H (Ω) satisfies the Hermitian

symmetry property stated in Eq. (11.39a). Since the impulse response h[k] is a

real-valued function, the magnitude spectrum |H (Ω)| is an even function of Ω and is therefore symmetric about the y-axis in Fig. 11.18(a). On the other hand,

the phase spectrum <H (Ω) is an odd function of Ω and is therefore symmetric

about the origin in Fig. 11.18(b). In cases where the impulse response h[k] is a

real-valued function, the plots in the rangeΩ = [0, π ] are sufficient to represent the frequency response completely. The frequency response within the range

Ω = [−π, 0] can then be obtained using the Hermitian symmetry property.

Example 11.20

Derive and plot the frequency responses of the LTID systems with the following

impulse responses:

(i) h[k] = sin(πk/6)

πk ; (11.61)

(ii) g[k] = δ[k] − sin(πk/6)

πk . (11.62)

Solution

(i) We express h[k] as a sinc function as h [k] = 1

sin (πk/6)

πk/6 =

6 sinc(k/6).

Using Table 11.2, the transfer function is given by

H (Ω) = {

1 |Ω| ≤ π/6 0 π/6 < |Ω| ≤ π. (11.63)

The impulse response h[k] and its magnitude spectrum |H (Ω)| are plotted in Figs. 11.19(a) and (b) within the frequency rangeΩ = [0, π ]. Since the transfer function H (Ω) is real-valued, the phase spectrum is zero. The transfer function,

Eq. (11.63), or equivalently the impulse response, Eq. (11.61), represents an

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511 11 Discrete-time Fourier series and transform

−25 −20 −15 −10 −5 0 5 10 15 20 25 −0.05

0.05

0.1

0.15

0.2

h [k

]

H(W)

0 p/2 p−p −p/2

stop bandpass band(a) (b)

Fig. 11.19. (a) Impulse response

h[k ] and (b) magnitude

spectrum |H(Ω)| of an ideal lowpass filter specified in

Example 11.20(i). The phase

response is zero for all

frequencies.

ideal lowpass filter since the low-frequency components in the input sequence,

which lie within the range 0 ≤ Ω ≤ π/6, are passed through the system without attenuation. On the other hand, the higher-frequency components within the

range π/6 ≤ Ω ≤ π are completely blocked. Lowpass filters are widely used in digital signal processing and will be considered in more detail in Chapter 14.

(ii) Expressing the impulse response g[k] in terms of the impulse response

h[k] given in part (i), we obtain

g[k] = δ[k] − h[k].

Using the linearity property, the transfer function of g[k] is given by

G(Ω) = 1 − H (Ω).

Substituting the value of H (Ω) from Eq. (11.63) yields

G(Ω) = {

0 |Ω| ≤ π/6 1 π/6 < |Ω| ≤ π. (11.64)

The impulse response g[k] and its magnitude spectrum |G(Ω)| are plotted in Figs. 11.20(a) and (b). It is observed that the low-frequency components within

the range 0 ≤ Ω ≤ π/6 are completely blocked from the output, while the high- frequency components within the range π/6 < Ω ≤ π are passed through the system without any attenuation. Such a system is referred to as an ideal highpass

filter. Like lowpass filters, the highpass filters are also widely used in digital

signal processing, and will be considered in more detail in Chapter 14.

In the previous example, we considered calculating the magnitude and phase

spectra of an LTID system. The following example illustrates how the spectra

may be used to calculate the output of an LTID system for elementary sinusoidal

sequences.

Example 11.21

A continuous-time audio signal x(t) = 3 cos(1000π t) + 5 cos(2000π t) is sam- pled at a sampling rate of 8000 samples/s to produce the DT sequence x[k].

Calculate the output signals if the DT signal x[k] is applied at the input of an

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512 Part III Discrete-time signals and systems

≈

−25 −20 −15 −10 −5 0

0.8333

5 10 15 20 25 −0.2

−0.1

0.1

0.2

0.3

h [k

]

G(W)

0 p/2 p−p −p/2

pass bandstop band(a) (b)

Fig. 11.20. (a) Impulse response

h[k ] and (b) magnitude

spectrum |H (Ω)| of an ideal highpass filter specified in

Example 11.20(ii). The phase

response is zero for all

frequencies.

LTID systems with the following transfer functions:

(i) H1(Ω) = 2

1 − 3

4 e−jΩ +

8 e−j2Ω

; (11.65)

(ii) H2(Ω) =







1 |Ω| ≤ π

0 π

6 < |Ω| ≤ π,

(11.66)

(iii) H3(Ω) =







0 |Ω| ≤ π

1 π

6 < |Ω| ≤ π.

(11.67)

Solution

The DT sequence x[k] is given by

x[k] = x(kTs) = 3 cos(1000πkTs) + 5 cos(2000πkTs).

Substituting Ts = 1/8000, we obtain

x[k] = 3 cos (

πk

)

+ 5 cos (

πk

)

which implies that x[k] consist of two frequency components, Ω1 = π/8 and Ω2 = π/4. This is also apparent from the DTFT of x[k], given by

X (Ω) = 3π [

δ (

Ω− π

)

+ δ (

Ω+ π

)]

+ 5π [

δ (

Ω− π

)

+ δ (

Ω+ π

)]

which consists of impulses at frequencies Ω1 = ±π/8 and Ω2 = ±π/4. As the DTFT is 2π -periodic, in the above equation we showed X (Ω) only

in the frequency range −π ≤ Ω ≤ π . This simplifies the analysis, and hence we will use the same approach to express the DTFTs in the following.

If the transfer function of an LTID system is H (Ω), the DTFT Y (Ω) of the

output sequence is given by

Y (Ω) = H (Ω)X (Ω)

= H (Ω) 3π [

δ (

Ω− π

)

+ δ (

Ω+ π

)]

+ 5π [

δ (

Ω− π

)

+ δ (

Ω+ π

)]

= 3π [

δ (

Ω− π

)

H (π

)

+ δ (

Ω+ π

)

H (

− π

)]

+ 5π [

δ (

Ω− π

)

H (π

)

+ δ (

Ω+ π

)

H (

− π

)]

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513 11 Discrete-time Fourier series and transform

The DTFT Y (Ω) is obtained by substituting the values of the transfer function

H(Ω) at frequencies Ω1 = ±π/8 and Ω2 = ±π/4. (i) For the transfer function in Eq. (11.65), the values of H1(Ω) are given by

Ω = π/8 H1(π/8) = 4.04 − j2.03, |H1(π/8)| = 4.52, < H1(π/8) = −0.465 radians;

Ω = −π/8 H1(−π/8) = 4.04 + j2.03, |H1(−π/8)| = 4.52, < H1(−π/8) = 0.465 radians;

Ω = π/4 H1(π/4) = 2.44 − j2.11, |H1(π/4)| = 3.22, < H1(π/4) = −0.71 radians;

Ω = −π/4 H1(−π/4) = 2.44 + j2.11, |H1(−π/4)| = 3.22, < H1(−π/4) = 0.71 radians.

The DTFT Y1(Ω) of the output sequence is therefore given by

Y1(Ω) = 3π [

δ (

Ω− π

)

4.52e−j0.465 + δ (

Ω+ π

)

· 4.52e j0.465 ]

+ π5 [

δ (

Ω− π

)

3.22e−j0.71 + δ (

Ω+ π

)

3.22ej0.71 ]

= 13.56π [

δ (

Ω− π

)

e−j0.465 + δ (

Ω+ π

)

e j0.465 ]

+ 16.10π [

δ (

Ω− π

)

e−j0.71 + δ (

Ω+ π

)

ej0.71 ]

Calculating the inverse DTFT, the output sequence is obtained as

y1[k] = 13.56 cos (π

8 k − 0.465

)

+ 16.10 cos (π

4 k − 0.71

)

where we have expressed the constant phase in radians. Expressing the constant

phase in degrees yields

y1[k] = 13.56 cos (π

8 k − 26.67◦

)

+ 16.10 cos (π

4 k − 40.80◦

)

The LTID system H1(Ω) acts like an amplifier as the sinusoidal component

3 cos(πk/8) with fundamental frequency Ω1 = π/8 is amplified by a factor

of 4.52, while the sinusoidal component 3 cos(πk/4) with fundamental fre-

quency Ω1 = π/4 is amplified by a factor of 3.22. The difference in the gains

is also apparent in the magnitude spectrum plotted in Fig. 11.18, where the

low-frequency components have a higher amplification factor than that of the

higher-frequency components.

(ii) For the transfer function in Eq. (11.65), the values of the transfer function

H2(Ω) at frequencies Ω1 = ±π/8 and Ω2 = ±π/4 are given by

Ω = π/8 H2(π/8) = 1, |H2(π/8)| = 1, <H2(π/8) = 0 radians;

Ω = −π/8 H2(−π/8) = 1, |H2(−π/8)| = 1, <H2(−π/8) = 0 radians;

Ω = π/4 H2(π/4) = 0, |H2(π/4)| = 0, <H2(π/4) = 0 radians;

Ω = −π/4 H2(−π/4) = 0, |H2(−π/4)| = 0, <H2(−π/4) = 0 radians.

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514 Part III Discrete-time signals and systems

The DTFT Y2(Ω) of the output sequence is therefore given by

Y2(Ω) = 3π [

δ (

Ω− π

)

· 1 + δ (

Ω+ π

)

· 1 ]

+ 5π [

δ (

Ω− π

)

· 0 + δ (

Ω+ π

)

· 0 ]

= 3π [

δ (

Ω− π

)

+ δ (

Ω+ π

)]

Calculating the inverse DTFT, the output sequence is obtained as

y2[k] = 3 cos (π

8 k )

The LTID system H2(Ω) acts like an ideal lowpass filter as the sinusoidal

component 3 cos(πk/8) with low fundamental frequency Ω1 = π/8 is not attenuated, while the sinusoidal component 3 cos(πk/4) with high fundamental

frequency Ω1 = π/4 is blocked from the output. (iii) For the transfer function in Eq. (11.67), the values of the transfer

function H3(Ω) at frequencies Ω1 = ± π/8 and Ω2 = ±π/4 are given by Ω = π/8, H3(π/8) = 0, |H3(π/8)| = 0, <H3(π/8) = 0 radians; Ω = −π/8, H3(−π/8) = 0, |H3(−π/8)| = 0, <H3(−π/8) = 0 radians; Ω = π/4, H3(π/4) = 1, |H3(π/4)| = 1, <H3(π/4) = 0 radians; Ω = −π/4, H3(−π/4) = 1, |H3(−π/4)| = 1, <H3(−π/4) = 0 radians.

The DTFT Y3(Ω) of the output sequence is therefore given by

Y3(Ω) = 3π [

δ (

Ω− π

)

· 0 + δ (

Ω+ π

)

· 0 ]

+ 5π [

δ (

Ω− π

)

· 1 + δ (

Ω+ π

)

· 1 ]

= 5π [

δ (

Ω− π

)

+ δ (

Ω− π

)]

Calculating the inverse DTFT, the output sequence is obtained as

y3[k] = 5 cos (π

4 k )

The LTID system H3(Ω) acts like an ideal highpass filter as the sinusoidal com-

ponent 3 cos(πk/8) with lower fundamental frequency Ω1 = π /8 is blocked, while the sinusoidal component 3 cos(πk/8) with higher fundamental frequency

Ω1 = π/4 is unattenuated in the output sequence.

11.8 Continuous- and discrete-time Fourier transforms

In Chapters 4, 5, and 10, we derived frequency representations for CT and DT

waveforms. In particular, we considered the following four frequency represen-

tations:

(1) CTFT for CT periodic signals;

(2) CTFT for CT aperiodic signals;

(3) DTFT for DT periodic sequences;

(4) DTFT for DT aperiodic sequences.

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515 11 Discrete-time Fourier series and transform

In this section, we compare the Fourier transforms for different types of signals.

The CT periodic signals are typically represented by the CTFS,

x̃(t) = ∞∑

n=−∞

Dne jnω0t ,

where Dn denotes the CTFS coefficients and ω0 is the fundamental frequency

of the CT periodic signal. By exploiting the CTFT pair

e jnω0t CTFT

←−−→ 2πδ(ω − nω0),

the CTFT for the CT periodic signals is given by

x̃(t) CTFT←−−→ X (ω) = 2π

∞∑

n=−∞ Dnδ(ω − nω0)

and consists of a train of time-shifted impulse functions. In other words, the

CTFT of CT periodic signals is discrete in nature.

For CT aperiodic signals, the CTFT X (ω) is given by

x(t) CTFT←−−→ X (ω) =

∞∫

−∞

x(t)e−jωt dt,

which is generally aperiodic and continuous in the frequency domain.

Similar to the CT periodic signal, the frequency representation for a DT

periodic sequence is obtained by using the following DTFS:

x̃[k] = ∑

n=〈K0〉 Dne

jnΩ0k,

where Dn denotes the DTFS coefficients and Ω0 is the fundamental frequency

of the DT periodic signal. We observed that the DTFS is periodic with period

K0 = 2π/Ω0 such that

Dn = Dn+mK0

for −∞ < m < ∞. By exploiting the DTFT pair

e jnΩ0k DTFT←−−→ 2πδ(Ω− nΩ0),

the DTFT for a DT periodic sequence is given by

x̃[k] DTFT←−−→ X (Ω) = 2π

∞∑

n=−∞ Dnδ(Ω− nΩ0).

We showed that the DTFT of a DT periodic sequence is discrete as it consists of

several discrete time-shifted impulse functions. In addition, the DTFT of a DT

periodic sequence is itself periodic in the frequency domain, with a fundamental

period Ω0 = 2π . Finally, the DTFT of a DT aperiodic sequence is given by

X (Ω) = ∞∑

k=−∞ x[k]e−jΩk .

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516 Part III Discrete-time signals and systems

Table 11.7. Fourier transforms for different types of waveforms

Time domain Frequency domain

x(t): continuous and periodic signals X (Ω): discrete and aperiodic CTFT

−2T0 −T0 0 T0 2T0

(a)

CTFT ←−−→ w

0 2p T0

2p T0

−

x(t): continuous and aperiodic signals X (Ω): Continuous and aperiodic CTFT

W − p

W p

(b)

CTFT←−−→ w

0 W

−W

x[k]: discrete and periodic signals X (Ω): discrete and periodic DTFT

0 2 4−6 −4 6 8−2−8

(c)

DTFT←−−→ W 0 p−2p 2p−p

x[k]: discrete and aperiodic signals X (Ω): continuous and periodic DTFT

0 2 4−6 −4

1 11

6 8−2−8

(d)

DTFT←−−→

0 2p 4p−4p −2p W

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We observed that the DTFT of a DT aperiodic sequence is continuous as it is

defined for all frequencies Ω. Like the DTFT of a DT periodic sequence, the

DTFT of a DT aperiodic sequence is periodic in the frequency domain, with a

fundamental period Ω0 = 2π . The aforementioned discussion on the four types of Fourier transforms is

summarized in Table 11.7, where we observe that periodicity in the time

domain corresponds to discreteness in the frequency domain. The CTFT

for the CT periodic signals, illustrated in row (a) of Table 11.7, and the

DTFT for the DT periodic signals, illustrated in row (c), are both dis-

crete in the frequency domain. The converse of the observation is also true,

as discreteness in the time domain corresponds to periodicity in the fre-

quency domain. The converse statement is illustrated in rows (c) and (d),

where periodic and aperiodic DT sequences are considered. The DTFT for

both the periodic and aperiodic DT sequences is periodic with period Ω0 = 2π .

When a signal is both discrete and periodic in the time domain, such as the DT

periodic sequence illustrated in row (c) of Table 11.7, the DTFT is also both

periodic and discrete in the frequency domain. This observation is exploited

in digital signal processing. To compute the DTFT on digital computers, it

is always assumed that the waveform is discrete and periodic, even when the

original waveform is neither discrete nor periodic. Chapter 12 presents the

theory of the discrete Fourier transform (DFT), which is a very powerful tool

for computing the CTFT and DTFT.

11.9 Summary

In this chapter, we presented the frequency representation for DT sequences.

For aperiodic sequences, we derived the DTFS, which is defined as

x̃[k] = ∑

n=〈K0〉 Dne

jnΩ0k,

whereΩ0 is the fundamental frequency, given byΩ0 = 2π/K0, and the discrete- time Fourier series (DTFS) coefficients Dn , for 1 ≤ n ≤ K0, are given by

Dn = 1

∑

k=〈K0〉 x[k]e−jnΩ0k .

The DTFS coefficients of periodic sequences are themselves periodic with a

period K0 such that

Dn = Dn+mK0 for m ∈ Z .

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518 Part III Discrete-time signals and systems

Section 11.2 derived the DTFT for an aperiodic sequence x[k] as follows:

DTFT synthesis equation x[k] = 1

2π

∫

〈2π〉

X (Ω)e jkΩdΩ;

DTFT analysis equation X (Ω) = ∞∑

k=−∞ x[k]e−jΩk,

and showed that the DTFT is periodic in the frequency domain with a period

Ω0 = 2π . As such, the frequencies Ω = 0, 2π, 4π, . . . are considered as the same frequencies and are referred to as the lowest possible frequency for

the DTFT. Similarly, the frequencies Ω = π, 3π, 5π, . . . are the same and are referred to as the highest possible frequency for the DTFT.

Section 11.3 derived a sufficient condition for the existence of the DTFT for

aperiodic DT sequences as follows:

∞∑

k=−∞ |x2[k]| < ∞.

The periodic DT sequences do not satisfy the above condition for the existence

of the DTFT. Instead the DTFT of a periodic sequence is obtained by calcu-

lating the DTFT of its DTFS representation, which results in the following

DTFT:

X (Ω) = 2π ∞∑

n=−∞ Dnδ

(

Ω− 2nπ

)

where Dn are the DTFS coefficients of the periodic sequence x[k].

Section 11.4 covered the properties of the DTFT. In particular, we covered

the following properties.

(1) The periodicity property states that the DTFT of any DT sequence is

periodic with period 2π .

(2) The Hermitian symmetry property states that the DTFT of a real-valued

sequence is Hermitian. In other words, the real component of the DTFT

of a real-valued sequence is even, while the imaginary component is

odd.

(3) The linearity property states that the overall DTFT of a linear combination

of DT sequences is given by the same linear combination of the individual

DTFTs.

(4) The time-scaling property is only applicable for time-expanded (or inter-

polated) sequences. It states that interpolating a sequence in the time

domain compresses its DTFT in the frequency domain.

(5) The time-shifting property states that shifting a sequence in the time

domain towards the right-hand side by an integer constant m is equiv-

alent to multiplying the DTFT of the original sequence by a complex

exponential exp(−jΩm). Similarly, shifting towards the left-hand side by

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519 11 Discrete-time Fourier series and transform

integer m is equivalent to multiplying the DTFT of the original sequence

by a complex exponential exp(jΩm).

(6) The frequency-shifting property is the converse of the time-shifting prop-

erty. It states that shifting the DTFT in the frequency domain towards the

right-hand side byΩ0 is equivalent to multiplying the original sequence by

a complex exponential exp(jΩ0m). Similarly, shifting the DTFT towards

the left-hand side by Ω0 is equivalent to multiplying the DTFT of the

original sequence by a complex exponential exp(−jΩ0m). (7) The frequency-differentiation property states that differentiating the

DTFT with respect to the frequency Ω is equivalent to multiplying the

original sequence by a factor of −jk. (8) Time differencing is defined as the difference between the original

sequence and its time-shifted version with a shift of one sample towards

the right-hand side. The time-differencing property states that time differ-

encing a signal in the time domain is equivalent to multiplying its DTFT

by a factor of (1 − exp(−jΩm)). (9) The time-summation property is the converse of the time-differencing

property. The time-summation property states that the DTFT of the run-

ning sum of a sequence is obtained by dividing the DTFT of the original

sequence by a factor of (1 − exp(−jΩm)) and adding DT impulses located at multiples of 2π .

(10) The time-convolution property states that the convolution of two DT

sequences is equivalent to the multiplication of the DTFTs of the two

sequences in the time domain.

(11) Periodic convolution is an extension of time convolution to periodic

sequences, where only single periods of the two periodic sequences are

convolved. The periodic-convolution property states that the periodic con-

volution in the time domain is equivalent to multiplying the DTFS coef-

ficients of the two periodic sequences by each other in the frequency

domain.

(12) The frequency-convolution property states that periodic convolution of

two DTFTs with period 2π is equivalent to multiplication of their

sequences in the time domain.

The DTFT of the impulse response of an LTID system is referred to as the Fourier

transfer function, which is generally complex-valued. The plot of the magnitude

of the Fourier transfer function with respect to frequencyΩ is referred to as the

magnitude spectrum, while the plot of the phase of the Fourier transfer function

with respect to frequency Ω is referred to as the phase spectrum. Sections 11.6

and 11.7 illustrated how the magnitude and phase spectra provide meaningful

insights into the analysis of the LTID systems. In particular, we covered the ideal

lowpass filter, which blocks all frequency components above a certain cut-off

frequency Ω > Ωc in the applied input sequence. All frequency components

Ω ≤ Ωc are left unattenuated in the output response of an ideal lowpass filter.

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520 Part III Discrete-time signals and systems

The magnitude spectrum of an ideal lowpass filter is unity within its pass band

(Ω ≤ Ωc) and zero within its stop band (Ωc < Ω ≤ π ). The converse of the ideal lowpass filter is the ideal highpass filter, which

blocks all frequency components below a certain cut-off frequency Ω > Ωc in

the applied input sequence. All frequency componentsΩ ≥ Ωc are left unatten- uated in the output response of an ideal highpass filter. The magnitude spectrum

of an ideal highpass filter is unity within the pass band (Ωc ≤ Ω ≤ π ) and zero within the stop band (0 ≤ Ω < Ωc).

Section 11.8 compared the Fourier representations of CT and DT periodic

and aperiodic waveforms. We showed that the Fourier representations of peri-

odic waveforms are discrete, whereas the Fourier representations of discrete

waveforms are periodic.

Problems

11.1 Determine the DTFS representation for each of the following DT peri- odic sequences. In each case, plot the magnitude and phase of the DTFS

coefficients.

(i) x[k] = k for 0 ≤ k ≤ 5 and x[k + 6] = x[k];

(ii) x[k] =







1 (0 ≤ k ≤ 2) 0.5 (3 ≤ k ≤ 5) 0 (6 ≤ k ≤ 8)

and x[k + 9] = x[k];

(iii) x[k] = 3 sin (

2π

7 k +

)

;

(iv) x[k] = 2e j( 5π 3

k+ π 4 );

(v) x[k] = ∞∑

m=−∞ δ(k − 5m);

(vi) x[k] = cos(10πk/3) cos(2πk/5); (vii) x[k] = |cos(2πk/3)|.

11.2 Given the following DTFS coefficients, determine the DT periodic sequence in the time domain:

(i) Dn =







1 (0 ≤ k ≤ 2) 0.5 (3 ≤ k ≤ 5) 0 (6 ≤ k ≤ 8)

and Dn+9 = Dn;

(ii) Dn =



  

  

1 − j0.5 (n = −1) 1 (n = 0) 1 + j0.5 (n = 1) 0 (2 ≤ n ≤ 5)

and Dn+7 = Dn;

(iii) Dn = 1 + 3

4 sin

(πn

)

(0 ≤ n ≤ 6) and Dn+7 = Dn;

(iv) Dn = (−1)n (0 ≤ n ≤ 7) and Dn+8 = Dn; (v) Dn = e jnπ/4 (0 ≤ n ≤ 7) and Dn+8 = Dn.

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521 11 Discrete-time Fourier series and transform

11.3 Determine if the following DT sequences satisfy the DTFT existence property:

(i) x[k] − 2;

(ii) x[k] = {

3 − |k| |k| < 3 0 otherwise;

(iii) x[k] = k3−|k|; (iv) x[k] = αk cos(ω0k)u[k], |α| < 1; (v) x[k] = αk sin(ω0k + φ)u[k], |α| < 1;

(vi) x[k] = sin(πk/5) sin(πk/7)

π2k2 ;

(vii) x[k] = ∞∑

m=−∞

δ(k − 5m − 3);

(viii) x[k] =

{

3 − |k| |k| < 3

0 |k| = 3 and x[k + 7] = x[k];

(ix) x[k] = e j(0.2πk+45 ◦);

(x) x[k] = k3−ku[k] + e j(0.2πk+45 ◦).

11.4 (a) Calculate the DTFT of the DT sequences specified in Problem 11.3. (b) Calculate the DTFT of the periodic DT sequences specified in

Problem 11.1.

11.5 Given the following transform pair:

x1[k] DTFT

←−−→ X1(Ω) and x2[k] DTFT←−−→ X2(Ω) ,

express the DTFT of the following DT sequences in terms of the DTFTs

X1(Ω) and X2(Ω):

(i) (−1)k x1[k]; (ii) (k − 5)2x2[k − 4];

(iii) ke−j4k x1[3 − k];

(iv)

∞∑

m=−∞ [x1[k − 4m] + x2[k − 6m]];

(v) x1[5 − k]x2[7 − k].

11.6 Calculate the DT sequences with the following DTFT representations defined over the frequency range −π ≤ Ω ≤ π :

(i) X (Ω) = 4e−jΩ

1 − 5e−jΩ + 6e−j2Ω ;

(ii) X (Ω) = 2e−j2Ω

(1 − 4e−jΩ)2(1 − 2e−jΩ) ;

(iii) X (Ω) = 8 sin(7Ω) cos(9Ω);

(iv) X (Ω) = 4e−j4Ω

10 − 6 cosΩ ;

(v) X (Ω) = {

1 0.25π ≤ |Ω| < 0.75π 0 |Ω| ≤ 0.25π and 0.75π ≤ |Ω| < π.

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522 Part III Discrete-time signals and systems

11.7 (a) Prove the Hermitian symmetry property, Eq. (11.39a), for a real- valued DT sequence.

(b) Problem 11.6 lists the DTFTs of several sequences. Applying the

Hermitian property, determine whether these sequences are real-

valued.

11.8 Prove the frequency-differentiation property of the DTFT.

11.9 Prove the time-convolution property of the DTFT.

11.10 Prove the time-shifting property of the DTFT.

11.11 Given the following transfer function:

H (Ω) = 1

(1 − 0.3e−jΩ)(1 − 0.5e−jΩ)(1 − 0.7e−jΩ) ,

(i) determine the impulse response of the LTID system;

(ii) determine the difference equation representation of the LTID

system;

(iii) determine the unit step response of the LTID system by using the

time-convolution property of the DTFT;

(iv) determine the unit step response of the LTID system by convolv-

ing the unit step sequence with the impulse response obtained in

part (i).

11.12 Given the following difference equation:

y[k] + y[k − 1] + 1

4 y[k − 2] = x[k] − x[k − 2],

(i) determine the transfer function representing the LTID system;

(ii) determine the impulse response of the LTID system;

(iii) determine the output of the LTID system for the input x[k] = (1/2)ku[k] using the time-convolution property;

(iv) determine the output of the LTID system by convolving the input

x[k] = (1/2)ku[k] with the impulse response obtained in part (ii).

11.13 Determine the output response of the LTID systems with the specified inputs and impulse responses using Fourier transform approach:

(i) x[k] = u[k] and h[k] = 4−|k|; (ii) x[k] = 2−ku[k] and h[k] = 2ku[−k − 1];

(iii) x[k] = u[k] − u[k − 9] and h[k] = 3ku[−k + 4]; (iv) x[k] = k5−ku[k] and h[k] = 5ku[−k]; (v) x[k] = u[k + 2] − u[−k − 3] and h[k] = u[k − 5] − u[−k − 6].

11.14 Given that the transfer function of an LTID system is given by

H (Ω) = 1

(1 + 3e−jΩ) ,

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523 11 Discrete-time Fourier series and transform

determine and sketch the following as a function of frequency Ω over

the range −π ≤ Ω ≤ π : (i) Re{H (Ω)};

(ii) Im{H (Ω)}; (iii) |H (Ω)|; (iv) <H (Ω).

11.15 Calculate and plot the magnitude and phase spectra of the LTID systems specified in Problem 11.13.

11.16 The impulse response of an LTID system is given by

h[k] = 3δ[k + 3] − 2δ[k + 2] + δ[k + 1] + 5δ[k] − δ[k − 1] − 2δ[k − 2] − 3δ[k − 3] + 4δ[k − 4].

Without explicitly determining the transfer function H (Ω), evaluate the

following using the properties of the DTFT:

(i) H (Ω)|Ω=0; (ii) H (Ω)|Ω=π ;

(iii) <H (Ω);

(iv)

∫ π

−π H (Ω)dΩ.

(v) Determine and sketch the DT sequence with the DTFT H (−Ω). (vi) Determine and sketch the DT sequence with the DTFT Re{H (Ω)}.

11.17 Using Parseval’s theorem, determine the following sum:

∞∑

k=−∞

sin(πk/5) sin(πk/7)

k2 .

11.18 Consider an LTID system with the following impulse response:

h[k] = sinc(3k/4).

Determine the output responses of the LTID system for the following

inputs:

(i) x[k] = cos(11πk/16) cos(3πk/16); (ii) x[k] = k for 0 ≤ k ≤ 5 and x[k + 6] = x[k];

(iii) x[k] =







1 (0 ≤ k ≤ 2) 0.5 (3 ≤ k ≤ 5) 0 (6 ≤ k ≤ 8)

and x[k + 9] = x[k];

(iv) x[k] = ∞∑

m=−∞ δ(k − 5m);

(v) x[k] = sinc(k/3).

11.19 When the DT sequence

x[k] = 4−ku[k] + 3−ku[k]

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524 Part III Discrete-time signals and systems

is applied at the input of an LTID system, the output response is given

y[k] = 2 (

u[k] − 4 (

u[k].

(i) Determine the Fourier transfer function H(Ω) of the LTID system.

(ii) Determine the impulse response h[k] of the LTID system.

(iii) Determine the difference equation representing the LTID

system.

(iv) Determine if the system is causal.

11.20 Repeat Example 11.21 for each of the following signals, assuming that the sampling rate to discretize the CT signals is 8000 samples/s:

(i) x1(t) = 2 + 3 cos(400π t) + 7 cos(800π t); (ii) x2(t) = 2 cos(4000π t) + 5 cos(6000π t);

(iii) x3(t) = 5 cos(600π t) + 9 cos(900π t) + 2 cos(3000π t); (iv) x4(t) = 4 cos(600π t) + 6 cos(12000π t).

11.21 Repeat Example 11.21 for each of the following signals, assuming that the sampling rate to discretize the CT signals is 22 000 samples/s:

(i) x1(t) = 2 + 3 cos(8000π t) + 7 cos(18 000π t); (ii) x2(t) = 2 cos(10 000π t) + 5 cos(30 000π t);

(iii) x3(t) = 5 cos(600π t) + 9 cos(900π t) + 2 cos(3000π t); (iv) x4(t) = 4 cos(28 000π t) + 6 cos(18 000π t).

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C H A P T E R

12 Discrete Fourier transform

In Chapter 11, we introduced the discrete-time Fourier transform (DTFT) that

provides us with an alternative representation for DT sequences. The DTFT

transforms a DT sequence x[k] into a function X (Ω) in the DTFT frequency

domain Ω. The independent variable Ω is continuous and is confined to the

range –π ≤ Ω < π . With the increased use of digital computers and special-

ized hardware in digital signal processing (DSP), interest has focused around

transforms that are suitable for digital computations. Because of the continuous

nature of Ω, direct implementation of the DTFT is not suitable on such digital

devices. This chapter introduces the discrete Fourier transform (DFT), which

can be computed efficiently on digital computers and other DSP boards.

The DFT is an extension of the DTFT for time-limited sequences with an

additional restriction that the frequency Ω is discretized to a finite set of values

given by Ω = 2πr/M , for 0 ≤ r ≤ (M − 1). The number M of the frequency

samples can have any value, but is typically set equal to the length N of the time-

limited sequence x[k]. If M is chosen to be a power of 2, then it is possible to

derive highly efficient implementations of the DFT. These implementations are

collectively referred to as the fast Fourier transform (FFT) and, for an M-point

DFT, have a computational complexity of O(M log2 M). This chapter discusses

a popular FFT implementation and extends the theoretical DTFT results derived

in Chapter 11 to the DFT.

The organization of this chapter is as follows. Section 12.1 motivates the

discussion of the DFT by expressing it as a special case of the continuous-time

Fourier transform (CTFT). The formal definition of the DFT is presented in

Section 12.2, including its matrix-vector representation. Section 12.3 applies the

DFT to estimation of the spectra of both DT and CT signals. Section 12.4 derives

important properties of the DFT, while Section 12.5 uses the DFT as a tool to

convolve two DT sequences in the frequency domain. A fast implementation of

the DFT based on the decimation-in-time algorithm is presented in Section 12.6.

Finally, Section 12.7 concludes the chapter with a summary of the important

concepts.

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526 Part III Discrete-time signals and systems

12.1 Continuous to discrete Fourier transform

In order to motivate the discussion of the DFT, let us assume that we are

interested in computing the CTFT of a CT signal x(t) using a digital computer.

The three main steps involved in the digital computation of the CTFT are

illustrated in Fig. 12.1. The waveforms for the CT signal x(t) and its CTFT

X (ω), shown in Figs. 12.1(a) and (b), are arbitrarily chosen, and hence the

following procedure applies to any CT signal. A brief explanation of each of

the three steps is provided below.

Step 1: Analog-to-digital conversion In order to store a CT signal into a digital computer, the CT signal is digitized. This is achieved through two pro-

cesses known as sampling and quantization, collectively referred to as analog-

to-digital (A/D) conversion by convention. In this discussion, we only consider

sampling, ignoring the distortion introduced by quantization. The CT signal

x(t) is sampled by multiplying it by an impulse train:

s1(t) = ∞∑

m=−∞

δ(t − mT1), (12.1)

illustrated in Fig. 12.1(c). The sampled waveform is given by

x1(t) = x(t)s1(t) = x(t) ×

∞∑

m=−∞

δ(t − mT1) (12.2)

and is shown in Fig. 12.1(e). Since multiplication in the time domain is equiv-

alent to convolution in the frequency domain, the CTFT X1(ω) of the sampled

signal x1(t) is given by

X1(ω) = ℑ

[

x(t)×

∞∑

m=−∞

δ(t − mT1)

]

= 1

2π

[

X (ω) ∗ 2π

∞∑

m=−∞

(

δ − 2mπ

)]

= 1

∞∑

m=−∞

(

ω − 2mπ

)

(12.3)

The above result was also derived in Eq. (9.5) of Chapter 9, and is graphically

illustrated in Figs. 12.1(b), (d), and (f), where we note that the spacing between

adjacent replicas of X (ω) in X1(ω) is given by 2π/T1. Since no restriction is

imposed on the bandwidth of the CT signal x(t), limited aliasing may also be

introduced in X1(ω).

To derive the discretized representation of x(t) from Eq. (12.3), sampling is

followed by an additional step (shown in Fig. 12.1(g)), where the CT impulses

are converted to the DT impulses. Equation (12.3) can now be extended to

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527 12 Discrete Fourier transform

X(w)1

x1[k]A

1 2 3

x1(t)

x(t)A

s1(t)

T10

S1(w)

X1(w)

X1(W)

2p T1

1 T1

04p T1

−

4p T1

− 2p T1

4p T1

2p T1

− 2p T1

4p T1

0 2p 4p−2p−4p

w[k]

0 1 2 … N−1

W(W)

0 2p 4p−2p−4p

(a)

(c)

(e)

(d)

(f)

(g) (h)

(i) (j)

(b)

Fig. 12.1. Graphical derivation of the discrete Fourier transform pair. (a) Original CT signal. (b) CTFT of the

original CT signal. (c) Impulse train sampling of CT signal. (d) CTFT of the impulse train in part (c). (e) CT

sampled signal. (f) CTFT of the sampled signal in part (e). (g) DT representation of CT signal in part (a).

(h) DTFT of the DT representation in part (g). (i) Rectangular windowing sequence. (j) DTFT of the

rectangular window. (k) Time-limited sequence representing part (g). (l) DTFT of time-limited sequence in

part (k). (m) Inverse DTFT of frequency-domain impulse train in part (n). (n) Frequency-domain impulse

train. (o) Inverse DTFT of part (p). (p) DTFT representation of CT signal in part (a). (q) Inverse DFT of part

(r). (r) DFT representation of CT signal in part (a).

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528 Part III Discrete-time signals and systems

Xw(W)

xw[k]A

0 1 2 … N−1 −4p −2p 2p W

2pT1 N

0 4p

s2[k]

M−M

x2[k]

0 k

0 1 2 … N−1 M

00 1 2 … N−1 M

S2(W)

M 2p

0 2p 4p−2p−4p

−M

Xw(W)

MT1

0 2p 4p−2p−4p

X2[r]

0 M 2M−M−2M

MT1

N x2[k]

(k) (l)

(m) (n)

(o) (p)

(q) (r)

Fig. 12.1. (cont.) derive the DTFT of the DT sequence x1[k] as follows:

x1[k] = ∞∑

m=−∞

x(mT1)δ(t − mT1). (12.4)

Calculating the CTFT of both sides of Eq. (12.4) yields

X1(ω) =

∞∑

m=−∞

x(mT1)e −jωmT1 . (12.5)

Substituting x1[m] = x(mT 1) and Ω = ωT1 in Eq. (12.5) leads to

X1(Ω) = X1(ω)|ω=Ω/T1 =

∞∑

m=−∞

x1[m]e −jmΩ,

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529 12 Discrete Fourier transform

which is the standard definition of the DTFT introduced in Chapter 11. The

DTFT spectrum X1(Ω) of x1[k] is obtained by changing the frequency axis

ω of the CTFT spectrum X1(ω) according to the relationship Ω = ωT1. The DTFT spectrum X1(Ω) is illustrated in Fig. 12.1(h).

Step 2: Time limitation The discretized signal x1[k] can possibly be of infi- nite length. Therefore, it is important to truncate the length of the discretized

signal x1[k] to a finite number of samples. This is achieved by multiplying the

discretized signal by a rectangular window,

w[k] = {

1 0 ≤ k ≤ (N − 1)

0 elsewhere, (12.6)

of length N . The DTFT Xw (Ω) of the time-limited signal xw [k] = x1[k]w[k]

is obtained by convolving the DTFT X1(Ω) with the DTFT W (Ω) of the rect-

angular window, which is a sinc function. In terms of X1(Ω), the DTFT Xw (Ω)

of the time-limited signal is given by

Xw (Ω) = 1

2π

[

X1(Ω) ⊗ sin(0.5NΩ)

sin(0.5Ω) e−j(N−1)/2

]

, (12.7)

which is shown in Fig. 12.1(l) with its time-limited representation xw [k] plotted

in Fig. 12.1(k). Symbol ⊗ in Eq. (12.7) denotes the circular convolution.

Step 3: Frequency sampling The DTFT Xw (Ω) of the time-limited signal xw [k] is a continuous function of Ω and must be discretized to be stored on a

digital computer. This is achieved by multiplying Xw (Ω) by a frequency-domain

impulse train, whose DTFT is given by

S2(Ω) = 2π

∞∑

m=−∞

(

Ω− 2πm

)

. (12.8)

The discretized version of the DTFT Xw (Ω) is therefore expressed as follows:

X2(Ω) = Xw (Ω)S2(Ω) = 1

[

X1(Ω) ⊗ sin(0.5NΩ)

sin(0.5Ω) e−j(N−1)/2

]

× ∞∑

m=−∞

(

Ω− 2πm

)

. (12.9)

The DTFT X2(Ω) is shown in Fig. 12.1(p), where the number M of frequency

samples within one period (−π ≤ Ω ≤ π ) of X2(Ω) depends upon the funda-

mental frequency Ω2 = 2π/M of the impulse train S2(Ω). Taking the inverse

DTFT of Eq. (12.9), the time-domain representation x2[k] of the frequency-

sampled signal X2(Ω) is given by

x2[k] = [xw [k] ∗ s2[k]] = [x1[k] · w[k]] ∗ ∞∑

m=−∞

δ(k − mM), (12.10)

and is shown in Fig. 12.1(o).

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530 Part III Discrete-time signals and systems

The discretized version of the DTFT Xw (Ω) is referred to as the discrete

Fourier transform (DFT) and is generally represented as a function of the fre-

quency index r corresponding to DTFT frequency Ωr = 2rπ/M , for 0 ≤ r ≤ (M− 1). To derive the expression for the DFT, we substitute Ω = 2rπ/M in

the following definition of the DTFT:

X2(Ω) =

N−1∑

k=0

x2[k]e −jkΩ, (12.11)

where we have assumed x2[k] to be a time-limited sequence of length N .

Equation (12.11) reduces as follows:

X2(Ωr ) =

N−1∑

k=0

x2[k]e −j(2πkr/M), (12.12)

for 0 ≤ r ≤ (M−1). Equation (12.12) defines the DFT and can easily be imple-

mented on a digital device since it converts a discrete number N of input samples

in x2[k] to a discrete number M of DFT samples in X2(Ωr ). To illustrate the

discrete nature of the DFT, the DFT X2(Ωr ) is also denoted asX2[r ]. The DFT

spectrum X2[r ] is plotted in Fig. 12.1(r).

Let us now return to the original problem of determining the CTFT X (ω) of

the original CT signal x(t) on a digital device. Given X2[r ] = X2(Ωr ), it is

straightforward to derive the CTFT X (ω) of the original CT signal x(t) by

comparing the CTFT spectrum, shown in Fig. 12.1(b), with the DFT spectrum,

shown in Fig. 12.1(r). We note that one period of the DFT spectrum within the

range −(M − 1)/2 ≤ r ≤ (M − 1)/2 (assuming M to be odd) is a fairly good

approximation of the CTFT spectrum. This observation leads to the following

relationship:

X (ωr ) ≈ MT1

N X2[r ] =

MT1

N−1∑

k=0 x2[k]e

−j(2πkr/M), (12.13)

where the CT frequencies ωr = Ωr/T1 = 2πr/(M × T1) for −(M − 1)/2 ≤ r ≤ (M − 1)/2.

Although Fig. 12.1 illustrates the validity of Eq. (12.13) by showing that the

CTFT X (ω) and the DFT X2[r ] are similar, there are slight variations in the two

spectra. These variations result from aliasing in Step 1 and loss of samples in

Step 2. If the CT signal x(t) is sampled at a sampling rate less than the Nyquist

limit, aliasing between adjacent replicas distorts the signal. A second distortion

is introduced when the sampled sequence x1[k] is multiplied by the rectangular

window w[k] to limit its length to N samples. Some samples of x1[k] are lost in

the process. To eliminate aliasing, the CT signal x(t) should be band-limited,

whereas elimination of the time-limited distortion requires x(t) to be of finite

length. These are contradictory requirements since a CT signal cannot be both

time-limited and band-limited at the same time. As a result, at least one of the

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531 12 Discrete Fourier transform

aforementioned distortions would always be present when approximating the

CTFT with the DFT. This implies that Eq. (12.12) is an approximation for the

CTFT X (ω) that, even at its best, only leads to a near-optimal estimation of the

spectral content of the CT signal.

On the other hand, the DFT representation provides an accurate estimate of

the DTFT of a time-limited sequence x[k] of length N . By comparing the DFT

spectrum, Fig. 12.1(h), with the DFT spectrum, Fig. 12.1(r), the relationship

between the DTFT X2(Ω) and the DFT X2[r ] is derived. Except for a factor of

K/M , we note that X2[r ] provides samples of the DTFT at discrete frequencies

Ωr = 2πr/M , for 0 ≤ r ≤ (M−1). The relationship between the DTFT and DFT is therefore given by

X2(Ωr ) = N

M X2[r ] =

N−1∑

k=0

x2[k]e −j(2πkr/M) (12.14)

forΩr = 2πr/M , 0 ≤ r ≤ (M−1). We now proceed with the formal definitions

for the DFT.

12.2 Discrete Fourier transform

Based on our discussion in Section 12.1, the M-point DFT and inverse DFT for a

time-limited sequence x[k], which is non-zero within the limits 0 ≤ k ≤ N − 1,

is given by

Forward DFT X [r ] =

N−1∑

k=0

x[k]e−j(2πkr/M) for 0 ≤ r ≤ M − 1; (12.15)

Inverse DFT x[k] = 1

M−1∑

r=0

X [r ]e j(2πkr/M) for 0 ≤ k ≤ N − 1.

(12.16)

Equations (12.15) and (12.16) are also, respectively, known as DFT analysis

and synthesis equations. Equation (12.15) was derived in Section 12.1. By

substituting the expression for X [r ] from Eq. (12.15), the analysis equation,

Eq. (12.16), can be formally proved, and vice versa. The formal proofs of the

DFT pair are left as an exercise for the reader. In Eqs. (12.15) and (12.16), the

length M of the DFT is typically set to be greater or equal to the length N of

the aperiodic sequence x[k]. Unless otherwise stated, we assume M = N in the

discussion that follows. Collectively, the DFT pair is denoted as

x[k] DFT

←−−→ X [r ]. (12.17)

Examples 12.1 and 12.2 illustrate the steps involved in calculating the DFTs of

aperiodic sequences.

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532 Part III Discrete-time signals and systems

0 1 2

−1

0 1 2 0

0 1 2 −0.5p

−p

0.5p

3 3 3

(a) (b) (c)

Fig. 12.2. (a) DT sequence x[k ];

(b) magnitude spectrum and

X[r ] computed in Example 12.1.

Example 12.1

Calculate the four-point DFT of the aperiodic sequence x[k] of length N = 4, which is defined as follows:

x[k] =



  

  

2 k = 0 3 k = 1

−1 k = 2 1 k = 3.

Solution

Using Eq. (12.15), the four-point DFT of x[k] is given by

X [r ] = 3∑

k=0 x[k]e−j(2πkr/4)

= 2 + 3 × e−j(2πr/4) − 1 × e−j(2π (2)r/4) + 1 × e−j(2π (3)r/4),

for 0 ≤ r ≤ 3. Substituting different values of r , we obtain

r = 0 X [0] = 2 + 3 − 1 + 1 = 5;

r = 1 X [1] = 2 + 3e−j(2π/4) − e−j(2π (2)/4) + e−j(2π (3)/4)

= 2 + 3(−j) − 1(−1) + 1(j) = 3 − 2j;

r = 2 X [2] = 2 + 3e−j(2π (2)/4) − e−j(2π (2)(2)/4) + e−j(2π (3)(2)/4)

= 2 + 3(−1) − 1(1) + 1(−1) = −3;

r = 3 X [3] = 2 + 3e−j(2π (3)/4) − e−j(2π (2)(3)/4) + e−j(2π (3)(3)/4)

= 2 + 3(j) − 1(−1) + 1(−j) = 3 + j2.

The magnitude and phase spectra of the DFT are plotted in Figs. 12.2(b) and

(c), respectively.

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533 12 Discrete Fourier transform

Example 12.2

Calculate the inverse DFT of

X [r ] =



  

  

5 r = 0 3 − j2 r = 1

−3 r = 2 3 + j2 r = 3.

Solution

Using Eq. (12.13), the inverse DFT of X [r ] is given by

x[k] = 1

3∑

r=0 X [r ]e j(2πkr/4) =

[

5 + (3 − j2) × e j(2πk/4) − 3 × e j(2π (2)k/4)

+ (3 + j2) × e j(2π (3)k/4) ]

for 0 ≤ k ≤ 3. On substituting different values of k, we obtain

x[0] = 1

4 [5 + (3 − j2) − 3 + (3 + j2)] = 2;

x[1] = 1

[

5 + (3 − j2)e j(2π/4) − 3e j(2π (2)/4) + (3 + j2)e j(2π (3)/4) ]

= 1

4 [5 + (3 − j2)( j) − 3(−1) + (3 + j2)(−j)] = 3;

x[2] = 1

[

5 + (3 − j2)e j(2π (2)/4) − 3e j(2π (2)(2)/4) + (3 + j2)e j(2π (3)(2)/4) ]

= 1

4 [5 + (3 − j2)(−1) − 3(1) + (3 + j2)(−1)] = −1;

x[3] = 1

[

5 + (3 − j2)e j(2π (3)/4) − 3e j(2π (2)(3)/4) + (3 + j2)e j(2π (3)(3)/4) ]

= 1

4 [5 + (3 − j2)(−j) − 3(−1) + (3 + j2)( j)] = 1.

Examples 12.1 and 12.2 prove the following DFT pair:

x[k] =



  

  

2 k = 0

3 k = 1

−1 k = 2

1 k = 3

DFT ←−−→ X [r ] =



  

  

5 r = 0

3 − j2 r = 1

−3 r = 2

3 + j2 r = 3,

where both the DT sequence x[k] and its DFT X [r ] have length N = 4.

Example 12.3

Calculate the N -point DFT of the aperiodic sequence x[k] of length N , which

is defined as follows:

x[k] =

{

1 0 ≤ k ≤ N1 − 1

0 N1 ≤ k ≤ N .

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534 Part III Discrete-time signals and systems

0 2 4 6 8 10 12 14 16 18 20 0

0.2

0.4

0.6

0.8

0 5 10 15 20 25 −p

−0.5p

0.5p

0 5 10 15 20 25 0

(a) (b)

(c)

Fig. 12.3. (a) Gate function x[k ]

in Example 12.3;

(b) magnitude spectrum and

Solution

Using Eq. (12.15), the DFT of x[k] is given by

X [r ] = N−1∑

k=0 x[k]e−j(2πkr/N ) =

N1−1∑

k=0 1 · e−j(2πkr/N )

+ N−1∑

k=N1

0 · e−j(2πkr/N ) = N1−1∑

k=0 e−j(2πkr/N ),

for 0 ≤ r ≤ (N−1). The right-hand side of this equation represents a GP series,

which can be simplified as follows:

X [r ] =

N1−1∑

k=0

e−j(2πkr/N ) =







N1 r = 0

1 − e−j(2πr N1/N )

1 − e−j(2πr/N ) r = 0







N1 r = 0

e−j(πr (N1−1)/N ) sin(πr N1/N )

sin(πr/N ) r = 0.

Since X [r ] is a complex-valued function, its magnitude and phase components

are given by

r = 0 |X [r ]| = N1 and <X [r ] = 0;

r = 0 |X [r ]| = sin(πr N1/N )

sin(πr/N )

<X [r ] = − πr (N1 − 1)

N + <sin(πr N1/N ) − <sin(πr/N ).

The magnitude and phase spectra for N1 = 7 and length N = 30 are shown in

Figs. 12.3(b) and (c).

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535 12 Discrete Fourier transform

12.2.1 DFT as matrix multiplication

An alternative representation for computing the DFT is obtained by expanding

Eq. (12.15) in terms of the time and frequency indices (k, r ). For N = M , the resulting equations are expressed as follows:

X [0] = x[0] + x[1] + x[2] + · · · + x[N − 1], X [1] = x[0] + x[1]e−j(2π/N ) + x[2]e−j(4π/N )

+ · · · + x[N − 1]e−j(2(N−1)π/N ), X [2] = x[0] + x[1]e−j(4π/N ) + x[2]e−j(8π/N )

+ · · · + x[N − 1]e−j(4(N−1)π/N ), ...

X [N − 1] = x[0] + x[1]e−j(2(N−1)π/N ) + x[2]e−j(4(N−1)π/N )

+ · · · + x[N − 1]e−j(2(N−1)(N−1)π/N ).



              

              

(12.18)

In the matrix-vector format they are given by



       

X [0]

X [1]

X [2]

X [N − 1]



       

︸︷︷︸

DFT vector �X



       

1 1 1 · · · 1

1 e−j(2π/N ) e−j(4π/N ) · · · e−j(2(N−1)π/N )

1 e−j(4π/N ) e−j(8π/N ) · · · e−j(4(N−1)π/N )

. . . .

1 e−j(2(N−1)π/N ) e−j(4(N−1)π/N ) · · · e−j(2(N−1)(N−1)π/N )



       

︸︷︷︸

DFT matrix F



       

x[0]

x[1]

x[2]

x[N − 1]



       

︸︷︷︸

signal vector �x

(12.19)

Equation (12.19) shows that the DFT coefficients X [r ] can be computed by left-

multiplying the DT sequence x[k], arranged in a column vector �x in ascending

order with respect to the time index k, by the DFT matrix F .

Similarly, the expression for the inverse DFT given in Eq. (12.16) can be

expressed as follows:



       

x[0]

x[1]

x[2]

x[N − 1]



       

︸︷︷︸

signal vector x

= 1



       

1 1 1 · · · 1

1 e j(2π/N ) e j(4π/N ) · · · e j(2(N−1)π/N )

1 e j(4π/N ) e j(8π/N ) · · · e j(4(N−1)π/N )

. . . .

1 e j(2(N−1)π/N ) e j(4(N−1)π/N ) · · · e j(2(N−1)(N−1)π/N )



       

︸︷︷︸

DFT matrix G=F−1



       

X [0]

X [1]

X [2]

X [N − 1]



       

︸︷︷︸

DFT vector X

(12.20)

which implies that the DT sequence x[k] can be obtained by left-multiplying

the DFT coefficients X [r ], arranged in a column vector �X in ascending order

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536 Part III Discrete-time signals and systems

with respect to the DFT coefficient index r , by the inverse DFT matrix G. It is

straightforward to show that G × F = F × G = IN , where IN is the identity matrix of order N .

Example 12.4 repeats Example 12.1 using the matrix-vector representation

for the DFT.

Example 12.4

Calculate the four-point DFT of the aperiodic signal x[k] considered in

Example 12.1.

Solution

Arranging the values of the DT sequence in the signal vector x , we obtain

x = [2 3 −1 1]T,

where superscript T represents the transpose operation for a vector. Using

Eq. (12.19), we obtain



  

X [0]

X [1]

X [2]

X [3]



  



  

1 1 1 1

1 e−j(2π/N ) e−j(4π/N ) e−j(6π/N )

1 e−j(4π/N ) e−j(8π/N ) e−j(12π/N )

1 e−j(6π/N ) e−j(12π/N ) e−j(18π/N )



  

︸︷︷︸

DFT matrix: F



  

x[0]

x[1]

x[2]

x[3]



  



  

1 1 1 1

1 e−j(2π/4) e−j(4π/4) e−j(6π/4)

1 e−j(4π/4) e−j(8π/4) e−j(12π/4)

1 e−j(6π/4) e−j(12π/4) e−j(18π/4)



  

︸︷︷︸

DFT matrix: F



  

−1 1



  



  

3 − j2 −3

3 + j2



  

The above values for the DFT coefficients are the same as the ones obtained in

Example 12.1.

Example 12.5

Calculate the inverse DFT of X [r ] considered in Example 12.2.

Solution

Arranging the values of the DFT coefficients in the DFT vector X , we obtain

X = [5 3 − j2 −3 3 + j2]T.

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537 12 Discrete Fourier transform

Using Eq. (12.20), the DFT vector X is given by



  

x[0]

x[1]

x[2]

x[3]



  

= 1



  

1 1 1 1

1 e j(2π/N ) e j(4π/N ) e j(6π/N )

1 e j(4π/N ) e j(8π/N ) e j(12π/N )

1 e j(6π/N ) e j(12π/N ) e j(18π/N )



  



   

X [0]

X [1]

X [2]

X [3]



   

= 1



  

1 1 1 1

1 e j(2π/4) e j(4π/4) e j(6π/4)

1 e j(4π/4) e j(8π/4) e j(12π/4)

1 e j(6π/4) e j(12π/4) e j(18π/4)



  



  

3 − j2 −3

3 + j2



  

= 1



  

−4 4



  



  

−1 1



  

The above values for the DT sequence x[k] are the same as the ones obtained

in Example 12.2.

12.2.2 DFT basis functions

The matrix-vector representation of the DFT derived in Section 12.2.1 can

be used to determine the set of basis functions for the DFT representation.

Expressing Eq. (12.20) in the following format: 

     

x[0]

x[1]

x[2] ...

x[N − 1]



     

= 1

N X [0]



     

1 ...



     

+ 1

N X [1]



     

e j(2π/N )

e j(4π/N )

...

e j(2(N−1)π/N )



     

+ 1

N X [2]



     

e j(4π/N )

e j(8π/N )

...

e j(4(N−1)π/N )



     

+ · · · 1

N X [N − 1]



     

e j(2(N−1)π/N )

e j(4(N−1)π/N )

...

e j(2(N−1)(N−1)π/N )



     

, (12.21)

it is clear that the basis functions for the N -point DFT are given by the following

set of N vectors:

Fr = 1

[

1 exp

( j2πr

)

exp

( j4πr

)

· · · exp (

j2(N − 1)πr N

)]T

for 0 ≤ r ≤ (N−1). Equation (12.21) illustrates that the DFT represents a DT

sequence as a linear combination of complex exponentials, which are weighted

by the corresponding DFT coefficients. Such a representation is useful for the

analysis of LTID systems.

As an example, Fig. 12.4 plots the real and imaginary components of the basis

vectors for the eight-point DFT of length N = 8. From Fig. 12.4(a), we observe

that the real components of the basis vectors correspond to a cosine function

sampled at different sampling rates. Similarly, the imaginary components of

the basis vectors correspond to a sine function sampled at different sampling

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538 Part III Discrete-time signals and systems

−0.2

0.2

0 1 2 3 4 5 6 7

0.2

0 1 2 3 4 5 6 7

0.2

0 1 2 3 4 5 6 7

0.2

0 1 2 3 4 5 6 7

0.2

0 1 2 3 4 5 6 7

0.2

0 1 2 3 4 5 6 7

0.2

0 1 2 3 4 5 6 7

0.2

0 1 2 3 4 5 6 7

0.2

0 1 2 3 4 5 6 7

0.2

0 1 2 3 4 5 6 7

0.2

0 1 2 3 4 5 6 7

0.2

0 1 2 3 4 5 6 7

0.2

0 1 2 3 4 5 6 7

0.2

0 1 2 3 4 5 6 7

0.2

4 54 5

1 2 3 51 2 3 5

(a)(a) (b)

Fig. 12.4. Basis vectors for an

eight-point DFT. (a) Real

components; (b) imaginary

components.

rates. This should not be surprising, since Euler’s identity expands a complex

exponential as a complex sum of cosine and sine terms.

We now proceed with the estimation of the spectral content of both DT and

CT signals using the DFT.

12.3 Spectrum analysis using the DFT

In this section, we illustrate how the DFT can be used to estimate the spectral

content of the CT and DT signals. Examples 12.6–12.8 deal with the CT signals,

while Examples 12.9 and 12.10 deal with the DT sequences.

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539 12 Discrete Fourier transform

Example 12.6

Using the DFT, estimate the frequency characteristics of the decaying expo-

nential signal g(t) = exp(−0.5t)u(t). Plot the magnitude and phase spectra.

Solution

Following the procedure outlined in Section 12.1, the three steps involved in

computing the CTFT are listed below.

Step 1: Impulse-train sampling Based on Table 5.1, the CTFT of the decay- ing exponential is given by

g(t) = e−0.5t u(t) CTFT←−−→ G(ω) = 1

0.5 + jω .

This CTFT pair implies that the bandwidth of g(t) is infinite. Ideally speak-

ing, the sampling theorem can never be satisfied for the decaying exponential

signal. However, we exploit the fact that the magnitude |G(ω)| of the CTFT

decreases monotonically with higher frequencies and we neglect any frequency

components at which the magnitude falls below a certain threshold η. Selecting

the value of η = 0.01 × |G(ω)|max, the threshold frequency B is given by ∣ ∣ ∣

0.5 + j2π B

∣ ∣ ∣ ≤ 0.01 × |G(ω)|max.

Since the maximum value of the magnitude |G(ω)| is 2 at ω = 0, the above

expression reduces to √

0.25 + (2π B)2 ≥ 50,

or B ≥ 7.95 Hz. The Nyquist sampling rate f1 is therefore given by

f1 ≥ 2 × 7.95 = 15.90 samples/s.

Selecting a sampling rate of f1 = 20 samples/s, or a sampling interval T1 =

1/20 = 0.05 s, the DT approximation of the decaying exponential is given by

g[k] = g(kT1) = e −0.5kT1 u[k] = e−0.025ku[k].

Since there is a discontinuity in the CT signal g(t) at t = 0 with g(0−) = 0 and

g(0+) = 1, the value of g[k] at k = 0 is set to g[0] = 0.5. based on Eq. (1.1).

Step 2: Time-limitation To truncate the length of g[k], we apply a rectangular window of length N = 201 samples. The truncated sequence is given by

gw [k] = e −0.025k(u[k] − u[k − 201]) =

{

e−0.025k 0 ≤ k ≤ 200

0 elsewhere.

The subscript w in gw [k] denotes the truncated version of g[k] obtained by

multiplying by the window function w[k]. Note that the truncated sequence

gw [k] is a fairly good approximation of g[k], as the peak magnitude of the

truncated samples is given by 0.0066 and occurs at k = 201. This is only 0.66%

of the peak value of the complex exponential g[k].

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Step 3: DFT computation The DFT of the truncated DT sequence gw [k] can now be computed directly from Eq. (12.16). M A T L A B provides a built-in

function fft, which has the calling syntax of

>> G = fft(g);

where g is the signal vector containing the values of the DT sequence gw [k]

and G is the computed DFT. Both g and G have a length of N , implying that

an N -point DFT is being taken. The built-in function fft computes the DFT

within the frequency range 0 ≤ r ≤ (N−1). Since the DFT is periodic, we can

obtain the DFT within the frequency range −(N − 1)/2 ≤ r ≤ (N − 1)/2 by

a circular shift of the DFT coefficients. In M A T L A B , this is accomplished by

the fftshift function.

Having computed the DFT, we use Eq. (12.12) to estimate the CTFT of the

original CT decaying exponential signal g(t). The M A T L A B code for comput-

ing the CTFT is as follows:

>> f1 = 20; % set sampling rate

>> t1 = 1/f1; % set sampling interval

>> N = 201; k = 0:N-1; % set length of DT sequence to

% N = 201

>> g = exp(-0.025*k); % compute the DT sequence

>> g(1) = 0.5; % initialize the first sample

>> G = fft(g); % determine the 201-point DFT

>> G = fftshift(G); % shift the DFT coefficients

>> G = t1*G; % scale DFT such that

% DFT = CTFT

>> dw = 2*pi*f1/N; % CTFT frequency resolution

>> w = -pi*f1:dw:pi*f1-dw; % compute CTFT frequencies

>> stem(w,abs(G)); % plot CTFT magnitude spectrum

>> stem(w,angle(G)); % plot CTFT phase spectrum

The resulting plots are shown in Fig. 12.5, where we have limited the frequency

axis to the range −5π ≤ ω ≤ 5π . The magnitude and phase spectra plotted

in Fig. 12.5 are fairly good estimates of the frequency characteristics of the

decaying exponential signal listed in Table 5.3.

In Example 12.6, we used the CTFT G(ω) to determine the appropriate sampling

rate. In most practical situations, however, the CTFTs are not known and one

−5p −4p −3p −2p −p 0 p 2p 3p 4p 5p 0

0.5

1.5

−5p −4p −3p −2p −p 0 p 2p 3p 4p 5p

−0.5p

−0.25p

0.25p

0.5p

(a) (b)

Fig. 12.5. Spectral estimation of

decaying exponential signal

g(t ) = exp(−0.5t )u(t ) using

the DFT in Example 12.6.

(a) Estimated magnitude

spectrum; (b) estimated phase

spectrum.

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541 12 Discrete Fourier transform

is forced to make an intelligent estimate of the bandwidth of the signal. If

the frequency and time characteristics of the signal are not known, a high

sampling rate and a large time window are arbitrarily chosen. In such cases, it

is advised that a number of sampling rates and lengths be tried before finalizing

the estimates.

Example 12.7

Using the DFT, estimate the frequency characteristics of the CT signal h(t) = 2 exp(j18π t) + exp(−j8π t).

Solution

Following the procedure outlined in Section 12.1, the three steps involved in

computing the CTFT are as follows.

Step 1: Impulse-train sampling The CT signal h(t) consists of two com- plex exponentials with fundamental frequencies of 9 Hz and 4 Hz. The Nyquist

sampling rate f1 is therefore given by

f1 ≥ 2 × 9 = 18 samples/s.

We select a sampling rate of f1 = 32 samples/s, or a sampling interval T1 = 1/32 s. The DT approximation of h(t) is given by

h[k] = h(kT1) = 2e j18πk/32 + e−j8πk/32.

Step 2: Time-limitation The DT sequence h[k] is a periodic signal with fun- damental period K0 = 32. For periodic signals, it is sufficient to select the length of the rectangular window equal to the fundamental period. Therefore,

N is set to 32.

Step 3: DFT computation The M A T L A B code for computing the DFT of the truncated DT sequence is as follows.

>> f1 = 32; % set sampling rate

>> t1 = 1/f1; % set sampling interval

>> N = 32; k = 0:N-1; % set length of DT sequence

>> h = 2*exp(j*18*pi*k/32) + exp(-j*8*pi*k/32);

% compute the DT sequence

>> H = fft(h); % determine the 32-point DFT

>> H = fftshift(H); % shift the DFT coefficients

>> H = t1*H; % scale DFT such that

% DFT = CTFT

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542 Part III Discrete-time signals and systems

−30p −20p −10p 0 10p 20p 30p 0

0.5

1.5

−30p −20p −10p 0 10p 20p 30p

−0.5p

0.5p

(a) (b)

Fig. 12.6. Spectral estimation of

decaying exponential signal h(t )

= 2 exp(j18π t ) + exp(−j8π t ) using the DFT in Example 12.7.

(a) Estimated magnitude

spectrum; (b) estimated

phase spectrum.

>> dw = 2*pi*f1/N; % CTFT frequency resolution

>> w = -pi*f1:dw:pi*f1-dw; % compute CTFT frequencies

>> stem(w,abs(H)); % plot CTFT magnitude spectrum

>> stem(w,angle(H)); % plot CTFT phase spectrum

The resulting plots are shown in Fig. 12.6, and they have a frequency resolution

of �ω = 2π . We know that the CTFT for h(t) is given by

2e j18π t + e−j8π t CTFT←−−→ 2δ(ω − 18π ) + δ(ω + 8π ).

We observe that the two impulses at ω = −8π and 18π radians/s are accurately

estimated in the magnitude spectrum plotted in Fig. 12.6(a). Also, the relative

amplitude of the two impulses corresponds correctly to the area enclosed by

these impulses in the CTFT for h(t).

The phase spectrum plotted in Fig. 12.6(b) is unreliable except for the two

frequencies ω = −8π and 18π radians/s. At all other frequencies, the magni-

tude |H (ω)| is zero, therefore the phase <H (ω) carries no information. This

is because the phase is computed as the inverse tangent of the ratio between

the imaginary and real components of H (ω). When |H (ω)| is close to zero,

the argument of the inverse tangent is given by ε1/ε2, with ε1and ε2 approach-

ing zero. In such cases, incorrect results are obtained for the phase. The phase

<H (ω) is therefore ignored when |H (ω)| is close to zero.

Example 12.8

Using the DFT, estimate the frequency characteristics of the CT signal x(t) =

2 exp(j19π t).

Solution

The three steps involved in computing the CTFT are as follows.

Step 1: Impulse-train sampling The CT signal x(t) constitutes a complex exponential with fundamental frequency 9.5 Hz. The Nyquist sampling rate f1

is therefore given by

f1 ≥ 2 × 9.5 = 19 samples/s.

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543 12 Discrete Fourier transform

As in Example 12.7, we select a sampling rate of f1 = 32 samples/s, or a sam- pling interval T1 = 1/32 s. The DT approximation of h(t) is given by

x[k] = x(kT1) = 2e j19πk/32.

Step 2: Time-limitation As in Example 12.7, we keep the length N of the rectangular window equal to 32.

Step 3: DFT computation The M A T L A B code for computing the DFT of the truncated DT sequence is as follows:

>> f1 = 32; % set sampling rate

>> t1 = 1/f1; % set sampling interval

>> N = 32; k = 0:N-1; % set length of DT sequence

% to N = 32

>> x = 2*exp(j*19*pi*k/32); % compute the DT sequence

>> X = fft(x); % determine the 32-point DFT

>> X = fftshift(X); % shift the DFT coefficients

>> X = t1*X; % scale DFT such that

% DFT = CTFT

>> dw = 2*pi*f1/N; % CTFT frequency resolution

>> w = -pi*f1:dw:pi*f1-dw; % compute CTFT frequencies

>> stem(w,abs(X)); % plot CTFT magnitude spectrum

The resulting magnitude spectrum is shown in Fig. 12.7(a), which has a fre-

quency resolution of �ω = 2π radians/s. Comparing with the CTFT for x(t), which is given by

2e j19π t CTFT

←−−→ 2δ(ω − 19π ),

we observe that Fig. 12.7(a) provides us with an erroneous result. This error

is attributed to the poor resolution �ω chosen to frequency-sample the CTFT.

Since �ω = 2π , the frequency component of 19π present in x(t) cannot be

displayed accurately at the selected resolution. In such cases, the strength of

the frequency component of 19π radians/s leaks into the adjacent frequencies,

leading to non-zero values at these frequencies. This phenomenon is referred

to as the leakage or picket fence effect.

Figure 12.7(b) plots the magnitude spectrum when the number N of sam-

ples in the discretized sequence is increased to 64. Since fft uses the same

number M of samples to discretize the CTFT, the resolution �ω = 2πT1/M =

π radians/s. The M A T L A B code for estimating the CTFT is as follows:

>> f1 = 32; t1 = 1/f1; % set sampling rate and interval

>> N = 64; k = 0:N-1; % set sequence length to N = 64

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−30p −20p −10p 0 10p 20p 30p 0

0.5

1.5

−30p −20p −10p 0 10p 20p 30p 0

0.5

1.5

(a) (b)

Fig. 12.7. Spectral estimation of

complex exponential signal

x (t ) = 2 exp( j19π t ) using the DFT in Example 12.8.

(a) Estimated magnitude

spectrum, with a 32-point DFT.

(b) Same as part (a) except that

a 64-point DFT is computed.

>> x = 2*exp(j*19*pi*k/32); % compute the DT sequence

>> X = fft(x); % determine the 64-point DFT

>> X = fftshift(X); % shift the DFT coefficients

>> X = 0.5*t1*X; % scale DFT so DFT = CTFT

>> w = 2*pi*f1/N; % CTFT frequency resolution

>> w = -pi*f1:dw:pi*f1-dw; % compute CTFT frequencies

>> stem(w,abs(X)); % plot CTFT magnitude spectrum

In the above code, we have highlighted the instructions that have been changed

from the original version. In addition to setting the length N to 64 in the above

code, we also note that the magnitude of the CTFT X is now being scaled by

a factor of 0.5 × T1. The additional factor of 0.5 is introduced because we are now computing the DFT over two consecutive periods of the periodic sequence

x[k]. Doubling the time duration doubles the values of the DFT coefficients,

and therefore a factor of 0.5 is introduced to compensate for the increase.

Figure 12.7(b), obtained using a 64-point DFT, is a better estimate for the

magnitude spectrum of x(t) than Fig. 12.7(a), obtained using a 32-point DFT.

The DFT can also be used to estimate the DTFT of DT sequences. Examples

12.9 and 12.10 compute the DTFT of two aperiodic sequences.

Example 12.9

Using the DFT, calculate the DTFT of the DT decaying exponential sequence

x[k] = 0.6k u[k].

Solution

Estimating the DTFT involves only Steps 2 and 3 outlined in Section 12.1.

Step 2: Time-limitation Applying a rectangular window of length N = 10, the truncated sequence is given by

xw[k] = {

0.6k 0 ≤ k ≤ 9

0 elsewhere.

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545 12 Discrete Fourier transform

Table 12.1. Comparison between the DFT and DTFT coefficients in Example 12.9

DTFT frequency,

DFT index, r Ωr = 2πr/N DFT coefficients, X [r ] DTFT coefficients, X (Ω)

−5 −π 0.6212 0.6250 −4 −0.8π 0.6334 + j0.1504 0.6373 + j0.1513 −3 −0.6π 0.6807 + j0.3277 0.6849 + j0.3297 −2 −0.4π 0.8185 + 0.5734 0.8235 + j0.5769 −1 −0.2π 1.3142 + j0.9007 1.3222 + j0.9062

0 0 2.4848 2.5000

1 0.2π 1.3142 − j0.9007 1.3222 − j0.9062 2 0.4π 0.8185 − j0.5734 0.8235 − j0.5769 3 0.6π 0.6807 − j0.3277 0.6849 − j0.3297 4 0.8π 0.6334 − j0.1504 0.6373 − j0.1513

Step 3: DFT computation The M A T L A B code for computing the DFT is as follows:

>> N = 10; k = 0:N-1; % set sequence length

% to N = 10

>> x = 0.6.ˆk; % compute the DT sequence

>> X = fft(x); % calculate the 10-point DFT

>> X = fftshift(X); % shift the DFT coefficients

>> w = -pi:2*pi/N:pi-2*pi/N; % compute DTFT frequencies

Table 12.1 compares the computed DFT coefficients with the corresponding

DTFT coefficients obtained from the following DTFT pair:

0.6ku[k] DTFT

←−−→ 1

1 − 0.6e−jΩ .

We observe that the values of the DFT coefficients are fairly close to the DTFT

values.

Example 12.10

Calculate the DTFT of the aperiodic sequence x[k] = [2, 1, 0, 1] for 0 ≤ k ≤

Solution

Using Eq. (12.6), the DFT coefficients are given by

X [r ] = [4, 2, 0, 2] for 0 ≤ r ≤ 3.

Mapping in the DTFT domain, the corresponding DTFT coefficients are given

X (Ωr ) = [4, 2, 0, 2] for Ωr = [0, 0.5π, π, 1.5π ] radians/s.

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−p −0.75p −0.5p −0.25p 0 0.25p 0.5p 0.75p 0

−p −0.75p −0.5p −0.25p 0 0.25p 0.5p 0.75p p −0.5p

−0.25p

0.25p

0.5p

(a) (b)

Fig. 12.8. Spectral estimation of

DT sequences using the DFT in

Example 12.10. (a) Estimated

magnitude spectrum;

(b) estimated phase spectrum.

The dashed lines show the

continuous spectrum obtained

from the DTFT.

If instead the DTFT is to be plotted within the range −π ≤ Ω ≤ π , then the DTFT coefficients can be rearranged as follows:

X (Ωr ) = [4, 2, 0, 2] for Ωr = [−π, −0.5π, 0, 0.5π ] radians/s.

The magnitude and phase spectra obtained from the DTFT coefficients are

sketched using stem plots in Figs. 12.8(a) and (b). For comparison, we use Eq.

(11.28b) to derive the DTFT for x[k]. The DTFT is given by

X (Ω) =

3∑

k=0

x[k]e−jΩk = 2 + e−jΩ + e−j3Ω.

The actual magnitude and phase spectra based on the above DTFT expression

are plotted in Figs. 12.8(a) and (b) respectively (see dashed lines). Although

the DFT coefficients provide exact values of the DTFT at the discrete fre-

quencies Ωr = [0, 0.5π , π, 1.5π ] radians/s, no information is available on

the characteristics of the magnitude and phase spectra for the intermediate

frequencies. This is a consequence of the low resolution used by the DFT

to discretize the DTFT frequency Ω. Section 12.3.1 introduces the concept

of zero padding, which allows us to improve the resolution used by the

DFT.

12.3.1 Zero padding

To improve the resolution of the frequency axis Ω in the DFT domain, a com-

monly used approach is to append the DT sequences with additional zero-valued

samples. This process is called zero padding, and for an aperiodic sequence x[k]

of length N is defined as follows:

xzp[k] =

{

x[k] 0 ≤ k ≤ N − 1

0 N ≤ k ≤ M − 1.

The zero-padded sequence xzp[k] has an increased length of M . The frequency

resolution �Ω of the zero-padded sequence is improved from 2π/N to 2π/M .

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547 12 Discrete Fourier transform

−p −0.75p −0.5p −0.25p 0 0.25p 0.5p 0.75p 0

−p −0.75p −0.5p −0.25p 0 0.25p 0.5p 0.75p

−0.5p

−0.25p

0.25p

0.5p

p p

(a) (b)

Fig. 12.9. Spectral estimation of

zero-padded DT sequences

using the DFT in Example 12.11.

(a) Estimated magnitude

spectrum; (b) estimated phase

spectrum.

Example 12.11 illustrates the improvement in the DTFT achieved with the

zero-padding approach.

Example 12.11

Compute the DTFT of the aperiodic sequence x[k] = [2, 1, 0, 1] for 0 ≤ k ≤ 3 by padding 60 zero-valued samples at the end of the sequence.

Solution

The M A T L A B code for computing the DTFT of the zero-padded sequence is

as follows:

>> N = 64; k = 0:N-1; % set sequence length

% to N = 64

>> x = [2 1 0 1 zeros(1,60)]; % zero-padded sequence

>> X = fft(x); % determine the 64-point DFT

>> X = fftshift(X); % shift the DFT coefficients

>> w = -pi:2*pi/N:pi-2*pi/N; % compute DTFT frequencies

>> stem(w,abs(X)); % plot magnitude spectrum

>> stem(w,angle(X)); % plot the phase spectrum

The magnitude and phase spectra of the zero-padded sequence are plotted in

Figs. 12.9(a) and (b), respectively. Compared with Fig. 12.8, we observe that

the estimated spectra in Fig. 12.9 provide an improved resolution and better

estimates for the frequency characteristics of the DT sequence.

12.4 Properties of the DFT

In this section, we present the properties of the M-point DFT. The length of

the DT sequence is assumed to be N ≤ M . For N < M , the DT sequence is

zero-padded with M − N zero-valued samples. The DFT properties presented

below are similar to the corresponding properties for the DTFT discussed in

Chapter 11.

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12.4.1 Periodicity

The M-point DFT of an aperiodic DT sequence with length N (M ≥ N ) is periodic with period M . In other words,

X [r ] = X [r + M], (12.22)

for 0 ≤ r ≤ M − 1.

12.4.2 Orthogonality

The column vectors Fr of the DFT matrix F , defined in Section 12.2.2, form

the basis vectors of the DFT. These vectors are orthogonal to each other and,

for the M-point DFT, satisfy the following:

FTp F ∗ q =

M∑

m=1

Fp(m, 1)[Fq (m, 1)] ∗ =

{

1/M for p = q

0 for p = q,

where the matrix FTp is the transpose of Fp and the matrix F ∗ q is the complex

conjugate of Fq .

12.4.3 Linearity

If x1[k] and x2[k] are two DT sequences with the following M-point DFT pairs:

x1[k] DFT

←−−→ X1[r ] and x2[k] DFT

←−−→ X2[r ],

then the linearity property states that

a1x1[k] + a2x2[k] DFT

←−−→ a1 X1[r ] + a2 X2[r ], (12.23)

for any arbitrary constants a1 and a2, which may be complex-valued.

12.4.4 Hermitian symmetry

The M-point DFT X [r ] of a real-valued aperiodic sequence x[k] is conjugate–

symmetric about r = M/2. Mathematically, the Hermitian symmetry implies

that

X [r ] = X∗[M − r ], (12.24)

where X∗[r ] denotes the complex conjugate of X [r ].

In terms of the magnitude and phase spectra of the DFT X [r ], the Hermitian

symmetry property can be expressed as follows:

|X [M − r ]| = |X [r ]| and <X [M − r ] = −<X [r ], (12.25)

implying that the magnitude spectrum is even and that the phase spectrum is

odd.

The validity of the Hermitian symmetry can be observed in the DFT plotted

for various aperiodic sequences in Examples 12.2–12.11.

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549 12 Discrete Fourier transform

12.4.5 Time shifting

If x[k] DFT

←−−→ X [r ], then

x[k − k0] DFT

←−−→ e−j2πk0r/M X [r ] (12.26)

for an M-point DFT and any arbitrary integer k0.

12.4.6 Circular convolution

If x1[k] and x2[k] are two DT sequences with the following M-point DFT pairs:

x1[k] DFT

←−−→ X1[r ] and x2[k] DFT

←−−→ X2[r ],

then the circular convolution property states that

x1[k] ⊗ x2[k] DFT

←−−→ X1[r ]X2[r ] (12.27)

and

x1[k]x2[k] DFT

←−−→ 1

M [X1[r ] ⊗ X2[r ]], (12.28)

where ⊗ denotes the circular convolution operation. Note that the two sequences

must have the same length in order to compute the circular convolution.

Example 12.12

In Example 10.11, we calculated the circular convolution y[k] of the aperiodic

sequences x[k] = [0, 1, 2, 3] and h[k] = [5, 5, 0, 0] defined over 0 ≤ k ≤ 3.

Recalculate the result of the circular convolution using the DFT convolution

property.

Solution

The four-point DFTs of the aperiodic sequences x[k] and h[k] are given by

X [r ] = [6, −2 + j2, −2, −2 − j2]

and

H [r ] = [10, 5 − j5, 0, 5 + j5]

for 0 ≤ r ≤ 3. Using Eq. (12.27), the four-point DFT of the circular convolu-

tion between x[k] and h[k] is given by

x1[k] ⊗ x2[k] DFT

←−−→ [60, j20, 0 − j20].

Taking the inverse DFT, we obtain

x1[k] ⊗ x2[k] = [15, 5, 15, 25],

which is identical to the answer obtained in Example 10.11.

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550 Part III Discrete-time signals and systems

12.4.7 Parseval’s theorem

If x[k] DFT

←−−→ X [r ], then the energy of the aperiodic sequence x[k] of length

N can be expressed in terms of its M-point DFT as follows:

Ex = N−1∑

k=0

|x[k]|2 = 1

M−1∑

k=0

|X [r ]|2. (12.29)

Parseval’s theorem shows that the DFT preserves the energy of the signal within

a scale factor of M .

12.5 Convolution using the DFT

In Section 10.6.1, we showed that the linear convolution x1[k] ∗ x2[k] between

two time-limited DT sequences x1[k] and x2[k] of lengths K1 and K2, respec-

tively, can be expressed in terms of the circular convolution x1[k] ⊗x2[k]. The

procedure requires zero padding both x1[k] and x2[k] to have individual lengths

of K ≥ (K1 + K2 – 1). It was shown that the result of the circular convolution

of the zero-padded sequences is the same as that of the linear convolution.

Since computationally efficient algorithms are available for computing the

DFT of a finite-duration sequence, the circular convolution property can be

exploited to implement the linear convolution of the two sequences x1[k] and

x2[k] using the following procedure.

(1) Compute the K -point DFTs X1[r ] and X2[r ] of the two time-limited

sequences x1[k] and x2[k]. The value of K is lower bounded by (K1 + K2 – 1), i.e. K ≥ (K1 + K2 – 1).

(2) Compute the product X3[r ] = X1[r ]X2[r ] for 0 ≤ r ≤ K − 1.

(3) Compute the sequence x3[k] as the inverse DFT of X3[r ]. The resulting

sequence x3[k] is the result of the linear convolution between x1[k] and

x2[k].

The above approach is explained in Example 12.13.

Example 12.13

Example 10.13 computed the linear convolution of the following DT sequences:

x[k] =







2 k = 0

−1 |k| = 1

0 otherwise

and h[k] =



  

  

2 k = 0

3 |k| = 1

−1 |k| = 2

0 otherwise,

using the circular convolution method outlined in Algorithm 10.4 in

Section 10.6.1. Repeat Example 10.13 using the DFT-based approach described

above.

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551 12 Discrete Fourier transform

Table 12.2. Values of X ′[r ], H ′[r ] and Y [r ] for 0 ≤ r ≤ 6 in Example 12.13

r X ′[r ] H ′[r ] Y [r ]

0 0 6 0

1 0.470 − j0.589 −1.377 − j6.031 −4.199 − j2.024

2 −0.544 − j2.384 −2.223 + j1.070 3.760 + j4.178

3 −3.425 − j1.650 −2.901 − j3.638 3.933 + j17.247

4 −3.425 + j1.650 −2.901 + j3.638 3.933 − j17.247

5 −0.544 + j2.384 −2.223 − j1.070 3.760 − j4.178

6 0.470 + j0.589 −1.377 + j6.031 −4.199 + j2.024

Solution

Step 1 Since the sequences x[k] and h[k] have lengths Kx = 5 and K y = 3, the value of K ≥ (5 + 3 − 1) = 7. We set K = 7 in this example:

padding (K − Kx ) = 4 additional zeros to x[k], we obtain x ′[k] = [−1, 2, −1, 0, 0, 0, 0];

padding (K − Kh) = 2 additional zeros to h[k], we obtain h′[k] = [−1, 3, 2, 3, −1, 0, 0].

The DFTs of x ′[k] are shown in the second column of Table 12.2, where the

values for X ′[r ] have been rounded off to three decimal places. Similarly, the

DFTs of h′[k] are shown in the third column of Table 12.2.

Step 2 The value of Y [r ] = X ′[r ]H ′[r ], for 0 ≤ r ≤ 6, are shown in the fourth column of Table 12.2.

Step 3 Taking the inverse DFT of Y [r ] yields

y[k] = [0.998 −5 5.001 −1.999 5 −5.002 1.001].

Except for approximation errors caused by the numerical precision of the com-

puter, the above results are the same as those obtained from the direct compu-

tation of the linear convolution included in Example 10.13.

12.5.1 Computational complexity

We now compare the computational complexity of the time-domain and DFT-

based implementations of the linear convolution between the time-limited

sequences x1[k] and x2[k] with lengths K1 and K2, respectively. For simplicity,

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552 Part III Discrete-time signals and systems

we assume that x1[k] and x2[k] are real-valued sequences with lengths K1 and

K2, respectively.

Time-domain approach This is based on the direct computation of the con- volution sum

y[k] = x1[k] ∗ x1[k] = ∞∑

m=−∞

x1[m]x2[k − m],

which requires roughly K1 × K2 multiplications and K1 × K2 additions. The

total number of floating point operations (flops) required with the time-domain

approach is therefore given by 2K1 K2.

DFT-based approach Step 1 of the DFT-based approach computes two K = (K1 + K2 − 1)-point DFTs of the DT sequences x1[k] and x2[k]. In Section

12.6, we show that the total number of flops required to implement a K -point

DFT using fast Fourier transform (FFT) techniques is approximately 5K log2 K .

Therefore, Step 1 of the DFT-based approach requires a total of 10K log2 K

flops.

Step 2 multiplies DFTs for x1[k] and x2[k]. Each DFT has a length of

K = K1 + K2 − 1 points; therefore, a total of K complex multiplications and

K − 1 ≈ K complex additions are required. The total number of computations required in Step 2 is therefore given by 8K or 8(K1 + K2 – 1) flops.

Step 3 computes one inverse DFT based on the FFT implementation requiring

5K log2 K flops.

The total number of flops required with the DFT-based approach is therefore

given by

15K log2 K + 6K ≈ 15K log2 K flops,

where K = K1 + K2 − 1. Assuming K1 = K2, the DFT-based approach pro- vides a computational saving of O((log2 K )/K ) in comparison with the direct

computation of the convolution sum in the time domain. Table 12.3 compares

the computational complexity of the two approaches for a few selected values

of K1 and K2. The length K of the DFT should be equal to or greater than

(K1 + K2 − 1) depending on its value. Where (K1 + K2 − 1) is not a power of 2, we have rounded (K1 + K2 − 1) to the next higher integer that is a power of 2. In the second row, for example, K1 = 32 and K2 = 5, which implies that (K1 + K2 − 1) = 36. Since the radix-2 FFT algorithm, described in Section 12.6, can only be implemented for sequences with lengths that are powers of 2, K is set to

64. Based on the DFT-based approach, the number of flops required to compute

the convolution of the two sequences is given by (15 × 64 × log2 (64)) = 5760. In Table 12.3, we observe that for sequences with lengths greater than 1000 sam-

ples, the DFT-based approach provides significant savings over the direct com-

putation of the circular convolution in the time domain. If x1[k] and x2[k] are

real-valued sequences, significant further savings (about 50%) can be achieved

using the procedures mentioned in Problems 12.18 and 12.19.

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553 12 Discrete Fourier transform

Table 12.3. Comparison of the computational complexities of the

time-domain versus the DFT-based approaches used to compute

the linear convolution

Computational complexity, flops

Length K1 Length K2 Time domain DFT

of x1[k] of x2[k] (2K1 × K2 flops) (15K log2 K flops)

32 5 320 5760

32 32 2048 5760

1000 200 400 000 337 920

1000 1000 2 000 000 337 920

2000 2000 8 000 000 737 280

12.6 Fast Fourier transform

There are several well known techniques including the radix-2, radix-4, split

radix, Winograd, and prime factor algorithms that are used for computing the

DFT. These algorithms are referred to as the fast Fourier transform (FFT) algo-

rithms. In this section, we explain the radix-2 decimation-in-time FFT algo-

rithm.

To provide a general frame of reference, let us consider the computational

complexity of the direct implementation of the K -point DFT for a time-limited

complex-valued sequence x[k] with length K . Based on its definition,

X [r ] = K−1∑

k=0 x[k]e−j(2πkr/K ), (12.30)

K complex multiplications and K − 1 complex additions are required to com- pute a single DFT coefficient. Computation of all K DFT coefficients requires

approximately K 2 complex additions and K 2 complex multiplications, where

we have assumed K to be large such that K − 1 ≈ K . In terms of flops, each complex multiplication requires four scalar multi-

plications and two scalar additions, and each complex addition requires two

scalar additions. Computation of a single DFT coefficient, therefore, requires

8K flops. The total number of scalar operations for computing the complete

DFT is given by 8K 2 flops.

We now proceed with the radix-2 FFT decimation-in-time algorithm. The

radix-2 algorithm is based on the following principle.

Proposition 12.1 For even values of K, the K-point DFT of a complex-valued

sequence x[k] with length K can be computed from the DFT coefficients of two

subsequences: (i) x[2k], containing the even-numbered samples of x[k], and

(ii) x[2k + 1], containing the odd-numbered samples of x[k].

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554 Part III Discrete-time signals and systems

Proof

Expressing Eq. (12.30) in terms of even- and odd-numbered samples of x[k],

we obtain

X [r ] = K−1∑

k=0,2,4,... x[k]e−j(2πkr/K )

︸︷︷︸

Term I

+ K−1∑

k=1,3,5,... x[k]e−j(2πkr/K )

︸︷︷︸

Term II

, (12.31)

for 0 ≤ r ≤ (M − 1). Substituting k = 2m in Term I and k = 2m + 1 in Term II,

Eq. (12.31) can be expressed as follows:

X [r ] =

K/2−1∑

m=0,1,2,...

x[2m]e−j(2π (2m)r/K ) +

K/2−1∑

m=0,1,2,...

x[2m + 1]e−j(2π (2m+1)r/K )

X [r ] =

K/2−1∑

m=0,1,2,...

x[2m]e−j2πmr/(K/2)

+ e−j(2πr/K ) K/2−1∑

m=0,1,2,...

x[2m + 1]e−j2πmr/(K/2), (12.32)

where exp[−j2π (2m)r/K ] = exp[−j2πmr/(K/2)]. By expressing g[m] =

x[2m] and h[m] = x[2m + 1], we can rewrite Eq. (12.32) in terms of the DFTs

of g[m] and h[m]:

X [r ] =

K/2−1∑

m=0,1,2,...

g[m]e−j2πmr/(K/2)

︸︷︷︸

=G[r ]

+ e−j2πr/K K/2−1∑

m=0,1,2,...

h[m]e−j2πmr/(K/2)

︸︷︷︸

=H [r ]

(12.33)

X [r ] = G[r ] + W rK H [r ], (12.34)

where WK is defined as exp(−j2π/K ). In FFT literature, W r K is generally

referred to as the twiddle factor. Note that in Eqs. (12.33) and (12.34), G[r ]

represents the (K/2)-point DFT of g[k], the even-numbered samples of x[k].

Similarly, H [r ] represents the (K/2)-point DFT of h[k], the odd-numbered

samples of x[k]. Equation (12.34) thus proves Proposition 12.1.

Based on Eqs. (12.34), the procedure for determining the K -point DFT can be

summarized by the following steps.

(1) Determine the (K/2)-point DFT G[r ] for 0 ≤ r ≤ (K/2 − 1) of the even-

numbered samples of x[k].

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555 12 Discrete Fourier transform

W1K

W2K

W3K

W4K

W6K

W5K

W7K

W0K

x[0]

x[2]

x[4]

x[6]

x[1]

x[3]

x[5]

x[7]

G[0]

G[1]

G[3]

G[4]

H[1]

H[2]

H[3]

H[4]

X [0]

X [1]

X [2]

X [3]

X [4]

X[5]

X [6]

X [7]

K/2 point

DFT

K/2 point

DFT

Fig. 12.10. Flow graph of a

K -point DFT using two

(K/2)-point DFTs for K = 8.

(2) Determine the (K/2)-point DFT H [r ] for 0 ≤ r ≤ (K/2 − 1) of the odd-

numbered samples of x[k].

(3) The K -point DFT coefficients X [r ] for 0 ≤ r ≤ (K − 1) of x[k] are

obtained by combining the K/2 DFT coefficients G[r ] and H [r ] using

Eq. (12.34a). Although the index r varies from zero to K− 1, we only

compute G[r ] and H [r ] over the range 0 ≤ r ≤ (K/2 − 1). Any outside

value can be determined by exploiting the periodicity properties of G[r ]

and H [r ], which state that

G[r ] = G[r + K/2] and H [r ] = H [r + K/2].

Figure 12.10 illustrates the flow graph for the above procedure for K = 8-point

DFT. In comparison with the direct computation of DFT using Eq. (12.30),

Fig. 12.10 computes two (K/2)-point DFTs along with K complex addi-

tions and K complex multiplications. Consequently, (K/2)2 + K complex

additions and (K/2)2 + K complex multiplications are required with the

revised approach. For K > 2, it is easy to verify that (K/2)2 + K < K 2;

therefore, the revised approach provides considerable savings over the direct

approach.

Assuming that K is a power of 2, Proposition 12.1 can be applied on

Eq. (12.34) to compute the (K/2)-point DFTs G[r ] and H [r ] as follows:

G[r ] =

K/4−1∑

ℓ=0,1,2,...

g[2ℓ]e−j(2πℓr/(K/4))

︸︷︷︸

G ′[r ]

+ W rK/2

K/4−1∑

ℓ=0,1,2,...

g[2ℓ + 1]e−j(2πℓr/(K/4))

︸︷︷︸

G ′′[r ]

(12.35)

and

H [r ] =

K/4−1∑

ℓ=0,1,2,...

h[2ℓ]e−j(2πℓr/(K/4))

︸︷︷︸

H ′[r ]

+ W rK/2

K/4−1∑

ℓ=0,1,2,...

h[2ℓ + 1]e−j(2πℓr/(K/4))

︸︷︷︸

H ′′[r ]

(12.36)

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556 Part III Discrete-time signals and systems

x[0]

x[4]

x[2]

x[6]

G[0]

G[1]

G[2]

G[3]

0 2/KW

1 2/KW

2 2/KW

3 2/KW

K/4 point

DFT

K/4 point

DFT

K/4 point

DFT

K/4 point

DFT

G′[0]

G′′[0]

G′′[1]

G′[1]

[1]x

[5]x

[3]x

x[7]

H[0]

H [1]

H [2]

H [3]

0 K/2W

1 K/2W

2 K/2W

3 K/2W

H ′ [0]

H ′ [1]

H ′′ [0]

H ′′ [1]

(a) (b)

Fig. 12.11. Flow graphs of

(K /2)-point DFTs using

(K /4)-point DFTs. (a) G[r ];

(b) H[r ].

Equation (12.35) expresses the (K/2)-point DFT G[r ] in terms of two (K /4)-

point DFTs of the even- and odd-numbered samples of g[k]. Figure 12.11(a)

illustrates the flow graph for obtaining G[r ] using Eq. (12.35). Similarly, Eq.

(12.36) expresses the (K/2)-point DFT H [r ] in terms of two (K/4)-point DFTs

of the even- and odd-numbered samples of h[k], which can be implemented

using the flow graph shown in Fig. 12.11(b). If K is a power of 2, then the

above process can be continued until we are left with a 2-point DFT. For the

aforementioned example with K = 8, the (K/4)-point DFTs in Fig. 12.11 can be implemented directly using 2-point DFTs. Using the definition of the DFT,

the top left 2-point DFTs G ′[0] and G ′[1], for example, in Fig. 12.11(a) are

expressed as follows:

G ′[0] = x[0] e−j2πℓr/2 ∣ ∣ ℓ=0,r=0

+ x[4] e−j2πℓr/2 ∣ ∣ ℓ=1,r=0

= x[0] + x[4]

(12.37)

and

G ′[1] = x[0] e−j2πℓr/2 ∣ ∣ ℓ=0,r=1

+ x[4] e−j2πℓr/2 ∣ ∣ ℓ=1,r=1

= x[0] − x[4].

(12.38)

The flow graphs for Eqs. (12.37) and (12.38) are shown in Fig. 12.12(a). By

following this procedure, the flow diagrams for the remaining 2-point DFTs

required in Fig. 12.11 are derived and are shown in Figs. 12.12(b)–(d). Because

of their shape, the elementary flow graphs shown in Fig. 12.12 are generally

referred to as the butterfly structures.

Combining the individual flow graphs shown in Figs. 12.10, 12.11, and 12.12,

it is straightforward to derive the overall flow graph for the 8-point DFT, which

is shown in Fig. 12.13; in this flow diagram, we have further reduced the number

of operations for an 8-point DFT by noting that

W rK/2 = e −j2πr/(K/2) = e−j4πr/K = W 2rK ,

and by placing the common terms between the twiddle multipliers of the two

]1[x

]5[x

G]2[x

]6[x

G′[0]

G′[1]

G′′[0]

G′′[1]

H ′[0]

H ′[1]

H ′′[0]

H ′′[1]

W1 = −1

W02 = 1

]0[x

]4[x

]3[x

]7[x

(a)

(b)

(c)

(d)

Fig. 12.12. Flow graphs of

2-point DFTs required for

Fig. 12.11. (a) Top 2-point DFT

G ′[0] and G ′[1] for Fig. 12.11(a).

(b) Bottom 2-point DFT G ′′[0]

and G ′′[1] for Fig 12.11(a).

H ′[1] for Fig 12.11(b).

(d) Bottom 2-point DFT H ′′[0]

and H ′′[1] for Fig. 12.11(b). branches, which are originating from the same node, before the source node.

12.6.1 Computational complexity

To derive the computational complexity of the decimation-in-time algorithm,

we generalize the results obtained in Fig. 12.13, where K is set to 8. We

observe that Fig. 12.13 consists of log2 K = 3 stages and that each stage

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557 12 Discrete Fourier transform

stage 1 stage 3

x[0]

x[4]

x[2]

x[6]

x[1]

x[5]

x[3]

x[7]

]0[X

]1[X

]2[X

]3[X

]4[X

]5[X

]6[X

]7[X

4 KW

0 KW

2 KW

0 KW

2 KW

4 KW

0 KW

1 KW

2 KW

3 KW

4 KW

stage 2

Fig. 12.13. Decimation-in-time

implementation of an 8-point

DFT.

requires K = 8 complex multiplications and K = 8 complex additions. For example, stage 3 in Fig. 12.13 requires multiplications with twiddle factors

W 0K , W 1 K , W

2 K , W

3 K , and four W

4 K s. This is also obvious from Eq. (12.34),

where in order to calculate the K-point DFT from two (K/2)-point DFTs,

we need to perform K complex multiplications (with the twiddle factors)

and approximately K complex additions. Therefore, the decimation-in-time

FFT implementation for a K-point DFT requires a total of K log2 K complex

multiplications and K log2 K complex additions.

Further reduction in the complexity of the decimation-in-time FFT imple-

mentation is obtained by observing that

W K/2

K = e −jπ = −1. (12.39)

Note that multiplication by a factor of −1 can be performed by simply revers- ing the sign bit. It is observed from Fig. 12.13 that each stage contains four

such multiplications (by a factor of W 4K ). In general, for a K-point FFT, K/2

such multiplications exist in each stage. Ignoring these trivial multiplications,

the total number of complex multiplications for all K stages can be reduced to

0.5K log2 K complex multiplications. However, the number of complex addi-

tions stays the same at K log2 K . In other words, the complexity of a K-point

FFT can be expressed as 0.5K log2 K butterfly operations where a butterfly

operation includes one complex multiplication and two complex additions.

Note that each complex multiplication requires a total of six flops (for four

scalar multiplications and two scalar additions), and that each complex addi-

tion requires two flops (for two scalar additions). As each butterfly operation

requires a total of ten flops, the overall complexity of the decimation-in-time

FFT implementation is 5K log2 K flops.

Table 12.4 compares the number of computations for the direct implemen-

tation of Eq. (12.30) and the FFT implementation. As explained above, the

number of scalar operations for the direct implementation is assumed to be 8K 2

flops. whereas the number of scalar operations for the FFT implementation is

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Table 12.4. Complexity of DFT calculation (in flops) with FFT

and direct implementations

Number of flops Increase in

K FFT (5K log2 K ) direct (8K 2) Speed

32 800 8192 10.2

256 10 240 524 288 51.2

1024 51 200 8 388 608 163.8

8192 532 480 536 870 912 1 008.2

Table 12.5. Data reordering in radix-2 decimation-in-time FFT

implementation

Bit-reversed representation

Original order, x[k] Binary representation binary decimal

x[b2b1b0] x[b0b1b2] xre[k]

x[0] x[000] x[000] x[0]

x[1] x[001] x[100] x[4]

x[2] x[010] x[010] x[2]

x[3] x[011] x[110] x[6]

x[4] x[100] x[001] x[1]

x[5] x[101] x[101] x[5]

x[6] x[110] x[011] x[3]

x[7] x[111] x[111] x[7]

assumed to be 5K log2 K flops. For large values of K , say 8192, Table 12.4

illustrates a speed-up by up to a factor of 1000 with the FFT implementation. For

real-valued sequences, the number of flops can be further reduced by exploit-

ing the symmetry properties of the DFT. Further reduction in the complexity

of the decimation-in-time FFT implementation can be obtained by ignoring

multiplications by the twiddle factor W 0K as W 0 K = 1.

12.6.2 Reordering of the input sequence

In Fig. 12.13, we observe that the input sequence x[k] with length K has been

arranged in an order that is considerably different from the natural order of

occurrence. This arrangement is referred to as the bit-reversed order and is

obtained by expressing the index k in terms of log2 K bits and then reversing

the order of bits such that the most significant bit becomes the least significant

bit, and vice versa. For K = 8, the reordering of the input sequence is illustrated in Table 12.5.

The function myfft, available in the accompanying CD, implements the

radix-2 decimation-in-time FFT algorithm. Direct computation of the DFT

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559 12 Discrete Fourier transform

coefficients using Eq. (12.16) is also implemented and provided as a second

function, mydft. The reader should confirm that the two functions compute

the same result, with the exception that the implementation of myfft is com-

putationally efficient

As mentioned earlier, M A T L A B also provides a built-in function fft to

compute the DFT of a sequence. Depending on the length of the sequence, the

fft function chooses the most efficient algorithm to compute the DFT. For

example, when the length of the sequence is a power of 2, it uses the radix-2

algorithm. On the other hand, if the length is such that a font method is not

possible, it uses the direct method based on Eq. (12.15).

12.7 Summary

This chapter introduces the discrete Fourier transform (DFT) for time-limited

sequences as an extension of the DTFT where the DTFT frequency Ω is dis-

cretized to a finite set of values Ω = 2πr/M , for 0 ≤ r ≤ (M − 1). The M- point DFT pair for a causal, aperiodic sequence x[k] of length N is defined as

follows:

DFT analysis equation X [r ] =

N−1∑

k=0

x[k]e−j(2πkr/M) for 0 ≤ r ≤ M − 1;

DFT synthesis equation x[k] = 1

M−1∑

r=0

X [r ]e j(2πkr/M) for 0 ≤ k ≤ N − 1.

For M = N , Section 12.2 implements the synthesis and analysis equations of

the DFT in the matrix-vector format as follows:

DFT synthesis equation x = FX;

DFT analysis equation X = F−1x,

where F is defined as the DFT matrix given by

F =



     

1 1 1 · · · 1

1 e−j(2π/N ) e−j(4π/N ) · · · e−j(2(N−1)π/N )

1 e−j(4π/N ) e−j(8π/N ) · · · e−j(4(N−1)π/N )

... ...

... . . .

...

1 e−j(2(N−1)π/N ) e−j(4(N−1)π/N ) · · · e−j(2(N−1)(N−1)π/N )



     

The columns (or equivalently the rows) of the DFT matrix define the basis

functions for the DFT.

Section 12.3 used the M-point DFT X [r ] to estimate the CTFT spectrum

X (ω) of an aperiodic signal x(t) using the following relationship:

X (ωr ) ≈ MT1

N X2[r ],

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560 Part III Discrete-time signals and systems

where T1 is the sampling interval used to discretize x(t), ωr are the CTFT

frequencies that are given by 2πr/(MT1) for −0.5(M − 1) ≤ r ≤ 0.5(M − 1), and N is the number of samples obtained from the CT signal. Similarly, the

DFT X [r ] can be used to determine the DTFT X (Ω) of a time-limited sequence

x[k] of length Nas

X (Ωr ) = N

M X [r ]

at discrete frequencies Ωr = 2πr/M , for 0 ≤ r ≤ M − 1.

Section 12.4 covered the following properties of the DFT.

(1) The periodicity property states that the M-point DFT of a sequence is

periodic with period M .

(2) The orthogonality property states that the basis functions of the DFTs are

orthogonal to each other.

(3) The linearity property states that the overall DFT of a linear combination

of DT sequences is given by the same linear combination of the individual

DFTs.

(4) The Hermitian symmetry property states that the DFT of a real-valued

sequence is Hermitian. In other words, the real component of the DFT

of a real-valued sequence is even, while the imaginary component is

odd.

(5) The time-shifting property states that shifting a sequence in the time domain

towards the right-hand side by an integer constant m is equivalent to mul-

tiplying the DFT of the original sequence by the complex exponential

exp(−j2πm/M). Similarly, shifting towards the left-hand side by an inte-

ger m is equivalent to multiplying the DTFT of the original sequence by

the complex exponential exp(j2πm/M).

(6) The time-convolution property states that the periodic convolution of two

DT sequences is equivalent to the multiplication of the individual DFTs of

the two sequences in the frequency domain.

(7) Parseval’s theorem states that the energy of a DT sequence is preserved in

the DFT domain.

Section 12.5 used the convolution property to derive alternative procedures

for computing the convolution sum. Depending on the sequence lengths, these

procedures may provide considerable savings over the direct implementation

of the convolution sum.

Section 12.6 covers the decimation-in-time FFT implementation of the DFT.

In deriving the FFT algorithm, we assume that the length N of the sequence

equals the number M of samples in the DFT, i.e. N = M = K . We showed

that if K is a power of 2, then the FFT implementations have a computational

complexity of O(K log2 K ).

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561 12 Discrete Fourier transform

Problems

12.1 Calculate analytically the DFT of the following sequences, with length 0 ≤ k ≤ N − 1:

(i) x[k] =

{

1 k = 0, 3

0 k = 1, 2 with length N = 4;

(ii) x[k] =

{

1 k even

−1 k odd with length N = 8;

(iii) x[k] = 0.6k with length N = 8;

(iv) x[k] = u[k] − u[k − 8] with length N = 8;

(v) x[k] = cos(ω0k) with ω0 = 2πm/N , m ∈ Z .

12.2 Calculate the DFT of the time-limited sequences specified in Examples 12.1(i)–(iv) using the matrix-vector approach.

12.3 Determine the time-limited sequence, with length 0 ≤ k ≤ N − 1, cor- responding to the following DFTs X [r ], which are defined for the DFT

index 0 ≤ r ≤ N − 1:

(i) X [r ] = [1 + j4, −2 − j3, −2 + j3, 1 − j4] with N = 4;

(ii) X [r ] = [1, 0, 0, 1] with N = 4;

(iii) X [r ] = exp −j(2πk0r/N ), where k0 is a constant;

(iv) X [r ] =

{

0.5N r = k0, N − k0

0 elsewhere where k0 is a constant;

(v) X [r ]=e−jπr (m−1)/N sin (πrm/N )

sin(πr/N ) where m ∈ Z and 0 < m < N ;

(vi) X [r ] = ( r

)

for 0 ≤ r ≤ N − 1.

12.4 In Problem 11.1, we determined the DTFT representation for each of the following DT periodic sequences using the DTFS. Using M A T L A B ,

compute the DTFT representation based on the FFT algorithm. Plot the

frequency characteristics and compare the computed results with the ana-

lytical results derived in Chapter 11.

(i) x[k] = k, for 0 ≤ k ≤ 5 and x[k + 6] = x[k];

(ii) x[k] =



 

 

1 0 ≤ k ≤ 2

0.5 3 ≤ k ≤ 5

0 6 ≤ k ≤ 8

and x[k + 9] = x[k] ;

(iii) x[k] = 3 sin

( 2π

7 k +

)

;

(iv) x[k] = 2 exp

(

j 5π

3 k +

)

;

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562 Part III Discrete-time signals and systems

(v) x[k] = ∞∑

m=−∞

δ[k − 5m];

(vi) x[k] = cos(10πk/3) cos(2πk/5);

(vii) x[k] = |cos(2πk/3)|.

12.5 (a) Using the FFT algorithm in M A T L A B , determine the DTFT rep- resentation for the following sequences. Plot the magnitude and phase

spectra in each case.

(i) x[k] = 2;

(ii) x[k] =

{

3 − |k| |k| < 3

0 otherwise;

(iii) x[k] = k3−|k|;

(iv) x[k] = αkcos(ω0k)u[k], |α| < 1;

(v) x[k] = αksin(ω0k + φ)u[k], |α| < 1;

(vi) x[k] = sin(πk/5) sin(πk/7)

π2k2 ;

(vii) x[k] =

∞∑

m=−∞

δ[k − 5m − 3];

(viii) x[k] =

{

3 − |k| |k| < 3

0 |k| = 3 and x[k + 7] = x[k];

(ix) x[k] = ej(0.2πk+45 ◦);

(x) x[k] = k3−ku[k] + ej(0.2πk+45 ◦).

(b) Compare the obtained results with the analytical results derived in

Problem 11.4(a).

12.6 Using the FFT algorithm in M A T L A B , determine the CTFT represen- tation for each of the following CT functions. Plot the frequency char-

acteristics and compare the results with the analytical results presented

in Table 5.1.

(i) x(t) = e−2t u(t);

(ii) x(t) = e−4|t |;

(iii) x(t) = t4e−4t u(t);

(iv) x(t) = e−4t cos(10π t)u(t);

(v) x(t) = e−t 2/2.

12.7 Prove the Hermitian property of the DFT.

12.8 Prove the time-shifting property of the DFT.

12.9 Prove the periodic-convolution property of the DFT.

12.10 Prove Parseval’s theorem for the DFT.

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563 12 Discrete Fourier transform

12.11 Without explicitly determining the DFT X [r ] of the time-limited sequence

x[k] = [6 8 −5 4 16 22 7 8 9 44 2],

compute the following functions of the DFT X [r ]:

(i) X [0];

(ii) X [10];

(iii) X [6];

(iv)

10∑

r=0 X [r ];

(v)

10∑

r=0 |X [r ]|2.

12.12 Without explicitly determining the the time-limited sequence x[k] for the following DFT:

X [r ] = [12, 8 + j4, −5, 4 + j1, 16, 16, 4−j1, −5, 8 −j4],

compute the following functions of the DFT X [r ]:

(i) x[0];

(ii) x[9];

(iii) x[6];

(iv)

9∑

r=0 x[k];

(v)

9∑

r=0 |x[k]|2;

12.13 Given the DFT pair

x[k] DFT

←−−→ X [r ],

for a sequence of length N , express the DFT of the following sequences

as a function of X [r ]:

(i) y[k] = x[2k];

(ii) y[k] =

{

x[0.5k] k even

0 elsewhere;

(iii) y[k] = x[N − 1 − k] for 0 ≤ k ≤ N − 1;

(iv) y[k] =

{

x[k] 0 ≤ k ≤ N − 1

0 N ≤ k ≤ 2N − 1;

(v) y[k] = (x[k] − x[k − 2])e j(10πk/N ).

12.14 Compute the linear convolution of the following pair of time-limited sequences using the DFT-based approach. Be careful with the time

indices of the result of the linear convolution.

(i) x1[k] =

{

k 0 ≤ k ≤ 3

0 otherwise and x2[k] =

{

2 −1 ≤ k ≤ 2

0 otherwise;

(ii) x1[k] = k for 0 ≤ k ≤ 3 and x2[k] =

{

5 k = 0, 1

0 otherwise;

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564 Part III Discrete-time signals and systems

(iii) x1[k] = {

2 0 ≤ k ≤ 2

0 otherwise and x2[k] =

{

k + 1 0 ≤ k ≤ 4

0 otherwise;

(iv) x1[k] =



  

  

−1 k = −1

1 k = 0

2 k = 1

0 otherwise

and x2[k] =



  

  

3 k = −1, 2

1 k = 0

−2 k = 1, 3

0 otherwise;

(v) x1[k] =

{

|k| |k| ≤ 2

0 otherwise and x2[k] =

{

2−k 0 ≤ k ≤ 3

0 otherwise.

12.15 Draw the flow graph for a 6-point DFT by subdividing into three 2- point DFTs that can be combined to compute X [r ]. Repeat for the

subdivision of two 3-point DFTs. Which flow graph provides more com-

putational savings?

12.16 Draw a flow graph for a 10-point decimation-in-time FFT algorithm using two DFTs of size 5 in the first stage of the flow graph and five DFTs

of size 2 in the second stage. Compare the computational complexity of

the algorithm with the direct approach based on the definition.

12.17 Assume that K = 33. Draw the flow graph for a K -point decimation- in-time FFT algorithm consisting of three stages by using radix-3 as

the basic building block. Compare the computational complexity of the

algorithm with the direct approach based on the definition.

12.18 Consider two real-valued N -point sequences x1[k] and x2[k] with DFTs X1[r ] and X2[r ], respectively. Let p[k] be an N -point complex-valued

sequence such that p[k] = x1[k] + jx2[k] and let the DFT of p[k] be

denoted by P[r ].

(a) Show that the DFTs X1[r ] and X2[r ] can be obtained from the DFT

P[r ].

(b) Assume that N = 2m and that the decimation-in-time FFT algorithm

discussed in Section 12.6 is used to calculate the DFT P[r ]. Estimate

the total number of flops required to calculate the DFTs X1[r ] and

X2[r ] using the procedure in part (a).

12.19 Consider a real-valued N -point sequence x[k], where N is a power of 2. Let x1[k] and x2[k] be two N /2-point real-valued sequences such that

x1[k] = x[2k] and x2[k] = x[2k + 1] for 0 ≤ k ≤ N/2 − 1. Let the N -

point DFT of x[k] be denoted by X [r ] and let the N /2-point DFT of

x1[k] and x2[k] be denoted by X1[r ] and X2[r ], respectively.

(a) Determine X [r ] in terms of X1[r ] and X2[r ].

(b) Estimate the total number of flops required to calculate X [r ] using

the procedure discussed in Problem 12.18.

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C H A P T E R

13 The z-transform

In Chapter 11, we introduced two frequency representations, namely the

discrete-time Fourier series (DTFS) and the discrete-time Fourier transform

(DTFT) for DT signals. These frequency representations are exploited to deter-

mine the output response of an LTID system. Unfortunately, the DTFT does

not exist for all signals (e.g., periodic signals). In situations where the DTFT

does not exist, an alternative transform, referred to as the z-transform, may be

used for the analysis of LTID systems. Even for DT sequences for which the

DTFT exists, the z-transforms are always real-valued, rational functions of the

independent variable z provided that the DT sequences are real. In comparison,

the DTFT is generally complex-valued. Therefore, using the z-transform sim-

plifies the algebraic manipulations and leads to flow diagram representations

of the DT systems, a pivotal step needed to fabricate the DT system in silicon.

Finally, the DTFT can only be applied to a stable LTID system for which the

impulse response is absolutely summable. Since the z-transform exists for both

stable and unstable LTID systems, the z-transform can be used to analyze a

broader range of LTID systems.

The difference between the DTFT and the z-transform lies in the choice of

the independent variable used in the transformed domain. The DTFT X (Ω) of a

DT sequence x[k] uses the complex exponentials ejkΩ as its basis function and

maps x[k] in terms of ejkΩ. The z-transform X (z) expresses x[k] in terms of

zk , where the independent variable z is given by z = e(σ+jΩ)k . The z-transform is, therefore, a generalization of the DTFT, just as the Laplace transform is a

generalization of the CTFT. In this chapter, we introduce the z-transform and

illustrate its applications in the analysis of LTID systems.

This chapter is organized as follows. Section 13.1 defines the bilateral, also

referred to as the two-sided, z-transform and illustrates the steps involved

in its computation through a series of examples. For causal signals, the

bilateral z-transform reduces to the one-sided, or unilateral, z-transform, which

is covered in Section 13.2. Section 13.3 presents inverse methods of calcu-

lating the time-domain representation of the z-transform. The properties of

the z-transform are derived in Section 13.4. Sections 13.5–13.9 cover various

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566 Part III Discrete-time signals and systems

applications of the z-transform. Section 13.5 applies the z-transform to calculate

the output of an LTID system from the input sequence and the impulse response

of the LTID system. The relationship between the Laplace transform and the

z-transform is discussed in Section 13.6. Stability analysis of the LTID system in

the z-domain is presented in Section 13.7, while graphical techniques to derive

the frequency response from the z-transform are discussed in Section 13.8.

Section 13.9 compares the DTFT and z-transform in calculating the steady state

and transient responses of an LTID system. Section 13.10 introduces important

M A T L A B library functions useful in computing the z-transform and in the

analysis of LTID systems. Finally, the chapter is concluded in Section 13.11

with a summary of important concepts.

13.1 Analytical development

Section 11.1 defines the synthesis and analysis equations of the DTFT pair

x[k] DTFT←−−→ X (Ω) as follows:

DTFT synthesis equation x[k] = 1

2π

∫

〈2π〉

X (Ω)ejΩkdΩ; (13.1)

DTFT analysis equation X (Ω) = ∞∑

k=−∞ x[k]e−jΩk . (13.2)

To derive the expression for the bilateral z-transform, we calculate the DTFT

of the modified version x[k]e−σk of the DT signal. Based on Eq. (13.2), the

DTFT of the modified signal is given by

ℑ {

x[k]e−σk }

= ∞∑

k=−∞ x[k]e−σke−jΩk =

∞∑

k=−∞ x[k]e−(σ+jΩ)k . (13.3)

Substituting eσ+jΩ = z in Eq. (13.3) leads to the following definition for the bilateral z-transform:

z-analysis equation X (z) = ℑ {

x[k]e−σk }

= ∞∑

k=−∞ x[k]z−k . (13.4)

It may be noted that the summation in Eq. (13.4) is absolutely summable only

for selected values of z. For other values of z, the infinite sum in Eq. (13.4)

may not converge to a finite value, and hence X (z) becomes infinite. The region

in the complex z-plane, where summation (13.4) is finite, is referred to as the

region of convergence (ROC) of the z-transform X (z).

By following a similar derivation for the DTFT synthesis equation, Eq. (13.1),

the expression for the inverse z-transform is given by

z-synthesis equation x[k] = 1

2π j

∮

X (z)zk−1 dz, (13.5)

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567 13 The z-transform

where C is a closed contour traversed in the counterclockwise direction within

the ROC. Solving Eq. (13.5) involves the application of contour integration

techniques and is, therefore, seldom used directly. In Section 13.3, we will

consider alternative approaches based on the look-up table, partial fraction

expansion, and power series expansion to evaluate the inverse z-transform.

Collectively, Eqs. (13.4) and (13.5) form the bilateral z-transform pair, which

is denoted by

x[k] z←→ X (z) or Z{x[k] } = X (z). (13.6)

To illustrate the steps involved in computing the z-transform, we consider the

following examples.

Example 13.1

Calculate the bilateral z-transform of the exponential sequence x[k] = αku[k].

Solution

Substituting x[k] = αku[k] in Eq. (13.4), we obtain

X (z) = ∞∑

k=−∞ αku[k]z−k =

∞∑

k=0 (αz−1)k







1 − αz−1 |αz−1| < 1

undefined elsewhere.

In the above expression, if |αz−1| ≥ 1 the bilateral z-transform has an infinite value. In such cases, we say that the z-transform is not defined. The set of values

of z over which the bilateral z-transform is defined is referred to as the region of

convergence (ROC) associated with the z-transform. In this example, the ROC

for the z-transform pair

αku[k] z←→

1 − αz−1 is given by

ROC: ∣ ∣αz−1

∣ ∣ < 1 or |z| > |α|.

Figure 13.1 highlights the ROC by shading the appropriate region in the complex

z-plane.

31 2 4 5 6 7 8−1 0−2

x[k] = aku[k]1

a2 a3 a4

Re{z}

Im{z}

(0,α)

(a) (b)

Fig. 13.1. (a) DT exponential

sequence x[k ] = αk u[k ]; (b) the ROC, |z| > |α|, associated with its bilateral z-transform. The ROC

is shown as the shaded area and

lies outside the circle of radius α.

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Example 13.1 derives the bilateral z-transform of the exponential sequence

x[k] = αku[k]:

αku[k] z←→

1 − αz−1 , with ROC |z| > |α|.

Since no restriction is imposed on the magnitude of α, the bilateral

z-transform of the exponential sequence exists for all values of α within the

specified ROC. Recall that the DTFT of an exponential sequence exists only

for α < 1. For α ≥ 1, the exponential sequence is not summable and its DTFT

does not exist. This is an important distinction between the DTFT and the bilat-

eral z-transform. While the DTFT exists for a limited number of absolutely

summable sequences, no such restrictions exist for the z-transform. By associ-

ating an ROC with the bilateral z-transform, we can evaluate the z-transform

for a much larger set of sequences.

Example 13.2

Calculate the bilateral z-transform of the left-hand-sided exponential sequence

x[k] = −αku[−k − 1].

Solution

For the DT sequence x[k] = −αku[−k − 1], Eq. (13.4) reduces to

X (z) =

∞∑

k=−∞ −αku[−k − 1]z−k = −

−1∑

k=−∞ (αz−1)k .

To make the limits of summation positive, we substitute m = −k in the above equation to obtain

X (z) = − ∞∑

m=1 (α−1z)m =







− α−1z

1 − α−1z |α−1z| < 1

undefined elsewhere,

which simplifies to

X (z) =







1 − αz−1 |z| < |α|

undefined elsewhere.

The DT sequence x[k] = −αku[−k − 1] and the ROC associated with its z-transform are illustrated in Fig. 13.2.

In Examples 13.1 and 13.2, we have proved the following z-transform pairs:

αku[k] z←→

1 − αz−1 , with ROC |z| > |α|,

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569 13 The z-transform

−3−5 −4 −2 −1 0 1 2−7 −6−8

x[k] = −aku[−k −1]

−a−4 −a−3

−a−2 −a−1

(a) (b)

Re{z} (0, a)

Im{z}Fig. 13.2. (a) Non-causal

function x[k ] = −αku[−k − 1]; (b) its associated ROC,

|z| < |α|, shown as the shaded area excluding the circle, over

which the bilateral z-transform

exists.

and

−αku[−k − 1] z←→ 1

1 − αz−1 , with ROC |z| < |α|.

Although the algebraic expressions for the bilateral z-transforms are the same

for the two functions, the ROCs are different. This implies that a bilateral z-

transform is completely specified only if both the algebraic expression and

the associated ROC are included in its specification.

13.2 Unilateral z-transform

In Section 13.1, we introduced the bilateral z-transform, which may be used to

analyze both causal and non-causal LTID systems. Since most physical systems

in signal processing are causal, a simplified version of the bilateral z-transform

exists in such cases. The simplified bilateral z-transform for causal signals

and systems is referred to as the unilateral z-transform, and it is obtained by

assuming x[k] = 0 for k < 0. The analysis equation, Eq. (13.4), simplifies as follows:

unilateral z-transform X (z) = ∞∑

k=0 x[k]z−k . (13.7)

Unless explicitly stated, we will, in subsequent discussion, assume the

“unilateral” z-transform when referring to the z-transform. If the bilateral

z-transform is being discussed, we will specifically state this. To clarify further

the differences between the unilateral and bilateral z-transforms, we summarize

the major points.

(1) The unilateral z-transform simplifies the analysis of causal LTID systems.

Since most physical systems are naturally causal, we will mostly use unilat-

eral z-transform in our computations. However, the unilateral z-transform

cannot be used to analyze non-causal systems directly.

(2) For causal signals and systems, the unilateral and bilateral z-transforms are

the same.

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570 Part III Discrete-time signals and systems

(3) The synthesis equation used for calculating the inverse of the unilateral

z-transform is the same as Eq. (13.5) used for evaluating the inverse of the

bilateral transform.

Example 13.3

Calculate the unilateral z-transform for the following sequences:

(i) unit impulse sequence, x1[k] = δ[k]; (ii) unit step sequence, x2[k] = u[k];

(iii) exponential sequence, x3[k] = αku[k]; (iv) first-order, time-rising, exponential sequence, x4[k] = kαku[k];

(v) time-limited sequence, x5[k] =







1 k = 0, 1 2 k = 2, 5 0 otherwise.

Solution

(i) By definition,

X1(z) = ∞∑

k=0 δ[k]z−k = δ[0]z0 = 1, ROC: entire z-plane.

The z-transform pair for an impulse sequence is given by

δ[k] z←→ 1, ROC: entire z-plane.

(ii) By definition,

X2(z) = ∞∑

k=0 u[k]z−k =

∞∑

k=0 z−k =







1 − z−1 for |z−1| < 1

undefined elsewhere.

The z-transform pair for a unit step sequence is given by

u[k] z←→

1 − z−1 , ROC: |z| > 1.

In the above transform pair, note that the ROC |z−1| < 1 is equivalent to |z| > 1 and consists of the region outside a circle of unit radius in the complex z-plane.

This circle of unit radius, with the origin of the z-plane as the center, is referred

to as the unit circle and plays an important role in the determination of the

stability of an LTID system. We will discuss stability issues in Section 13.7.

(iii) By definition,

X3(z) = ∞∑

k=0 αku[k]z−k =

∞∑

k=0 (αz−1)k =







1 − αz−1 for

∣ ∣αz−1

∣ ∣ < 1

undefined elsewhere.

The z-transform pair for an exponential sequence is therefore given by

αku[k] z←→

1 − αz−1 , ROC: |z| > |α|.

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571 13 The z-transform

In the above transform pair, the ROC |αz−1| < 1 is equivalent to |z| > α and consists of the region outside the circle of radius |z| = α in the complex z- plane. Example 13.1 derives the bilateral z-transform for the function x3[k] = αku[k]. Since the function is causal, the bilateral and unilateral z-transforms

are identical.

(iv) By definition,

X (z) = ∞∑

k=0 kαku[k]z−k =

∞∑

k=0 k(αz−1)k .

Using the following result:

∞∑

k=0 kr k =

(1 − r )2 , provided |r| < 1,

the above summation reduces to

X (z) = αz−1

(1 − αz−1)2 , ROC: |αz−1| < 1.

The z-transform pair for a time-rising, complex exponential is given by

kαku[k] z←→

αz−1

(1 − αz−1)2 or

αz

(z − α)2 , ROC: |z| > |α|.

(v) Since the input sequence x5[k] is zero outside the range 0 ≤ k ≤ 5, Eq. (13.4) reduces to

X (z)= ∞∑

k=0 x[k]z−k = x[0]+x[1]z−1+x[2]z−2+x[3]z−3+x[4]z−4+x[5]z−5.

Substituting the values of x5[k] for the range 0 ≤ k ≤ 5, we obtain

X (z) = 1 + z−1 + 2 z−2 + 2 z−5 ROC: entire z-plane, except z = 0.

For finite-duration sequences, the ROC is always the entire z-plane except for

the possible exclusion of z = 0 and z = ∞.

13.2.1 Relationship between the DTFT and the z-transform

Comparing Eq. (13.2) with Eq. (13.4), the DTFT can be expressed in terms of

the bilateral z-transform as follows:

X (Ω) = ∞∑

k=−∞ x[k]z−k = X (z)|z=ejΩ . (13.8)

Since, for causal functions, the bilateral and unilateral z-transforms are the

same, Eq. (13.8) is also valid for the unilateral z-transform for causal functions.

Equation (13.8) shows that the DTFT is a special case of the z-transform

with z = ejΩ. The equality z = ejΩ corresponds to the circle of unit radius (|z| = 1) in the complex z-plane. Equation (13.8) therefore implies that the

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572 Part III Discrete-time signals and systems

Table 13.1. Unilateral z-transform pairs for several causal DT sequences

DT sequence z-transform with ROC

x[k] = 1

2π j

∮

X (z)zk−1dz X (z) = ∞∑

k=−∞ x[k]z−k

(1) Unit impulse

x[k] = δ[k] 1, ROC: entire z-plane

(2) Delayed unit impulse

x[k] = δ[k − k0] z−k0 ,ROC: entire z-plane, except z = 0

(3) Unit step

x[k] = u[k]

1 − z−1 =

z − 1 , ROC: |z| > 1

(4) Exponential

x[k] = αku[k]

1 − αz−1 =

z − α , ROC: |z| > |α|

(5) Delayed exponential

x[k] = αk−1u[k − 1]

z−1

1 − αz−1 =

z − α , ROC: |z| > |α|

(6) Ramp

x[k] = ku[k]

z−1

(1 − z−1)2 =

(z − 1)2 , ROC: |z| > 1

(7) Time-rising exponential

x[k] = kαku[k]

αz−1

(1 − αz−1)2 =

αz

(z − α)2 , ROC: |z| > |α|

(8) Causal cosine

x[k] = cos(Ω0k)u[k]

1 − z−1 cosΩ0 1 − 2z−1 cosΩ0 + z−2

= z[z − cosΩ0]

z2 − 2z cosΩ0 + 1 , ROC: |z| > 1

(9) Causal sine

x[k] = sin(Ω0k)u[k]

z−1 sinΩ0

1 − 2z−1 cosΩ0 + z−2 =

z sinΩ0

z2 − 2z cosΩ0 + 1 , ROC: |z| > 1

(10) Exponentially modulated cosine

x[k] = αk cos(Ω0k)u[k]

1 − αz−1 cosΩ0 1 − 2αz−1 cosΩ0 + α2z−2

= z[z − α cosΩ0]

z2 − 2αz cosΩ0 + α2 , ROC: |z| > |α|

(11) Exponentially modulated sine I

x[k] = αk sin(Ω0k)u[k]

αz−1 sinΩ0

1 − 2αz−1 cosΩ0 + α2z−2 =

αz sinΩ0

z2 − 2αz cosΩ0 + α2 , ROC: |z| > α

(12) Exponentially modulated sine II

x[k] = rαk sin(Ω0k + θ )u[k], with α ∈ R.

A + Bz−1

1 + 2γ z−1 + α2z−2 =

z(Az + B) z2 + 2γ z + γ 2

, ROC: |z| ≤ |α|(a)

(a) Where r =

√

A2α2 + B2 − 2ABγ α2 − γ 2

, Ω0 = cos−1 (

−γ α

)

, and θ = tan−1 (

A √

α2 − γ 2 B − Aγ

)

DTFT is obtained by computing the z-transform along the unit circle in the

complex z-plane.

Table 13.1 lists the z-transforms for several commonly used sequences. Com-

paring Table 13.1 with Table 11.2, we observe that when the sequence is causal

and its DTFT exists, the DTFT can be obtained from the z-transform by sub-

stituting z = ejΩ. Since the substitution z = ejΩ can only be made if the ROC contains the unit circle, an alternative condition for the existence of the DTFT

is the inclusion of the unit circle within the ROC of the z-transform. If the ROC

of a z-transform does not include the unit circle, we cannot substitute z = ejΩ and we say that its DTFT cannot be obtained from Eq. (13.8). For example, the

ROC of the unit step function is given by |z| > 1, which does not contain the

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573 13 The z-transform

unit circle. Equation (13.8) is, therefore, not valid for the unit step function.

This may also be verified from Table 11.2, where the DTFT of the unit step

function is different from the value obtained by substituting z = ejΩ in its z- transform. The DTFT of the unit step function in Table 11.2 contains the Dirac

delta functions, which makes the amplitude of the DTFT infinite at certain

frequencies. No Dirac delta functions exist in the z-transform of the unit step

function. Likewise, the ROCs for the z-transforms of the sine and cosine waves

do not contain the unit circle, and Eq. (13.8) is also not valid in these cases.

13.2.2 Region of convergence

As a side note to our discussion, we observe that the z-transform is guaranteed

to exist at all points within the ROC. For example, consider the causal sinusoidal

sequence x[k] = cos(0.2πk)u[k], whose z-transform is given in Table 13.1 as follows:

X (z) = 1 − cos(Ω0)z−1

1 − cos(Ω0)z−1 + z−2 , ROC: |z| > 1,

with Ω0 = 0.2π . We are interested in calculating the values of its z-transform at two points z1 = 2 + j0.6 and z2 = 0.8 + j0.6. Since z1 lies within the ROC, |z| > 1, the value of the z-transform at z1 is given by

X (z) = 1 − cos(0.2π )z−1

1 − cos(0.2π )z−1 + z−2

∣ ∣ ∣ ∣ z=2+j0.6

= 1.39 − j0.05.

However, the point z2 = 0.8 + j0.6 lies outside the ROC, |z| > 1. Therefore, the z-transform of the causal sinusoidal sequence cannot be computed for z2. In

the following, we list the important properties of the ROC for the z-transform.

(1) The ROC consists of a 2D plane of concentric circles of the form |z| > z0 or |z| < z0. All entries in Table 13.1 have ROCs that are concentric circles.

(2) The ROC does not include any poles of the z-transform.

The poles of a z-transform are defined as the roots of its denominator poly-

nomial. Since the value of the z-transform is infinite at the location of a pole,

the ROC cannot include any pole. Property (2) can be verified for all entries in

Table 13.1. Consider, for example, the unit step function, which has a single

pole at z = 1. The ROC of the z-transform of the unit step function is given by |z| > 1 and does not include its pole (z = 1).

(3) The ROC of a right-hand-sided sequence (x[k] = 0 for k < k0) is defined by the region outside a circle. In other words, the ROC of a right-hand-sided

sequence has the form |z| > z0.

Entries (3)–(12) in Table 13.1 are right-hand-sided sequences. Consequently, it

is observed that the ROC for all these sequences is of the form |z| > z0.

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574 Part III Discrete-time signals and systems

(4) The ROC of a left-hand-sided sequence (x[k] = 0 for k > k0) is defined by the inside region of a circle. Mathematically, this implies that the ROC

of a left-sided sequence has the form |z| < z0.

In Example 13.2, we computed the ROC for the left-hand-sided exponential

sequence x[k] = −αku[−k − 1] as |z| < α, which satisfies Property (4).

(5) The ROC of a double-sided (or bilateral) sequence, which extends to infinite

values of k in both directions, is confined to a ring with a finite area and

has the form z1 < |z| < z2.

An example of a double-sided sequence is x[k] = βku[k] − αku[−k − 1]. By applying the linearity property, which is formally derived in

Section 13.4.1, it is observed that the z-transform is given by

βku[k] − αku[−k − 1] z←→ 1

1 − αz−1 +

1 − βz−1 , ROC: β < |z| < α,

which satisfies Property (5).

(6) The ROC of a finite-length sequence (x[k] = 0 for k < k1, k > k2) is the entire z-plane except for the possible exclusion of the points z = 0 and z = ∞.

As an example of Property (6), we consider entries (1) and (2) of Table 13.1.

Also, sequence x5[k] defined in Example 13.3 is a finite-length sequence. In

such cases, we note that the ROC consists of the entire z-plane except for the

possible exclusion of z = 0 and z = ∞.

13.3 Inverse z-transform

Evaluating the inverse of z-transform is an important step in the analysis of

LTID systems. There are four commonly used methods to evaluate the inverse

z-transform:

(i) table look-up method;

(ii) inversion formula method;

(iii) partial fraction expansion method;

(iv) power series method.

Evaluating the inverse z-transform using the inversion formula (method (ii))

involves contour integration, which is fairly complex and beyond the scope of

the text. In this section, we cover the remaining three methods in more detail.

13.3.1 Table look-up method

In this method, the z-transform function X (z) is matched with one of the entries

in Table 13.1. As the transform pairs are unique, the inverse transform is readily

obtained from the time-domain entry. For example, if the inverse z-transform

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575 13 The z-transform

of the function

X (z) = 1

1 − 0.3z−1 , ROC: |z| > 0.3

is required, we determine that the matching entry in Table 13.1 is given by the

transform pair

αku[k] z←→

1 − αz−1 , ROC: |z| > α.

Substituting α = 0.3, the inverse z-transform of X (z) is given by x[k] = 0.3ku[k]. The scope of the table look-up method is limited to the list of

z-transforms available in Table 13.1.

13.3.2 Inversion formula method

In this method, the inverse z-transform is calculated directly by solving the

complex contour integral specified in the synthesis equation, Eq. (13.5). This

approach involves contour integration, which is beyond the scope of the text.

13.3.3 Partial fraction method

In LTID signals and systems analysis, the z-transform of a function x[k] gen-

erally takes the following rational form:

X (z) = N (z)

D(z) =

bm z m + bm−1zm−1 + · · · + b1z + b0

zn + an−1zn−1 + · · · + a1z + a0 (13.9a)

or alternatively

X (z) = N ′(z)

D′(z) = zm−n

bm + bm−1z−1 + · · · + b1z−m+1 + b0z−m

1 + an−1z−1 + · · · + a1z−n+1 + a0z−n . (13.9b)

Note that the numerator N (z) and denominator D(z) in Eq. (13.9a) are polyno-

mials of the complex function z. In this case, the inverse z-transform of X (z) can

be calculated using the partial fraction expansion method. The method consists

of the following steps.

Step 1 Calculate the roots of the characteristic equation of the rational function Eq. (13.9a). The characteristic equation is obtained by equating the denominator

D(z) in Eq. (13.9a) to zero, i.e.

D(z) = zn + an−1zn−1 + · · · + a1z + a0 = 0. (13.10)

For an nth-order characteristic equation, there will be n first-order roots.

Depending on the value of the coefficients {bl}, 0 ≤ l ≤ n − 1, roots {pr}, 1 ≤ r ≤ n, of the characteristic equation may be real-valued and/or complex- valued. By expressing D(z) in the factorized form, the z-transform X (z) is

represented as follows:

X (z)

z ≡

N (z)

z(z − p1)(z − p2) · · · (z − pn−1)(z − pn) . (13.11)

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576 Part III Discrete-time signals and systems

It may be noted that in Eq. (13.11) we represent X (z)/z, not X (z), in terms of

its poles. The reason for this will become clear after step 3.

Step 2 Using Heaviside’s partial fraction expansion formula, explained in Appendix D, decompose X (z)/z into a summation of the first- or second-order

fractions. If no roots are repeated, X (z)/z is decomposed:

X (z)

z =

z +

z − p1 +

z − p2 + · · · +

kn−1

z − pn−1 +

z − pn , (13.12)

where the coefficients {kr}, 0 ≤ r ≤ n, are obtained from the following expres- sion:

kr = [

(z − pr ) N (z)

zD(z)

]

z=pr . (13.13)

It may be noted that Eq. (13.13) appends roots {pr}, 1 ≤ r ≤ n, of the char- acteristic equation, Eq. (13.10), with an additional root p0 = 0 such that n + 1 partial fraction coefficients are obtained by solving Heaviside’s expression.

If there are repeated roots, X (z) takes a slightly different form (see Appendix

D for more details). It is important to associate separate ROCs with each partial

fraction term in Eq. (13.12). The ROC for each partial fraction term is deter-

mined such that the intersection of these individual ROCs results in the overall

ROC specified for X (z).

Multiplying both sides of Eq. (13.12) by z, we obtain

X (z) ≡ k0 + k1 z

z − p1 + k2

z − p2 + · · · + kn−1

z − pn−1 + kn

z − pn (13.14a)

X (z) ≡ k0 + k1 1

1 − p1z−1 + k2

1 − p2z−1 + · · · + kn−1

1 − pn−1z−1

+ kn 1

1 − pnz−1 . (13.14b)

Step 3 The inverse transform of X (z) can now be calculated by calculating the inverse transform of each individual partial fraction in Eq. (13.14a) using the

following transform pair (see Table 13.1):

αku[k] z←→

1 − αz−1 =

z − α , ROC: |z| > α,

and is given by

x[k] = k0δ[k] + k1(p1)ku[k] + k2(p2)ku[k] + · · · + kn−1(pn−1)ku[k] + kn(pn)ku[k]. (13.14c)

The reason for performing a partial fraction expansion of X (z)/z and not of

X (z) should now be clear. It was done so that the transform pair in Eq. (13.14b)

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577 13 The z-transform

can readily be applied to calculate the inverse transform. Otherwise, we would

be missing the factor of z in the numerator of Eq. (13.14a), and application of

Eq. (13.14b) would have been more complicated.

To illustrate the aforementioned procedure (steps (1)–(3)) for evaluating the

inverse z-transform using the partial fraction expansion, we consider the

following example.

Example 13.4

The z-transform of three right-sided functions is given below. Calculate the

inverse z-transform in each case.

(i) X1(z) = z

z2 − 3z + 2 ;

(ii) X2(z) = 1

(z − 0.1)(z − 0.5)(z + 0.2) ;

(iii) X3(z) = 2z(3z + 17)

(z − 1)(z2 − 6z + 25) .

Solution

(i) The characteristic equation of X1(z) is given by z 2 − 3z + 2 = 0, which has

two roots, at z = 1 and 2. The z-transform X1(z) can therefore be expressed as follows:

X1(z)

z =

z2 − 3z + 2 ≡

z − 1 +

z − 2 .

Using Heaviside’s partial fraction expansion formula, the coefficients of the

partial fractions k1 and k2 are given by

k1 = [

(z − 1) 1

(z − 1)(z − 2)

]

z=1 =

[ 1

z − 2

]

z=1 = −1

and

k2 = [

(z − 2) 1

(z − 1)(z − 2)

]

z=2 =

[ 1

z − 1

]

z=2 = 1.

The partial fraction expansion of X1(z) is therefore given by

X1(z) = −z

(z − 1) ︸︷︷︸

ROC:|z|>1

+ z

(z − 2) ︸︷︷︸

ROC:|z|>2

= −1

(1 − z−1) ︸︷︷︸

ROC:|z|>1

+ 1

(1 − 2z−1) ︸︷︷︸

ROC:|z|>2

where the ROC is obtained by noting that each term in X1(z) corresponds

to a right-hand-sided sequence. This follows directly from knowing that

x[n] is right-hand-sided; hence, each term in X1(z) should also correspond

to a right-hand sequence. Calculating the inverse z-transform of X1(z), we

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578 Part III Discrete-time signals and systems

obtain

x1[k] = −u[k] + 2ku[k] = (2k − 1)u[k].

(ii) The characteristic equation of X2(z) has three roots, at z = 0.1, 0.5 and −0.2. Therefore, X2(z)/z can be expressed as follows:

X2(z)

z =

z(z − 0.1)(z − 0.5)(z + 0.2)

= k0

z +

z − 0.1 +

z − 0.5 +

z + 0.2 .

The partial fraction coefficients k0, k1, k2, and k3 are given by

k0 = [

z 1

z(z − 0.1)(z − 0.5)(z + 0.2)

]

z=0 = 100,

k1 = [

(z − 0.1) 1

z(z − 0.1)(z − 0.5)(z + 0.2)

]

z=0.1 = −

250

3 ,

k2 = [

(z − 0.5) 1

z(z − 0.1)(z − 0.5)(z + 0.2)

]

z=0.5 =

7 ,

and

k3 = [

(z + 0.2) 1

z(z − 0.1)(z − 0.5)(z + 0.2)

]

z=−0.2 = −

500

21 .

The partial fraction expansion of X2(z)/z is therefore given by

X2(z)

z =

100

z −

250

(z − 0.1) +

(z − 0.5) −

500

(z + 0.2) or

X2(z) = 100 − 250

(1 − 0.1z−1) +

(1 − 0.5z−1) −

500

(1 + 0.2z−1) .

Using the pairs in Table 13.1 and assuming a right-hand-sided sequence, the

inverse z-transform is given by

x2[k] = 100δ[k] + {

− 250

3 (0.1)k +

7 (0.5)k −

500

21 (0.2)k

}

u[k].

(iii) The characteristic equation of X3(z) has one real-valued root at z = 1 and two complex-conjugate roots at z = 3 ± j4. Combining the complex roots in a quadratic term, X3(z)/z can be expressed as follows:

X3(z)

z =

2(3z + 17) (z − 1)(z2 − 6z + 25)

≡ k1

z − 1 +

k2z + k3 z2 − 6z + 25

Using Heaviside’s partial fraction expansion formula, coefficient k1 is given by

k1 = [

(z − 1) 2(3z + 17)

(z − 1)(z2 − 6z + 25)

]

z=1 = 2.

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579 13 The z-transform

To determine the remaining partial fraction coefficients k2 and k3, we expand

2(3z + 17) (z − 1)(z2 − 6z + 25)

≡ 2

z − 1 +

k2z + k3 z2 − 6z + 25

by cross-multiplying and equating the numerators, we obtain

2(3z + 17) ≡ 2(z2 − 6z + 25) + (k2z + k3)(z − 1).

Comparing coefficients of z2 and z yields

coefficients of z2 0 ≡ 2 + k2 ⇒ k2 = −2; coefficients of z 6 ≡ −12 − k2 + k3 ⇒ k3 = 16.

The partial fraction expansion of X3(z) can therefore be expressed as follows:

X3(z)

z =

z − 1 +

−2z + 16 z2 − 6z + 25

X3(z) = 2 z

z − 1 − 2

z(z − 5 × 0.6) z2 − 2 × 5 × z × 0.6 + 52

+ 5

z(5 × 0.8) z2 − 2 × 5 × z × 0.6 + 52

where the final rearrangement makes the three terms in the above expres-

sion consistent with entries (4), (10), and (11) of Table 13.1, with α = 5, and cos(Ω0) = 0.6 and sin(Ω0) = 0.8. Assuming that the three terms repre- sent right-hand-sided sequences, the inverse z-transform for each term is given

term 1 2 z

z − 1 z−1←−−→ 2u[k];

term 2 − 2 [

z(z − 5 × 0.6) z2 − 2 × 5 × z × 0.6 + 52

]

z−1←−−→ −2 · cos(cos−1(0.6)k) · 5ku[k];

term 3 5

[ z(5 × 0.8)

z2 − 2 × 5 × z × 0.6 + 52

]

z−1←−−→ 5

2 · sin(cos−1(0.6)k) · 5ku[k].

Substituting cos−1(0.6) = 0.9273, the three terms are combined as follows:

x3[k] = 2u[k] − 2 · 5k cos(0.9273k)u[k] + 5

2 · 5k sin(0.9273k) u[k],

which can be simplified to

x3[k] = {

2 + 3.2016 × 5k cos(0.9273 k − 128.7◦) }

u[k].

The DT sequences x1[k], x2[k], and x3[k] are plotted in Fig. 13.3 for duration

0 ≤ k ≤ 6.

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580 Part III Discrete-time signals and systems

k 31 2 4 5 6−1 0−2

x1[k] = (2 k −1)u[k]

1 3

15 31

≈ ≈

≈

k 31 2 4 5 6−1 0−2

x2[k]

0.4 0.23

0.11

k 31 2 4

5 6

−1 0−2

x3[k]

346 219

≈

−7297

≈≈

−49239(a) (b) (c)

Fig. 13.3. DT sequences

obtained in Example 13.4.

13.3.4 Power series method

When X (z) is a rational function of the form in Eq. (13.9), the partial fraction

expansion is a convenient method of calculating the inverse z-transform. At

times, however, it may be difficult to expand X (z) as partial fractions, especially

when X (z) is not a rational function. In such cases, we use the power series

method. Alternatively, we may be interested in determining a few values of x[k]

for k ≥ 0. The power series method is easy to apply since it does not require

the evaluation of the complete inverse z-transform.

In the power series method, the transform X (z) is expanded by long division

as follows:

X (z) = N (z)

D(z) = a + bz−1 + cz−2 + dz−3 + · · · . (13.15a)

Taking the inverse z-transform of Eq. (13.15), we obtain

x[k] = aδ[k] + bδ[k − 1] + cδ[k − 2] + dδ[k − 3] + · · · , (13.15b)

which implies that x[0] = a, x[1] = b, x[2] = c, and x[3] = d . Additional

samples of x[k] can be obtained by determining additional terms in the quotient

of Eq. (13.15a). We now illustrate the application of the power series method

with an example.

Example 13.5

Calculate the first four non-zero values of the following right-sided sequences

using the power series approach:

(i) X1(z) = z

z2 − 3z + 2 ;

(ii) X2(z) = 1

(z − 0.1)(z − 0.5)(z + 0.2) ;

(iii) X3(z) = 2z(3z + 17)

(z − 1)(z2 − 6z + 25) .

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581 13 The z-transform

Solution

(i) Using long division, X1(z) can be expressed as follows:

z−1 + 3z−2 + 7z−3 + 15z−4

z2 − 3z + 2 ∣ ∣ ∣ ∣

z − 3 + 2z−1 3 − 2z−1 3 − 9z−1 + 6z−2

7z−1 − 6z−2 7z−1 − 21z−2 + 14z−3

15z−2 − 14z−3 15z−2 − 45z−3 + 30z−4.

In other words,

X1(z) = z

z2 − 3z + 2 = 0z0 + z−1 + 3z−2 + 7z−3 + 15z−4 + · · · .

Taking the inverse transform gives the following values for the first five samples

of x1[k]:

x1[0] = 0, x1[1] = 1, x1[2] = 3, x1[3] = 7, x1[4] = 15.

Note that the above values are consistent with Fig. 13.3(a) obtained in Example

13.4 (i).

(ii) Using long division, X2(z) can be expressed as follows:

z−3 + 0.4z−4 + 0.23z−5 + 0.11z−6

z3 − 0.4z2 − 0.07z + 0.01

∣ ∣ ∣ ∣ ∣

1 − 0.4z−1 − 0.07z−2 + 0.010z−3

0.4z−1 + 0.07z−2 − 0.010z−3

0.4z−1 − 0.16z−2 − 0.028z−3 + 0.0040z−4

0.23z−2 + 0.018z−3 − 0.0040z−4

0.23z−2 − 0.092z−3 − 0.0161z−4 + 0.0023z−5

0.11z−3 + 0.0121z−4 − 0.0023z−5

0.11z−3 + 0.0440 z−4 − 0.0077z−5 + 0.0011z−5.

In other words,

X2(z) = 1

(z − 0.1)(z − 0.5)(z + 0.2) = 0z0 + 0z−1 + 0z−2 + z−3 + 0.4z−4 + 0.23z−5 + 0.11z−6 + · · · .

Taking the inverse transform gives the following values for the first seven sam-

ples of x2[k]:

x2[0] = 0, x2[1] = 0, x2[2] = 0, x2[3] = 1, x2[4] = 0.4, x2[5] = 0.23, x2[6] = 0.11.

This result is consistent with Fig. 13.3(b) obtained in Example 13.4(ii).

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582 Part III Discrete-time signals and systems

(iii) Using long division, X3(z) can be expressed as follows:

6z−1 + 76z−2 + 346z−3 + 216z−4

z3 − 7z2 + 31z − 25

∣ ∣ ∣ ∣ ∣

6z2 + 34z 6z2 − 42z + 186 − 150z−1

76z − 186 + 150z−1

76z − 532 + 2356z−1 − 1900z−2

346 − 2206z−1 + 1900z−2

346 − 2422z−1 + 10726z−2 − 8650z−3

216z−1 − 8826z−2 + 8650z−3

216z−1 − 1512z−2 + 6696z−3 − 5400z−3.

In other words,

X3(z) = 2z(3z + 17)

(z − 1)(z2 − 6z + 25) = 0z0 + 6z−1 + 76z−2 + 346z−3 + 216z−4 + · · · .

Taking the inverse transform gives the following values for the first five samples

of x3[k]:

x3[0] = 0, x3[1] = 6, x3[2] = 76, x3[3] = 346, x3[4] = 216.

The result is consistent with Fig. 13.3(c) obtained in Example 13.4(iii).

13.4 Properties of the z-transform

The unilateral and bilateral z-transforms have several interesting properties,

which are used in the analysis of signals and systems. These properties are

similar to the properties of the DTFT, which were covered in Section 11.4. In

this section, we discuss several of these properties, including their proofs and

applications, through a series of examples. A complete list of the properties is

provided in Table 13.2. In most cases, we prove the properties for the unilateral

z-transform. The proof for the bilateral z-transform follows along similar lines

and is not included to avoid repetition.

13.4.1 Linearity

If x1[k] and x2[k] are two DT sequences with the following z-transform pairs:

x1[k] z←→ X1(z), ROC: R1

and

x2[k] z←→ X2(z), ROC: R2,

then

a1x1[k] + a2x2[k] z←→ a1 X1(z) + a2 X2(z), ROC: at least R1 ∩ R2. (13.16)

The linearity property is satisfied by both unilateral and bilateral z-transforms.

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583 13 The z-transform

Table 13.2. Properties of the z-transform for transform pairs x[k ] z←→X(z), ROC: R x ; x[k ]u[k ]

z←→X (c)(z), ROC: R x ; x1[k ]

z←→X 1(z), ROC: R1; x2[k ] z←→X 2(z), ROC: R 2

Properties Time domain z-domain ROC

Linearity a1x1[k] + a2x2[k] a1 X1(z) + a2 X2(z) at least R1 ∩ R2 Time scaling x (m)[k]

for m = 1, 2, 3, . . . X (zm) (Rx )

1/m

Time shifting

(non-causal)

x[k − m] zm X (z)

Time shifting

(causal)

x[k − m]u[k − m] zm X (c)(z) Rx , except for the possible deletion or

addition of z = 0 or z = ∞

x[k + m]u[k] zm X (c)(z) − zm m−1∑

k=0 x[k]z−k

x[k − m]u[k] z−m X (c)(z) + z−m m∑

k=1 x[−k]zk

Frequency shifting ejΩ0k x[k] X (e−jΩ0 z) Rx

Time differencing x[k] − x[k − 1] (1 − z−1)X (z) Rx , except for the possible deletion of

the origin

Time accumulation y[k] = k∑

m=0 x[m] (a)

z − 1 X (z) Rx ∩ ( |z| > 1)

z-domain

differentiation

kx[k] −z dX (z)

dz Rx

Time convolution x1[k] ∗ x2[k] X1(z)X2(z) at least R1 ∩ R2 Initial-value theorem x[0] = lim

z→∞ X (z) provided x[k] = 0

for k < 0

Final-value theorem x[∞] = lim k→∞

x[k]= lim z→1

(z − 1)X (z) provided x[∞] exists

(a) Provided that the sequence y[k] has a finite value for all k.

Proof

Using Eq. (13.7), the z-transform of a1x1[k] + a2x2[k] is calculated as follows:

Z{a1x1[k] + a2x2[k]} = ∞∑

k=0 {a1x1[k] + a2x2[k]} z−k

= a1 ∞∑

k=0 x1[k]z

−k

︸︷︷︸

X1(z)

+ a2 ∞∑

k=0 x2[k]z

−k

︸︷︷︸

X2(z)

which proves the algebraic expression, Eq. (13.16). To determine the ROC of

the linear combination, we note that the z-transform X1(z) is finite within the

specified ROC, R1. Similarly, X2(z) is finite within its ROC, R2. Therefore, the

linear combination a1 X1(z) + a2 X2(z) should be finite at least within region R

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that represents the intersection of the two regions, i.e. R = R1 ∩ R2. In certain cases, due to the interaction between x1[k] and x2[k], which may lead to cance-

lation of certain terms, the overall ROC R may be larger than the intersection

of the two regions. On the other hand, if there is no common region between

R1 and R2, the z-transform of {a1x1[k] + a1x2[k]} does not exist.

13.4.2 Time scaling

As mentioned in Chapter 1, there are two types of scaling in the DT domain:

decimation and interpolation.

13.4.2.1 Decimation

Because of the irreversible nature of the decimation operation, the z-transform

of x[k] and its decimated sequence y[k] = x[mk] are not related to each

other.

13.4.2.2 Interpolation

Section 1.3.2.2 defines the interpolation of x[k] as follows:

x (m)[k] =

{

x [k/m] if k is a multiple of integer m

0 otherwise.

The z-transform of an interpolated sequence is given by the following property.

If x[k] z←→X (z) with ROC Rx , then the z-transform X (m)(z) of x (m)[k] is given

x (m)[k] z←→ X (m)(z) = X (zm), ROC: (Rx )1/m (13.17)

for 2 ≤ m < ∞. The interpolation property is satisfied by both unilateral and bilateral z-transforms.

Proof

Z{x (m)[k]} = ∞∑

k=0 x (m)[k]z−k

= x (m)[0] + x (m)[1]z−1+· · · + x (m)[m]z−m + x (m)[m+1]z−(m+1)

+ · · · + x (m)[2m]z−2m + · · · .

Based on Eq. (13.17), the interpolated sequence x (m)[k] is zero everywhere

except when k is a multiple of m. This reduces the above transform as follows:

Z{x (m)[k]} = x (m)[0] + x (m)[m]z−m + x (m)[2m]z−2m + x (m)[3m]z−3m + · · · . = x[0] + x[1]z−m + x[2]z−2m + x[3]z−3m + · · ·

= ∞∑

k=0 x[k](zm)−k = X (zm).

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585 13 The z-transform

Because X (z) is finite-valued within the region z ∈ Rx , X (zm) will have a finite value when zm ∈ Rx or z ∈ (Rx )1/m .

13.4.3 Time shifting

The time-shifting property for a bilateral z-transform is as follows.

If x[k] bilateral z← → X (z) with ROC Rx , then

x[k − m] bilateral z← → z −m X (z), (13.18)

with ROC given by Rx except for the possible deletion or addition of z = 0 or z = ∞. The ROC is altered because of the inclusion of the zm or z−m term, which affects the roots of the denominator D(z) in X (z).

For causal sequences, the time-shifting property is more complicated. For

any causal sequence x[k]u[k] satisfying the DTFT pair

x[k]u[k] z←→X (z)

and having the ROC Rx , the unilateral z-transform of the following time-shifted

sequences are expressed as follows (for a positive integer m):

(a) x[k − m]u[k − m] z←→ z−m X (z); (13.19)

(b) x[k + m]u[k] z←→ zm X (z) − zm m−1∑

k=0 x[k]z−k ; (13.20)

k=1 x[−k]zk . (13.21)

In Eqs. (13.19)–(13.21), the ROC of the time-shifted sequences is given by Rx ,

except for the possible deletion or addition of z = 0 or z = ∞. To illustrate the three time-shifting operations, consider a two-sided sequence

x[k] = α|k| with |α| < 1, as illustrated in Fig. 13.4(a). Figures 13.4(b)–(d) illus- trate the three time-shifting operations defined above in Eqs. (13.19)–(13.21)

for m = 2.

Proof

We prove Eqs. (13.19)–(13.21) separately.

Equation (13.19)

Z{x[k − m]u[k − m]} = ∞∑

k=0 x[k − m]u[k − m]z−k =

∞∑

k=m x[k − m]z−k .

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586 Part III Discrete-time signals and systems

0−2 −1 21 3 4 5−4 −3−5

x[k−2]u[k−2]

a a2 a3

(b)

k 0−2 −1 21 3 4 5−4 −3−5

x[k] = a|k| 1

a a2

a3 a4 a5

a a2

a3a4a5

(a)

0−2 −1 21 3 4 5−4 −3−5

x[k+2]u[k]

a2 a3 a4 a5 a6 a7

(d)

0−2 −1 21 3 4 5−4 −3−5

x[k−2]u[k]

a a2 a3

a a2

(c)

Fig. 13.4. (a) Original DT

sequence x[k ] = α|k | . Parts (b)–(d) show sequences

obtained by time shifting the

sequence in part (a):

(b) x[k − 2]u[k − 2]; (c) x[k − 2]u[k ]; (d) x[k + 2]u[k ].

Substituting p = k − m, the above summation reduces to

Z{x[k − m]u[k − m]} = ∞∑

p=0 x[p]z−(p+m) = z−m

∞∑

p=0 x[p]z−p, = z−m X (z).

Equation (13.20)

Z{x[k + m]u[k]} = ∞∑

k=0 x[k + m]u[k] z−k =

∞∑

k=0 x[k + m]z−k .

Substituting p = k + m, the above summation reduces to

Z{x[k + m]u[k]} = ∞∑

p=m x[p]z−(p−m) = zm

∞∑

p=0 x[p]z−p − zm

m−1∑

p=0 x[p]z−p,

= zm X (z) − zm m−1∑

k=0 x[k]z−k .

Equation (13.21)

Z{x[k − m]u[k]} = ∞∑

k=0 x[k − m]u[k]z−k =

∞∑

k=0 x[k − m]z−k .

Substituting p = k − m, the above summation reduces to

Z{x[k − m]u[k]} = ∞∑

p=−m x[p]z−(p+m)

= z−m ∞∑

p=0 x[p]z−p + z−m

−1∑

p=−m x[p]z−p.

= z−m x(z) + z−m m∑

k=1 x[−k]zk .

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587 13 The z-transform

Example 13.6

Consider a non-causal DT sequence x[k] with initial values x[−1] = 11/6 and x[−2] = 37/36. Express the z-transform of the function

g[k] = (x[k] − 5x[k − 1] + 6x[k − 2])u[k]

in terms of the z-transform Z{x[k]u[k]} = X (z).

Solution

Applying the time-shifting property, Eq. (13.21), the z-transforms of

x[k − 1]u[k] and x[k − 2]u[k] are given by

Z{x[k − 1]u[k]} = z−1 X (z) + z−1x[−1]z = z−1 X (z) + 11

and

Z{x[k − 2]u[k]} = z−2 X (z) + z−2x[−1]z + z−2x[−2]z2

= z−2 X (z) + 11

6 z−1 +

36 .

Applying the linearity property, the z-transform of g[k] is given by

G(z) = X (z) − 5 [

z−1 X (z) + 11

]

+ 6 [

z−2 X (z) + 11

6 z−1 +

]

= (

1 − 5z−1 + 6z−2 )

X (z) + 11z−1 − 3.

13.4.4 Time differencing

If x[k] z←→X (z) with ROC Rx , then the z-transform of the time-difference

sequence x[k] − x[k − 1] is given by

x[k] − x[k − 1] z←→ (1 − z−1)X (z), (13.22)

with the ROC given by Rx except for the possible deletion of z = 0. The time- differencing property can be proved easily by applying the linearity and time-

shifting properties with m = 1. The time-differencing property is satisfied by both unilateral and bilateral z-transforms.

Example 13.7

Based on the z-transform pair

u[k] z←→

1 − z−1 , ROC: |z| > 1,

calculate the z-transform of the impulse function x[k] = δ[k] using the time- differencing property.

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588 Part III Discrete-time signals and systems

Solution

Using the time-differencing property, the z-transform of u[k] − u[k − 1] is given by

u[k] − u[k − 1] z←→ (1 − z−1) · Z{u[k]}, ROC: |z| > 1.

Substituting the value of Z{u[k]} = 1/(1 − z−1) and noting that u[k] − u[k − 1] = δ[k], we obtain

δ[k] z←→ 1.

Since the z-transform of the unit impulse function is finite for all values of z,

the ROC of the aforementioned z-transform pair is the entire z-plane.

13.4.5 z-domain differentiation

If x[k] z←→ X (z) with ROC Rx , then

kx[k] z←→ −z

dX (z)

dz , ROC: Rx . (13.23)

The z-domain differentiation property is satisfied by both unilateral and bilateral

z-transforms.

Proof

By definition,

X (z) = ∞∑

k=0 x[k]z−k .

Differentiating both sides with respect to z yields

dX (z)

dz =

∞∑

k=0 x[k]

dz−k

dz =

∞∑

k=0 x[k](−k)z−k−1.

Multiplying both sides by −z, we obtain

−z dX (z)

dz =

∞∑

k=0 kx[k]z−k,

which proves Eq. (13.23).

Example 13.8

Given the z-transform pair

αku[k] z←→

1 − αz−1 , ROC: |z| > |α|,

calculate the z-transform of the function kαku[k].

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589 13 The z-transform

Solution

We use the frequency-differentiation property,

kαk x[k] z←→ − z

[ 1

1 − αz−1

]

which reduces to

kαk x[k] z←→

αz−1

(1 − αz−1)2 ROC: |z| > |α|.

13.4.6 Time convolution

If x1[k] and x2[k] are two arbitrary functions with the following z-transform

pairs:

x1[k] z←→ X1(z), ROC: R1

and

x2[k] z←→ X2(z), ROC: R2,

then the convolution property states that

x1[k] ∗ x2[k] z←→ X1(z)X2(z), ROC: at least R1 ∩ R2. (13.24)

The convolution property is valid for both unilateral and bilateral z-transforms.

The overall ROC of the convolved signals may be larger than the intersection

of regions R1 and R2 because of the possible cancelation of some poles of the

convolved sequences.

Proof

By definition, the convolution of two sequences is given by

x1[k] ∗ x2[k] = ∞∑

m=−∞ x1[m]x2[k − m].

Taking the z-transform of both sides yields

x1[k] ∗ x2[k] z←→

∞∑

k=−∞

∞∑

m=−∞ x1[m]x2[k − m]z−k .

By interchanging the order of the two summations on the right-hand side of the

transform pair, we obtain

x1[k] ∗ x2[k] z←→

∞∑

m=−∞ x1[m]

∞∑

k=−∞ x2[k − m]z−k .

Substituting p = k − m in the inner summation leads to

x1[k] ∗ x2[k] z←→

∞∑

m=−∞ x1[m]

∞∑

p=−∞ x2[p]z

−(p+m)

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x1[k] ∗ x2[k] z←→

∞∑

m=−∞ x1[m] z

−m ∞∑

p=−∞ x2[p]z

−p,

which proves Eq. (13.24).

Like the DTFT convolution property discussed in Chapter 11, the time-

convolution property of the z-transform provides us with an alternative approach

to calculate the output y[k] when a DT sequence x[k] is applied at the input of

an LTID system with the impulse response h[k]. The procedure for calculating

the output y[k] of an LTID system in the complex z-domain consists of the

following four steps.

(1) Calculate the z-transform X (z) of the input sequence x[k]. If the input

sequence and the impulse response are both causal functions, then the

unilateral z-transform is used. If either of the two functions is non-causal,

the bilateral z-transform must be used.

(2) Calculate the z-transform H (z) of the impulse response h[k] of the LTID

system. The z-transform H (z) is referred to as the z-transfer function of

the LTID system and provides a meaningful insight into the behavior of the

system.

(3) Based on the convolution property, the z-transform Y (z) of the resulting

output y[k] is given by the product of the z-transforms of the input signal

and the impulse response of the LTID system. Mathematically, this implies

that Y (z) = X (z)H (z). (4) Calculate the output response y[k] in the time domain by taking the inverse

z-transform of Y (z) obtained in step (3).

Example 13.9

The exponential decaying sequence x[k] = aku[k], 0 ≤ a ≤ 1, is applied at the input of an LTID system with the impulse response h[k] = bku[k], 0 ≤ b ≤ 1. Using the z-transform approach, calculate the output of the system.

Solution

Based on Table 13.1, the z-transforms for the input sequence and the impulse

response are given by

X (z) = 1

1 − az−1 and H (z) =

1 − bz−1 .

The z-transform of the output signal is, therefore, calculated as follows:

Y (z) = H (z)X (z) = 1

(1 − az−1)(1 − bz−1) .

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591 13 The z-transform

The inverse of Y (z) takes two different forms depending on the values of a and

Y (z) =



  

  

(1 − az−1)2 a = b

(1 − az−1)(1 − bz−1) a �= b.

We consider the two cases separately while calculating the inverse z-transform

of Y (z).

Case 1 (a = b) From Table 13.1, we know that

kaku[k] z←→

az−1

(1 − az−1)2 .

Applying the time-shifting property, we obtain

(k + 1)ak+1u[k + 1] z←→ a

(1 − az−1)2 .

The output response is therefore given by

y[k] = Z−1 {

(1 − az−1)2

}

= (k + 1)aku[k + 1],

which is the same as

y[k] = (k + 1) aku[k].

Case 2 (a �= b) Using partial fraction expansion, the function Y (z) is expressed as follows:

Y (z) = 1

(1 − az−1)(1 − bz−1) ≡

1 − az−1 +

1 − bz−1 , (13.25)

where the partial fraction coefficients are given by

A = 1

1 − bz−1

∣ ∣ ∣ ∣ az−1=1

= a

a − b and

B = 1

1 − az−1

∣ ∣ ∣ ∣ bz−1=1

= − b

a − b .

Substituting the values of A and B into Eq. (13.25) and taking the inverse DTFT

yields

y[k] = a

a − b × aku[k] −

a − b × bku[k] =

a − b [

ak+1 − bk+1 ]

u[k].

Combining case 1 with case 2, we obtain

y[k] =







(k + 1)aku[k] a = b 1

a − b [

ak+1 − bk+1 ]

u[k] a �= b, (13.26)

which is identical to the result of Example 11.15.

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592 Part III Discrete-time signals and systems

13.4.7 Time accumulation

If x[k] z←→X (z) with ROC Rx , then

k∑

m=0 x[m]

z←→ z

z − 1 X (z), ROC: Rx ∩ ( |z| > 1). (13.27)

Proof

To prove the time-accumulation property, we make use of the following con-

volution result:

k∑

m=0 x[m] = x[k] ∗ u[k].

Taking the z-transform of both sides and applying the time-convolution property

yields

k∑

m=0 x[m]

z←→ X (z)Z{u[k]}.

In the above equation, we substitute the z-transform of the unit step function,

u[k] z←→

1 − z−1 , ROC: |z| > 1,

to obtain

k∑

m=0 x[m]

z←→ X (z) 1

1 − z−1 ,

which proves Eq. (13.27).

Example 13.10

Given the z-transform pair

u[k] z←→

1 − z−1 , ROC: |z| > 1 ,

calculate the z-transform of the function ku[k] using the time-accumulation

property.

Solution

Note that

ku[k] = k∑

m=0 u[m] − u[k].

Calculating the z-transform of both sides and applying the time-accumulation

property, we obtain

ku[k] z←→

(z − 1) 1

(1 − z−1) −

1 − z−1 ,

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593 13 The z-transform

which reduces to

ku[k] z←→

z−1

(

1 − z−1 )2

which can be expressed in the following alternative form:

ku[k] z←→

(z − 1)2 .

Note that the ROC for ku[k] is the same as that for u[k].

13.4.8 Initial- and final-value theorems

If x[k] z←→ X (z) with ROC Rx , then

initial-value theorem x[0] = lim z→∞

X (z), provided x[k] = 0 for k < 0;

(13.28)

final-value theorem x[∞] = lim k→∞

x[k] = lim z→1

(z − 1)X (z),

provided x[∞]exists. (13.29)

Note that the initial-value theorem is valid only for the unilateral z-transform as

it requires the reference signal x[k] to be zero for k < 0. The final-value theorem,

however, may be used with either the unilateral or bilateral z-transform. It is

possible to get a finite value from Eq. (13.29) even though x[∞] is undefined or equal to infinity. Readers are advised to check that x[∞] indeed converges to a finite value before using the final-value theorem. This generally happens if

all poles of (z − 1)X (z) lie inside the unit circle.

Example 13.11

Given the following z-transforms of right-sided sequences, determine the initial

and final values:

(i) X1(z) = z

z2 − 3z + 2 ;

(ii) X2(z) = 1

(z − 0.1)(z − 0.5)(z + 0.2) ;

(iii) X3(z) = z2(2z − 1.5)

(z − 1)(z − 0.5)2 .

Solution

(i) Using the initial-value theorem,

x1[0] = lim z→∞

X1(z) = lim z→∞

[ z

z2 − 3z + 2

]

= lim z→∞

[ 1

z − 3 + 2z−1

]

= 0.

Using the final-value theorem, we obtain

x1[∞] = lim z→1

(z − 1)X1(z) = lim z→1

[ z(z − 1)

z2 − 3z + 2

]

= lim z→∞

[ z

z − 2

]

= −1.

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From Example 13.4 part (i), where we determined x1[k], it can be verified that

x1[0] = 0. However, we obtain x1[∞] = ∞ from the result in Example 13.4, which is different from the result obtained above using the final-value theorem.

Actually, in this case the final-value theorem cannot be applied as x1[∞] is not finite. This can be guessed from the fact that (z − 1)X (z) has a pole at z = 2, which is not inside the unit circle.

(ii) Using the initial-value theorem,

x2[0] = lim z→∞

X2(z) = lim z→∞

[ 1

(z − 0.1)(z − 0.5)(z + 0.2)

]

= 0.

Using the final-value theorem,

x2[∞] = lim z→1

(z − 1)X2(z) = lim z→1

[ (z − 1)

(z − 0.1)(z − 0.5)(z + 0.2)

]

= 0.

From the expression of x2[k] derived in Example 13.4 part (ii), it can be verified

that the above values are indeed correct.

(iii) Using the initial-value theorem,

x3[0] = lim z→∞

X3(z) = lim z→∞

[ z2(2z − 1.5)

(z − 1)(z − 0.5)2

]

= lim z→∞

[ 2 − 1.5z−1

(1 − z−1)(1 − 0.5z−1)2

]

= 2.

Using the final-value theorem,

x3[∞] = lim z→1

(z − 1)X3(z) = lim z→1

[ (z − 1)z2(2z − 1.5) (z − 1)(z − 0.5)2

]

= lim z→1

[ z2(2z − 1.5) (z − 0.5)2

]

= 2.

By calculating the inverse z-transform of X3(z), we obtain

x3[k] = (2 + k × 2−k)u[k].

Based on the above expression, x3[0] = 2 and x3[∞] = 2, which are indeed the values obtained using the initial- and final-value theorems.

13.5 Solution of difference equations

An important application of the z-transform is to solve linear, constant-

coefficient difference equations. In Section 10.1, we used a time-domain

approach to obtain the zero-input, zero-state, and overall solutions of differ-

ence equations. In this section, we discuss an alternative approach based on the

z-transform. We illustrate the steps involved in the z-transform-based approach

through Example 13.12.

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595 13 The z-transform

Example 13.12

A causal system is represented by the following difference equation:

y[k + 2] − 5y[k + 1] + 6y[k] = 3x[k + 1] + 5x[k]. (13.30)

Calculate the output y[k] for the input x[k] = 2−ku[k] and the initial conditions y[−1] = 11/6, y[−2] = 37/36.

Solution

Substituting k − 2 for k in Eq. (13.30), we obtain

y[k] − 5y[k − 1] + 6y[k − 2] = 3x[k − 1] + 5x[k − 2]. (13.31)

Note that the input sequence x[k] = 2−ku[k] is causal, hence x[−2] = x[−1] = 0. Using the time-shifting property, Eq. (13.19), the z-transform of the right-hand side of Eq. (13.31) is given by

3x[k − 1] + 5x[k − 2] z←→ 3z−1 X (z) + 5z−2 X (z).

Using the z-transform pair,

x[k] = 2−ku[k] = 0.5ku[k] z←→ X (z) = 1

1 − 0.5z−1 ,

the z-transform of the right-hand side of Eq. (13.31) is given by

3x[k − 1] + 5x[k − 2] z←→ 3z−1

1 − 0.5z−1 +

5z−2

1 − 0.5z−1 =

3z−1 + 5z−1

1 − 0.5z−1 .

The output response is not causal as the initial conditions y[−1] and y[−2] are not zero. We are interested in determining the causal component y[k]u[k] of

the response y[k]. Let us denote the z-transform of y[k]u[k] by Y (z). Using the

results in Example 13.6, the z-transform of the left-hand side of Eq. (13.31) is

given by

y[k] − 5y[k − 1] + 6y[k − 2] z←→ (1 − 5z−1 + 6z−2)Y (z) + (11z−1 − 3).

Equating the z-transforms of both sides of Eq. (13.31), we obtain

(1 − 5z−1 + 6z−2)Y (z) + (11z−1 − 3) = 3z−1 + 5z−2

1 − 0.5z−1 ,

which reduces to

Y (z) = 3 − 11z−1

1 − 5z−1 + 6z−2 +

3z−1 + 5z−2

(1 − 0.5z−1)(1 − 5z−1 + 6z−2)

= (3 − 11z−1)(1 − 0.5z−1) + 3z−1 + 5z−2

(1 − 0.5z−1)(1 − 5z−1 + 6z−2)

= 3 − 9.5z−1 + 10.5−2

(1 − 0.5z−1)(1 − 2z−1)(1 − 3z−1) .

Using partial fraction expansion, Y (z) can be expressed as follows:

Y (z) = 26

15 ×

1 − 0.5z−1 −

3 ×

1 − 2z−1 +

5 ×

1 − 3z−1 .

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Taking the inverse transform, we obtain

y[k] = [

15 × 0.5k −

3 × 2k +

15 × 3k

]

for k > 0.

The output response is plotted in Fig. 13.5.k 2 30 1 4 5−2 −1

y[k]

3 7

23.5

78.75

254.38

≈ ≈

800.19

≈

1.831.03

Fig. 13.5. Output response of

the LTID system specified in

Example 13.12.

13.6 z-transfer function of LTID systems

In Chapters 10 and 11, we used the impulse response h[k] and Fourier transfer

function H (Ω) to represent an LTID system. An alternative representation for

an LTID system is obtained by taking the z-transform of the impulse response:

h[k] z←→ H (z).

The DTFT H (z) is referred to as the z-transfer function of the LTID system.

In conjunction with the linear convolution property, Eq. (13.24), the z-transfer

function H (z) may be used to determine the output response y[k] of an LTID

system when an input sequence x[k] is applied at its input. In the time domain,

the output response y[k] is given by

y[k] = x[k] ∗ h[k]. (13.32)

Taking the z-transform of both sides of Eq. (13.32), we obtain

Y (z) = X (z)H (z), (13.33)

where Y (z) and X (z) are, respectively, the z-transforms of the output response

y[k] and the input sequence x[k]. Equation (13.33) provides us with an alter-

native definition for the transfer function as the ratio of the z-transform of the

output response and the z-transform of the input signal. Mathematically, the

transfer function H (z) can be expressed as follows:

H (z) = Y (z)

X (z) . (13.34)

The z-transfer function of an LTID system can be obtained from its difference

equation representation, as described in the following.

Consider an LTID system whose input–output relationship is given by the

following difference equation:

y[k + n] + an−1 y[k + n − 1] + · · · + a0 y[k] = bm x[k + m] + bm−1x[k + m − 1] + · · · + b0x[k]. (13.35)

By taking the z-transform of both sides of the above equation, we obtain {

zn + an−1zn−1 + · · · + a0z }

Y (z) = {

bm z m + bm−1zm−1 + · · · + b0

}

X (z),

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which reduces to the following transfer function:

H (z) = Y (z)

X (z) =

bm z m + bm−1zm−1 + · · · + b0

zn + an−1zn−1 + · · · + a0 (13.36a)

or alternatively as

H (z) = zm−n bm + bm−1z−1 + · · · + b0z−m

1 + an−1z−1 + · · · + a0z−n . (13.36b)

13.6.1 Characteristic equation, poles, and zeros

The z-transfer function plays an important role in the analysis of LTID sys-

tems analysis. In this section, we will define a few key concepts related to the

z-transfer function.

Characteristic equation The characteristic equation for the transfer function, Eq. (13.36a), is defined as follows:

D(z) = anzn + an−1zn−1 + · · · + a0 = 0. (13.37)

Zeros The zeros of the transfer function H (z) of an LTID system are finite locations in the complex z-plane, where |H (z)| = 0. For the transfer function, Eq. (13.36a), the location of the zeros can be obtained by solving the following

equation:

N (z) = bm zm + bm−1zm−1 + · · · + b0 = 0. (13.38)

Since N (z) is an mth-order polynomial, it will have m roots leading to m zeros.

Poles The poles of the transfer function H (z) of an LTID system are defined as locations in the complex z-plane, where |H (z)| has an infinite value. The poles corresponding to the transfer function, Eq. (13.36a), can be obtained by solving

the characteristic equation, Eq. (13.37). Since D(z) is an nth-order polynomial,

it will have n roots leading to n poles.

Because D(z) is an nth-order polynomial and N (z) is an mth-order polyno-

mial, the transfer function will have a total of n poles and m zeros. However,

in some cases, the location of a pole may coincide with the location of a zero.

In that case, the pole and zero will cancel each other, and the actual number

of poles and zeros will be reduced. In order to calculate the zeros and poles, a

transfer function is factorized and typically represented as follows:

H (z) = N (z)

D(z) =

(z − z1)(z − z2) · · · (z − zm) (z − p1)(z − p2) · · · (z − pn)

, (13.39a)

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or alternatively as

H (z) = zm−n (1 − z1z−1)(1 − z2z−1) · · · (1 − zm z−1) (1 − p1z−1)(1 − p2z−1) · · · (1 − pnz−1)

. (13.39b)

Example 13.13

Determine the poles and zeros of the following LTID systems:

(i) H1(z) = z

z2 − 3z + 2 ;

(ii) H2(z) = 1

(z − 0.1)(z − 0.5)(z + 0.2) ;

(iii) H3(z) = z2(2z − 1.5)

(z + 0.4)(z − 0.5)2 ;

(iv) H4(z) = z2 + 0.7z + 1.6

(z2 − 1.2z + 1)(z + 0.3) .

Solution

(i) H1(z) = z

z2 − 3z + 2 =

(z − 1)(z − 2) .

There is one zero, at z = 0, and two poles, at z = 1 and 2.

(ii) H2(z) = 1

(z − 0.1)(z − 0.5)(z + 0.2) .

There is no zero, but there are three poles, at z = 0.1, 0.5, and −0.2.

(iii) H3(z) = z2(2z − 1.5)

(z + 0.4)(z − 0.5)2 .

There are three zeros, at z = 0, 0, and 0.75. There are three poles, at z = −0.4, 0.5, and 0.5.

(iv) H4(z) = (z − 0.5)(z + 1.2)

((z − 0.6)2 + 0.82)(z + 0.3)

= (z − 0.5)(z + 1.2)

(z − 0.6 + j0.8)(z − 0.6 − j0.8)(z + 0.3) .

There are two zeros, at z = 0.5 and −1.2. There are three poles, at z = 0.6 − j0.8, 0.6 + j0.8, and −0.3.

The poles and zeros of the above four systems are shown in Fig. 13.6. In the

plot, × marks the position of a pole and • marks the position of a zero.

Im{z}

Re{z} 1 2−1−2

−1

−2

Im{z}

Re{z} 0.5 1−0.5−1

0.5

−0.5

−1

Im{z}

Re{z} 0.5 1−0.5−1

0.5

−0.5

−1

Im{z}

Re{z} 0.5 1−0.5−1

0.5

−0.5

−1

(a)

(b)

(c)

(d)

Fig. 13.6. Pole and zero plots

for transfer functions in Example

13.13. Plot (a) corresponds to

part (i) of Example 13.13; plot

(b) corresponds to part (ii); plot

plot (d) corresponds to part

(iv). Also note that plot (c)

includes double zeros at z = 0 and double poles at z = 0.5.

13.6.2 Determination of impulse response

The impulse response h[k] of an LTID system can be obtained by calculating

the inverse z-transform of the transfer function H (z). Example 13.14 explains

the steps involved in determining the impulse response.

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599 13 The z-transform

Example 13.14

The input–output relationship of an LTID system is given by the following

difference equation:

y[k + 2] − 3

4 y[k + 1] +

8 y[k] = 2x[k + 2]. (13.40)

Determine the transfer function and the impulse response of the system.

Solution

Substituting m = k + 2, Eq. (13.40) can be written as follows:

y[m] − 3

4 y[m − 1] +

8 y[m] = 2x[m].

Calculating the z-transform on both sides of the equation yields

Y (z) − 3

4 z−1Y (z) +

8 z−2Y (z) = 2X (z),

which results in the following transfer function:

H (z) = Y (z)

X (z) =

1 − 3

4 z−1 +

8 z−2

To calculate the impulse response of the LTID system, consider the partial

fraction expansion of H (z) as

H (z) = 2

(

1 − 1

2 z−1

) (

1 − 1

4 z−1

) ≡ 4

1 − 1

2 z−1

− 2

1 − 1

4 z−1

By calculating the inverse z-transform of both sides, the impulse response h[k]

is obtained:

h[k] = 4 (

u[k] − 2 (

u[k],

which is identical to the result obtained by Fourier technique in Example 11.18.

13.7 Relationship between Laplace and z-transforms

LTID signals and systems can be considered as special cases of LTIC signals

and systems. Therefore, the Laplace transform can also be used to analyze such

signals and systems. In this section, we derive the relationship between the

Laplace and z-transforms.

If a DT sequence x[k] is obtained by sampling a CT signal x(t) with a

sampling interval T , the CT sampled signal xs(t) may be expressed as follows:

xs(t) = ∞∑

k=−∞ x(kT )δ(t − kT ),

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600 Part III Discrete-time signals and systems

LTID system

H(z)x[k] y[k]

H(e sT)

LTIC system

∑ x(kT )d(t−kT ) ∞

k=−∞

∑ y(kT )d(t−kT ) ∞

k=−∞

(a)

(b)

Fig. 13.7. Using Laplace

transform techniques to analyze

LTID systems. (a) Reference LTID

system; (b) equivalent LTIC

system with CT input and output

signals.

where x(kT) are the sampled values of x(t) which equals the DT sequence x[k].

Calculating the Laplace transform of xs(t), we obtain

X (s) = L{xs(t)} = ∞∑

k=−∞ x(kT )L{δ(t − kT )} =

∞∑

k=−∞ x(kT )e−kT s .

Comparing X (s) with the z-analysis equation,

X (z) = ∞∑

k=−∞ x[k]z−k,

it is clear that

X (s) = X (z)|z=esT (13.41a)

since x[k] = x(kT ). Equation (13.41a) illustrates the relationship between the Laplace transform X (s) of a sampled function and the z-transform X (z) of the

DT sequence obtained from the samples. As illustrated in Fig. 13.7, an LTID

system can be analyzed using an equivalent LTIC system. Figure 13.7(a) shows

an LTID system with the z-transfer function H (z) and sequence x[k] applied

at its input. The analysis of the LTID system can be completed in the s-domain

with the LTIC system shown in Fig. 13.7(b). The transfer function of the LTIC

system is given by

H (s) = H (z)|z=esT (13.41b)

and the DT input is transformed to an equivalent CT input of the form

xs(t) = ∞∑

k=−∞ x(kT )δ(t − kT ).

The output in Fig. 13.7(b) can be calculated using CT analysis techniques. The

resulting output y(t) can then be transformed back into the DT domain using

the relationship y[k] = x(t) at t = kT .

Example 13.15

A DT system is represented by the following impulse response function:

h[k] = 0.55u[k]. (13.42)

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601 13 The z-transform

(i) Determine the z-transfer function of the system.

(ii) Determine the equivalent Laplace transfer function of the system.

(iii) Using the Laplace domain approach, determine if the system is stable.

sIm

sRe 4 848

j20p

j40p

j20p

j40p

Fig. 13.8. Location of poles in

the s-plane for the system in

Example 13.15 with T = 0.1.

Solution

(i) H (z) = Z {

0.5ku[k] }

= 1

1 − 0.5z−1 , or

z − 0.5 , ROC: |z| > 0.5.

(ii) Using Eq. (13.41b), the Laplace transfer function is given by

H (s) = H (z)|z=esT = esT

esT − 0.5 , (13.43)

where T is the sampling interval.

(iii) A causal LTIC system is stable if all the poles corresponding to the

Laplace transfer function lie in the left-hand half of the s-plane. Therefore, we

will first calculate the pole locations in the s-plane, and then determine if the

system is stable. The poles of the transfer function, Eq (13.43), are calculated

from the characteristic equation as follows:

esT − 0.5 = 0 ⇒ esT = 0.5 ⇒ e(sT ±j2πm) = 0.5,

where m = 0, 1, 2, . . . Solving for the roots of this equation yields

s = 1

T [ln 0.5 ± j2πm] ≈

T [−0.693 ± j2πm].

It is observed that an LTID system has an infinite number of poles in the

s-domain. The locations of these poles for T = 0.1 are shown in Fig. 13.8. It is clear that these poles would lie in the left-half of the s-plane, irrespective of

the value of the sampling interval T . The LTID system is, therefore, causal and

stable.

Alternatively, the stability of the LTID system can be determined from its

impulse response by noting that

∞∑

k=−∞ |h[k]| =

∞∑

k=−∞ 0.5k = 2 < ∞,

which satisfies the BIBO stability requirement derived in Chapter 10.

13.8 Stabilty analysis in the z-domain

In Example 13.15, the stability of an LTID system was determined by trans-

forming its z-transfer function H (z) to the Laplace transfer function H (s) of an

equivalent LTIC system and observing if the poles of H (s) lie in the left-half

s-plane. In this section, we derive a z-domain condition to check the stability

of a system directly from its z-transfer function.

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Consider a pole z = pz of an LTID system with the z-transfer function given by H (z). Based on Eq. (13.41b), the location of the corresponding s-domain

pole, s = ps , of its equivalent LTIC system H (esT ) is related to the location of the z-domain pole, z = pz , of H (z) by the following relationship:

pz = eps T or pz = eRe{ps T }.e j Im{ps T }, (13.44)

where ps T is decomposed into real and imaginary components as Re{ps T } + j Im{ps T }.

We consider two different cases. Case 1 refers to a stable system, which is

not necessarily causal, while case 2 refers to a stable and causal system.

Case 1 Stable (not necessarily causal) LTID system The LTIC stability condition for a stable system H (s) is that the ROC of H (s) must contain the

vertical imaginary jω-axis in the complex s-plane. Since the ROC cannot contain

any pole, in terms of the pole s = ps T , this implies that Re{ps T } �= 0 such that no pole exists on the imaginary jω-axis. Substituting Re{ps T } �= 0 into

Eq. (13.44) and calculating its magnitude yields

|pz| = ∣ ∣eRe{ps T }

∣ ∣

︸︷︷︸

term I �=1 if Re{psT}�=0

× ∣ ∣ej Im{ps T }

∣ ∣

︸︷︷︸

term II=1

�= 1, (13.45)

which implies that an LTID system H (z) is stable if there is no pole on the unit

circle of the z-plane. In terms of the ROC, it implies that the ROC must contain

the unit circle for the system to be stable. The above condition does not assume

the system to be causal, which is considered next.

Case 2 Stable and causal LTID system The LTIC stability condition for a stable and causal system H (s) is that all poles of H (s) must lie in the left-

half of the complex s-plane. In terms of the pole s = ps T , this implies that

Re{ps T } < 0. Substituting Re{ps T } < 0 into Eq. (13.44) and taking the mag-

nitude yields

|pz| = ∣ ∣eRe{ps T }

∣ ∣

︸︷︷︸

term I<1 if Re {psT} < 0

× ∣ ∣ej Im{ps T }

∣ ∣

︸︷︷︸

term II=1

< 1. (13.46)

Equation (13.46) states that an LTID system H(z) is stable if all poles lie within

the unit circle. Alternatively, the requirement for a causal and stable LTID

system is stated as follows.

An LTID system will be absolutely BIBO stable and causal if and only if the

ROC occupies the region outside and inclusive of the unit circle. In other words,

the ROC for a stable and causal system is given by |z| > z0, with z0 < 1.

Example 13.16

Consider the LTID systems in Example 13.13. Considering various possibilities

of the ROC, determine if the systems are absolutely BIBO state.

Determine if the systems are absolutely BIBO stable.

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603 13 The z-transform

Solution

(i) Since

H1(z) = z

z2 − 3z + 2 =

(z − 1)(z − 2) ,

there are two poles of the LTID system, at z = 1 and 2. Since one pole lies on the unit circle, the ROC cannot contain the unit circle. The LTID system H1(z)

is therefore not absolutely BIBO stable.

(ii) Since

H2(z) = 1

(z − 0.1)(z − 0.5)(z + 0.2) ,

there are three poles, at z = 0.1, 0.5, and −0.2. There are three choices for the ROC, which are given by

ROC 1: |z| < 0.1. Such an implementation of the LTID system is not abso- lutely stable since the ROC does not contain the unit circle.

ROC 2: 0.1 < |z| < 0.2. Such an implementation is not absolutely stable since the ROC does not contain the unit circle.

ROC 3: 0.2 < |z| < 0.5. Such an implementation is not absolutely stable since the ROC does not contain the unit circle.

ROC 4: |z| > 0.5. Such an implementation is absolutely stable since the ROC contains the unit circle.

(iii) Since

H3(z) = z2(2z − 1.5)

(z + 0.4)(z − 0.5)2 ,

there are three poles, at z = −0.4, 0.5, and 0.5. There are three choices for the ROC, which are given by

ROC 1: |z| < 0.4. Such an implementation of the LTID system is not abso- lutely stable since the ROC does not contain the unit circle.

ROC 2: 0.4 < |z| < 0.5. Such an implementation of the LTID system is not absolutely stable since the ROC does not contain the unit circle.

ROC 3: |z| > 0.5. Such an implementation of the LTID system is absolutely stable since the ROC contains the unit circle.

(iv) Since

H4(z) = z2 + 0.7z + 1.6

(z2 − 1.2z + 1)(z + 0.3) =

(z − 0.5)(z + 1.2) (z − 0.6 + j0.8)(z − 0.6 − j0.8)(z + 0.3)

there are three poles, at z = 0.6 − j0.8, 0.6 + j0.8, and −0.3. The three choices of the ROC are given by

ROC 1: |z| < 0.3. Such an implementation of the LTID system is not abso- lutely stable since the ROC does not contain the unit circle.

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ROC 2: 0.3 < |z| < |0.6 ± j0.8| or 0.3 < |z| < 1. Such an implementation of the LTID system is not absolutely stable since the ROC does not contain

the unit circle.

ROC 3: |z| > |0.6 ± j0.8| or |z| > 1. Such an implementation of the LTID system is not absolutely stable since the ROC does not contain the unit

circle.

13.8.1 Marginal stability

Equation (13.46) can be used to determine if a causal LTID system is absolutely

stable. An absolutely stable and causal system has all poles inside the unit circle

in the complex z-plane. On the contrary, if a causal system has one or more

poles outside the unit circle then the system will not be absolutely stable. The

impulse response of such a system includes a growing exponential function,

making the system unstable. An intermediate case arises when a causal system

has unrepeated poles on the unit circle and the remaining poles are inside the

circle in the complex z-plane. Such a system is referred to as a marginally

stable system. The condition for marginally stable and causal system is stated

below.

A causal system with M unrepeated poles pm = am + jbm, 1 ≤ m ≤ M, on the unit circle (such that |pm | = 1) and all the remaining poles inside the unit circle in the z-plane is stable for all bounded input signals that do not include

complex exponential terms of the form {exp(jΩmk)}, withΩm = tan−1(bm/am), for 1 ≤ m ≤ M. If any of the poles on the unit circle are repeated then the LTID system is unstable.

The following example demonstrates that a marginally stable system becomes

unstable if the input signal includes a complex exponential exp(jΩm) with

frequency Ωm = tan−1(bm/am) corresponding to the location of the pole at pm = am + jbm on the unit circle in the complex z-plane.

Example 13.17

A causal LTID system with transfer function given by

H (z) = 1

z2 − z + 1 =

(z − 0.5 − j( √

3/2))(z − 0.5 + j( √

3/2))

is a marginally stable system because of two unrepeated poles, at z = 0.5 ± j0.866, on the unit circle. We will demonstrate the marginal stability by calcu-

lating the output for the following bounded input sequences:

(i) x1[k] = u[k]; (ii) x2[k] = sin(πk/3)u[k].

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605 13 The z-transform

Solution

(i) Taking the z-transform of the input sequence, we obtain

X1(z) = z

z − 1 .

Applying the convolution property, the z-transform Y1(z) of the output response

is given by

Y1(z) = H (z)X1(z) = z

(z − 1)(z2 − z + 1) =

z−2

(1 − z−1)(1 − z−1 + z−2) .

Taking the partial fraction expansion of Y1(z) yields

Y1(z) = 1

1 − z−1 −

1 − z−1 + z−2 .

Using entries (3) and (12) of Table 13.1 (see Problem 13.5), we obtain

u[k] z←→

1 − z−1

and

2 √

3 sin

( πk

3 +

)

u[k] z←→

1 − z−1 + z−2 .

Using the linearity property, the output y1[k] is given by

y1[k] = [

1 − 2

√ 3

sin

( πk

3 +

)]

u[k].

Note that the output response contains a unit step function and a sinusoidal term

and is, therefore, bounded.

(ii) Taking the z-transform of the input sequence, we obtain

X2(z) = ( √

3/2)z−1

1 − z−1 + z−2 .

Applying the convolution property, the z-transform Y2(z) of the output response

is given by

Y2(z) = H (z)X2(z) = ( √

3/2)z−1

1 − z−1 + z−2 ·

z−2

1 − z−1 + z−2 =

( √

3/2)z−3

(1 − z−1 + z−2)2 .

Using the frequency-differentiation property (see Problem 13.6), it can be

shown that the following is a z-transform pair: [

3 sin

(π

3 k )

− k

√ 3

sin (π

3 k +

) ]

u[k] z←→ =

( √

3/2)z−3

(1 − z−1 + z−2)2 .

Therefore, the output response is given by

y2[k] = [

3 sin

(π

3 k )

− k

√ 3

sin (π

3 k +

) ]

u[k].

Note that the output is a growing sinusoid function because of the k/ √

3 scaling

factor. Therefore, as k increases, the |y2[k]| increases without bound, leading to an unstable situation.

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Table 13.3. Discrete frequencies corresponding to a few selected points along the unit circle in the z-domain

z-coordinates 1 + j0 1

√ 2

+ j 1

√ 2

0 + j1 − 1

√ 2

+ j 1

√ 2

−1 + j0 − 1

√ 2

− j 1

√ 2

0 − j1 1

√ 2

− j 1

√ 2

Frequency, Ω 0 π/4 π/2 3π/4 π 5π /4 3π/2 7π /4

In this example, we observe that the output response for the first input signal

x1[k] = u[k] is bounded. On the other hand, the output produced by the second input, x2[k] = sin(πk/3)u[k], is unbounded. Note that the second input is a sinusoidal sequence, which contains two complex exponentials:

sin

( πk

)

u[k] = 1

[

ejπk/3 − e−jπk/3 ]

with discrete frequencies Ωm = ±π/3. Since the frequencies of the complex exponentials are the same as the value of tan−1(bm/am) = tan−1(±

√ 3/4) =

±π/3, determined from the poles, at z = 0.5 ± j √

3/2, on the unit circle, the

output response is unbounded. This is consistent with the marginal stability

condition mentioned above.

13.9 Frequency-response calculation in the z-domain

Based on Eq. (13.8), the DTFT transfer function is related to the z-transfer

function by the following relationship:

H (Ω) = ∞∑

k=−∞ h[k]z−k = H (z)|z=ejΩ , (13.47)

which may be used to derive the DTFT transfer function from the z-transfer

function. Equation (13.47) has wider implications, as we discuss in the follow-

ing.

(1) Taking the magnitude of both sides of the relationship z = exp(jΩ) gives |z| = 1; therefore, Eq. (13.47) is only valid if the ROC of the z-transfer function contains the unit circle. Otherwise, the substitution z = exp(jΩ) cannot be made and the DTFT transfer function does not exist.

(2) Equation (13.47) can also be used to compute the magnitude and phase

spectra of the LTID system by evaluating the z-transfer function at dif-

ferent frequencies (0 ≤ Ω ≤ 2π ) along the unit circle. The correspon- dence between the discrete frequency Ω and the z-coordinates is shown in

Fig. 13.9. A selected subset of the discrete frequencies along the unit circle

is shown in Table 13.3.

The computation of the magnitude and phase spectra from the z-transfer

function is illustrated in the following example.

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607 13 The z-transform

Im{z}

Re{z}

( ) ⇒ W = 3p1 42

,− 1

2 ( ) ⇒ W = p1

2 , 1

( ) ⇒ W = 5p1 42

,− − 1

2 ( ) ⇒ W = 7p1

42 , −

(−1, 0) ⇒ W = p

(1, 0) ⇒ W = 0

⇒ W =

2 ⇒ W =

(0, −1)

(0, 1)

Fig. 13.9. Determination of the

magnitude and phase spectra

from the z-transfer function.

0 0p/2 p−p −p/2 p/2 p−p −p/2 W

.0.245p

−0.245p

H(W) <H(W)

(a) (b)

Fig. 13.10. (a) Magnitude

spectrum and (b) phase

spectrum of the LTID system

considered in Example 13.18.

The responses are shown in the

frequency rangeΩ = [−π, π ].

Example 13.18

Consider the system with z-transfer function given by

H (z) = 2z2

z2 − (3/4)z + (1/8) =

1 − (3/4)z−1 + (1/8)z−2 .

Calculate and plot the amplitude and phase spectra of the system.

Solution

The DTFT transfer function is given by

H (Ω) = H (z)|z=ejΩ = 2

1 − (3/4)e−jΩ + (1/8)e−j2Ω .

The magnitude spectrum |H (Ω)| and the phase spectrum <H (Ω) are plotted in Fig. 13.10, which are identical to the spectra shown in Fig. 11.18.

13.10 DTFT and the z-transform

In Chapter 11 and in this chapter, we presented two different frequency-domain

approaches to analyze DT signals and systems. The DTFT-based approach,

introduced in Chapter 11, uses the real frequency Ω, whereas the z-transform-

based approach uses the complex frequency σ + jΩ. The output response of

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608 Part III Discrete-time signals and systems

an LTID system can be computed using the convolution property of either the

DTFT or the z-transform. In addition, the frequency-domain approach offers

insight about the system characteristics, which is not readily available from the

time-domain approach. However, an important issue is to determine which of

the two transforms should be used to analyze the LTID system. Both approaches

have their own advantages. Depending upon the application under considera-

tion, the appropriate transform should be selected.

Example 13.19

Consider an LTID system represented by the unit impulse response h[k] = 0.8ku[k]. Calculate the overall output and steady state output of the LTID system

for the input sequence x[k] = cos(πk/3)u[k].

Solution

z-transform method Using Table 13.1, the z-transforms of the impulse response h[k] and the input x[k] are given by

H (z) = 1

1 − 0.8z−1

and

X (z) = 1 − z−1 cos(π/3)

1 − 2z−1 cos(π/3) + z−2 =

1 − 0.5z−1

1 − z−1 + z−2 .

Using the convolution property, the z-transform of the output response is given

Y (z) = H (z)X (z) = 1 − 0.5z−1

(1 − 0.8z−1)(1 − z−1 + z−2) .

By partial fraction expansion, the above expression becomes

Y (z) = 2

7 ×

1 − 0.8z−1 +

7 ×

1 + 0.5z−1

1 − z−1 + z−2

= 2

7 ×

1 − 0.8z−1 +

7 ×

1 − 0.5z−1

1 − z−1 + z−2 +

7 ×

z−1

1 − z−1 + z−2 .

Taking the inverse z-transform, the output response is given by

y[k] = 2

7 × 0.8ku[k] +

7 × cos

( πk

)

u[k] + 10

7 √

3 × sin

( πk

)

u[k]

= [

0.287(0.8)k + 1.091 cos (

πk

3 − 0.857r

) ]

u[k]

where the superscript r indicates that the angle is expressed in radians.

The steady state output yss[k] is computed by neglecting the transient term

(0.8)k , which decays to zero with time. The steady state output response is,

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609 13 The z-transform

therefore, given by

yss[k] = 1.091 cos (

πk

3 − 0.857r

)

u[k].

DTFT method As in the CT case, the calculation of the actual output is difficult using the DTFT. However, the steady state value of the output can be

easily calculated using DTFT. We have

H (Ω) = 1

1 − 0.8e−jΩ .

The value of the DTFT transfer function atΩ = π/3, the fundamental frequency of the sinusoidal input, is given by

H (Ω)|Ω=π/3 = 1

1 − 0.8e−j(π/3) = 0.714 − j0.285 = 1.091e−j0.857,

implying that |H (Ω)| = 1.091 and <H (Ω) = −0.857 radians. Therefore, the steady state output response is given by

yss[k] = |H (Ω)| × cos (

πk

3 + <H (Ω)

)

u[k]

= 1.091 cos (

πk

3 − 0.857r

)

u[k].

Example 13.19 shows that the z-transform is a more convenient tool for tran-

sient analysis. For the steady state analysis, the z-transform does not offer much

advantage over the DTFT. In signal processing applications, such as audio,

image and video processing, the transients are generally ignored. In such appli-

cations, the DTFT is sufficient to analyze the steady state response. On the

other hand, the transient analysis is important for applications such as control

systems and process control. This is precisely the reason for the widespread use

of the z-transform in digital control and system design, whereas the DTFT is

preferred in signal processing applications.

13.11 Experiments with M A T L A B

M A T L A B provides several M-files for working with z-transforms. In this sec-

tion, we explore five important functions, residuez, residue, tf2zp, zp2tf, andzplane. To illustrate the application of these M-files, we consider the following linear, constant-coefficient difference equation representation:

an y[k] + an−1 y[k − 1] + · · · + a0 y[k − n] = bm x[k] + bm−1x[k − 1] + · · · + b0x[k − m],

for modeling the relationship between the input sequence x[k] and output

response y[k] of an LTID system. The above equation is a more general case

of Eq. (13.35), where an was set to 1. Recall that Section 10.9 covered the

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610 Part III Discrete-time signals and systems

M A T L A B file filter used to compute the output response y[k] from spec- ified sample values of the input sequence x[k] and the ancillary conditions. In

this section, we focus on the z-transfer function representation,

H (z) = Y (z)

X (z) =

bm + bm−1z−1 + · · · + b0z−m

an + an−1z−1 + · · · + a0z−n , (13.48)

which can also be factorized as follows:

H (z) = Y (z)

X (z) = K

(1 − z0z−1)(1 − z1z−1) · · · (1 − zM z−1) (1 − p0z−1)(1 − p1z−1) · · · (1 − pN z−1)

. (13.49)

Since M A T L A B assumes that the numerator and denominator of the z-transfer

function are expressed in increasing powers of z−1, we prefer the aforemen-

tioned format for the z-transfer function.

13.11.1 Partial fraction expansion

To calculate the partial fraction expansion of a rational z-transfer function,

M A T L A B provides the residuez function, which has the following syntax:

>> [R,P,K] = residuez(B,A);

In terms of the transfer function in Eq. (13.48), the input variables B and A are defined as follows:

A = [an an−1 . . . a0] and B = [bm bm−1 . . . b0].

The output parameter R returns the values of the partial fraction coefficients, P returns the location of the poles, while K contains the direct term in the row vector.

Example 13.20

To illustrate the usage of the built-in function residuez, let us calculate the partial fraction expansion of the z-transfer function,

H (z) = 2z(3z + 17)

(z − 1)(z2 − 6z + 25) ,

considered in Example 13.4(iii).

Solution

Expressing the z-transfer function in increasing powers of z−1 yields

H (z) = 6z−1 + 34z−2

1 − 7z−1 + 31z−2 − 25z−3 .

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The M A T L A B code to determine the partial fraction expansion is given below.

The explanation follows each instruction in the form of comments.

>> B = [0; 6; 34; 0]; % Coeff. of the numerator N(z) >> A = [1; -7; 31; -25]; % Coeff. of the denominator D(z) >> [R,P,K] = residuez(B,A) % Calc. partial fraction expansion

The returned values are given by

R = [-1.0000-1.2500j, -1.0000+1.2500j, 2.0000]

P = [3.0000+4.0000j, 3.0000-4.0000j, 1.0000] and K=[].

The transfer function H (z) can therefore be expressed as follows:

H (z) = −1 − j1.25

1 − (3 + j4)z−1 +

−1 + j1.25 1 − (3 − j4)z−1

+ 2

1 − z−1 .

Alternative partial fraction expansion Sometimes, it is desirable to perform the partial fraction in terms of the polynomials of z, instead of the polynomials

of z−1. In such cases, the M A T L A B function residue is used. We solve Example 13.20 in terms of the alternative expression for the transfer function,

H (z)

z =

6z + 34 z3 − 7z2 + 31z − 25

The M A T L A B code to determine the partial fraction expansion of the alternative

expression is given below. As before, the explanation follows each instruction

in the form of comments.

>> B = [0; 0; 6; 34]; % Coeff. of the numerator N(z) >> A = [1; -7; 31; -25]; % Coeff. of the D(z) >> [R,P,K] = residue(B,A) % Calc. partial fraction expansion

The returned values are given by

R = [-1.0000-1.2500j, -1.0000+1.2500j, 2.0000]

P = [3.0000+4.0000j, 3.0000-4.0000j, 1.0000] and K = [].

The transfer function H (z) can therefore be expressed as follows:

H (z)

z =

−1 − j1.25 z − (3 + j4)

+ −1 + j1.25 z − (3 − j4)

+ 2

z − 1 ,

which is the same as result obtained in Example 13.4(iii).

13.11.2 Computing poles and zeros from the z-transfer function

M A T L A B provides the built-in function tf2zp to calculate the location of the poles and zeros from the z-transfer function. Another function zplane can be

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612 Part III Discrete-time signals and systems

used to plot the poles and zeros in the complex z-plane. In terms of Eq. (13.48),

the syntaxes for these functions are given by

>> [Z,P,K] = tf2zp(B,A); % Calculate poles and zeros

>> zplane(Z,P); % plot poles and zeros,

where the input variables B and A are defined as follows:

A = [an an−1 . . . a0] and B = [bm bm−1 . . . b0].

They are obtained from the transfer function given in Eq. (13.48). The vector

Z contains the location of the zeros, vector P contains the location of the poles, while K returns a scalar providing the gain of the numerator.

Example 13.21

For the z-transfer function

H (z) = 2z(3z + 17)

(z − 1)(z2 − 6z + 25) ,

compute the poles and zeros and give a sketch of their locations in the complex

z-plane.

Solution

The M A T L A B code to determine the location of zeros and poles is listed below.

The explanation follows each instruction in the form of comments.

>> B = [0, 6, 34, 0]; % Coefficients of the numerator N(z) >> A = [1, -7, 31, -25]; % Coefficients of the denominator D(z) >> [Z,P,K] = tf2zp(B,A) % Calculate poles and zeros >> zplane(Z,P) % plot poles and zeros

The returned values are given by

Z = [0, -5.6667],

P = [3.0000+4.0000j 3.0000-4.0000j 1.0000] and K = 6.

The transfer function H (z) can therefore be expressed as follows:

H (z) = 6 z(z + 5.6667)

(z − (3 + j4))(z − (3 − j4))(z − 1)

= 6 z−1(1 + 5.6667z−1)

(1 − (3 + j4)z−1)(1 − (3 − j4)z−1)(1 − z−1) .

The pole–zero plot for H (z) is shown in Fig. 13.11.

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−6 −4 −2 0 real part

im ag

in ar

y p

ar t

2 4

−4

−3

−2

−1

4Fig. 13.11. Location of poles and

zeros obtained in Example 13.21

using MATLAB

13.11.3 Computing the z-transfer function from poles and zeros

M A T L A B provides the built-in function zp2tf to calculate the z-transfer function from poles and zeros. In terms of Eq. (13.49), the syntax for zp2tf is given by

>> [B,A] = zp2tf(Z,P,K); % Calculate poles and zeros

where vector Z contains the location of the zeros, vector P contains the location of the poles, andK is a scalar providing the gain of the numerator. The numerator coefficients are returned in B and the denominator coefficients in A.

Example 13.22

Consider the poles and zeros calculated in Example 13.21. Using the values of

the poles and, zeros and the gain factor, determine the transfer function H (z).

Solution

The M A T L A B code to determine the coefficients of the transfer function is

listed below. The explanation follows each instruction in the form of comments.

>> Z = [0; -5.666667]; % Zeros in a column vector

>> P = [3+4 * j; 3-4 * j; 1]; % Poles in a column vector

>> K = 6; % Gain of the numerator

>> [B,A] = zp2tf(Z,P,K); % Calculate poles and zeros

The returned values are given by

B = [0 6 34 0] and A = [1 -7 31 -25],

which implies that the transfer function is given by

H (z) = 6z2 + 34z

z3 − 7z2 + 31z − 25 .

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614 Part III Discrete-time signals and systems

The aforementioned expression is identical to the transfer function specified in

Example 13.21.

13.12 Summary

In this chapter, we defined the bilateral z-transform for DT sequences as follows:

z-analysis equation X (z) = ℑ {

x[k]e−σk }

= ∞∑

k=−∞ x[k]z−k .

z-synthesis equation x[k] = 1

2π j

∮

X (z)zk−1dz.

Unlike the DTFT, which requires the DT sequences to be absolutely summable

for the DTFT to exist, the z-transform exists for a much larger set of DT

sequences. Associated with the bilateral z-transform is a region of convergence

(ROC) in the complex z-plane over which the z-transform is defined.

For causal sequences, the bilateral z-transform simplifies to the unilateral

z-transform, defined in Section 13.2 as follows:

unilateral z-transform X (z) = ∞∑

k=0 x[k]z−k .

Section 13.3 introduced the look-up table, the partial fraction expansion, and

the power-series-based approaches for determining the inverse z-transform of

a rational function.

Section 13.4 presented the properties of the z-transform, which are summa-

rized in the following.

(1) The linearity property states that the overall z-transform of a linear com-

bination of DT sequences is given by the same linear combination of the

individual z-transforms.

(2) The time-scaling property is only applicable for time-expanded (or inter-

polated) sequences. The time-scaling property states that interpolating a

sequence in the time domain compresses its z-transform in the complex

z-domain.

(3) The time-shifting property states that shifting a sequence in the time domain

towards the right-hand side by an integer constant m is equivalent to mul-

tiplying the z-transform of the original sequence by a complex term z−m .

Similarly, shifting towards the left-hand side by integer m is equivalent to

multiplying the z-transform of the original sequence with a complex term

zm .

(4) Time differencing is defined as the difference between an original sequence

and its time-shifted version with a shift of one sample towards the right-

hand side. The time-differencing property states that time-differencing a

signal in the time domain is equivalent to multiplying its DTFT by a factor

of (1 − z−1).

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615 13 The z-transform

(5) The z-domain-differentiation property states that differentiating the z-

transform with respect to z and then multiplying with the variable −z is equivalent to multiplying the original sequence by a factor of k.

(6) The time-convolution property states that the convolution of two DT

sequences is equivalent to the multiplication of the z-transforms of the

two sequences in the z-domain.

(7) The time-accumulation property is the converse of the time-differencing

property. The accumulation property states that the z-transform of the run-

ning sum of a sequence is obtained by multiplying the z-transform of the

original sequence by a factor of z/(z − 1). (8) The initial- and final-value theorems can be used to determine the initial

value at k = 0 and final value at k → ∞ directly from the z-transform of a DT sequence.

Section 13.5 covered the application of the z-transform in solving finite-

difference equations and showed how the z-transfer function can be obtained

from a difference equation of the following form:

H (z) = Y (z)

X (z) =

bm + bm−1z−1 + · · · + b0z−m

an + an−1z−1 + · · · + a0z−n .

Section 13.6 defined the characteristic equation, poles, and zeros of an LTID

system from the above rational expression of the z-transfer function. The char-

acteristic equation for the transfer function is based on the denominator D(z)

of the z-transfer function H (z) and is defined as follows:

D(z) = anzn + an−1zn−1 + · · · + a0 = 0.

The roots of the characteristic equation define the poles of the LTID system as

locations in the complex z-plane, where |H (z)| has an infinite value. Similarly, the zeros of the transfer function H (z) of an LTID system are finite locations

in the complex z-plane where |H (z)| approaches zero. If N (z) is the numerator of H (z), the zeros can be obtained by calculating the roots of the following

equation:

N (z) = bm zm + bm−1zm−1 + · · · + b0 = 0.

Sections 13.7 and 13.8 exploited the relationship between the z-transfer function

H (z) of an LTID system and the Laplace transfer function H (s) of an equivalent

LTIC system to derive the stability conditions for a causal and stable LTID

system.

Section 13.9 showed how the magnitude and phase spectra can be obtained

directly from the z-transform, while Section 13.10 compared the z-transfer-

function-based analysis techniques with those based on the DTFT transfer

function. We showed that the z-transform is a more convenient tool for transient

analysis, while the DTFT is more appropriate for steady state analysis.

Finally, Section 13.11 illustrated some M A T L A B library functions used to

analyze the LTID systems in the complex z-domain.

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616 Part III Discrete-time signals and systems

Problems

13.1 Calculate the bilateral z-transform of the following non-causal functions: (i) x1[k] = 0.5k+1u[k + 5];

(ii) x2[k] = (k + 2)0.5|k|; (iii) x3[k] = |k + 2| × 0.5|k+2|; (iv) x4[k] = 3k+1 cos

(π

3 k −

)

u[−k + 5].

13.2 Calculate the unilateral z-transform of the following causal functions:

(i) x1[k] =







1 k = 10, 11 2 k = 12, 15 0 otherwise;

(ii) x2[k] = 3−k+2u[k] + 4∑

m=1 mδ[k − m];

(iii) x3[k] = sin (

πk

5 +

)

u[k];

(iv) x4[k] = 2−k sin (

πk

5 +

)

u[k];

(v) x5[k] = ku[k].

13.3 Using the partial fraction method, calculate the inverse z-transform of the following DT causal sequences:

(i) X1(z) = z

z2 − 0.9z + 0.2 ;

(ii) X2(z) = z

z2 − 2.1z + 0.2 ;

(iii) X3(z) = z2 + 2

(z − 0.3)(z + 0.4)(z − 0.7) ;

(iv) X4(z) = z2 + 2

(z − 0.3)(z + 0.4)2 ;

(v) X5(z) = 4z−1

1 − 5z−1 + 6z−2 ;

(vi) X6(z) = 4z−2

10 − 6(z1 + z−1) ;

(vii) X7(z) = 2z−2

(1 − 4z−1)2(1 − 2z−1) .

13.4 Using the power series expansion method, calculate the inverse z-transform of the DT causal sequences in Problem 13.3 for the first

five non-zero values.

13.5 (a) Prove entry (12) of Table 13.1. (b) Using the proved result, derive the following z-transform pair used in Example 13.17(i):

2 √

3 sin

( πk

3 +

)

u[k] z←→

1 − z−1 + z−2 .

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617 13 The z-transform

13.6 (a) Using the z-domain-differentiation property and pairs (9) and (12) in Table 13.1, show that

(i) k sin(Ω0k)u[k] z←→

z(z2 − 1) sinΩ0 (z2 − 2z cosΩ0 + 1)2

, ROC: |z| > 1;

(ii) k sin(Ω0k + θ )u[k] z←→

z[sin(Ω0+θ )z2−2z sin θ − sin(Ω0 − θ )] (z2 − 2z cosΩ0 + 1)2

ROC: |z| > 1. (b) Using the above result, or otherwise, prove the following z-transform

pair used in Example 13.17 (ii):

[ 2

3 sin

(π

3 k )

− k

√ 3

sin (π

3 k +

) ]

u[k] z←−−→

( √

3/2)z

(z2 − z + 1)2

= ( √

3/2)z−3

(1 − z−1 + z−2)2 , ROC: |z| > 1.

13.7 Using the time-shifting property and the results in Example 13.3(v), calculate the z-transform of the following function:

g[k] =







1 k = 10, 11 2 k = 12, 15 0 otherwise.

13.8 Prove the initial-value theorem stated in Section 13.4.8.

13.9 Prove the final-value theorem stated in Section 13.4.8.

13.10 Determine the z-transform of the following sequences using the specified property:

(i) x[k] = (5/6)ku[k − 6], based on the z-transform pair (4) in Table 13.1 and the time-shifting property;

(ii) x[k] = k(2/9)ku[k], based on the z-transform pair (4) in Table 13.1 and the z-domain differentiation property;

(iii) x[k] = ku(k), based on the z-transform pair (3) in Table 13.1 and the accumulation property;

(iv) x[k] = ek sin(k)u[k], based on the z-transform pair (4) in Table 13.1 and the linearity property.

13.11 By selecting different ROCs, calculate four possible impulse responses of the transfer function

H (z) = 1 − z−1

(1 − 0.5z−1)(1 − 0.75z−1)(1 − 1.25z−1) .

Determine the impulse response of the system that is stable. Is it causal?

Why or why not?

13.12 You are given the unit impulse response of an LTID system,

h[k] = 5−ku[k].

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618 Part III Discrete-time signals and systems

(i) Determine the impulse response hinv[k] of the inverse system that

satisfies the property

hinv[k] ∗ h[k] = δ[k].

(ii) Using any method, obtain the output y[k] of the original system

h[k] for each of the following inputs: (a) x1[k] = u[k]; (b) x2[k] =

5δ[k − 4] − 2δ[k + 4]; and (c) x3[k] = e (k+2)u[−k + 2].

13.13 You are hired by a signal processing firm and you are hoping to impress them with the skills that you have acquired in this course. The firm asks

you to design an LTID system that has the property that if the input is

given by

x[k] = (1/3)k u[k] − (1/4)k−1 u[k],

the output is given by

y[k] = (1/4)k u[k].

(i) Determine the z-transfer function of the LTID system.

(ii) Determine the impulse response of the LTID system.

(iii) Determine the difference-equation representation of the LTID sys-

tem.

13.14 The transfer function of a physically realizable system is as follows:

H (z) = 1

(1 − 0.3z−1)(1 − 0.5z−1)(1 − 0.7z−1) .

(i) Determine the impulse response of the LTID system.

(ii) Determine the difference-equation representation of the LTID sys-

tem.

(iii) Determine the unit step response of the LTID system by using the

time-convolution property of the z-transform.

(iv) Determine the unit step response of the LTID system by convolv-

ing the unit step sequence with the impulse response obtained in

part (i).

13.15 Given the difference equation

y[k] + y[k − 1] + 1

4 y[k − 2] = x[k] − x[k − 2],

(i) determine the transfer function representing the LTID system;

(ii) determine the impulse response of the LTID system;

(iii) determine the output of the LTID system to the input x[k] =

(1/2)ku[k] using the time-convolution property;

(iv) determine the output of the LTID system by convolving the input

x[k] = (1/2)ku[k] with the impulse response obtained in part (ii).

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619 13 The z-transform

13.16 Determine the output response of the following LTID systems with the specified inputs and impulse responses:

(i) x[k] = u[k + 2] − u[−k − 3] and h[k] = u[k − 5] − u[k − 6]; (ii) x[k] = u[k] = u[k − 9] and h[k] = 3−ku[k − 4];

(iii) x[k] = 2−ku[k] and h[k] = k(u[k] − u[k − 4]); (iv) x[k] = u[k] and h[k] = 4−|k|; (v) x[k] = 2−ku[k] and h[k] = 2ku[−k − 1].

13.17 When the DT sequence

x[k] = (1/4)k u[k] + (1/3)ku[k]

is applied at the input of a causal LTID system, the output response is

given by

y[k] = 2 (1/4)k u[k] − 4 (3/4)k u[k].

(i) Determine the z-transfer function H(z) of the LTID system.

(ii) Determine the impulse response h[k] of the LTID system.

(iii) Determine the difference-equation representation of the LTID sys-

tem.

13.18 Consider an LTIC system with the following transfer function:

H (s) = esT

esT − 0.3 .

Calculate the output response y(t) of the LTIC system for the following

input sequence:

f (t) = ∞∑

k=0 (0.2)kT δ(t − kT ).

13.19 Plot the poles and zeros of the following LTID systems. Assuming that the systems are causal, determine if the systems are BIBO stable.

(i) H (z) = z − 2

(z − 0.6 + j0.8)(z2 + 0.25) ;

(ii) H (z) = (z − 2)(z − 1)

(z2 − 2.5z + 1)(z2 + 0.25) ;

(iii) H (z) = z − 0.2

(z + 0.1)(z2 + 4) ;

(iv) H (z) = z−1 − 2z−2 + z−3;

(v) H (z) = (z2 + 2.5z + 0.9 + j0.15)z

z3 + (1.8 + j0.3)z2 + (0.6 + j0.6)z − 0.2 + j0.3 ;

(vi) H (z) = z3 − 1.2z2 + 2.5z + 0.8

z6 + 0.3z5 + 0.23z4 + 0.209z3 + 0.1066z2 − 0.04162z − 0.07134 .

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620 Part III Discrete-time signals and systems

13.20 Consider an LTID system with the following transfer function:

H (z) = z

z + 0.1 .

(i) Using M A T L A B , calculate the frequency response H (ejΩ) forΩ = [−π :π/20:π ]. Plot the amplitude and phase spectra.

(ii) If the DT signal x[k] = 5 cos(πk/10) is passed through the system, what will be the steady state output of the system?

13.21 (a) Using M A T L A B , determine the poles and zeros of the z-transfer functions specified in Problem 13.19. (b) Plot the location of poles and

zeros in the complex z-plane using M A T L A B .

13.22 (a) Using M A T L A B , determine the partial fraction expansion of the z-transfer functions specified in Problem 13.19. (b) From the par-

tial fraction expansion, calculate the impulse response function of the

systems.

13.23 Assume that the functions in Problem 13.3 are z-transfer functions of some causal LTID systems. (a) Using M A T L A B , determine the impulse

responses of these systems. (b) Plot the impulse responses.

C H A P T E R

14 Digital filters

A digital filter is defined as a system that transforms a sequence, applied at

the input of the filter, by changing its frequency characteristics in a predefined

manner. A convenient classification of digital filters is obtained by specifying the

shape of their magnitude and phase spectra in the frequency domain. Based on

the magnitude response, digital filters are classified in four important categories:

lowpass, highpass, bandpass, and bandstop. A lowpass filter removes the higher-

frequency components from an input sequence and is widely used to smooth out

any sharp changes present in the sequence. An example of lowpass filtering is the

elimination of the hissing noise present in magnetic audio cassettes. Since the

background hissing noise contains higher-frequency components than the music

itself, a lowpass filter removes the hissing noise. A highpass filter eliminates the

lower-frequency components and tends to emphasize sharp transitions in the

input sequence. An application of highpass filtering is the detection of edges of

different objects present in still images. While eliminating the smooth regions,

represented by low frequencies, within each object, a highpass filter retains

the boundaries between the objects. A bandpass filter allows a selected range

of frequencies, referred to as the pass band, within the input sequence to be

preserved at the output of the filter. All frequencies outside the pass band are

eliminated from the input sequence. Bandpass filters are used, for example,

in detecting the dual-tone multifrequency (DTMF) signals in digital telephone

systems. As shown in Fig. 14.1, each DTMF key is represented by a pair of

frequencies. At the receiver, a bank of bandpass filters, each tuned to one of the

seven frequencies specified in Fig. 14.1, is used to determine the pressed key

by isolating the pair of frequencies present in the transmitted signal. Bandstop

filters are the converse of bandpass filters and allow all frequencies, except those

in a specified stop band, to be retained at the output. An application of bandstop

filters is to eliminate narrow-band noise, seen as bright and dark blotches in

digital videos.

This chapter focuses on digital filters and introduces the basic filtering

concepts and implementations useful in the design of digital filters. Sec-

tion 14.1 describes four categories of frequency-selective filters, based on the

621

622 Part III Discrete-time signals and systems

magnitude characteristics of the transfer function H (Ω). A second classifica-

tion of digital filters is made on the basis of the length of the impulse response

h[k] and is covered in Section 14.2. Yet another classification of digital filters

is made on the basis of the linearity of the phase <H (Ω), which is presented

in Section 14.3. The impulse response of the ideal frequency-selective filters,

considered in Section 14.1, is infinite, which makes them physically unreal-

izable. Section 14.4 describes realizable implementations of the ideal filters,

which are causal. Sections 14.5–14.7 cover physical implementations of digital

filters using special-purpose hardware confined to delays, adders, and scalar

multipliers. During the actual implementation of digital filters in software or

hardware, the filter coefficients can only be represented with finite precision.

The impact of finite-precision arithmetic on the performance of digital filters

is covered in Section 14.8. Important M A T L A B library functions used in the

analysis of digital filters are presented in Section 14.9. Finally, Section 14.9

concludes the chapter with summary of the important concepts.

2ABC 3DEF1697 Hz

4GHI 5JKL 6MNO770 Hz

7PQRS 8TUVW 9WXYZ852 Hz

1477 Hz1336 Hz1209 Hz

7PQRS 8TUVW 9WXYZ941 Hz

Fig. 14.1. Dual-tone

multifrequency (DTMF) signals

used in digital telephone

systems.

14.1 Filter classification

A digital filter is often classified on the basis of the magnitude and phase spectra

derived from its transfer function. In this section, we consider a classification

based on the shape of the magnitude spectrum of the filter. In the case of ideal

filters, the shape of the magnitude spectrum is rectangular with a sharp transition

between the range of frequencies passed and the range of frequencies blocked

by the filter. The range of frequencies passed by the filter is referred to as the

pass band of the filter, while the range of blocked frequencies is referred to as

the stop band.

14.1.1 Ideal lowpass filter

The transfer function Hilp(Ω) of an ideal lowpass filter, with a cut-off frequency

of Ωc, is given by

Hilp(Ω) = {

1 |Ω| ≤ Ωc 0 Ωc < |Ω| ≤ π,

(14.1a)

which has a pass band of |Ω| ≤ Ωc and a stop band ofΩc ≤ |Ω| ≤ π . Since the

frequencyΩ = π is the highest frequency present in the DTFT, the lowpass filter

removes the higher frequencies in the range of Ωc < |Ω| ≤ π . The magnitude

response of the lowpass filter is shown in Fig. 14.2(a). It is observed that the

lowpass filter has a unity gain in the pass band and zero gain in the stop band.

Sometimes, a lowpass filter has a pass band gain different from unity. If the

gain is greater than one, the pass band signal is amplified, if the gain is less than

one, the pass band signal is attenuated.

623 14 Digital filters

W 0

Hilp(W)

Wc−Wc p−p

0 Wc1 Wc2−Wc1−Wc2 p−p W

Hibp(W)

0 Wc1 Wc2−Wc1−Wc2 p−p W

Hibs(W)

Hihp(W)

0 Wc−Wc p−p W

(a) (b)

Fig. 14.2. Magnitude response of ideal filters. (a) Lowpass filter; (b) highpass filter; (c) bandpass filter;

(d) bandstop filter.

The impulse response hilp[k] of the ideal lowpass filter is obtained by calcu-

lating the inverse DTFT of Eq. (14.1a) and is given by

hilp[k] = sin(kΩc)

kπ = Ωc

π sinc

(

kΩc

)

. (14.1b)

14.1.2 Ideal highpass filter

The transfer function Hihp(Ω) of an ideal highpass filter, with a cut-off frequency

of Ωc, is given by

Hihp(Ω) = {

0 |Ω| < Ωc 1 Ωc ≤ |Ω| ≤ π,

(14.2a)

which has a pass band of Ωc ≤ |Ω| ≤ π and a stop band of |Ω| < Ωc. From

the magnitude response of the highpass filter shown in Fig. 14.2(b), it is clear

that the highpass filter blocks the lower frequencies |Ω| < Ωc, while the higher

frequencies Ωc ≤ |Ω| ≤ π are passed with a unity gain.

The transfer function Hihp(Ω) of an ideal highpass filter is related to the

transfer function Hilp(Ω) of an ideal lowpass filter as follows:

Hihp(Ω) = 1 − Hilp(Ω), (14.2b)

provided that the cut-off frequenciesΩc of both filters are the same. Calculating

the inverse DTFT of Eq. (14.2b), the impulse response hihp[k] of the ideal

highpass filter is obtained:

hihp[k] = δ[k] − hilp[k]. (14.3a)

624 Part III Discrete-time signals and systems

Substituting the expression for hilp[k] given in Eq. (14.1b) into the above equa-

tion, the impulse response hihp[k] can be expressed as follows:

hihp[k] = δ[k] − hilp[k] = δ[k] − Ωc

π sinc

(

kΩc

)

. (14.3b)

14.1.3 Ideal bandpass filter

The transfer function Hibp(Ω) of an ideal bandpass filter, with cut-off frequencies

of Ωc1 and Ωc2, is given by

Hibp(Ω) = {

1 Ωc1 ≤ |Ω| ≤ Ωc2

0 Ωc1 < |Ω| and Ωc2 < |Ω|≤ π, (14.4a)

which has a pass band of Ωc1 ≤ |Ω| ≤ Ωc2 and a stop band of |Ω| ≤ Ωc1 and

Ωc2 ≤ |Ω| ≤ π . The magnitude response of the ideal bandpass filter is shown

in Fig. 14.2(c).

The transfer function Hibp(Ω) is expressed in terms of the transfer functions

of two ideal lowpass filters:

Hibp(Ω) = Hilp1(Ω)

∣

cut-off freq=Ωc2 − Hilp2(Ω)

∣

cut-off freq=Ωc1 . (14.4b)

Calculating the inverse DTFT of Eq. (14.4a), the impulse response hibp[k] of

the ideal bandpass filter can be expressed as follows:

hibp[k] = hilp1[k] ∣

∣

Ωc=Ωc2 − hilp2[k]

∣

Ωc=Ωc1 . (14.4c)

Substituting the expression for hilp[k] given in Eq. (14.1b) into the above equa-

tion, the impulse response hibp[k] of the ideal bandpass filter can be expressed

as follows:

hibp[k] = Ωc2

π sinc

(

kΩc2

)

− Ωc1

π sinc

(

kΩc1

)

. (14.4d)

Equation. (14.4b) shows that a bandpass filter can be formed by a parallel

configuration of two lowpass filters. The first lowpass filter in the parallel con-

figuration should have a cut-off frequency of Ωc2, while the second lowpass

filter has a cut-off frequency of Ωc1. Other configurations of bandpass filters

are also possible, such as a series combination of a lowpass and a highpass

filter.

14.1.4 Ideal bandstop filter

The transfer function Hibs(Ω) of an ideal bandstop filter, with cut-off frequencies

Ωc1 and Ωc2, is given by

Hibs(Ω) =

{

0 Ωc1 ≤ |Ω| ≤ Ωc2

1 |Ω| < Ωc1 and Ωc2 < |Ω| ≤ π, (14.5a)

which has a pass band of |Ω| < Ωc1 and Ωc2 < |Ω| ≤ π and a stop band of

Ωc1 ≤ |Ω| ≤ Ωc2. The magnitude response of the ideal bandstop filter is shown

in Fig. 14.2(d).

625 14 Digital filters

Table 14.1. Impulse response of ideal lowpass, highpass, bandpass, and bandstop

filters in terms of normalized cut-off frequencies, Ωn = Ωc/π The pass-band gain is assumed to be unity. For bandpass and bandstop filters, there

are two cut-off frequencies, and Ωn2 > Ωn1

Filter Normalized cut-

Type off frequency Ideal filter impulse response

Lowpass Ωn hilp[k] = Ωn sinc[kΩn] Highpass Ωn hilp[k] = δ[k] − Ωn sinc[kΩn] Bandpass Ωn1,Ωn2 hibp[k] = Ωn2 sinc[kΩn2] − Ωn1 sinc[kΩn1] Bandstop Ωn1,Ωn2 hibs[k] = δ[k] − Ωn2 sinc[kΩn2] + Ωn1 sinc[kΩn1]

The transfer function Hibs(Ω) of an ideal bandstop filter is related to the

transfer function Hibp(Ω) of an ideal bandpass filter by

Hibs(Ω) = 1 − Hibp(Ω), (14.5b)

provided that the the cut-off frequenciesΩc1 andΩc2 of both filters are the same.

Calculating the inverse DTFT of Eq. (14.5b), the impulse response hibs[k] of

the ideal bandstop filter is obtained:

hibs[k] = δ[k] − hibp[k] ∣

∣

Ωc=Ωc2,Ωc1

= δ[k] = hilp1[k] ∣

∣

Ωc=Ωc2 − hilp2[k]

∣

Ωc=Ωc1 (14.6)

= δ[k] − Ωc2

π sinc

(

kΩc2

)

− Ωc1

π sinc

(

kΩc1

)

Equation (14.6) shows that a bandstop filter can be formed by a parallel con-

figuration of two lowpass filters having cut-off frequencies Ωc2 and Ωc1.

The impulse responses of the four types of frequency-selective ideal filters

discussed above are summarized in Table 14.1 in terms of the normalized cut-

off frequencies. It is observed that the impulse responses primarily include one

or two sinc functions and that all four types of ideal filters are non-causal.

14.2 FIR and IIR filters

A second classification of digital filters is made on the length of their impulse

response h[k]. The length (or width) of a digital filter is the number N of samples

k beyond which the impulse response h[k] is zero in both directions along the

k-axis. A filter of length N is also referred to as an N -tap filter.

A finite impulse response (FIR) filter is defined as a filter whose length N

is finite. On the other hand, if the length N of the filter is infinite, the filter is

called an infinite impulse response (IIR) filter. Below, we provide examples of

FIR and IIR filters with length N specified in the parentheses.

626 Part III Discrete-time signals and systems

h[k] = 0.6k u[k]

0.6 0.36

0 1 2 3 4 5−4 −3 −2 −1−5

> 3

≤ 3

1 − 3

|k|

|k| h[k] =

|k|

0 1 2 3 4 5−4 −3 −2 −1−5

0.67

0.33

0.67

0.33

(a) (b)

Fig. 14.3. (a) FIR filter; (b) IIR

filter.

FIR filters

Triangular sequence h[k] =







1 − |k| 3

|k| ≤ 3 0 elsewhere

(N = 5);

shifted impulse sequence h[k] = 0.1δ[k − 2] + δ[k] + 0.2δ[k − 2]

(N = 5);

exponentially decaying triangular sequence h[k] =

5 ∑

m=−5

0.4|k|δ[k − m]

(N = 11);

decaying impulses h[k] =

10 000 ∑

m=0

m + 1 δ[k − m] (N = 10 001).

IIR filters

Causal decaying exponential h[k] = 0.6ku[k] (N = ∞);

causal decaying sinusoidal h[k] = 0.5k sin(0.2πk)u[k] (N = ∞).

Other examples of IIR filters include non-causal ideal filters as shown in

Table 14.1.

Figure 14.3(a) plots the triangular sequence with length N = 5 as an example

of the FIR filter. Likewise, Fig. 14.3(b) plots the causal decaying exponential

sequence with infinite length as an example of the IIR filter. An important

consequence of a finite-length impulse response h[k] is observed during the

determination of the output response of an FIR filter resulting from a finite-

length input sequence. Since the output response is obtained by the convolution

of the impulse response and the input sequence, the output of an FIR filter is

finite in length if the input sequence itself is finite in length. On the other hand,

an IIR filter produces an output response that is always infinite in length.

A second consequence of the finite length of the FIR filters is observed in the

stability characteristics of such filters. Recall that an LTID system with impulse

response function h[k] is BIBO stable if

∞ ∑

k=−∞

|h[k]| < ∞.

627 14 Digital filters

Since the FIR filter is non-zero for only a limited number of samples k, the

stability criterion is always satisfied by an FIR filter. As IIR filters contain an

infinite number of impulse functions, even if the amplitudes of the constituent

impulse functions are finite, the summation ∑

h[k] in an IIR filter may not be

finite. In other words, it is not guaranteed that an IIR filter will always be stable.

Therefore, care should be taken when designing IIR filters so that the filter is

stable.

The implementation cost, typically measured by the number of delay ele-

ments used, is another important criterion in the design of filters. IIR filters are

implemented using a feedback loop, in which the number of delay elements is

determined by the order of the IIR filter. The number of delay elements used in

FIR filters depends on its length, and so the implementation cost of such filters

increases with the number of filter taps. An FIR filter with a large number of

taps may therefore be computationally infeasible.

14.3 Phase of a digital filter

In Section 14.1, we introduced ideal frequency-selective filters as having rect-

angular magnitude response with sharp transitions between the pass band and

stop band. The phase of ideal filters is assumed to be zero at all frequencies. An

ideal filter is physically unrealizable because of the sharp transitions between

the pass bands and stop bands and also because of the zero phase. In this sec-

tion, we illustrate the effect of the phase on the performance of digital filters.

In particular, we show that distortionless transmission within the pass band can

be achieved by using a filter having a linear phase within the pass band.

Consider the following sinusoidal sequence:

x[k] = A1 cos(Ω1k) + A2 cos(Ω2k) + A3 cos(Ω3k),

consisting of three tone frequencies Ω1 < Ω2 < Ω3 applied at the input of a

physically realizable lowpass filter with the frequency response H (Ω) illustrated

in Fig. 14.4. The magnitude spectrum |H (Ω)| of the filter is shown by a solid line, while the phase spectrum <H (Ω) is shown by a dashed line. The filter

has a cut-off frequency Ωc, such that Ω2 < Ωc < Ω3, and the cut-off frequency

lies within the transition band. Based on the frequency response H (Ω) shown

H(W)

−W3

stop band pass band

stop band

transition

band

transition

band

−W2 −W1 W1 W2 W3 p−p Fig. 14.4. Physically realizable

lowpass filter with transition

bands and non-zero phase.

628 Part III Discrete-time signals and systems

in Fig. 14.4, the magnitudes and phases of the transfer function at the tone

frequencies are given by

frequency Ω = ±Ω1 |H (Ω)| = 1, <H (Ω) = ∓m1Ω1; frequency Ω = ±Ω2 |H (Ω)| = 1, <H (Ω) = ∓m2Ω2; frequency Ω = ±Ω3 |H (Ω)| = 0, <H (Ω) = ∓m3Ω3;

where m1, m2, and m3 are the slopes of the phase response at Ω = Ω1,Ω2 and Ω3, respectively.

Using the convolution property, the DTFT of the output of the filter is given

Y (Ω) = A1π [δ(Ω−Ω1)ejm1Ω + δ(Ω−Ω1)e−jm1Ω] + A2π [δ(Ω+Ω2)ejm2Ω + δ(Ω−Ω2)e−jm2Ω] + A3π [δ(Ω+Ω3) + δ(Ω−Ω3)] · 0.

Taking the inverse DTFT of the above equation, we obtain

y[k] = A1 cos(Ω1(k − m1)) + A2 cos(Ω2(k − m2)).

For m1 �= m2, the input tones A1 cos(Ω1k) and A2 cos(Ω2k) are delayed une- qually and the output sequence y[k] is a distorted version of the sinusoidal

components present within the pass band of the filter. To retain the shape of the

pass-band components, each sinusoidal term A1 cos(Ω1k) and A2 cos(Ω2k) in

y[k] should be delayed equally, i.e. m1 = m2. In signal processing, the following two types of delays are defined:

phase delay dp = −φ(Ω)/Ω;

roup delay dg = − dφ(Ω)

dΩ ;

where φ(Ω) is the phase of the filter transfer function, i.e. φ(ω) = � H (ω). In other words, the phase delay (dp) is defined as the phase divided by the

frequency, whereas the group delay (dg) is defined as the derivative of the phase

with respect to frequency. From the above definitions, it is observed that the

delay of a filter will be constant if the phase φ(Ω) of the filter is a linear function

of frequency. A filter is said to have a linear phase response if it satisfies the

following relationships.

φ(Ω) = −αΩ, or φ(Ω) = −αΩ+ β.

The first condition ensures that the filter has constant phase and group delay,

whereas the second condition ensures only constant group delay. Although it is

desirable to have both constant group and phase delays, a constant group delay

is generally sufficient in many applications.

Based on the above discussion, the conditions for distortionless filtering,

where the pass-band components are retained precisely at the filter output, are

enlisted as follows.

629 14 Digital filters

(1) The pass-band gain of the filter should be the same for all frequency com-

ponents present in the input signal that lie within the pass-band of the

filter.

(2) The phase <H (Ω) of the filter should be linear for all input frequency

components that lie within the pass band of the filter.

(3) The stop-band gain of the filter should be zero within the stop band of the

filter.

Conditions (1)–(3) are valid for distortionless transmission within the pass bands

of both FIR and IIR filters and are checked by plotting the magnitude and phase

spectra of the filters. For FIR filters, the linear phase condition (condition (2))

can also be checked directly from the impulse response h[k] as explained

next.

14.3.1 Linear-phase FIR filters

Consider an FIR filter with impulse response h[k], which is non-zero within

the range 0 ≤ k ≤ N − 1. The z-transform of the FIR filter is expressed as

follows:

H (z) =

N−1 ∑

k=0

h[k]z−k = h[0] + h[1] z−1 + h[2] z−2 + · · · + h[N − 1] z−(N−1).

(14.7)

The following proposition provides sufficient conditions for the phase linearity

of an FIR filter.

Proposition 14.1 If the impulse response function of an N-tap filter, with

z-transfer function given by Eq. (14.7), satisfies either of the following

relationships:

symmetrical impulse response h[k] = h[N − 1 − k]; (14.8a)

antisymmetrical impulse response h[k] = −h[N − 1 − k], (14.8b)

then the frequency response function can be represented as follows:

H (Ω) = G(Ω)ej(−αΩ+β), (14.9)

where G(Ω) is a real-valued function ofΩ, α = (N − 1)/2, and β is a constant

that can be either zero or π/2. Depending on the symmetry/anti-symmetry and

even/odd length of h[k], the FIR filters can be divided into four types: type 1,

type 2, type 3 and type 4. Table 14.2 defines these four types of filters and the

corresponding G(Ω) and β values. It is observed that type 1 and type 2 filters

have constant phase and group delays, whereas type 3 and type 4 filters only

have constant group delay.

630 Part III Discrete-time signals and systems

Table 14.2. Linear-phase FIR filter types and the corresponding G(Ω) and β values

The coefficients a[k ] and b[k ] in column 4 are defined as follows: a[0] = h[(N − 1)/2], a[k ] = 2h[(N − 1)/ 2 − k ], b[k ] = 2h[N /2 − k ]

Type of FIR filter Length, N Symmetry G(Ω) β

Type 1 odd h[k] = h[N − 1 − k] (N−1)/2

∑

k=0 a[k] cos(Ωk) 0

Type 2 even h[k] = h[N − 1 − k] N/2 ∑

k=1 b[k] cos[Ω(k − 0.5)] 0

Type 3 odd h[k] = −h[N − 1 − k] (N−1)/2

∑

k=1 a[k] sin(Ωk) π/2

Type 4 even h[k] = −h[N − 1 − k] N/2 ∑

k=1 b[k] sin[Ω(k − 0.5)] π/2

Proof

We prove Proposition 14.1 for a type 1 filters. The proof for type 2, type 3, and

type 4 filters follows along the same lines.

By substituting z = exp(jΩ) in Eq. (14.7), we get

H (Ω) = h[0] + h[1] e−jΩ + · · · + h[N − 2] e−j(N−2)Ω + h[N − 1] e−j(N−1)Ω.

Taking exp(j(N − 1)Ω/2) common from the left-hand side of the above equa- tion yields

H (Ω) = e−j(N−1)Ω/2 [

h[0] e j(N−1)Ω/2 + h[1] e j(N−3)Ω/2

+ · · · + h[N − 2] e−j(N−3)Ω/2 + h[N − 1] e−j(N−1)Ω/2 ]

. (14.10)

We now pair the first term with the last term, the second term with the second

last term, and so on for the remaining terms. Note that for a type 1 filter, N has an

odd value and h[k] = h[N – 1 – k]. By pairing terms in Eq. (14.10), we obtain

H (Ω) = e−j(N−1)Ω/2 [(

h[0] e j(N−1)Ω/2 + h[N − 1] e−j(N−1)Ω/2 )

+ (

h[1] e j(N−3)Ω/2 + h[N − 2]e−j(N−3)Ω/2 )

+ · · ·

+ (

[

N −1 2

−1 ]

e jΩ + h [

N −1 2

+1 ]

e−jΩ )

+ h [

N − 1 2

]]

Because h[k] = h[N − 1 − k], the above equation reduces as follows:

H (Ω) = e−j(N−1)Ω/2 [

2h[0] cos

(

(N − 1)Ω 2

)

+ 2h[1] cos (

(N − 3)Ω 2

)

+ · · · + 2h [

N − 1 2

− 1 ]

cos(Ω) + h [

N − 1 2

]]

= e−j(N−1)Ω/2 {

[

N − 1 2

]

+ (N−3)/2 ∑

k=0 2h[k] cos

[

Ω

(

N − 1 2

− k )]

}

631 14 Digital filters

Table 14.3. Examples of FIR filters with linear and non-linear phase

Number Phase

of taps, N z-transfer function, H (z) (linear or non-linear) Phase value

4 1 − 2z−1 + 2z−2 − z−3 type 4, linear −1.5Ω+ π/2 3 1 − z−2 type 3, linear −Ω+ π/2 3 1 + 2z−1 + 2z−2 non-linear 4 1 + 2z−1 + 2z−2 + z−3 type 2, linear −1.5Ω 4 1 + 2z−1 − 2z−2 + z−3 non-linear 5 1 + 2z−1 + 3z−2 + 2z−3 + z−4 type 1, linear −2Ω 5 1 + 2z−1 + 3z−2 + 2z−3 − z−4 non-linear

Substituting m = (N−1) 2

− k in the above equation, we obtain

H (Ω) = e−jN−1)Ω/2 {

[

N − 1 2

]

+ (N−1)/2 ∑

m−1 2h

[

N − 1 2

− m ]

cos(mΩ)

}

= e−j(N−1)Ω/2 {

[

N − 1 2

]

+ (N−1)/2 ∑

k=1 2h

[

N − 1 2

− k ]

cos(kΩ)

}

= e−j(N−1)Ω/2 {

(N−1)/2 ∑

k=0 a[k] cos(kΩ)

}

where a[0] = h[(N − 1)/2] and a[k] = 2h[(N − 1)/2 − k]. It is observed that the derived H (Ω) matches with Eq. (14.9), with α = (N − 1)/2 and G(Ω) given in Table 14.2.

Example 14.1

Determine if the FIR filters specified in column 2 of Table 14.3 have linear

phase or not. Also determine the value of the phase.

Solution

The phase linearity can be determined using the conditions given in Eq. (14.8).

The third column of Table 14.3 shows whether a filter is linear phase and the

type of linear-phase filter. The phase function, i.e. (−αΩ+ β) in Eq. (14.9), is shown in the fourth column.

To confirm the results of the last two entries of Table 14.3, Fig. 14.5 plots

the magnitude and phase spectra of the FIR filter specified in the second to last

row of Table 14.3. The phase plot in Fig. 14.5(b) confirms that the FIR filter

has a linear phase. Since a phase of π is the same as that of −π , the sharp transitions at Ω = ±0.5π are not discontinuities but correspond to the same value. The magnitude spectrum illustrates non-uniform gains within the pass

band and stop bands, implying that the FIR filter is not an ideal lowpass filter

despite having a linear phase.

632 Part III Discrete-time signals and systems

W 0 p−p −0.5p 0.5p

H (W) 9

W 0 p−p −0.5p 0.5p

< H(W)−p

−0.5p

0.5p

(a) (b)

Fig. 14.5. Example of an FIR

filter H(z) = 1 + 2z−1 + 3z−2 + 2z−3 + z−4 with linear phase. (a) Magnitude spectrum;

(b) phase spectrum.

W 0 p−p −0.5p 0.5p

H(W) 7

W 0 p−p −0.5p 0.5p

< H(W)−p

−0.5p

0.5p

(a) (b)

Fig. 14.6. Example of an FIR

filter H(z) = 1 + 2z−1 + 3z−2 + 2z−3 − z−4) with non-linear phase. (a) Magnitude

spectrum; (b) phase spectrum.

Likewise, Fig. 14.6 plots the magnitude and phase spectra of the FIR filter

specified in the last row of Table 14.3. The phase plot shown in Fig. 14.6(b)

confirms that the FIR filter has a non-linear phase.

14.4 Ideal versus non-ideal filters

Table 14.1 shows the impulse response of four types of frequency-selective

ideal filters. It is observed that the ideal impulse responses are non-zero for

k < 0. Therefore, these ideal filters are non-causal and hence physically non-

realizable. It is, however, possible to realize a non-causal filter by applying an

appropriate delay. To elaborate, let us consider the transfer function of an ideal

lowpass filter shown in Eq. (14.1a) in a slightly different form as follows:

Hilp(Ω) = {

e−jmΩ |Ω| ≤ Ωc 0 Ωc < |Ω| ≤ π,

(14.11)

where a linear-phase component of exp(−jmΩ), is included within the pass

band. The variable m is a constant that corresponds to the delay of the filter.

The impulse response hilp[k] of the ideal lowpass filter is obtained by taking

the inverse DTFT of Eq. (14.11), and is given by

hilp[k] = sin((k − m)Ωc)

(k − m)π = Ωc

π sinc

(

(k − m)Ωc

)

. (14.12)

Figure 14.7 plots the impulse response hlp[k]. As illustrated in Fig. 14.7, the

impulse response hlp[k] of the ideal lowpass filter has an infinite length and is

still non-causal. The ideal lowpass filter is therefore not physically realizable,

irrespective of the value of delay m. Since the magnitude of the impulse response

decays in both directions from its origin, k = m, a simple method to derive a

633 14 Digital filters

hilp[k]

m +

m+2

p Wc

Ωc

π− Ωc

m−2

Fig. 14.7. Impulse response of

an ideal lowpass filter with a

cut-off frequency ofΩc and

delay m.

causal implementation of the ideal lowpass filter is to truncate its impulse

response on either side of its origin. We consider two such implementations:

FIR implementation I

h1[k] =











Ωc

π sinc

(

(k − m)Ωc π

)

m − 70 ≤ k ≤ m + 70

0 elsewhere;

(14.13a)

FIR implementation II

h2[k] =











Ωc

π sinc

(

(k − m)Ωc

)

m − 10 ≤ k ≤ m + 10

0 elsewhere.

(14.13b)

The length of the truncated FIR approximation is 141 in Eq. (14.13a) and 21 in

Eq. (14.13b). The magnitude spectra for the two implementations are shown in

Fig. 14.8. Compared with the ideal lowpass filter, we observe three significant

changes in the causal implementations.

(1) The gain within the pass band of the causal implementations is no

longer constant but includes several oscillating ripples, referred to as the

pass-band ripples. The distortion caused by the pass-band ripples is sig-

nificantly higher when the truncated length is small. Compared with Eq.

(14.13a) with a truncated length of 141, Eq. (14.13b) has a length of 21

and results in a higher ripple distortion.

W 0 p

H1(W)

H2(W)

−p −0.5p

Fig. 14.8. Magnitude spectrum

of FIR implementations h1[k ]

and h2[k ] obtained by

truncating the impulse response

h ilp[k ] of an ideal lowpass filter.

634 Part III Discrete-time signals and systems

H(W)

Wp Ws

1−dp

1+dp

pass band stop bandtransition

band

0 p

Fig. 14.9. Specifications of a

practical lowpass filter with three

modifications from an ideal

lowpass filter. First, a pass-band

ripple of δp is included about the

unity pass-band gain. Then a

stop-band ripple of δs is

included. Finally, a transition

band ofΩs −Ωp allows for smooth transition between the

stop and pass bands.

(2) Unlike the ideal lowpass filter, the FIR implementations have a significant

transition band between the pass band and the stop band. The width of

the transition band depends upon the length of the FIR implementations.

The smaller the truncated length, the larger the width of the transition

region.

(3) The gain within the stop band of the causal implementations is no longer

zero but contains ripples, referred to as the stop-band ripple. As in the pass

band, the distortion produced by the stop-band ripples is higher when the

truncated length is smaller.

Since ideal filters are not physically realizable, a practical implementation of

these filters is obtained by allowing acceptable variations in the magnitude

response within the pass band and stop band. In addition, a transition band is

included between the pass band and the stop band so that the magnitude response

of the filter can drop off smoothly. Figure 14.9 specifies the magnitude response

of a practical lowpass filter with the following characteristics:

pass band (1 − δp) ≤ |H (Ω)| ≤ (1 + δp) for |Ω| ≤ Ωp; transition band Ωp < |Ω| ≤ Ωs;

stop band 0 ≤ |H (Ω)| ≤ δs for Ωs ≤ |Ω| ≤ π.

The objective of a good design is to obtain a filter with limited ripples within

the pass band and stop band, narrow transition bandwidth, and a linear phase

at a reasonable implementation cost. Such an objective is self-contradictory.

For example, a smaller transition band requires a relatively longer FIR filter or,

alternatively, a higher-order IIR filter. In the case of FIR filters, the complexity

of the filter is directly proportional to its length. Keeping the transition band

small therefore results in a higher cost. Likewise, for IIR filters, the complexity

depends upon the order of the filter. Increasing the order of the IIR filter to

reduce the transition bandwidth increases the implementation cost. The design

635 14 Digital filters

Table 14.4. Impulse response of a 21-tap FIR filter

k 0, 20 1, 19 2, 18 3, 17 4, 16 5, 15 6, 14 7, 13 8, 12 9, 11 10

h[k] −0.0014 0.0015 0.0066 0.0081 −0.0059 −0.0330 −0.0411 0.121 0.1320 0.2619 0.3183

process generally involves some trade-offs between the desired characteristics

of the specified digital filter. We will revisit this issue in Chapters 15 and 16,

where we introduce several design techniques for the FIR and IIR filters.

Examples 14.2 and 14.3 consider the FIR and IIR filters.

Example 14.2

Calculate the transfer function of a causal DT FIR filter whose impulse response

h[k] is specified in Table 14.4. Determine and plot the magnitude spectrum of

the FIR filter. What are the values of the stop-band ripple δs and the transition

bandwidth?

Solution

The impulse response of the FIR filter is plotted in Fig. 14.10(a). To determine

the frequency characteristics of the filter, we determine the z-transfer function

of the FIR filter:

H (z) = 20

∑

k=0 h[k]z−k .

h[k]

k 7

5 6

8 9 10 11 12

14 16

17 18 19200 2 3

4 W

0 p−p −0.5p 0.5p

H(W) 1

(a) (b)

0 p−p −0.5p 0.5p

20 × log10( |H(W)| )0

−40

−60

−20

W 0 p−p −0.5p 0.5p

< H(W)−p

−0.5p

0.5p

Fig. 14.10. FIR filter in Example

14.2. (a) Impulse response h[k ];

(b) magnitude spectrum

|H(Ω)|; (c) phase spectrum <H(Ω); (d) magnitude spectrum

|H(Ω)| in decibels.

636 Part III Discrete-time signals and systems

Substituting z = exp(jΩ), the Fourier transfer function of the FIR filter is given by

H (Ω) = 20

∑

k=0 h[k]e−jkΩ,

which is used to plot the magnitude and phase spectra of the FIR filter in

Figs. 14.10(b) and (c). It is observed that the gain of the filter is close to

unity at low frequencies (Ω ≈ 0), while the gain is zero at high frequencies

(Ω ≈ π ). Therefore, the impulse response h[k] represents a lowpass filter. Also,

Fig. 14.10(c) illustrates that the phase of the FIR is piecewise linear.

Without knowing the exact values of the pass and stop bands, it is difficult

to determine the exact values of the stop-band ripple δ2 and the transition

bandwidth. An intelligent guess can be made by looking at the Bode plot of

the FIR filter. Recall that the Bode plot is the same as the magnitude spectrum

except that the magnitude |H (Ω)| of the filter is expressed in decibels (dB) as

follows:

gain in dB = 20 log10(|H (Ω)|).

From the Bode plot shown in Fig. 14.10(d), we observe that the maximum value

of |H (Ω)| within the stop band is approximately –52 dB. Expressed on a linear

scale, the stop-band ripple δ2 is given by

δs (dB) = 20 log10(δ2) = −52 ⇒ δs = 10 −2.6 = 0.0025.

Figure 14.10(d) also provides approximate estimations of the pass band and

stop band as follows:

pass band (0 ≤ |Ωp| ≤ 0.5) and stop band (1.5 ≤ |Ωs| ≤ π ).

The transition band is therefore given by 0.5 < |Ω| < 1.5.

Example 14.3

The transfer function of a DT IIR filter is given by

H (z) = 0.12z

z2 − 1.2z + 0.32 .

Determine and sketch the impulse response h[k] of the filter. Determine and

plot the magnitude response of the IIR filter.

Solution

The characteristic equation of H (z) is given by z2 − 1.2z + 0.32 = 0, which

has two roots, at z = 0.8 and 0.4. The z-transfer function H (z) can therefore

be expressed as follows:

H (z)

z =

0.12

z2 − 1.2z + 0.32 ≡

z − 0.8 +

z − 0.4 .

637 14 Digital filters

h[k]

k 75 6 8 9 1011 12 1513 14 16 171819200 2 3 4

0.12 0.144

0.134

W 0 p−p −0.5p 0.5p

|H(W)| 1

W 0 p−p −0.5p 0.5p

<H(W)−p

−0.5p

0.5p

W 0 p−p −0.5p 0.5p

20 × log10( |H(W)| )0

−40

−60

−20

(a) (b)

Fig. 14.11. IIR filter in Example

14.3. (a) Impulse response h[k ];

(b) magnitude spectrum

|H(Ω)|; (c) phase spectrum <H(Ω); (d) magnitude spectrum

|H(Ω)| in decibels.

Using Heaviside’s partial fraction formula the coefficients of the partial fractions

k1 and k2 are given by

k1 = [

(z − 0.8) 0.12

(z − 0.8)(z − 0.4)

]

z=0.8 =

[

0.12

z − 0.4

]

z=0.8 = 0.3

and

k2 = [

(z − 0.4) 0.12

(z − 0.8)(z − 0.4)

]

z=0.4 =

[

0.12

z − 0.8

]

z=0.4 = −0.3.

The partial fraction expansion of H (z) is therefore given by

H (z) = 0.3z

z − 0.8 +

−0.3z z − 0.4

Taking the inverse z-transform of H (z) yields

h[k] = 0.3[(0.8)k − (0.4)k]u[k].

which is plotted in Fig. 14.11(a). Note that the IIR filter has infinite length, as

expected.

The Fourier transfer function of the IIR filter is obtained by substituting

z = exp(jΩ):

H (Ω) = 0.12e−jΩ

1 − 1.2e−jΩ + 0.32e−j2Ω .

The magnitude spectrum of the IIR filter is plotted in Figs. 14.11(b) and (d).

Since the gain of the filter is unity at low frequencies (around Ω ≈ 0) and

close to zero at high frequencies (around Ω ≈ π ), the impulse response h[k]

638 Part III Discrete-time signals and systems

represents a lowpass filter. Figure 14.11(c) illustrates that the phase of the IIR

filter is non-linear; therefore, the IIR filter introduces distortion within the pass

band.

14.5 Filter realization

In the preceding chapters, we presented several different techniques to calculate

the output of a DT system. In the time domain, the output response y[k] can be

determined from its input x[k] either by solving a linear, constant-coefficient,

difference equation of the following form:

a0 y[k] + a1 y[k − 1] + · · · + aN y[k − N ] = b0x[k] + b1x[k − 1] + · · · + bM x[k − M]

or, alternatively, by calculating the convolution sum between the input x[k] and

the impulse response h[k]. The convolution sum is given by

y[k] = x[k] ∗ h[k] = ∞

∑

m=−∞

x[m]h[k − m].

In the frequency domain, the convolution property is used to express the con-

volution sum in terms of the transfer function H (Ω) and the CTFT X (Ω) of the

input as follows:

Y (Ω) = X (Ω)H (Ω),

from which the output y[k] can be determined by calculating the inverse CTFT

of Y (Ω). On digital computers and specialized DSP boards, the output of a dig-

ital filter is generally obtained by iteratively evaluating the recurrence formula,

y[k] = − 1

a0 (+a1 y[k − 1] + · · · + aN y[k − N ])

+ 1

a0 (b0x[k] + b1x[k − 1] + · · · + bM x[k − M]),

z−1x[k] x[k−1]

x1[k] x1[k]+x2[k]

x2[k]

x[k] ax[k] a

(a)

(b)

(c)

Fig. 14.12. Fundamental

elements for building digital

implementations for FIR and IIR

filters. (a) Unit delay element;

(b) adder; (c) constant-

coefficient multiplier.

derived from the difference equation. Implementing the recurrence formula

requires delaying the samples of the input and output sequences, multiplying

the sample values with constant coefficients, and adding the resulting prod-

ucts. In other words, we require three mathematical operations, shift or delay,

multiplication, and addition, to solve a difference equation iteratively. In the fol-

lowing, we introduce the schematic representation of these three fundamental

operations.

14.5.1 Shift or delay operator

On digital computers and specialized DSP boards, the shift operation is

implemented using a cascaded combination of delay elements. The schematic

639 14 Digital filters

representation of a unit delay element is illustrated in Fig. 14.12(a), where the

transfer function of the block is given by H (z) = z−1. The impulse response h[k] of the unit delay element is given by h[k] = δ[k – 1]. The output is there-

fore given by x[k] ∗ δ[k − 1] = x[k − 1]. If a delay of more than one sample

is required, several unit delay elements may be cascaded together in a series

configuration.

14.5.2 Adder

On digital devices, adders are typically implemented using combinational or

sequential circuits consisting of registers and logic gates. The schematic repre-

sentation of an adder is illustrated in Fig. 14.12(b), where the input sequences

x1[k] and x2[k] produce an output x1[k] + x2[k].

14.5.3 Multiplication by a constant

On digital devices, multipliers are typically implemented using sequential cir-

cuits consisting of registers, shift delays, and logic gates. The schematic rep-

resentation of a constant multiplier is shown in Fig. 14.12(c), where the input

sequence x[k] is multiplied with a constant a, producing an output ax[k].

In the following sections, we sketch signal flow graphs for efficient implemen-

tations of both FIR and IIR digital filters using the aforementioned elements,

referred to as the fundamental elements. By manipulating the signal flow graphs,

we present several different but equivalent structures for the same transfer func-

tion. We also demonstrate the effect of finite-precision arithmetic on the gain–

frequency characteristics of digital filters, and provide several design tips to

alleviate the problems arising from finite-precision arithmetic.

14.6 FIR filters

A causal FIR filter, of finite length N and having non-zero values in the range

0 ≤ k ≤ (N − 1), is represented by the following transfer function:

H (z) =

N−1 ∑

k=0

h[k]z−k = h[0] + h[1]z−1 + h[2]z−2 + · · · + h[N − 1]z−(N−1)

(14.14)

or, alternatively by a difference equation obtained by solving the convolution

sum:

y[k] =

N−1 ∑

m=0

h[k]x[k − m]

= h[0]x[k] + h[1]x[k − 1] + · · · + h[N − 1]x[k − (N − 1)]. (14.15)

640 Part III Discrete-time signals and systems

z−1 z−1 z−1 z−1x[k] …x [k

− 1

]

x[ k

− 2

]

x[ k

− 3

]

x[ k

– (

N –

2 )]

x[ k

– (

N –

1 )]

h[0]

+ h[1]

+ + + + h[2] h[3] h[N−2] h[N−1]

y[k]

Fig. 14.13. Direct form for

causal FIR filters of length N .

There are several flow graph representations of the FIR filter. In the following,

we discuss some of them.

14.6.1 Direct form

The flow graph for direct form is achieved by implementing Eq. (14.15) directly.

In direct form, the constant multipliers are the same as the coefficients of the

difference equation, Eq. (14.15). The direct form of the flow graph for a causal

FIR filter is shown in Fig. 14.13. Since the cost of implementation of a filter is

directly proportional to the number of fundamental elements used, we include

a count of these elements for each flow graph. The number of the fundamental

elements used in Fig. 14.13 is shown in the second row of Table 14.5.

The flow graph for the direct form resembles a tapped delay line used fre-

quently in communication systems for channel equalization. The filter shown

in Fig. 14.13 is therefore referred to as a tapped delay line filter or sometimes

as a transversal filter.

14.6.2 Cascaded form

The flow graph for the cascaded form is achieved by expressing Eq. (14.14) in

terms of a product of quadratic terms:

H (z) = h[0] ⌈ N+12 ⌉∏

n=1

(1 + b1nz −1 + b2nz

−2). (14.16)

Factorizing H (z) in terms of quadratic terms ensures coefficients b1n and b2n

to be real-valued provided that the impulse response h[k] is also real-valued.

Had linear factors been considered in Eq. (14.16) there would be no guarantee

for the coefficients of the linear factors to be real-valued, even with real-valued

h[k]. The upper limit ⌈(N − 1)/2⌉ in the summation in Eq. (14.16) represents

a ceiling operation, which equals (N − 1)/2 if N is odd. If N is even, the upper

limit equals N/2 with b2n = 0 for the last product term.

The flow graph of the cascaded form is achieved by considering ⌈(N − 1)/2⌉

substructures and cascading the substructures together in a series configuration.

The resulting flow graph is shown in Fig. 14.14. The number of fundamental

elements used in Fig. 14.14 is shown in the third row of Table 14.5.

641 14 Digital filters

Table 14.5. Number of elements required to implement different types

of FIR filter structures

Structure two-input address Unit delays Constant multipliers

Direct form N − 1 N − 1 N Cascaded form N − 1 N − 1 N Linear phase N − 1 N − 1 N/2 (N even)

filters (N + 1)/2(N odd)

b11

b21 b22

b12+

b1(N−1)/2

b2(N−1)/2

+][kx y[k]

z−1 z−1 z−1

z−1z−1z−1

Fig. 14.14. Cascaded form for

causal FIR filters of length N .

14.6.3 Linear-phase FIR filters

As proved in Proposition 14.1, an N -tap linear phase FIR filter satisfies the

following symmetry condition:

h[k] = h[N − 1 − k] or h[k] = −h[N − 1 − k].

For the symmetry condition h[k] = h[N − 1 − k], we show that the condition can be used to reduce the number of constant multipliers. The derivation for the

antisymmetry condition, h[k] = −h[N − 1 − k], follows along similar lines. If the length N of the filter is even, Eq. (14.14) is rearranged as follows:

H (z) = h[0] (

1 + z−(N−1) )

+ h[1] (

z−1 + z−(N−2) )

+ · · · + h [

2 − 1

]

(

z−(N/2−1) + z−(N/2) )

On the other hand, if the length N of the filter is odd, Eq. (14.14) is rearranged

as follows:

H (z) = h[0] (

1 + z−(N−1) )

+ h[1] (

z−1 + z−(N−2) )

+ · · · + h [

N − 1 2

− 1 ]

(

z−((N−1)/2−1) + z−((N−1)/2+1) )

+ h [

N − 1 2

]

z−(N−1)/2.

Using the above equations, the flow graphs of the linear-phase FIR filter satis-

fying the symmetry condition is shown in Fig. 14.15. Both even and odd values

of length N are considered. The numbers of fundamental elements required are

shown in the fourth row of Table 14.5. It is observed that the number of constant

642 Part III Discrete-time signals and systems

h[0] h[1] h[2] h[3]

x[k] z−1 z−1 z−1 z−1…

z−1

z−1 z−1 z−1 z−1

z−1 z−1 z−1

z−1

z−1 z−1 z−1…

…+

y[k]

2 − 1]N−1h[

2 ] N−1

2 − 2] − 1N−1h[ 2

N hh[0] h[1] h[2] h[3]h

x[k] …

…

+y[k]

(a)

(b)

[

][

Fig. 14.15. Flow graphs for

linear-phase FIR filters.

(a) Length N is odd; (b) length

N is even.

multipliers is roughly half that required in direct form or cascaded form. The

number of unit delay and addition elements, however, stays the same.

14.6.4 Transposed forms

Alternative flow graphs for implementations in Sections 14.6.1–14.6.3 can be

realized by applying the transpose operation. Transposition of a flow graph

is achieved by (i) interchanging the role of the input and output; (ii) revers-

ing the directions of all branches within a flow graph; and (iii) replacing the

source nodes by adders, and vice versa. Note that the number of fundamental

elements required to implement a filter does not change if the transposed form

is used for implementation. We explain the principle of transposition with an

example.

Example 14.4

Implement direct form and cascaded configurations of the flow graph for the

FIR filter with transfer function given by

H (z) = −0.3 − 0.4z−1 + 1.4z−2 − 0.4z−3 − 0.8z−4.

643 14 Digital filters

z−1 z−1 z−1 z−1x[k]

−0.3

+ −0.4

+ + + 1.4 −0.4 −0.8

y[k]

z−1

3.5633 +

1.7868

−2.23 +

1.4924

x[k] y[k] −0.3

(a) (b)

Fig. 14.16. (a) Direct form I and

(b) cascaded configurations for

the FIR filter in Example 14.4.

Using transposition, derive an alternative configuration from the cascaded

implementation.

Solution

The flow graph for the direct form is shown in Fig. 14.16(a). For the cascaded

configuration, we factorize H (z) as follows:

H (z) = −0.3 (1 + 2.9595z−1)(1 + 0.6038z−1)(1 − (1.1150 − j0.4992)z−1) ×(1 − (1.1150 + j0.0.4992)z−1).

Expressing H (z) as a product of quadratic terms, we obtain

H (z) = −0.3(1 + 3.5633z−1 + 1.7868z−2)(1 − 2.23z−1 + 1.4924z−2),

which has the flow graph illustrated in Fig. 14.16(b).

The alternative configuration for the cascaded form, obtained by applying

the transposition principle, is shown in Fig. 14.17 using two steps. Step 1

interchanges the role of the input and output, reverses the directions of all

branches, and replaces the source nodes with adders. Similarly, the adders are

replaced by source nodes. The resulting configuration is shown in Fig. 14.17(a),

where the input is on the right-hand side of the flow graph and the output is on

the left-hand side. Figure 14.17(b) is a reordered version of Fig. 14.17(a) with

the input and output, rearranged to the standard right-hand and left-hand sides,

respectively.

y[k] x[k]

−2.23

1.4924

z−1

3.5633

1.7868

−0.3 x[k]

3.5633

1.7868

+ −2.23

1.4924

y[k] −0.3

(a) (b)

Fig. 14.17. Transpose

configurations of flow graph in

Fig. 14.16(b).

644 Part III Discrete-time signals and systems

Table 14.6. Number of elements required to implement different types of IIR filter

structures

M and N are, respectively, the degree of the numerator and denominator

polynomials in H(z), as shown in Eq. (14.17)

Structure Unit delays Two-input adders Constant multipliers

Direct form I M + N M + N M + N + 1 Direct form II max(M, N ) M + N M + N + 1 Cascaded form max(M, N ) M + N M + N + 1 Parallel form max(M, N ) M + N (M ≥ N ) M + N + 1 (M ≥ N )

2N (M < N ) 2N + 1 (M < N )

Finally, it should be noted that H (z) does not represent a linear-phase FIR

filter. As such, the linear-phase configuration cannot be derived for this filter.

The direct form and cascaded implementations of the FIR filters can be extended

to the IIR filters, which are discussed in Section 14.7.

14.7 IIR filters

The transfer function of an IIR filter is given by

H (z) = b0 + b1z−1 + · · · + bM z−M

1 + a1z−1 + · · · + aN z−N , (14.17)

where the coefficient a0 of the constant term in the denominator is normalized

to one. Based on Eq. (14.17), an IIR filter can alternatively be modeled by the

linear, constant-coefficient difference equation given by

y[k] + a1 y[k − 1] + · · · + aN y[k − N ] = b0x[k] + b1x[k − 1] + · · · + bM x[k − M]. (14.18)

There are four major architectures to implement the IIR filters, which are con-

sidered in the following.

14.7.1 Direct form I

To derive the IIR realization of the transfer function, Eq. (14.17), we implement

the numerator and denominator functions, defined as follows:

numerator N (z) = b0 + b1z−1 + · · · + bM z−M ; denominator D(z) = 1 + a1z−1 + · · · + aN z−N ,

separately. The resulting flow graph is shown in Fig. 14.18, where the first

structure represents N (z) and the second structure represents D(z). The numbers

of fundamental elements required in direct form I are shown in the second row

of Table 14.6.

645 14 Digital filters

D(z)N(z)

x[k] y[k]

x[k−1]

x[k−2]

x[k −M ]

y[k−1]

y[k−2]

y[k − (N−1)]

y[k −N]

+ z−1

z−1

+ b0

bM−1

−a1

−a2

−aN−1

−aN

x[k − (M−1)]

Fig. 14.18. Direct form I for IIR

filters where numerator

polynomial N (z) and

denominator polynomial D (z)

are implemented as cascaded

systems. The degree M of the

numerator is assumed to the

same as the degree N of the

denominator.

D(z)

N(z)

x[k] y[k]

x[k−1]

x[k−2]

z−1

y[k−1]

y[k−3]

y[k−2]

−2

0.1

0.07

0.065

Fig. 14.19. Direct form I

realization for the IIR filter in

Example 14.5.

Example 14.5

Implement the direct form I realization of an IIR filter with the following transfer

function:

H (z) = z3 − 2z2 + z

z3 − 0.1z2 − 0.07z − 0.065 . (14.19)

Solution

The transfer function H (z) can be represented as follows:

H (z) = 1 − 2z−1 + z−2

1 − 0.1z−1 − 0.07z−2 − 0.065z−3 ,

with the difference equation given by

y[k] = x[k] − 2x[k − 1] + x[k − 2] − {−0.1y[k − 1] − 0.07y[k − 2] − 0.065y[k − 3]}

= x[k] − 2x[k − 1] + x[k − 2] + 0.1y[k − 1] + 0.07y[k − 2] + 0.065y[k − 3].

The flow graph using direct form I is illustrated in Fig. 14.19.

646 Part III Discrete-time signals and systems

D(z)

x[k] y[k]

N(z)

a′

b′

m′

n′

bM−1

z−1

−a2

−a1

−aN−1

−aN

bM−1

x[k] y[k]

z−1

−a2

−a1

−aN−1

−aN

(a) (b)

Fig. 14.20. Direct form II for IIR

filters where degrees of the

numerator (M ) and

denominator (N ) are assumed

to be the same.

14.7.2 Direct form II

Direct form II is realized by noting that the order of structures N (z) and D(z) can

be interchanged as for any two systems in a series combination. The resulting

flow graph is shown in Fig. 14.20(a). Since nodes α and α′ have the same

polarity, these nodes can be merged by replacing the top two delay elements

by one delay element. Similarly, nodes β and β ′ can be merged, and so on for

the rest of the adjacent nodes below the delays in structures D(z) and N (z).

The resulting flow diagram is referred to as direct form II and is illustrated in

Fig. 14.20(b). The number of fundamental elements required in direct form II

is shown in the third row of Table 14.6.

A flow graph that requires the minimum number of delay elements, multi-

pliers, and adders to implement a filter is referred to as a canonical structure.

It can be shown that the implementation complexity of an arbitrary IIR filter

with a numerator of degree M and a denominator of degree N cannot be less

than the complexity of the flow graph for direct form II shown in Fig. 14.20(b).

Therefore, direct form II with the flow graph shown in Fig. 14.20(b), is a canon-

ical architecture. On the other hand, direct form II with the flow graph shown

in Fig. 14.20(a) is a non-canonical architecture.

647 14 Digital filters

x[k] y[k]

+ 1

−2

0.1

0.07

0.065

z−1

Fig. 14.21. Direct form II

architecture for the IIR filter in

Example 14.6.

Example 14.6

Implement the filter in Example 14.5 using the direct form II realization.

Solution

The flow graph for direct form II realization is shown in Fig. 14.21.

14.7.3 Cascaded form

The flow graph for the cascaded form is achieved by expressing the numerator

and denominator polynomials in Eq. (14.17) in terms of a product of quadratic

terms:

H (z) = b0

⌈ M2 ⌉∏

m=1

(1 + b1m z −1 + b2m z

−2)

⌈ N2 ⌉∏

n=1

(1 + a1nz −1 + a2nz

−2)

. (14.20)

Factorizing H (z) in terms of quadratic terms ensures coefficients b1n and b2n

to be real-valued provided that the impulse response h[k] is also real-valued.

y[k]x[k]

b11

b21

a11

a21

b1q

b2q

a1q

a2q

Q(z)

z−1 z−1

z−1z−1

Fig. 14.22. Cascaded form

architecture for IIR filters.

648 Part III Discrete-time signals and systems

y[k]x[k]

+ +

z−1

z−1 z−1

−0.5−2

−0.13

−0.4

Fig. 14.23. Cascaded form

architecture for the IIR filter in

Example 14.7.

In general, the quadratic terms may be coupled together in the following

form:

H (z) = b0 (1 + b11z−1 + b21z−2) (1 + a11z−1 + a21z−2)

× (1 + b12z−1 + b22z−2) (1 + a12z−1 + a22z−2)

× · · · × (1 + b1q z−1 + b2q z−2) (1 + a1q z−1 + a2q z−2)

× Q(z), (14.21)

where q = min(⌈N/2⌉, ⌈(M/2⌉), and Q(z) represents the uncoupled terms arising from unequal values of degree N and M . The first q quadratic terms in

Eq. (14.21) are implemented using a cascaded configuration of the direct form

II realization, while Q(z) may be implemented in either direct form I or direct

form II realization. The flow graph for Eq. (14.21) is shown in Fig. 14.22. The

numbers of fundamental elements required in cascaded form are shown in the

fourth row of Table 14.6.

Example 14.7

Implement the filter in Example 14.5 using the cascaded form.

Solution

The transfer function H (z) is expressed as follows:

H (z) = 1 − 2z−1 + z−2

1 − 0.1z−1 − 0.07z−2 − 0.065z−3

= (1 − z−1)(1 − z−1)

(1 − 0.5z−1)[1 − (−0.2 + j0.3)z−1][1 − (−0.2 − j0.3)z−1] .

Note that if the filter is implemented using only first-order filters, the filter

coefficients will be complex. In order to avoid complex values for the filter

coefficients, the complex roots are combined into a quadratic term as follows:

H (z) = 1 − 2z−1 + z−2

1 + 0.4z−1 + 0.13z−2 ×

1 + 0.5z−1 . (14.22)

The flow diagram for Eq. (14.22) is shown in Fig. 14.23, where we have omitted

scalar multiplications where the multiplier is unity.

649 14 Digital filters

Table 14.7. Comparison of the number of fundamental elements in flow graphs

obtained from different forms for the IIR filter implemented in Examples 14.4–14.7

Number of

Form unit delays scalar multipliers dual-input adders

Direct form I 5 4 5

Direct form II 3 4 5

Cascaded form 3 4 5

Parallel form 3 6 5

14.7.4 Parallel form

In this form, IIR filters are implemented as a parallel combination of first- and/or

second-order filters. To derive the parallel realization, the transfer function H (z)

is expressed in terms of its partial fractions:

H (z) ≡ Q(z) + k1

1 − p1z−1 +

1 − p2z−1 + · · · +

1 − pN z−1 , (14.23)

where k1, k2, . . . , kN are partial fraction coefficients, obtained from Heaviside’s

formula, and p1, p2, . . . , pN are the poles of H (z). To prevent complex-valued

coefficients, Eq. (14.23) is expressed in terms of quadratic terms as follows:

H (z) = Q(z) + ⌊ N+12 ⌋ ∑

n=1

b1n + b2nz−1 1 + a1nz−1 + a2nz−2

. (14.24)

If the degree N of the denominator in H (z) is odd, a2n = b2n = 0 (for n = ⌈N/2⌉). The parallel form of the IIR filter is illustrated in Fig. 14.24. The

number of fundamental elements required in the parallel form are shown in

the fifth row of Table 14.6. Note that the parallel architecture has the same

complexity as the direct form II and cascade architectures when N = M . If the

numerator and the denominator are not of the same degree, a larger number of

scalar multipliers and two-input adders are required.

Example 14.8

Implement the IIR filter in Example 14.5 using the parallel form.

Solution

Using partial fraction expansion, the transfer function H (z) is expressed as

follows:

H (z) ≡ k1

1 − 0.5z−1 +

k21 + k22z −1

1 + 0.4z−1 + 0.13z−2 ,

where the partial fraction coefficients are determined as k1 = 0.431, k21 =

0.569, and k22 = −1.8879. Figure 14.25 shows the parallel form of the IIR

filter.

650 Part III Discrete-time signals and systems

y[k]

Q(z) +x[k]

z−1

−a11

b11

b12

b22

b1N

b2N

b21

−a21

−a12

−a22

−a1N

−a2N

Fig. 14.24. Parallel form

architecture for IIR filters.

y[k]x[k]

+ 0.569

0.5

0.431

z−1

−0.4

−0.13

−1.8879

Fig. 14.25. Parallel form

architecture for the IIR filter in

Example 14.8.

In Table 14.7, we compare the different realizations of the IIR filter specified in

Example 14.5. Trivial scalar multiplications, where the scalar multiplier is unity,

are ignored. The cascaded form yields the minimum number of fundamental

elements used. This, however, is valid only for Example 14.5 and is not true in

general.

14.7.5 Transposed forms

As was the case for FIR filters, alternative flow graphs for the implementations

in Sections 14.7.1–14.7.4 can be realized by applying the transpose operation.

651 14 Digital filters

14.7.6 Choice of structures

The direct form II, cascaded and parallel forms are referred to as canonical

structures and have roughly the same implementation complexity. The actual

complexity of each form of realization depends on the transfer function under

consideration. Table 14.7 compares the four structures in terms of the number

of unit delays, scalar multipliers, and dual-input adders for the filter consid-

ered in Examples 14.4–14.7. It is observed that the direct form I requires the

largest number of delay elements. The direct form II, cascaded, and parallel

structures require an identical number of delay elements and adders. However,

the cascaded form needs to implement the lowest number of multipliers. This

is because there are two multipliers that perform multiplication by a factor of

one. These unity multipliers need not be implemented. The parallel structure

requires the largest number of multipliers.

Irrespective of the arithmetic complexity, all of these realizations should

provide identical outputs for the same input. As we shall see in the following

section, the filter coefficients are implemented using finite precision. The impact

of finite-precision arithmetic on the performance of digital filters is the focus of

our discussion in Section 14.8. The following are some empirical observations

that should be kept in mind when choosing a particular realization.

(i) When the poles of the transfer function lie close to each other or close to

the unit circle in the complex z-plane, direct form realizations, with filter

coefficients represented using finite precision, produce large deviations

from the output of an exact filter.

(ii) The order in which the first- and second-order systems are implemented

in cascaded forms affects the output of the filter in finite-precision imple-

mentations. Changing the order may reduce the deviation from the output

of an exact filter.

(iii) Pairing of complex poles and zeros is important for all cascaded and

parallel realizations.

(iv) In cascaded realizations, scalar multipliers between different systems may

be required to prevent the partial fraction coefficients from becoming too

large or too small.

14.8 Finite precision effect

Figure 14.26 illustrates the processing of analog signals with digital systems.

The analog signal y(t) produced by such a system contains distortions from

several sources, including

(i) analog-to-digital conversion (ADC) noise;

(ii) finite-precision approximation of filter coefficients;

652 Part III Discrete-time signals and systems

x(t) y(t) x[k]

sampling DT

system reconstruction

y[k]

Fig. 14.26. Processing of analog

signals with digital filters (iii) round-off errors;

(iv) register overflow.

These effects are considered in the following.

14.8.1 Analog-to-digital conversion (ADC) noise

The process of encoding the analog signal x(t) into a DT signal x[k], quantized

to a fixed number of bits, involves discarding the higher resolution information

of the analog signal. The resulting distortion is referred to as the analog-to-

digital (ADC) noise. The amount of ADC noise is inversely proportional to the

number of bits used in the quantization process. For example, assume that the

true value of a sample is given by 0.875 364 573 894 562 234 5. If the sample

is quantized by a 3-bit uniform quantizer with a peak-to-peak range of ±1 V, the sample value would be quantized to 0.9375, leading to an ADC noise of

−0.062 135 426 105 44. If instead an 8-bit uniform quantizer is used, the sample value would be approximated to 0.878 906 25 with an ADC noise of −0.003 541 676 105 44. The ADC noise can, therefore, be reduced by using a higher-

resolution quantizer with a larger number of reconstruction levels, but it can

never be eliminated. The ADC noise causes the analog signal y(t), recovered

from the processed digital sequence y[k], to deviate from the output signal

produced by a completely analog system, which is equivalent to the schematic

representation of Fig. 14.26.

A second error introduced by the quantizer is referred to as the saturation

noise, which occurs when the input signal x(t) exceeds the peak-to-peak operat-

ing range for which the quantizer is designed. Since the range of the saturation

noise is unlimited, the saturation noise is more objectionable than the ADC

noise.

14.8.2 Finite-precision approximation of filter coefficients

The filter coefficients designed from a given specification are analog and have

infinite precision. When the filter coefficients are represented using a finite

number of bits, quantization noise is introduced. As a result, the characteristics

of the digital filter may change considerably from the design specifications.

A common standard used for representing floating point numbers on a digital

computer is the IEEE 754 floating point standard, which uses 32 bits in the

single-precision mode. The representation for the 32-bit IEEE standard is shown

653 14 Digital filters

Table 14.8. Representation used in the 32-bit IEEE 754 floating point single-precision standard

31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0

s exponent significand

(1bit) (8 bits) (23 bits)

Table 14.9. IEEE 754 floating point representation for the decimal number for −0.75ten

31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0

1 0 1 1 1 1 1 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

s exponent significand

in Table 14.8, where a single-precision, floating point number in IEEE 754

standard is represented in scientific notation as follows:

(−1)s × (1 + 0.significand) × 2(exponent−127). (14.25)

Note that the s-bit represents the sign of the floating point. The s-bit is set to

unity for negative numbers and to zero for positive numbers. The significand

specifies the decimal fraction, while the exponent represents the power in terms

of 2. As an example, consider the IEEE 754 binary representation of the decimal

number −0.75, represented by −0.75ten. The binary representation of −0.75ten is given by

−0.75ten = −0.11two,

which in scientific notation is represented by

−0.75ten = −1.1two × 2−1.

Comparing with Eq. (14.25), the values of the exponent and significand are

given by

0.significand = 0.1two and exponent = 126 or 011 111 10two.

The single-precision representation for −0.75ten is specified in Table 14.9. To derive the resolution of the 32-bit single-precision arithmetic, we calculate

the two smallest numbers that can be represented by Eq. (14.25). The smallest

number is given by

(−1)1 × (1 + 0.111 111 11two) × 2–127

= −1 × 1.996 093 75 × 2−127 = −1.173 198 463 418 338 × 10−38.

The next smallest number represented by the 32-bit single-precision arithmetic

(−1)1 × (1 + 0.111 111 10two) × 2−127

= −1 × 1.992 187 50 × 2−127 = −1.170 902 576 014 388 × 10−38.

654 Part III Discrete-time signals and systems

The resolution of the 32-bit single-precision arithmetic is therefore the differ-

ence of these numbers:

−1.173 198 463 418 338 × 10−38 − (−1.170 902 576 014 388 × 10−38) = −2.295 887 403 950 041 × 10−41.

If the hardware allows for IEEE 754 single precision, then the quantization error

is proportional to −2.295 887 403 950 041 × 10−41. Generally, specialized DSP boards are restricted to a smaller number of bits than the IEEE 32-bit single-

precision representation.

The addition of quantization noise into the filter is a non-linear process. The

detailed analysis of the effect of the quantization noise on the performance

of the filter is beyond the scope of this text. In the following, Example 14.9

illustrates the effect of finite-precision arithmetic on the magnitude response of

a 21-tap FIR filter.

14.8.3 Round-off errors

Because of the limited resolution of DSP boards, the output response of a filter

cannot be accurately represented. In the 32-bit signed IEEE 754 floating point

standard, the resolution of each sample of the output response is restricted

to 2.295 887 403 950 041 × 10−41. The distortion due to rounding off sample values to the resolution allowed by the DSP board is less damaging than the

finite-precision representation of the filter coefficients. In the latter case, the

distortion is substantially magnified. Still, the round-off errors in the sample

values should be considered in the analysis of a filter performance.

14.8.4 Arithmetic overflow

Arithmetic overflow occurs during multiplication, division, or addition, when

the final answer falls outside the range of the DSP board. For example, the

dynamic range of the 32-bit signed IEEE 754 floating point standard is restricted

to a maximum value of 2.0ten × 1038 and a minimum value of −2.0ten × 1038. If the result of any mathematical operation between the two floating point numbers

falls outside this range, then an overflow occurs.

Example 14.9

Consider the 21-tap FIR filter with impulse response as shown in Table 14.10,

where each coefficient is represented by 14 decimal digits. The FIR filter is

implemented on a DSP board, which uses finite-precision arithmetic given by

(−1)s × (0 + 0.significand),

where the significand represents the decimal fraction of the number and is

limited to a fixed number of bits. There are no bits allocated for the exponent.

655 14 Digital filters

Table 14.10. Finite impulse response

h[k ] of the 21-tap FIR filter specified in

Example 14.9

k h[k]

10 0.318 348 783 765 15

9,11 0.261 850 185 125 51

8,12 0.132 021 415 468 16

7,13 0.012 135 562 150 39

6,14 −0.041 086 983 052 48 5,15 −0.032 969 416 668 68 4,16 −0.005 898 263 640 95 3,17 0.008 055 858 168 72

2,18 0.006 608 361 295 03

1,19 0.001 494 396 943 68

0, 20 −0.001 385 507 671 95

Table 14.11. Impulse response of the FIR filter in Example 14.9 with 4-bit and 8-bit finite precisions

h[k]

k Exact 8-bit binary representation 4-bit precision 8-bit precision

10 0.318 348 783 765 15 0.010 100 01 0.3125 0.316 406 25

9, 11 0.261 850 185 125 51 0.010 000 11 0.25 0.261 718 75

8, 12 0.132 021 415 468 16 0.001 000 01 0.125 0.128 906 25

7, 13 0.012 135 562 150 39 0.000 000 11 0 0.011 718 75

6, 14 −0.041 086 983 052 48 −0.000 010 10 0 −0.039 062 5 5, 15 −0.032 969 416 668 68 −0.000 010 00 0 −0.031 25 4, 16 −0.005 898 263 640 95 −0.000 000 01 0 −0.003 906 25 3, 17 0.008 055 858 168 72 0.000 000 10 0 0.007 812 5

2, 18 0.006 608 361 295 03 0.000 000 01 0 0.003 906 25

1, 19 0.001 494 396 943 68 0.000 000 00 0 0

0, 20 −0.001 385 507 671 95 0.000 000 00 0 0

Calculate the filter coefficients with the significand restricted to a total of 7 bits

and where 1 bit is allocated for the sign. Plot the magnitude response of the

filter. Repeat for a 3-bit significand with 1 bit allocated for the sign.

Solution

The filter coefficients with the 4-bit and 8-bit finite-precision arithmetic are

shown in Table 14.11. We illustrate how we derived the result for the filter

coefficient h[10] = 0.318 348 783 765 15. The remaining entries can be derived by following the procedure specified for h[10].

656 Part III Discrete-time signals and systems

W 0 p−p −0.5p 0.5p

20 × log10 (|H(W)|)0

−40

−60

−20

exact

4-bit

precision

8-bit

precision

Fig. 14.27. Frequency

characteristics of the filter with

quantized coefficients in

Example 14.9.

The binary representation for h[10] = 0.318 348 783 765 15 is given by

0.31834878376515ten = 0.010 100 010 111 111 1 . . .two.

For 4-bit precision, the finite-precision representation of h[10] is given by

(−1)0 × (0 + 0.0101)two = 2−2 + 2−4 = 0.3125.

For 8-bit precision, the finite-precision representation of h[10] is given by

(−1)0 × (0 + 0.010 100 01)two = 2−2 + 2−4 + 2−8 = 0.316 406 25.

In deriving the above values, the finite-precision representations are truncated to

the available number of bits. Alternatively, the numerical values can be rounded

off to the nearest available level in each representation. The latter reduces the

quantization noise.

In Table 14.7, we observe that several filter coefficients are reduced to zero.

With 8-bit precision, the values of h[0], h[1], h[19], and h[20] are all represented

by zero. With 4-bit precision, a total of 16 values within the ranges 0 ≤ k ≤ 7

and 13 ≤ k ≤ 20 are reduced to zero. In other words, the FIR filter becomes a

17-tap filter with 8-bit precision and a 5-tap filter with 4-bit precision.

A comparison of the frequency characteristics for the three filters, with coef-

ficients listed in Table 14.11, is shown in Fig. 14.27. Noticeable differences in

the magnitude spectrum are observed in the three implementations. The width

of the transition band increases substantially for the FIR filter represented with

4-bit precision. The stop-band ripple also increases with the finite-precision

filters. The original filter has a minimum attenuation of 50 dB in the stop band.

The minimum attenuation is decreased to 40 dB with 8-bit finite precision and

to 20 dB with 4-bit precision. In fact, it is difficult to describe the 4-bit finite-

precision filter as a lowpass filter since the higher-frequency components pass

through the system with comparatively little attenuation.

Increasing the number of bits used in the finite-precision representation gen-

erally improves the approximation of the original filter characteristics. However,

the increase in precision also increases the implementation cost.

657 14 Digital filters

14.9 M A T L A B examples

In Chapter 13, we introduced a M A T L A B M-file residuez for the partial fraction expansion of a given rational function. Similarly, the M-file tf2zp was introduced to calculate the location of poles and zeros for a given transfer

function. These M-files can also be used to derive the cascaded and parallel

forms of the transfer function. We illustrate the application of these M-files by

deriving the cascaded and parallel forms for the transfer function,

H (z) = z3 − 2z2 + z

z3 − 0.1z2 − 0.07z − 0.065 =

1 − 2z−1 + z−2

1 − 0.1z−1 − 0.07z−2 − 0.065z−3 ,

considered in Example 14.5.

14.9.1 Parallel form

The M A T L A B code to determine the partial fraction expansion is given below.

The explanation follows each instruction in the form of comments.

>> B = [1 −2 1 0]; % Coefficients of the % numerator of H(z)

>> A = [1 −0.1 -0.07 −0.065]; % Coefficients of the % denominator of H(z)

>> [R, P, K] = residuez(B, A); % Calculate partial

% fraction expansion

The returned values are given by

R = [0.4310 0.2845+3.3362j 0.2845−3.3362j] P = [0.5000 −0.2000+0.3000j −0.2000−0.3000j] and K = 0.

The transfer function H (z) can therefore be expressed as follows:

H (z) = 0.4310

1 − 0.5z−1 +

0.2845 + j3.3362 1 − (−0.2 + j0.3)z−1

+ 0.2845 − j3.3362

1 − (−0.2 − j0.3)z−1 .

To eliminate complex-valued coefficients, we combine the complex poles as

follows:

H (z) = 0.4310

1 − 0.5z−1 +

0.5690 − 1.8879z−1

1 + 0.4z−1 + 0.13z−2 .

The partial fraction expansion is then implemented using the parallel form as

shown in Fig. 14.25.

658 Part III Discrete-time signals and systems

14.9.2 Series form

The M A T L A B code to determine the poles and zeros of H(z) is given by

>> B = [0 1 −2 1]; % The numerator of H(z) >> A = [1 −0.1 −0.07 −0.065]; % The denominator of H(z) >> [Z, P, K] = tf2zp(B, A); % Calculate poles and

% zeros

The locations of the poles and zeros are given by

Z = [0 1 1]

P = [0.5000 −0.2000+0.3000j −0.2000−0.3000j] and K = 1.

The transfer function H (z) can therefore be expressed as follows:

H (z) = 1 (1 − 0z−1)(1 − 1z−1)(1 − 1z−1)

(1 − 0.5z−1)(1 − (−0.2 + j0.3)z−1)(1 − (−0.2 − j0.3)z−1)

= (1 − z−1)2

(1 − 0.5z−1)(1 − (−0.2 + j0.3)z−1)(1 − (−0.2 − j0.3)z−1) .

Combining the complex roots in the denominator, the cascaded configuration

is given by

H (z) = 1 − z−1

1 − 0.5z−1 ×

1 − z−1

1 + 0.4z−1 + 0.13z−2 .

The cascaded configuration is then implemented using the series form as shown

in Fig. 14.23.

14.10 Summary

Chapter 14 defined digital filters as systems used to transform the frequency

characteristics of the DT sequences, applied at the input of the filter, in a prede-

fined manner. Based on the magnitude spectrum |H (Ω)|, Section 14.1 classifies filters in four different categories. A lowpass filter removes the higher-frequency

components above a cut-off frequencyΩc from an input sequence, while retain-

ing the lower-frequency components Ω ≤ Ωc. A highpass filter is the converse

of the lowpass filter and removes the lower-frequency components below a cut-

off frequencyΩc from an input sequence, while retaining the higher-frequency

components Ω ≥ Ωc. A bandpass filter retains a selected range of frequency components between the lower cut-off frequency Ωc1 and the upper cut-off

frequency Ωc2 of the filter. A bandstop filter is the converse of the bandpass

filter, which rejects the frequency components between the lower cut-off fre-

quencyΩc1 and the upper cut-off frequencyΩc2 of the filter. All other frequency

components are retained at the output of the bandstop filter.

Section 14.2 introduces a second classification of digital filters based on the

length of the impulse response h[k] of the digital filter. Finite impulse response

659 14 Digital filters

(FIR) filters have a finite length impulse response, while the length of infinite

impulse response (IIR) filters is infinite. The ideal frequency-selective filters,

introduced in Section 14.1, are not practically realizable because of constant

gains within the pass band and stop band, the sharp transitions between the

pass band and the stop band, and because of the zero phase. Sections 14.3 and

14.4 explore practical realizations of the ideal filter obtained by allowing some

variations in the pass-band and stop-band gains, introducing a linear-phase

within the pass band, and by leaving some transitional bandwidth between

the pass band and the stop band. The transition bandwidth allows the filter

characteristics to change gradually. Section 14.3 also proved the following

sufficient condition for ensuring a linear phase for FIR filters. If the impulse

response function of an N -tap filter, with z-transfer function given by Eq. (14.7),

satisfies either of the following relationships:

symmetrical impulse response h[k] = h[N − 1 − k]; antisymmetrical impulse response h[k] = −h[N − 1 − k],

then the phase <H (Ω) of the filter is linearly proportional to the frequency.

When implementing digital filters on digital computers or specialized DSP

boards, the output of a digital filter is obtained by solving the following recursive

formula:

y[k] = − 1

a0 (a1 y[k − 1] + · · · + aN y[k − N ])

+ 1

a0 (b0x[k] + b1x[k − 1] + · · · + bM x[k − M]).

Based on the aforementioned formula, Sections 14.5–14.7 derived physical real-

izations of digital filters using three fundamental elements: a two-input adder,

a scalar multiplier, and a unit delay. For FIR filters, direct form, series form,

and parallel forms are derived in Section 14.6. The series form is obtained by

factorizing the transfer function in terms of a product of quadratic polynomials

and then cascading the transfer function for the individual quadratic polynomi-

als. The parallel form is obtained by partial fraction expansion of the transfer

function. Section 14.7 derived similar realizations for IIR filters. For both FIR

and IIR filters, alternative flow diagrams are obtained by applying the transpose

operation. Transposition of a flow graph is achieved by (i) interchanging the

role of the input and output; (ii) reversing the directions of all branches within

a flow graph; and (iii) replacing the source nodes with adders and the adders

with source nodes.

Direct form II, the series form, and the cascaded form are defined as canon-

ical representations since there forms, in general, use the minimal number of

fundamental elements, whereas direct form I is referred to as a non-canonical

representation. Irrespective of the arithmetic complexity, all of these realizations

provide identical outputs for the same input sequence.

660 Part III Discrete-time signals and systems

During the actual realization of digital filters in software or hardware, the

filter coefficients are implemented with finite precision. Section 14.8 discussed

the impact of finite-precision arithmetic on the performance of digital fil-

ters. It is observed that the effect of finite-precision arithmetic varies from

one realization of the filter to another. We list in the following some empir-

ical observations that should be kept in mind while choosing a particular

realization.

(i) When the poles of the transfer function lie close to each other or close to

the unit circle in the complex z-plane, direct form realizations, with filter

coefficients represented using finite precision, produce large deviations

from the output of an exact filter.

(ii) The order in which the first- and second-order systems are implemented

in cascaded forms affects the output of the filter in finite-precision imple-

mentations. Changing the order may reduce the deviation from the output

of an exact filter.

(iii) Pairing of complex poles and zeros is important for all cascaded and

parallel realizations.

(iv) In cascaded realizations, scalar multipliers between different systems may

be required to prevent the partial fraction coefficients from becoming too

large or too small.

Section 14.9 presented two M A T L A B functions, residuez and tf2zp, for deriving the physical realizations of digital filters.

Problems

14.1 Determine if the filters represented by the following transfer functions are

(a) FIR or IIR, and (b) causal or non-causal. If a filter is FIR, determine

if its phase is linear.

(i) H (z) = 0.7 + 0.2z−1 + 0.8z−2;

(ii) H (z) = 1

3 (z + 1 + z−1)

(iii) H (z) = 0.7 + 0.2z−1 + 0.8z−2

1 + 0.5z−1 − 0.24z−2 ;

(iv) H (z) = 1 − 0.1z−1 − 0.06z−2

1 + 0.2z−1 .

14.2 Consider two filters with transfer functions given by

H1(z) = 1 + 2z−1 + 3z−2 + 2z−3 + z−4 and H2(z) = 1 + 2z−1 + 3z−2 + 2z−3 − z−4.

661 14 Digital filters

x[k]

1.0

+ 2.0

+ + + 0.4 0.2 0.1

y[k]

z−1z−1z−1z−1 Fig. P14.5. The FIR system for

Problem 14.5.

(i) Determine and plot the frequency characteristics of the filters.

(ii) If the sequence x[k] cos(0.5k) + cos(k) is applied at the input of filter H1(z), determine the output of the filter from the frequency

characteristics obtained in (i).

(iii) Repeat (ii) for filter H2(z). What advantage do you see with the

linear-phase filter?

14.3 Consider a digital filter with impulse response given by

h[k] = {

1/3 −1 ≤ k ≤ 1 0 otherwise.

(i) Calculate the transfer function of the filter.

(ii) Sketch the amplitude and phase responses of the filter with respect

to frequency.

(iii) How will you classify this filter – lowpass, bandpass, bandstop, or

highpass?

(iv) Does it have a linear phase?

14.4 Consider a digital filter with transfer function given by

H (z) = 0.7 + 0.2z−1 + 0.8z−2

1 + 0.5z−1 − 0.24z−2 .

(i) Plot the impulse response and the frequency characteristics of the

filter.

(ii) From the frequency characteristics, determine the maximum magni-

tude of the pass-band and stop-band ripples and the transition band-

width.

14.5 Given the flow graph in Fig. P14.5, calculate the transfer function and the

impulse response of the LTI system of the realization. From the transfer

function, calculate the magnitude and phase spectra for the filter.

14.6 The flow graph of Fig. P14.5 can be implemented by using only three

scalar multipliers. Sketch the flow graph which uses three scalar multipli-

ers without increasing the number of delay elements or two-input adders.

14.7 Repeat Problem 14.5 for the flow graph shown in Fig. P14.7.

14.8 Draw the flow graphs for (i) the direct form and (ii) the cascaded form

for an FIR filter with a transfer function given by

H (z) = 0.4 − 0.8z−1 + 0.4z−2.

662 Part III Discrete-time signals and systems

x[k] y[k]

+ 0.7

0.2

0.8

z−1

−0.5

0.24

Fig. P14.7. The IIR system for

Problem 14.7.

14.9 Using the principle of transposition for the flow graphs, derive two alter-

native representations for the FIR filter specified in Problem 14.8.

14.10 Draw the linear-phase flow graph for the FIR filter specified in Problem

14.8.

14.11 The transfer function of an IIR filter is given by

H (z) = (1 − 0.25z−1)8.

Draw the flow graphs based on the following forms: (i) cascade of eight

first-order FIR systems; (ii) cascade of four second-order FIR systems;

(iii) cascade of two third-order FIR systems and one second-order FIR

system; (iv) cascade of two fourth-order FIR systems; (v) cascade of one

sixth-order FIR system and one second-order FIR system. Compare the

computational complexity of each realization.

14.12 The transfer function of an IIR filter, with impulse response given by

h[k] = 0.5k sin (

4 k

)

u[k],

is given by the following expression:

H (z) = 0.5z sin

(

)

z2 − 2 × 0.5 cos (

)

z + 0.25 ≈

0.3536z

z2 − 0.7071z + 0.25 .

Draw the flow graphs for (i) direct form I, (ii) direct form II, (iii) the

cascaded form, and (iv) the parallel form realizations of the IIR filter.

14.13 Using the principle of transposition for the flow graphs, derive four

alternative flow graph representations for the IIR filter specified in

Problem 14.12.

14.14 The transfer function of a digital system is given by

H (z) = 1 − 0.8z−1 + 0.15z−2

1 − 0.7z−1 − 0.18z−2 .

Draw the flow graphs for (i) direct form I, (ii) direct form II, (iii) the

cascaded form, and (iv) the parallel form realizations of the IIR filter.

663 14 Digital filters

14.15 Using the principle of transposition for the flow graphs, derive four alter-

native flow graph representations for the IIR filter specified in Problem

14.14.

14.16 An allpass filter has a constant gain for all frequencies, i.e. |H (Ω)| = 1. (i) Show that the transfer functions

H1(z) = α1 + z−1

1 + α1z−1 and H2(z) =

α1α2 + α1z−1 + z−2

1 + α1z−1 + α2z−2

represent allpass filters.

(ii) Sketch the flow graph for the first-order allpass filter H1(z), which

uses a single scalar multiplier.

(iii) Sketch the flow graph for the second-order allpass filter H2(z) with

only two scalar multipliers. There is no restriction on the number

of unit delay elements or two-input adders in each case.

14.17 The impulse response of an LTID system is given by

h[k] = {

αk 0 ≤ k ≤ 9

0 elsewhere.

(i) Draw the flow graph for the above LTID system with no feedback

paths.

(ii) The z-transfer function for the above impulse response is given by

H (z) = 1 − α10z−10

1 − αz−1 .

Draw the flow graph of the IIR system specified by this transfer

function.

(iii) Compare the two implementations with respect to the number of

delays, scalar multipliers and two-input adders.

14.18 Implement the filter with transfer function given by

H (z) = 0.4 − 0.8z−1 + 0.4z−2

with finite-precision arithmetic given by

(−1)s × (0 + 0.significand),

where the significand represents the decimal fraction of the coefficients

and is limited to 3 bits with 1 bit allocated for the sign. Compare the

magnitude response of the original filter with the magnitude response of

the filter implemented with finite-precision representation.

664 Part III Discrete-time signals and systems

14.19 Repeat Problem 14.14 for the following transfer function:

H (z) = 1 − 0.8z−1 + 0.15z−2

1 − 0.7z−1 − 0.18z−2 .

14.20 Repeat Problem 14.14 for the following transfer function:

H (z) = 0.5z sin

(

)

z2 − cos (

)

z + 0.25 .

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CUUK852-Mandal & Asif May 28, 2007 14:9

C H A P T E R

15 FIR filter design

In Chapter 14, we defined frequency-selective filters as systems that modify

the frequency components of the input signals in a predefined manner. Further

classification of frequency-selective filters is based on the length N of their impulse responses h[k]. If the length N of the impulse response of a frequency- selective filter is finite, the filter is referred to as a finite impulse response (FIR)

filter. If the length N is infinite, the frequency-selective filter is referred to as an infinite impulse response (IIR) filter. In this chapter, we consider the design

of frequency-selective FIR filters.

The design of digital filters involves three distinct stages. Stage 1 describes

the desired specifications of the frequency characteristics of the filter. Based

on the specified frequency characteristics, stage 2 derives the transfer function

H (z), or the impulse response h[k], of the filter. Finally, stage 3 develops the canonical realization of the filter using one of the several forms presented in

Chapter 14. While deriving the impulse response h[k] of an FIR filter in stage 2, the following two conditions must also be satisfied.

(1) Causality condition. This implies that the impulse response h[k] of an FIR filter is zero for k < 0. This will ensure a causal, and hence a physically realizable, filter.

(2) Linear-phase condition. This implies that the impulse response h[k] of an FIR filter of length N is symmetrical or anti-symmetrical, i.e. h[k] = ±h[N − 1 − k]. The linear-phase condition ensures that no distortion is introduced in the input frequency components lying within the pass band

of the FIR filter.

Generally, FIR filters are designed directly from the impulse response of an

ideal lowpass filter. Section 15.1 describes the windowing approach, where an

appropriate window function w[k] is used to truncate the impulse response of an ideal lowpass filter to a finite length N . The specifications of the FIR filter, along with the characteristics of the window function, are used to calculate

the length N of the FIR filter. Sections 15.2 and 15.4 extend the windowing

665

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CUUK852-Mandal & Asif May 28, 2007 14:9

666 Part III Discrete-time signals and systems

approach to the design of highpass, bandpass, and bandstop FIR filters. The FIR

filter design techniques, based on the windowing function, can result in several

alternative designs, all of which satisfy the given specifications. Section 15.5

presents the Parks–McClellan method, which recursively computes the opti-

mal filter for a given length N . Section 15.6 presents several library functions available in M A T L A B to design FIR filters. Finally, the chapter is concluded in

Section 15.7.

15.1 Lowpass filter design using windowing method

In Section 14.1, it was shown that the impulse response of an ideal lowpass

filter is a sinc function, and therefore that an ideal lowpass filter is non-causal

and IIR. In Section 14.4, it was shown that a causal lowpass FIR filter can

be obtained by delaying the ideal impulse response by m time units (see Fig.

15.1a) and truncating the impulse response. To generate an N-tap FIR filter, the

truncation of the ideal impulse response is performed as follows:

hlp[k] =







hilp[k] = Ωc

π sinc

( (k − m)Ωc

)

0 ≤ k ≤ N − 1

0 elsewhere;

(15.1)

where the value of m in Eq. (15.1) is selected to be (N − 1)/2. This approach of designing an FIR filter is referred to as the windowing method, and is shown

in Fig. 15.1. Note that the impulse response hlp[k] of the resulting FIR filter is non-zero only within the range 0 ≤ k ≤ N–1. In addition, the impulse response hlp[k] is symmetrical about k = (N − 1)/2, i.e.

h[k] = h[N − 1 − k], (15.2)

and satisfies the linear-phase condition given in Eq. (14.8a). If N is an odd- valued integer, the resulting FIR filter is a type 1 linear-phase filter with an

integer delay m. On the other hand, if N is an even-valued integer, the resulting FIR filter is a type 2 linear-phase filter with a fractional delay m.

Truncating the impulse response of an ideal lowpass filter affects the fre-

quency characteristics of the ideal lowpass filter. In addition to introducing

ripples within the pass and stop bands, the truncation leads to a transition band

between the pass band and the stop band. In the following subsection, we ana-

lyze the effect of truncating the impulse response hlp[k] of the ideal lowpass filter with a rectangular window of length N .

15.1.1 Rectangular window

The rectangular window of length N , centered at k = (N − 1)/2, is defined as follows:

wrect [k] =

{

1 0 ≤ k ≤ N − 1 0 otherwise,

(15.3)

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667 15 FIR filter design

hilp[k]

k m m+2

Wc/π

Wc hlp[k]

m−p/Wc m+p/Wc

m−2

m m+2m−2

0 p

Hilp(W)

−p −0.5p 0.5p W

Wc−Wc

k W 0 p

Hlp(W)

−p −0.5p 0.5p

0 p

Wrect(W)

−p −0.5p 0.5p

wrect[k]

0 N−1

(a) (b)

(e) ( f )

m−p/Wc m+p/Wc

Fig. 15.1. Windowing operation to derive a truncated FIR filter from an ideal lowpass filter. The left-hand

column (plots (a), (c), and (e)) represents the windowing operation in the time domain, and the right-hand

column (plots (b), (d), and (f)) represents the windowing operation in the frequency domain. (a) Impulse

response h ilp[k ] of an ideal lowpass filter. (b) Magnitude spectrum |H ilp(Ω)| of an ideal lowpass filter. (c) Rectangular window wrect[k ]. (d) DTFT Wrect(Ω) of the rectangular window. (e) Impulse response

h lp[k ] = h ilp[k ]wrect[k ] of the truncated lowpass filter. (f) Magnitude spectrum |H lp(Ω)| = |(1/2π )H ilp(Ω) ∗ Wrect(Ω)| of the truncated lowpass filter.

where we have assumed that the length N of the windowing function is odd. Taking the DTFT of Eq. (15.3) results in the following frequency characteristics

for the rectangular window:

Wrect (Ω) = e−j(N−1)Ω/2 × sin(NΩ/2)

sin(Ω/2) . (15.4)

The rectangular window wrect[k] and its magnitude spectrum |Wrect(Ω)| are illustrated in Figs. 15.1(c) and (d), respectively. The narrow lobe, centered at

Ω = 0, in Wrect(Ω) is referred to as the main lobe, while the lobes on each side of the main lobe are referred to as the side lobes of the rectangular window.

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668 Part III Discrete-time signals and systems

Truncating the impulse response hilp[k] of the ideal lowpass filter to length N is the same as multiplying the impulse response hilp[k] by the rectangular window in the time domain. The truncation operation is, therefore, modeled as

follows:

hlp[k] = hilp[k]wrect[k], (15.5)

with m = (N − 1)/2. The result of the truncation step is illustrated in Fig. 15.1(e). Since multiplication in the time domain is equivalent to convolution in

the frequency domain, the transfer function of the truncated FIR filter is given

Hlp(Ω) = 1

2π [Wrect(Ω) ∗ Hilp(Ω)] =

2π

∫

〈2π〉

Hilp(θ )Wrect(θ − Ω) dθ, (15.6)

which results in the magnitude spectrum shown in Fig. 15.1(f). Comparing the

magnitude spectrum |Hilp(Ω)| of the ideal lowpass filter with the magnitude spectrum |Hlp(Ω)| of the truncated lowpass filter, we note three major differ- ences. First, there are significant ripples within the pass band of the truncated

lowpass filter. Secondly, the magnitude spectrum of the truncated lowpass fil-

ter does not change abruptly in between the pass band and stop band. In fact,

a transition band of finite width appears. Thirdly, there are additional ripples

within the stop band of the truncated lowpass filter. The appearance of ripples in

the pass band and stop band is referred to as the Gibbs phenomenon. In order to reduce the ripples and eliminate the transition band, the DTFT Wrect(Ω) should be a narrow impulse function. This would imply that the length N of the win- dowing function is very large, increasing the implementation complexity of the

truncated lowpass filter.

In Fig. 15.1(c), we observe that the rectangular window has abrupt truncations

outside the range 0 ≤ k ≤ N − 1. The pass-band and stop-band ripples, as well as the transition band, can be decreased by selecting alternative windows that

taper smoothly to zero from the peak value of 1 at k = (N − 1)/2. Section 15.1.2 discusses several alternatives to the rectangular window.

15.1.2 Commonly used windows

There are a number of alternatives to the rectangular window. A few popular

choices are defined in the following.

Bartlett (triangular) window

wbart[k] =



    

    

N − 1 0 ≤ k ≤ (N − 1)/2

2 − 2k

N − 1 (N − 1)/2 < k ≤ N − 1

0 otherwise.

(15.7)

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669 15 FIR filter design

N−1 k

N−10

Blackman

Hamming

Hanning

Bartlett

rectangular

1 w[k]Fig. 15.2. Commonly used

windows of length N .

Generalized Hamming window For 0 < α < 1,

wgene[k] =







α − (1 − α) cos [

2πk

N − 1

]

0 ≤ k ≤ N − 1

0 otherwise.

(15.8)

Hamming window

whamm[k] =







0.54 − 0.46 cos

[ 2πk

N − 1

]

0 ≤ k ≤ N − 1

0 otherwise.

(15.9)

Hanning window

whann[k] =







0.5 − 0.5 cos

[ 2πk

N − 1

]

0 ≤ k ≤ (N − 1)

0 otherwise.

(15.10)

Blackman window

wblac[k] =







0.42 − 0.5 cos

[ 2πk

N − 1

]

+ 0.08 cos

[ 4πk

N − 1

]

0 ≤ k ≤ N − 1

0 otherwise.

(15.11)

The shapes of the windows are shown in Fig. 15.2, where, for convenience of

illustration, continuous plots are used. In reality, the windows are a function of

the DT variable k. It may be noted that the Hamming and Hanning windows are special cases of the generalized Hamming window. For the Hamming window,

variable α in Eq. (15.8) of the generalized Hamming window equals 0.54.

Similarly, for the Hanning window, variable α in Eq. (15.8) equals 0.5.

The DTFTs of the aforementioned windows are shown in Fig. 15.3, where

the vertical axis represents the magnitude of the DTFTs based on the decibel

(dB) scale. The two important parameters used in the FIR filter design are (i) the

width of the main lobes of the DTFT of the windows; (ii) the relative strength

of the highest value side lobe with respect to the main lobe. The width of the

main lobe is defined as the distance between the nearest zero crossings of the

main lobe, while the relative side lobe strength is defined as the difference in

dB between the magnitudes of the highest value side lobe and the main lobe.

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670 Part III Discrete-time signals and systems

0 0.2p 0.4p 0.6p 0.8p p −100

−80

−60

−40

−20

0 0.25p 0.5p 0.75p p −100

−80

−60

−40

−20

0 0.2p 0.4p 0.6p 0.8p p −100

−80

−60

−40

−20

Hanning window

0 0.2p 0.4p 0.6p 0.8p p −100

−80

−60

−40

−20

Hamming window

0 0.2p 0.4p 0.6p 0.8p p −100

−80

−60

−40

−20

Blackman window

(a)

(e)

(b)

rectangular window Bartlett window

Fig. 15.3. DTFTs of commonly

used windows with length

N = 75. (a) Rectangular window; (b) Bartlett window;

(d) Hamming window;

(e) Blackman window.

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671 15 FIR filter design

Table 15.1. Comparison of the properties of the commonly used windows

Kaiser window(b)

Window

Width of

main lobe

Peak side lobe

amplitude(a) (dB)

Max. stop/pass-band

error 20log10(δ) β transition width

Rectangular 4π /N −13.3 −21 0 1.81π /(N − 1) Bartlett 8π /(N− 1) −26.5 −25 1.33 2.37π /(N − 1) Hanning 8π /(N− 1) −31.4 −44 3.86 5.01π /(N − 1) Hamming 8π /(N− 1) −42.6 −53 4.86 6.27π /(N − 1) Blackman 12π /(N− 1) −58.0 −74 7.04 9.19π /(N − 1)

aThe peak side lobe magnitude in column 3 is relative to the magnitude of the main lobe. bThe last two columns for the Kaiser window are explained in Section 15.1.5.

The second and third columns of Table 15.1 compare these two parameters for

the commonly used windows as a function of the length N of the window. The fourth column of Table 15.1 quantifies the maximum difference between the

magnitude spectra within the pass and stop bands of the ideal lowpass filter and

the causal FIR filter obtained from the windowing method. In other words, it

provides an upper bound on the values of the ripples in the pass and stop bands

of the causal FIR filter. For example, the maximum pass- and stop-band error of

–21dB for the rectangular window implies that the pass- and stop-band ripples

are confined to –21dB in the FIR filter obtained with the rectangular window.

In filter design, we prefer to minimize the transition band and reduce the

strength of the ripples. These are conflicting requirements, as we see next.

To minimize the transition band in the FIR filter, the main lobe width of the

windows should be as small as possible. To reduce the pass-band and stop-

band ripples in the FIR filter, the area enclosed by the side lobes (in other

words, the relative strength of the side lobes) of the windows should be small.

Table 15.1 illustrates that these two requirements are contradictory. The rect-

angular window has the smallest width main lobe, but the relative strength of

its highest side lobe with respect to the main lobe is the largest. As a result,

for the rectangular window, the transition bandwidth is small, but the ripple

magnitude is large. On the contrary, the relative strength of the side lobe for the

Blackman window is the smallest, but the width of its main lobe is the largest.

In other words, for the Blackman window, the transition bandwidth is large, but

the ripple magnitude is small.

In the following example, we illustrate the effect of the rectangular and

Hamming windows on the frequency characteristics of an ideal lowpass filter

truncated with these windows.

Example 15.1

Calculate the impulse response of an ideal DT lowpass filter with radian cut-

off frequency Ωc = 1. From the ideal filter, design two 21-tap FIR filters with Ωc = 1 using the rectangular and Hamming windows.

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672 Part III Discrete-time signals and systems

hrect[k]

0.3183

10 12 14 16 18 200 6 8

hhamm[k]

0.3183

10 12 14 16 18 200 4 6 82 4 2

(a) (b)

Fig. 15.4. Impulse response of

FIR filters obtained by truncating

the impulse response of the

ideal lowpass filter with (a) a

rectangular window and (b) a

Hamming window.

Solution

Substituting Ωc = 1 in Eq. (14.12), the impulse response of an ideal lowpass filter is given by

hilp[k] = sin(k − m) (k − m)π

= 1

π sinc

( k − m

)

, (15.12)

where m = (N − 1)/2 = 10. The expressions for the rectangular and Hamming windows with 21 taps are as follows:

rectangular window wrect[k] = {

1 0 ≤ k ≤ 20 0 otherwise;

(15.13)

Hamming window whamm[k] =



 

 

0.54 − 0.46 cos

( 2πk

)

0 ≤ k ≤ 20

0 otherwise.

(15.14)

The FIR filters are obtained by multiplying the impulse response of the ideal

lowpass filter by the expressions for the rectangular and Hamming windows.

The resulting impulse responses are as follows:

rectangular window hrect[k] =



 

 

π sinc

( k − 10

)

0 ≤ k ≤ 20

0 otherwise; (15.15)

Hamming window hhamm[k]







π sinc

( (k − 10)

) (

0.54 − 0.46 cos

[ 2πk

])

0 ≤ k ≤ 20

0 otherwise.

(15.16)

The impulse responses for FIR filters obtained by truncating the ideal lowpass

filter impulse response with the rectangular and Hamming windows are shown in

Fig. 15.4. Although the two impulse responses have the same value at k = 10, the impulse response hhamm[k], shown in Fig. 15.4(b), decays more rapidly as we move away from the central point (k = 10) and is different from the impulse response hrect[k], shown in Fig. 15.4(a). Typically, the pass-band gain of the truncated FIR filters, obtained from the ideal lowpass filters using the

windowing method, is not unity, as desired. To prove this, we calculate the value

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673 15 FIR filter design

|H (W)|

W 0 p0.75p0.5p0.25p

0.8

0.6

0.4

0.2 Hamming

rectangular

|H(W)|20 log10

W 0 p0.75p0.5p0.25p

−20

−40

−60

Hamming rectangular

(a) (b)

Fig. 15.5. Magnitude spectra of the FIR filters obtained by truncating the impulse response of the ideal

lowpass filter with the rectangular and Hamming windows. The magnitude spectrum of the FIR filter

obtained from the rectangular window is plotted as a solid line, and the magnitude spectrum of the FIR

filter obtained from the Hamming window is plotted as a dashed line. (a) Plotted using a linear scale for

the gain. (b) Plotted using a dB scale for the gain.

of Hlp(0) by substituting Ω = 0 in the DTFT Hlp(Ω):

Hlp(0) = ∞∑

k=−∞ hlp[k]e

−jΩk

∣ ∣ ∣ ∣ ∣ Ω=0

= ∞∑

k=−∞ hlp[k]. (15.17)

Equation (15.17) can therefore be used to calculate the pass-band gain atΩ = 0. Using the values of the samples plotted in Figs. 15.4(a) and (b), the values of

the gain of the two truncated filters at Ω = 0 are given by

rectangular window Hrect(0) = ∞∑

k=−∞ hrect[k] = 0.9754;

Hamming window Hhamm(0) = ∞∑

k=−∞ hhamm[k] = 0.9982.

To ensure a unity gain in the pass band, the impulse response corresponding

to the rectangular window in Eq. (15.15) is normalized by a factor of 0.9754.

Similarly, the impulse response corresponding to the Hamming window is nor-

malized by a factor of 0.9982. The resulting magnitude spectra of the two

normalized FIR filters are shown in Fig. 15.5, where the gains of the filter are

plotted on a linear scale in Fig. 15.5(a) and on a logarithmic scale in Fig. 15.5(b).

It is observed that the dc gain, defined as the gain of the filter atΩ = 0, for both filters is unity. The rectangular window results in higher pass-band and stop-

band ripples. However, the rectangular window provides a smaller transition

band than the Hamming window.

From Fig. 15.5(b), we quantify the gain in the stop band for the FIR filter

obtained using the Hamming window and compare its value with the stop-band

gain for the FIR filter obtained using the rectangular window. The maximum

gain in |Hhamm(Ω)| is less than −50 dB in the stop band (Ω > 0.49π ). Equiv- alently, we can also say that the minimum attenuation in the stop band of the

Hamming window is greater than 50 dB. The maximum gain in |Hhamm(Ω)|,

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674 Part III Discrete-time signals and systems

|Hlp(W)|

Wp Ws

1−dp

1+dp

pass band stop bandtransition

band

W 0 p

Fig. 15.6. Desired specifications

of a lowpass filter.

obtained from the rectangular window, is about −22 dB forΩ > 0.37π . In other words, the Hamming window attenuates the higher-frequency components of

the input signals more strongly than the rectangular window. As discussed ear-

lier, this improvement in the stop-band attenuation is at the expense of a higher

transitional bandwidth in the truncated FIR filter obtained from the Hamming

window.

15.1.3 Design of FIR lowpass filters

We now list the main steps involved in the design of FIR filters using the

windowing method. The design specifications for a lowpass filter are illustrated

in Fig. 15.6 and are given by

pass band (0 ≤ Ω ≤ Ωp) (1 − δp) ≤ |Hlp(Ω)| ≤ (1 − δp);

stop band (Ωs < Ω ≤ π ) 0 ≤ |Hlp(Ω)| ≤ δs.

Expressed in decibels (dB), 20 log10(δp) is referred to as the pass-band ripple

or the peak approximation error within the pass band. Similarly, 20 log10(δs) is

referred to as the stop-band ripple or the peak approximation error in the stop

band. The stop-band ripple can also be expressed in terms of the stop-band

attenuation as −20 log10(δs) dB.

For digital filters, the pass and stop bands are generally specified in the DT

frequency Ω domain, which is limited to the range 0 ≤ Ω ≤ 2π . A DT system

may also be used to process a CT signal. The schematic representation of such

a system was shown in Fig. 9.1. In such cases, it is possible that the pass and

stop bands of the overall system are specified in the CT frequency ω domain

and we are required to compute the transfer function of the DT system shown as

the central block in Fig. 9.1. We assume that the sampling frequency f0 used in the analog to digital (A/D) converter is known. The following nine steps design

an FIR filter using the windowing method.

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675 15 FIR filter design

Step 1 Calculate the normalized cut-off frequency of the filter based on the following expressions:

DT frequency Ω specifications

cut-off frequency, Ωc = 0.5(Ωp + Ωs);

normalized cut-off frequency, Ωn = Ωc/π ;

CT frequency ω (or f ) specifications

cut-off frequency, ωc = 0.5(ωp + ωs) or fc = 0.5( fp + fs);

normalized cut-off frequency, Ωn = ωc0.5ω0 �= fc

0.5 f0 .

Note that the for CT specifications, ωp and ωs denote the pass-band and stop-

band edge frequencies in radians/s, and fp and fs denote the pass-band and stop- band edge frequencies in Hz, respectively. The above frequency normalization

scales the DT frequency range [0, π ] to [0, 1]. For CT, the frequency range

[0, 0.5ω0] (in radians/s) or [0, 0.5 f0] (in Hz) is scaled to [0, 1]. The normalized cut-off frequency Ωn can have a value in the range [0, 1].

Step 2 The impulse response of an ideal lowpass filter is given by

hilp[k] = sin((k − m)Ωc)

(k − m)π = Ωc

π sinc

( (k − m)Ωc

)

= Ωn sinc((k − m)Ωn),

where Ωc = πΩn and m = (N − 1)/2, where N is the filter length to be cal- culated in step 6. Note that the DT filter impulse response hilp[k] primarily depends on the normalized frequency Ωn. If the DT filter is used to process DT

signals obtained using different sampling rates, the CT cut-off frequency will

change depending on the sampling frequency, but the Ωn will remain same.

Step 3 Calculate the minimum attenuation A using A = min(δp, δs) and then convert it to the dB scale.

Because of the nature of the windowing method and the inherent symmetry in

the window functions, the resulting FIR filter has identical attenuations of A dB in both the pass and stop bands. If δp > δs, the designed filter will satisfy

the pass-band attenuation requirement and exceed the stop-band attenuation

requirement. Conversely, if δs > δp, then the filter will satisfy the stop-band

attenuation requirement and exceed the pass-band attenuation requirement.

Step 4 Use the first three columns of Table 15.2 to choose the window type for the specified attenuation A.

In Table 15.2, the attenuation A, specified in the first two columns, is relative to the pass-band gain. For a given value of A, more than one choice of the window type is possible. With a minimum attenuation requirement of 20 dB,

for example, any of the four windows may be selected. Although the higher

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676 Part III Discrete-time signals and systems

Table 15.2. Selection of the type of window based on the attenuation values

obtained from step 3

Minimum attenuation (A)

dB linear scale Type of window

Transition bandwidth,

�Ωn

≤20 ≤0.1 rectangular 1.8/N ≤40 ≤0.01 Hanning 6.2/N ≤50 ≤0.003 Hamming 6.6/N ≤70 ≤0.000 03 Blackman 11/N

attenuation windows (Hanning, Hamming, or Blackman) reduce the pass- and

stop-band ripples, the transition bands of the resulting FIR filters obtained with

these windows are larger than the transition band of the FIR filter obtained with

the rectangular window.

The first two columns of Table 15.2 are approximated directly from the fourth

column of Table 15.1, which lists the stop-band attenuation. The last column of

Table 15.2 is based on empirical observations.

Step 5 Calculate the normalized transition bandwidth for the FIR filter using the following expressions:

DT frequency Ω specifications

transition BW, �Ωc = Ωs − Ωp;

normalized transition BW, �Ωn = �Ωc/π ;

CT frequency specifications

transition BW, �ωc = ωs − ωp or � fc = fs − fp;

normalized transition BW, �Ωn = �ωc

0.5ω0 =

� fc 0.5 f0

Step 6 Using the last column of Table 15.2, determine the minimum length N of the filter for the computed transitional bandwidth �Ωn obtained in step 5 and the window function selected in step 4.

Step 7 Determine the expression w[k] for the window function using the window type selected in step 4 for length N obtained in step 6. The expression for the rectangular window is given in Eq. (15.3), while the expressions for the

remaining window functions are specified in Eqs. (15.7)–(15.11).

Step 8 Derive the impulse response hlp[k] of the FIR filter:

hlp[k] = hilp[k]w[k].

If the pass-band gain |Hlp(0)| at Ω = 0, given by ∑

hlp[k], is not equal to one, we normalize hlp[k] with

∑

hlp[k], where ∑

denotes the summation operation.

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677 15 FIR filter design

Step 9 Confirm that the impulse response hlp[k] satisfies the initial specifica- tions by plotting the magnitude spectrum |hlp(k)| of the FIR filter obtained in step 8.

We illustrate the working of the aforementioned FIR filter design algorithm in

Example 15.2.

Example 15.2

Figure 9.1 is used to process a CT signal with a digital filter. The overall

characteristics of the CT system, modeled with Fig. 9.1, are specified below:

(i) pass-band edge frequency (ωp) = 3π kradians/s (or 1500 Hz); (ii) stop-band edge frequency (ωs) = 4π kradians/s (or 2000 Hz);

(iii) minimum stop-band attenuation, −20 log10(δs) = 50 dB; (iv) sampling frequency ( f0) = 8 ksamples/s.

Design the DT system in Fig. 9.1 based on the aforementioned CT specifications.

Solution

Step 1 suggests that the cut-off frequency of the filter is given by

ωc = 0.5(ωp + ωs) = 3.5π kradians/s.

Using ω0 = 2π f0 = 16π × 103, the normalized cut-off frequency is given by

Ωn = ωc/(0.5ω0) = (

3.5π × 103 )

/ (

0.5 × 2π × 8 × 103 )

= 0.4375.

Based on step 2, the impulse response of the ideal lowpass filter with the

normalized cut-off frequency Ωn = 0.4375 is given by

hilp[k] = 0.4375 sinc(0.4375(k − m))

with m set to (N − 1)/2. The value of N is determined in step 6. Step 3 determines the minimum attenuation A to be 50 dB. Step 4 determines the type of window. For the minimum stop-band attenua-

tion of 50 dB, Table 15.2 limits our choice to either the Hamming or Blackman

window. We select the Hamming window because its length N will be lower than that of the Blackman window.

Step 5 computes the normalized transition bandwidth:

�Ωn = �ωc/(0.5ω0) = (4π − 3π ) × 103/(0.5 × 2π × 8 × 103) = 0.1250.

Step 6 evaluates the length N of the Hamming window:

6.6/N = 0.1250 ⇒ N = 8 × 6.6 = 52.8.

Ceiling off the length of the window to the nearest larger odd integer, we obtain

N = 53.

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678 Part III Discrete-time signals and systems

|H(W)|20 log10

W 0 p0.75p0.5p0.25p

−20

−40

−60

Fig. 15.7. Magnitude spectrum

of the FIR filter designed in

Example 15.2.

Step 7 derives the expression for the Hamming window of length N = 53:

whamm[k] =







0.54 − 0.46 cos [

2πk

]

0 ≤ k ≤ 52

0 otherwise.

Step 8 gives the impulse response of the FIR filter:

hlp[k] =







0.4375 sinc(0.4375(k − 26))

{

0.54 − 0.46 cos

[ 2πk

]}

0 ≤ k ≤ 52

0 otherwise.

Since ∑

hlp[k] = 0.9998 ≈ 1, the impulse response hlp[k] of the FIR filter is not normalized with

∑

hlp[k]. The magnitude spectrum of the FIR filter is plotted in Fig. 15.7 using a dB scale. We observe that the pass-band frequency com-

ponents below Ω= 1.5 kHz are passed without any attenuation. The minimum

attenuation in the stop band is also observed to be less than 50 dB.

15.1.4 Kaiser window

As shown in Table 15.1, the minimum stop-band attenuation δ in the FIR

filter obtained using either the rectangular, Bartlett, Hamming, Hanning, or

Blackman window is fixed. In most cases, the selected window surpasses the

required specifications for the attenuation δ. Consider, for example, the design

of an FIR filter with the minimum attenuation specified at 60 dB. Table 15.2

determines that only the Blackman window can be used, and it exceeds the

minimum attenuation requirement by 10 dB. There is no alternative choice

available, and the selection of the Blackman window is an overkill achieved

at the cost of a wider transition band. Several advanced windows, such as

Lanczos, Tukey, Dolph–Chebyshev and Kaiser windows have been proposed,

which provide control over the stop-band ripple δ by means of an additional

parameter characterizing the window. In this section, we introduce the Kaiser

window and outline the steps for designing FIR filters with the Kaiser window.

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25 500 k

1 w[k]

b = 0

b = 1

b = 4.86

b = 10

b = 20

b = 3

Fig. 15.8. Kaiser windows of

length N = 51 for different values of the shape control

parameter β .

The Kaiser window is based on the zeroth-order Bessel function of the first

kind and is defined as follows:

wkaiser[k] =



 

 

I0 [

(√

1 − [(k − m) /m]2 )]

I0[β] −

N − 1 2

≤ k ≤ N − 1

2 0 otherwise,

(15.18)

where m = (N − 1)/2, N is the length of the filter, and I0[·] represents the zeroth-order Bessel function of the first kind, which can be approximated by

I0[β] ≈ 1 + ∞∑

r=1

[ (β/2)

r !

. (15.19)

The parameter β is referred to as the shape control parameter. By varying β with respect to the window’s length N , the shape of the Kaiser window can be adjusted to trade the amplitude of the side lobe for the width of the main lobe

of the DTFT of the Kaiser window. Figure 15.8 illustrates the variations in the

shape of the Kaiser window as β varies from 0 to 20. The length N of the window is kept constant at 51. From Fig. 15.8, we observe that the Kaiser window can

be used to approximate any of the rectangular, Bartlett, Hamming, Hanning, or

Blackman windows by appropriately selecting the value of β. When β = 0, for

example, the shape of the Kaiser window is identical to the rectangular window.

Similarly, when β = 4.86, the shape of the Kaiser window is almost identical

to the Hamming window. Since the shape of the window also determines the

maximum ripples within the pass and stop bands, parameter β is also referred

to as the ripple control parameter. We now explain the last two columns included in Table 15.1 As explained

earlier, the Kaiser window can be used to approximate the five basic windows

covered in Section 15.1.2. The second to last column in Table 15.1 specifies the

value of the shape control parameter β for which the Kaiser window approaches

the basic windows. Setting β = 4.86, for example, will cause the shape of

the Kaiser window to be similar to that of the Hamming window. The last

column lists the width of the transition band of the FIR filter obtained by using

the Kaiser window. For β = 4.86, the Kaiser window would approach the

Hamming window. The transition band of the resulting FIR filter obtained by

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680 Part III Discrete-time signals and systems

truncating the ideal lowpass filter to length N with the Kaiser window is given by 6.27π/(N − 1). We can explain the remaining entries in the last two columns of Table 15.1 in a similar fashion.

15.1.5 Lowpass filter design steps using the Kaiser window

The steps involved in designing a lowpass FIR filter using the Kaiser window

are similar to those in the filter design outlined in Section 15.1.3, except for

steps 4, 6, and 7. Below, we only include a brief description of steps 1–3, which

are common to the two algorithms. The steps that are different are explained in

more detail.

Step 1 Calculate the normalized cut-off frequency Ωn of the filter. See step 1 of Section 15.1.3 for details.

Step 2 Determine the expression for the impulse response of an ideal lowpass filter:

hilp[k] = Ωn sinc(Ωn(k − m)),

where m = (N − 1)/2 and N is the length of the FIR filter, which is calculated in step 6.

Step 3 Calculate the minimum attenuation A on a dB scale using A = min(δp, δs).

Step 4 Based on the value of A obtained in step 3, calculate the shape parameter β from the following:

β =







0 A ≤ 21 dB 0.5842(A − 21)0.4 + 0.0789(A − 21) 21 dB < A < 50 dB 0.1102(A − 8.7) A ≥ 50 dB.

(15.20)

The above expression was derived empirically by J. F. Kaiser, who came up

with the specifications of the Kaiser window.

Step 5 Calculate the normalized transitional bandwidth �Ωn for the FIR filter. See step 5 of Section 15.1.3 for details.

Step 6 The length N of the Kaiser window is calculated from the following expression:

N ≥ A − 7.95

2.285π × �Ωn . (15.21)

Equation (15.21) was also derived by Kaiser from empirical observations. Select

an appropriate value of N and then calculate m = (N − 1)/2.

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681 15 FIR filter design

Step 7 Determine the Kaiser window by substituting the values of β (obtained in step 4) and m (obtained in step 6) into Eq. (15.18). Let the determined Kaiser window be denoted by wkaiser[k] .

Step 8 The impulse response of the FIR filter is given by:

hlp[k] = hilp[k]wkaiser[k]. (15.22)

If the pass-band gain |Hlp(0)| at Ω = 0, given by ∑

hlp[k], is not equal to one, we normalize hlp[k] with

∑

hlp[k].

Step 9 Confirm that the impulse response hlp[k] satisfies the initial specifica- tions by plotting the magnitude spectrum |Hlp(Ω)| of the FIR filter obtained in step 8.

Example 15.3 uses the above algorithm to design an FIR filter using the Kaiser

window.

Example 15.3

Using the Kaiser window, design the FIR filter specified in Example 15.2.

Solution

Following steps 1–3 of Example 15.2, we determine the following values for

the normalized cut-off frequency, impulse response of the ideal lowpass filter,

and minimum attenuation A:

Ωn = 0.4375; hlp[k] = 0.4375 sinc(0.4375(k − m)); A = 50 dB.

The value of m in the impulse response is set to (N − 1)/2. Step 4 of Section 15.1.5 determines the value of β:

β = 0.1102(A − 8.7) = 4.5513.

Step 5 computes the normalized transition bandwidth:

�Ωn = �ωc/(0.5ω0) = (4π − 3π ) × 103/ (

0.5 × 2π × 8 × 103 )

= 0.1250.

Using Step 6, the length of the Kaiser window is given by

N ≥ A − 7.95

2.285π × �Ωn =

50 − 7.95

2.285π × 0.125 = 46.8619,

which is rounded off to the closest higher odd number as 47.

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682 Part III Discrete-time signals and systems

0 5 10 15 20 25 30 35 40 45 0

0.2

0.4

0.6

0.8

1 w[k]

k 0 5 10 15 20 25 30 35 40 45

−0.1

0.1

0.2

0.3

0.4

h[k]

(a) (b)

Fig. 15.9. (a) Kaiser window

w[k ] of length N = 46 and β = 4.5513. (b) Impulse response h[k ] of the FIR filter

obtained by multiplying the

ideal lowpass filter impulse

response by the Kaiser window

in Example 15.3.

Substituting β = 4.5513 and N = 47 in Eq. (15.22), the expression for the Kaiser window is given by

wkaiser[k] =



 

 

I0 [

4.5513 (√

1 − [(k − 23) /23]2 )]

I0[4.5513] 0 ≤ k ≤ 46

0 otherwise.

The impulse response of the FIR filter is then given by h[k] = hilp[k] wkaiser[k]. Figure 15.9(a) plots the time-domain representation of the Kaiser window of

length N = 47 and shape control parameter β = 4.5513, and Fig. 15.9(b) plots the impulse response of the FIR filter.

The frequency characteristics of the FIR filter are shown in Fig. 15.10. Since ∑

h[k] = 0.9992 ≈ 1, the impulse response h[k] of the FIR filter is not normal- ized by

∑

h[k]. It is observed that the FIR filter meets the design specification. The stop-band attenuation is lower than 50 dB.

By comparing the results of Example 15.3 with those of Example 15.2, we

observe that the FIR filter obtained from the Kaiser window in Example 15.3 has

a smaller length N = 47 than the FIR filter obtained from the Hamming window

|H (W)|20 log10

0 p0.75p0.5p0.25p

−20

−40

−60

Fig. 15.10. Magnitude response

of the lowpass FIR filter

designed in Example 15.3.

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683 15 FIR filter design

in Example 15.2, which has a length of N = 53. Therefore, the Kaiser window provides an FIR filter with a lower implementational cost. This reduction in

cost is attributed to the flexibility in the Kaiser window of closely meeting the

stop-band attenuation of 50 dB. The stop-band attenuation in the Hamming

window is fixed to 60 dB and cannot be varied.

Example 15.4

Design a lowpass FIR filter based on the following specifications:

(i) cut-off frequency Ωc = 0.3636π radians/s; (ii) transition width �Ωc = 0.0727π radians/s;

(iii) pass-band ripple 20 log10(1 + δp) ≤ 0.07 dB; (iv) stop-band attenuation −20 log10(δs) ≥ 40 dB.

Solution

The specifications for the digital filter are specified in the DT frequency Ω

domain.

Step 1 suggests that the normalized cut-off frequency is given by

Ωn = (0.3636π )/π = 0.3636.

Step 2 determines the impulse response of the ideal lowpass filter with the

normalized cut-off frequency Ωn = 0.3636:

hilp[k] = 0.3636 sinc(0.3636(k − m)),

with m set to (N − 1)/2. Step 3 determines the value of the minimum attenuation A. The pass-band

ripple 20 log10(1 + δp) is limited to 0.07 dB. Expressed on a linear scale, we

obtain δp ≤ 0.0081. Similarly, the stop-band ripple 20 log10(δs) is limited to

−40 dB, which implies δs ≤ 0.01. The value of the minimum attenuation is

therefore given by A = min(0.0081, 0.01) = 0.0081. Expressed in decibels, the value of the minimum attenuation is −20 log10(A) = 41.83 dB.

Step 4 determines the value of the shape control parameter β from

Eq. (15.20):

β = 0.5842(A − 21)0.4 + 0.0789(A − 21) = 3.6115.

Step 5 computes the normalized transition bandwidth:

�Ωn = �Ωc/π = 0.0727.

Step 6 determines the length of the Kaiser window:

N ≥ A − 7.95

2.285π × �Ωn =

41.83 − 7.95

2.285π × 0.0727 = 64.92,

which is rounded off to the closest higher odd number as 65.

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684 Part III Discrete-time signals and systems

|H (W)|20 log10

W 0 p0.5p 0.75p0.25p

−20

−40

−60

Fig. 15.11. Magnitude response

of the lowpass FIR filter

designed in Example 15.4.

Substituting β = 3.6115 and N = 65 in Eq. (15.22), the expression for the Kaiser window is given by

wkaiser[k] =



  

  

I0 [

3.6115 (√

1 − [(k − 32) /32]2 )]

I0 [3.6115] 0 ≤ k ≤ 64

0 otherwise.

The impulse response of the FIR filter is then given by h[k] = hilp[k]wkaiser[k]. The magnitude response of the FIR filter is plotted in Fig. 15.11 using a dB

scale.

15.2 Design of highpass filters using windowing

The windowing method is not restricted to design of lowpass FIR filters; it

can be generalized to design other types of FIR filters. Section 15.2 considers

highpass FIR filters, and Sections 15.3 and 15.4 extend the windowing method

to bandpass and bandstop FIR filters, respectively.

The transfer function of an ideal highpass filter was defined in Section 14.1.2,

and is reproduced here for convenience:

Hihp(Ω) =

{

0 |Ω| < Ωc

1 Ωc ≤ |Ω| ≤ π. (15.23)

It was shown in Section 14.1.2 that the impulse response of a highpass filter

can be related to the impulse response of a lowpass filter with the same cut-off

frequency, it is given by Eqs. (14.3a) and (14.3b). As shown in Table 14.1, the

impulse response of the ideal highpass filter with a normalized cut-off frequency

Ωn is given by

hihp[k] = δ[k] − Ωn sinc[Ωnk]. (15.24)

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685 15 FIR filter design

|Hhp(W)|

Ws Wp

1−dp

1+dp

stop band pass bandtransition

band

W 0 p

Fig. 15.12. Desired

specifications of a highpass filter.

The filter with this impulse response is non-causal and hence non-realizable. By

applying a delay m, the impulse response of an ideal highpass filter is obtained:

hihp[k] = δ[k − m] − Ωn sinc[Ωn(k − m)]. (15.25)

Given the impulse response of an ideal highpass filter, we can use the windowing

method to design a highpass FIR filter. The specifications for the highpass FIR

filter are illustrated in Fig. 15.12 and are given by

stop band (0 ≤ Ω ≤ Ωs) 0 ≤ |Hhp(Ω)| ≤ δs;

pass band (Ωp < Ω ≤ π ) (1 − δp) ≤ |Hhp(Ω)| ≤ (1 + δp).

The steps involved in the design of a highpass FIR filter are given in the fol-

lowing.

Step 1 Calculate the normalized cut-off frequency Ωn of the filter:

cut-off frequency Ωc = 0.5 (

Ωp + Ωs )

;

normalized cut-off frequency Ωn = Ωc/π.

Step 2 Determine the expression for the impulse response of an ideal highpass filter:

hihp[k] = δ[k − m] − Ωn sin[Ωn(k − m)], (15.26)

where Ωc = πΩn and m = (N − 1)/2, where N is the length of the FIR filter.

Step 3 Calculate the minimum attenuation A on a dB scale using A = min(δp, δs).

Step 4 Calculate the normalized transition band �Ωn for the FIR filter:

transition BW �Ωc = (Ωp − Ωs);

normalized transition BW �Ωn = �Ωc/π.

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686 Part III Discrete-time signals and systems

Step 5 Design an appropriate window with parameters A and �Ωn using the procedures mentioned in Section 15.1.3 or Section 15.1.5. Denote this window

by w[k].

Step 6 Derive the impulse response of the FIR filter:

hhp[k] = hihp[k]w[k]. (15.27)

We now derive the condition for the gain |H (π )| to be equal to one. Substituting Ω = π in the DTFT H (Ω), we obtain

H (π ) = N−1∑

k=0 h[k]e−jkΩ

∣ ∣ ∣ ∣ ∣ Ω=π

= N−1∑

k=0,2,... h[k] −

N−1∑

k=1,3,... h[k]. (15.28)

In other words, the difference between the sum of the even-numbered samples

of h[k] and the sum of the odd-numbered samples of h[k] should equal one. If not, we calculate the normalized impulse response h′hp[k] = hhp[k]/H (π ).

Step 7 Confirm that the impulse response h[k] satisfies the initial specifications by plotting the magnitude spectrum |Hhp(Ω)| of the FIR filter obtained in step 6.

We observe that the above algorithm is similar to the design of a lowpass filter,

except that the impulse response of the ideal lowpass filter is replaced by the

impulse response of the ideal highpass filter. Example 15.5 uses the above

algorithm to design a highpass FIR filter using the Kaiser window.

Example 15.5

Design a highpass FIR filter, using the Kaiser window, with the following

specifications:

(i) pass-band edge frequency Ωp = 0.5π radians/s;

(ii) stop-band edge frequency Ωs = 0.125π radians/s;

(iii) pass-band ripple 20 log10(1 + δp) ≤ 0.01 dB;

(iv) stop-band attenuation −20 log10(δs) ≥ 60 dB.

Plot the frequency characteristics of the designed filter.

Solution

The cut-off frequency Ωc of the filter is given by Ωc = 0.5(0.125π + 0.5π ) =

0.3125π . The normalized cut-off frequency Ωn of the filter is Ωc/π = 0.3125.

The impulse response of the ideal high pass filter with a cut-off frequency of

0.3125 is given by

hihp[k] = δ[k − m] − 0.3125 sinc(0.3125(k − m)). (15.29)

To determine the minimum attenuation A, we calculate δp and δs. Since 20 log10(1 + δp) <= 0.01 dB, the pass-band ripple δp should be less than

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687 15 FIR filter design

|H (W)|20 log10

0 p0.25p 0.5p 0.75p

−20

−40

−60

Fig. 15.13. Magnitude response

of the highpass FIR filter

designed in Example 15.5.

100.01/20 − 1 = 0.0012, while δs should be less than 10−60/20 − 1 = 0.001. The minimum attenuation A is therefore given by A = min(0.0012, 0.001) = 0.001, or 60 dB.

The shape parameter is evaluated from Eq. (15.20) as follows:

β = 0.1102(A − 8.7) = 5.6533.

The transition band �Ωc for the FIR filter is Ωp − Ωs = 0.375π . The nor- malized transition band �Ωn is therefore given by �Ωc/π = 0.375. Using �Ωn = 0.375, the length N of the Kaiser window is given by

N ≥ 60 − 7.95

2.285π × 0.375 = 19.3354.

Rounding off to the higher closest odd number, we obtain N = 21. The expression for the Kaiser window is given by

wkaiser[k] =



 

 

I0 [

5.6533 (√

1 − [(k − 10) /10]2 )]

I0[5.6533] 0 ≤ k ≤ 20

0 otherwise.

(15.30)

The impulse response of the highpass FIR filter is given by

hhp[k] = hihp[k]wkaiser[k],

where hihp[k] is specified in Eq. (15.29) with m = 10 and wkaiser[k] is given in Eq. (15.30). The filter gain at Ω = π is given by

Hhp(π ) = N−1∑

k=0,2,...

hhp[k] − N−1∑

k=1,3,...

hhp[k] = 1.0002.

As H (π ) ≈ 1, the coefficients of h[k] need not be normalized. The magnitude response of the highpass FIR filter is plotted in Fig. 15.13,

which verifies that the initial specifications of the filter are satisfied.

In Example 15.5 we designed a highpass FIR filter directly from the given spec-

ifications. An alternative procedure to design a highpass FIR filter is to exploit

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688 Part III Discrete-time signals and systems

1+dp

1−dp

stop

band I

stop

band II

pass band

0 p

|Hbp(W)|

Ws1 Wp1 Wp2 Ws2

ds2 ds1

Fig. 15.14. Desired

specifications of a bandpass

filter.

Eq. (14.2b) and implement Hlp(Ω) instead. Based on the frequency character- istics of the highpass FIR filter illustrated in Fig. 15.12, the specifications of

the Hlp(Ω) in Eq. (14.2b) are given by

pass band (0 ≤ Ω ≤ Ωs) (1 − δs) ≤ |Hlp(Ω)| ≤ (1 + δs);

stop band (Ωp < Ω ≤ π ) 0 ≤ |Hlp(Ω)| ≤ δp.

The impulse response of the lowpass FIR filter ĥlp [k] is then transformed to the impulse response ĥhp [k] of the highpass FIR filter using the following equation:

ĥhp[k] = δ[k − m] − ĥlp[k].

15.3 Design of bandpass filters using windowing

The design specifications for bandpass filters are specified in Fig. 15.14 and are

given by

stop band I (0 ≤ Ω ≤ Ωs1) 0 ≤ |H (Ω)| ≤ δs1;

stop band II (Ωs2 ≤ Ω ≤ π ) 0 ≤ |H (Ω)| ≤ δs2;

pass band (Ωp1 < Ω ≤ Ωp2) (1 − δp) ≤ |H (Ω)| ≤ (1 + δp),

where we assume that the values of ripples δs1 and δs2 allowed in the two stop

bands are different. The algorithm used to design a bandpass FIR filter using

windowing is similar to the design for the highpass filter described in Section

15.2, except that the impulse response of an ideal bandpass filter is used in

step 2.

The transfer function of an ideal bandpass filter was defined in Section 14.1.3,

and is reproduced here for convenience:

Hibp(Ω) =

{

1 Ωc1 ≤ |Ω| ≤ Ωc2

0 |Ω| < Ωc1 and Ωc2 ≤ |Ω| ≤ π. (15.31)

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As shown in Table 14.1, the impulse response of the ideal bandpass filter with

normalized cut-off frequencies of Ωn1,Ωn2 (Ωn2 > Ωn1) is given by

hibp[k] = Ωn2 sinc[Ωn2k] − Ωn1 sinc[Ωn1k]. (15.32)

As the filter with this impulse response is non-causal, we apply a delay of m, and the modified impulse response is obtained:

hibp[k] = Ωn2 sinc[Ωn2(k − m)] − Ωn1 sinc[Ωn1(k − m)]. (15.33)

The steps for designing a bandpass filter using windowing are as follows.

Step 1 Calculate the two normalized cut-off frequencies Ωn1 and Ωn2 of the bandpass filter:

cut-off frequenciesΩc1 = 0.5(Ωp1 + Ωs1) and Ωc2 = 0.5(Ωp2 + Ωs2); normalized cut-off frequencies Ωn1 = Ωc1/π and Ωn2 = Ωc2/π.

Step 2 Determine the impulse response of the ideal bandpass filter by substi- tuting the values of Ωn1 and Ωn2 in Eq. (15.33).

Step 3 Calculate the minimum attenuation A on a dB scale using A = min(δp, δs1, δs2).

Step 4 Calculate the normalized transition bandwidth �Ωn for the FIR filter:

transitional BW �Ωc1 = (Ωp1 − Ωs1) and �Ωc2 = (Ωs2 − Ωp2); normalized transition BW �Ωn = min (�Ωc2, �Ωc1) /π.

Step 5 Design an appropriate window with parameters A and �Ωn using the procedures mentioned in Section 15.1.3 or Section 15.1.5. Denote this window

by w[k].

Step 6 Derive the impulse response of the FIR filter:

hbp[k] = hibp[k]w[k]. (15.34)

Step 7 Confirm that the impulse response hbp[k] satisfies the initial specifica- tions by plotting the magnitude spectrum |Hbp(Ω)| of the FIR filter obtained in step 6.

Example 15.6 illustrates the working of the above algorithm by designing a

bandpass FIR filter using the Kaiser window.

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690 Part III Discrete-time signals and systems

Example 15.6

Design a bandpass FIR filter with the following specifications:

(i) pass-band edge frequencies, Ωp1 = 0.375π and Ωp2 = 0.5π radians/s; (ii) stop-band edge frequencies, Ωs1 = 0.25π and Ωs2 = 0.625π radians/s;

(iii) stop-band attenuations, δs1 > 50 dB and δs2 > 50 dB.

Plot the gain–frequency characteristics of the designed bandpass filter.

Solution

The cut-off frequencies of the bandpass filter are given by

Ωc1 = 0.5 (0.25π + 0.375π ) = 0.3125π and

Ωc2 = 0.5 (0.5π + 0.625π ) = 0.5625π.

The normalized cut-off frequencies are given by Ωn1 = Ωc1/π = 0.3125 and Ωn2 = Ωc2/π = 0.5625. The impulse response of an ideal bandpass filter is given by

hibp[k] = 0.5625 sinc[0.5625(k − m)] − 0.3125 sinc[0.3125(k − m)]. (15.35)

Since only the stop-band attenuations are specified, and these are both equal to

50 dB, the minimum attenuation A = 50 dB. The shape parameter β of the Kaiser window is computed to be

β = 0.1102(50 − 8.7) = 4.5513.

The transition bands �Ωc1 and �Ωc2 for the bandpass FIR filter are given by

�Ωc1 = 0.375π − 0.25π = 0.125π and

�Ωc2 = 0.625π − 0.5π = 0.125π,

which lead to the normalized transition BW of �Ωn = 0.125. The length N of the Kaiser window is given by

N ≥ 50 − 7.95

2.285π × 0.125 = 46.8619.

Rounded to the closest higher odd number, N = 47, and the value of m in Eq. (15.35) is 23. The expression for the Kaiser window is as follows:

wkaiser[k] =



 

 

I0 [

4.5513 (√

1 − [(k − 23)/23]2 )]

I0[4.5513] 0 ≤ k ≤ 46

0 otherwise.

(15.36)

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691 15 FIR filter design

20 log10 |H(W)|

W 0 p0.25p 0.5p 0.75p

−20

−40

−60

Fig. 15.15. Magnitude response

of the bandpass FIR filter

designed in Example 15.6.

The impulse response of the bandpass FIR filter is given by

hbp[k] = hibp[k]wkaiser[k].

where hibp[k] is specified in Eq. (15.35) with m = 23 and wkaiser[k] is specified in Eq. (15.36).

The magnitude spectrum of the bandpass FIR filter is plotted in Fig. 15.15.

It is observed that the bandpass filter satisfies the design specifications.

In Example 15.6, we designed a bandpass FIR filter directly. As for the

highpass FIR filter, an alternative procedure to design a bandpass FIR filter

is to exploit Eq. (14.4e) and implement two lowpass FIR filters with impulse

responses Hlp1(k) and Hlp2(k). The specifications for the two lowpass filters should be carefully derived such that the pass- and stop-band ripples of the

combined system are limited to values allowed in the original bandpass filter’s

specifications.

15.4 Design of a bandstop filter using windowing

As illustrated in Fig. 15.16, the design specifications for a bandstop filter are

given by

pass band I (0 ≤ Ω ≤ Ωp1) (1 − δp1) ≤ |Hbs(Ω)| ≤ (1 + δp1);

pass band II (Ωp2 ≤ Ω ≤ π ) (1 − δp2) ≤ |Hbs(Ω)| ≤ (1 + δp2);

stop band (Ωs1 < Ω ≤ Ωs2) 0 ≤ |Hbs(Ω)| ≤ δs.

The steps involved in the design of a bandpass FIR filter using windowing are

similar to those specified for the bandpass filter in Section 15.3.

The transfer function of an ideal bandstop filter was defined in Section 14.1.4,

and is reproduced here for convenience:

Hibs(Ω) =

{

0 Ωc1 ≤ |Ω| ≤ Ωc2

1 |Ω| < Ωc1 and Ωc2 < |Ω| ≤ π, (15.37)

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pass

band I

|Hbs(W)|

1+dp1

1−dp1

pass

band II

stop

band

W 0 p

Wp1 Ws1 Ws2 Wp2

1+dp2

1−dp2

Fig. 15.16. Desired

specifications of a bandstop

filter.

As shown in Table 14.1, the impulse response of the ideal bandstop filter with

normalized cut-off frequencies of Ωn1,Ωn2 (Ωn2 > Ωn) is given by

hibs[k] = δ[k] − Ωn2 sinc[Ωn2k] + Ωn1 sinc[Ωn1k]. (15.38)

By applying a delay m, the modified impulse response of an ideal bandpass filter is obtained:

hibs[k] = δ[k − m] − Ωn2 sinc[Ωn2(k − m)] + Ωn1 sinc[Ωn1(k − m)]. (15.39)

In the following example, we illustrate the steps involved in designing a practical

bandstop filter using the windowing method.

Example 15.7

Design a bandstop FIR filter, using a Kaiser window, with the following speci-

fications:

(i) pass-band edge frequencies, Ωp1 = 0.25π and Ωp2 = 0.625π radians/s; (ii) stop-band edge frequencies, Ωs1 = 0.375π and Ωs2 = 0.5π radians/s;

(iii) stop-band attenuations, δs1 > 50 dB and δs2 > 50 dB.

Solution

The cut-off frequencies of the bandpass filter are given by

Ωc1 = 0.5 (0.25π + 0.375π ) = 0.3125π and

Ωc2 = 0.5 (0.5π + 0.625π ) = 0.5625π.

The normalized cut-off frequencies are given by Ωn1 = 0.3125 and Ωn2 = 0.5625. The impulse response of an ideal bandpass filter is given by

hibs[k] = δ[k − m] − 0.5625 sinc[0.5625(k − m)] + 0.3125 sinc[0.3125(k − m)]. (15.40)

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693 15 FIR filter design

The minimum attenuation A = 50 dB. Therefore, The shape parameter β of the Kaiser window is computed as

β = 0.1102(50 − 8.7) = 4.5513.

The transition bands �Ωc1 and �Ωc2 for the bandpass FIR filter are given by

�Ωc1 = (0.375π − 0.25π ) = 0.125π and �Ωc2 = (0.625π − 0.5π ) = 0.125π,

which leads to the normalized transition BW of �Ωn = 0.125. The length N of the Kaiser window is given by

N ≥ 50 − 7.95

2.285π × 0.125 = 46.8619.

Rounded to the closest higher odd number, N = 47, and the value of m in Eq. (15.40) is 23.

The expression for the Kaiser window is as follows:

wkaiser[k] =



 

 

I0 [

4.5513 (√

1 − [(k − 23) /23]2 )]

I0[4.5513] 0 ≤ k ≤ 46

0 otherwise.

(15.41)

The impulse response of the bandpass FIR filter is given by

hbs[k] = hibs[k]wkaiser[k],

where hibs[k] is specified in Eq. (15.40) with m = 23 and wkaiser[k] is as shown in Eq. (15.41).

The magnitude response of the bandstop FIR filter is plotted in Fig. 15.17. It

is observed that the bandstop filter satisfies the design specifications.

In the above example, a bandstop FIR filter was designed directly. As for the

highpass and bandpass FIR filters, an alternative design procedure (see Eq. 14.6)

is to express the transfer function of a bandstop FIR filter in terms of the transfer

functions of two lowpass filters as follows:

hibs[k] = δ[k − m] − hilp1[k]|Ωc=Ωc2 + hilp2[k]|Ωc=Ωc1 . (15.42)

The specifications for these two lowpass filters are derived from the given

design specifications of the bandpass filter. As for bandpass FIR filters, the

specifications of the lowpass filters should be carefully assigned such that the

pass- and stop-band ripples of the combined system satisfy the original bandstop

filter’s specifications.

15.5 Optimal FIR filters

Designing an FIR filter using the windowing approach is simple but suffers

from one severe limitation. The minimum attenuation obtained in the stop

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694 Part III Discrete-time signals and systems

20 log10 |H(W)|

W 0 p0.75p0.5p0.25p

−20

−40

−60

Fig. 15.17. Magnitude response

of the bandstop FIR filter

designed in Example 15.7.

band of the FIR filter is fixed for the elementary window types covered in

Section 15.1.2. The Kaiser window, introduced in Section 15.1.4, provides some

flexibility in controlling the stop-band attenuation by introducing an additional

design parameter β. However, there is no guarantee that the FIR filter, designed

with either the elementary windows or the Kaiser window, is optimal. In this

section, we introduce a computational optimization procedure for the design

of FIR filters. The procedure is commonly referred to as the Parks–McClellan

algorithm, which iteratively minimizes the absolute value of the error:

ε(Ω) = W (Ω) [Hd(Ω) − H (Ω)], (15.42)

where Hd(Ω) is the transfer function of the desired or ideal filter, whose fre- quency characteristics are to be approximated, H (Ω) is the transfer function of the approximated FIR filter, and W (Ω) is a weighting function introduced to emphasize the relative importance of various frequency components of the

filter. For a lowpass filter, for example, a logical way to select the values of the

weighting function is to set

lowpass filter W (Ω) =







1/δp 0 ≤ Ω ≤ Ωp

0 Ωp < Ω < Ωs

1/δs Ωs ≤ Ω ≤ π.

(15.43)

Equation (15.43) implies that if the condition for the pass-band ripple is more

stringent (i.e. smaller) than the condition for the stop-band ripple, the weighting

function allocates a higher weight to the pass band than to the stop band, and

vice versa. Zero weight is associated with the transition band, which means

that the weighting function does not care about the characteristics of the FIR

filter in the transition band as long as the filter’s gain changes steadily between

the pass and stop bands. Scaling Eq. (15.43) with δs, the normalized weighting

function is given by

lowpass filter W (Ω) =







δs/δp 0 ≤ Ω ≤ Ωp

0 Ωp < Ω < Ωs

1 Ωs ≤ Ω ≤ π.

(15.44)

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The weighting function for highpass, bandpass, and bandstop filters can be

derived in a similar fashion. Given a weighting function, the Parks–McClellan

algorithm seeks to solve the following optimization problem:

min

{h[k]}

[

max

Ω ∈ S |ε(Ω)|

]

, (15.45)

where S is defined as a set of discrete frequencies chosen within the pass and stop bands. For a lowpass filter, the set of frequencies that can be included in S should lie in the following range:

lowpass filter S = [

0 ≤ Ω ≤ Ωp ]

∪ [Ωs ≤ Ω ≤ π ] (15.46)

Similarly, the sets S of discrete frequencies are carefully selected over the pass and stop bands for other types of filters.

Because Eq. (15.45) minimizes a cost function, which is the maximum of the

error ε(Ω), Eq. (15.45) is also referred to as the minimax optimization problem.

The goal in solving Eq. (15.45) is to determine the set of coefficients for the

impulse response h[k] of the optimal FIR filter of length N . It was shown in Proposition 14.1 (see Section 14.3.1) that if the filter coe-

fficients of an FIR filter are symmetric or anti-symmetric, the phase response

of the filter is a linear function of frequency, and the transfer function can be

expressed as follows:

H (Ω) = G(Ω)ej(β−αΩ), (15.47)

where α = (N − 1)/2, β is a constant, and G(Ω) is a real-valued function. Table 14.2 shows the values of β and G(Ω) for four types of linear-phase FIR filters.

The Parks–McClellan algorithm exploits Proposition 14.1 to solve the mini-

max optimization problem, as explained in the following. For various types of

linear-phase FIR filter, G(Ω) is a summation of a finite number of sinusoidal terms of the form cos(Ωk) or sin(Ωk), which themselves can be expressed as polynomials of cos(Ω). For example, cos(Ωk) can be expressed as a kth-order polynomial of cos(Ω), which, for k = 2 and 3, can be expressed as follows:

cos (2Ω) = 2 cos2(Ω− 1);

cos (3Ω) = 4 cos3(Ω) − cos(Ω).

It is observed from Table 14.2 that the function G(Ω) can be expressed as a sum of several higher-order terms cos(Ωk) or sin(Ωk). Therefore, G(Ω) can also be expressed as a polynomial of cos(Ω). It can be shown that the error function ε(Ω)

in Eq. (15.42), corresponding to linear-phase FIR filters, can also be expressed

as a polynomial of cos(Ω). Parks and McClellan applied the alternation theorem

from the theory of polynomial approximation to solve the minimax optimization

problem. For convenience, we first express the alternation theorem in the context

of polynomial approximation, and later we show its adaptation to the minimax

optimization problem.

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15.5.1 Alternation theorem

Let S be a compact subset on the real axis x and let D(x) be a desired function of x which is continuous on S. Let D(x) be approximated by P(x), an Lth-order polynomial of x , which is given by

P(x) = L∑

m=0 cm x

m . (15.48a)

Define the approximation error ε(x) and the amplitude of the maximum error value εmax on S as follows:

ε(x) = W (x)[D(x) − P(x)] : (15.48b) εmax = arg max

x∈S |ε(x)|. (15.48c)

A necessary and sufficient condition for P(x) to be the unique Lth-order poly- nomial minimizing εmax is that ε(x) exhibits at least L + 2 alternations. In other words, there must exist L + 2 values of x , {x1 < x2 < · · · xL+2} ∈ S such that ε(xm) = −ε(xm+1) = ±εmax.

Note that the minimax optimization problem for optimal filter design fits

very well in the framework of the alternation theorem. In the filter design

problem, S is the subset of DT frequencies, D(x) is the desired filter response, P(x) is the approximated filter response, and εmax is the maximum deviation between the desired and approximated filter response. Therefore, the FIR filters

obtained using minimax optimization is also expected to exhibit alternations in

its frequency response. However, note that G(Ω) is a polynomial of cos(Ω) and not ofΩ. This issue can be addressed by using the mapping function x = cos(Ω). In this case, the frequency spaceΩ = [0, π ] can be mapped to x = [−1, 1], and the optimization problem can be reformulated around x to calculate the optimal filter coefficients. It can be shown that the alternation in the frequency response

of the optimal filters is still applicable.

Based on the above discussion, the alternation theorem can be restated for

the minimax optimization problem as follows. Consider the following minimax

optimization problem:

{h[k], 0 ≤ min

k ≤ (N − 1)}

[

max Ω∈S

|ε(Ω)|

]

; (15.49a)

ε(Ω) = W (Ω)



 Hd (Ω) − G(Ω)e

−jΩ(N−1/2)ejφ ︸︷︷︸

H (Ω)



  . (15.49b)

where S is a set of discrete extremal frequencies chosen within the pass and stop bands, W (Ω) is a positive weighting function, Hd(Ω) is the transfer function of the ideal filter with a unity gain within the pass band and a zero gain within the

stop band, and G(Ω) is a polynomial of cos(Ω) with degree L , which is uniquely

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697 15 FIR filter design

specified by the impulse response h[k]. Let εmax max denote the maximum value of the error |ε(Ω)|. The polynomial G(Ω), which best approximates Hd(Ω) (i.e. minimizes εmax), produces the error function ε(Ω) that must satisfy the

following property. There should be at least L + 2 discrete frequencies {Ω1 < Ω2 < · · · < ΩL+2} ∈ S at which the maximum and minimum peak values of the error alternate, i.e. ε(Ωm+1) = −ε(Ωm) = εmax.

Before presenting some examples of the application of the alternation theo-

rem, we briefly comment on the degree L of the error function ε(Ω) in the FIR filters. The value of L is determined by evaluating the highest power of cos(Ω) in the G(Ω) function of the filters. For the four types of FIR filters with length N , the value of L is specified as follows:

type I FIR filters L = N − 1

2 ;

type II FIR filters L = N − 2

2 ;

type III FIR filters L = N − 3

2 ;

type IV FIR filters L = N − 2

2 .

The alternation theorem states that the minimum number of alternations for the

optimal FIR filter should be at least L + 2. The actual number of alternations in an optimal FIR may, however, exceed the minimum number specified by the

alternation theorem. An optimally designed lowpass or highpass filter can have

up to L + 3 alternations, while an optimal bandpass or bandstop filter can have up to L + 5 alternations.

Example 15.8

The magnitude spectra of two lowpass FIR filters with lengths N = 13 and 20 are, respectively, shown in Figs. 15.18(a) and (b), where the filter gain

within the pass and stop bands is enclosed within a frame box. Determine if the

two filters satisfy the alternation theorem.

Solution

Figure 15.18(a) shows the frequency response of a type I FIR filter with

length N = 13. The degree L of cos(Ω) in the polynomial ε(Ω) is given by L = (13 − 1)/2 = 6. Based on the alternation theorem, there should be at least L + 2 = 8 alternations in polynomial ε(Ω). Note that the absolute value of error |ε(Ω)| is the difference |H (Ω) −Hd(Ω)|, where Hd(Ω) has a unity gain within the pass band and zero gain within the stop band. Therefore, counting the

number of alternations in ε(Ω) is the same as counting the number of alterna-

tions in H (Ω) with respect to the pass- and stop-band ripples. From Fig. 15.18 we observe that there are indeed eight alternations (shown by × symbols) in

H (Ω). One of these alternations occurs at the pass-band edge frequency Ωp,

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0 0.25p 0.5p 0.75p p 0

0.2

0.4

0.6

0.8

0 0.25p 0.5p 0.75p p 0

0.2

0.4

0.6

0.8

(a) (b)

Fig. 15.18. Magnitude spectrum

of lowpass FIR filters. (a) Type I

FIR filter of length N = 13. (b) Type II FIR filter of length

N = 20.

and two of these alternations occur at the stop-band edge frequencies Ωs and

π . In other words, Fig. 15.18(a) satisfies the alternation theorem.

Figure 15.18(b) shows the frequency response of a type II FIR filter with

length N = 20. The degree L of cos(Ω) in polynomial ε(Ω) is given by L = (20 − 2)/2 = 9. Based on the alternation theorem, there should be at least L + 2 = 11 alternations in polynomial ε(Ω). In Fig. 15.18(b), we observe 12 alternations in H (Ω), which exceed the minimum required number of alterna- tions. Therefore, Fig. 15.18(b) satisfies the alternation theorem.

15.5.2 Parks–McClellan algorithm

In this section we present steps of the Parks–McClellan algorithm for designing

optimal filters. In this discussion, we will consider only type I filters. Algorithms

for other types of filters can be obtained in the same manner. To derive the

Parks–McClellan algorithm, the approximated error function in Eq. (15.49b) is

expressed as follows:

G(Ω) + ε(Ω)

W (Ω) ≈ Hd(Ω). (15.50)

For type I filters, we obtain G(Ω) from Table 14.2 as follows:

G(Ω) = h

[ N − 1

]

+ 2

(N−1)/2∑

k=1

[ N − 1

2 − k

]

cos(Ωk).

Since we are interested in calculating (N − 1)/2 + 1, or L + 1, coefficients of h[k] in G(Ω) and the value of the maximum error εmax, we pick L + 2

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699 15 FIR filter design

discrete frequencies {Ω1 < Ω2 < · · · < ΩL+2} ∈ S, and solve Eq. (15.50) at the selected frequencies. Assuming that the selected frequencies are the extremal

frequencies at which the maximum error changes between its peak value of

±εmax, Eq. (15.50) reduces to

G(Ωm) + 1

W (Ωm) (−1)mεmax = Hd(Ωm). (15.51)

for 1 ≤ m ≤ (L + 2). The resulting set of (L + 2) simultaneous equations is as follows:



     

1 cos (

Ω1 )

· · · cos (

LΩ1 )

−1/W (Ω0) 1 cos

(

Ω2 )

· · · cos (

LΩ2 )

−1/W (Ω2) . . .

. . . .

1 cos (

ΩL+1 )

· · · cos (

LΩL+1 )

(−1)L+1 /W (ΩL+1) 1 cos

(

ΩL+2 )

· · · cos (

LΩL+2 )

(−1)L+2 /W (

ΩL+2 )



     

︸︷︷︸

�(cos(kΩ))



      

h [

N−1 2

]

2h [

N−1 2

− 1 ]

2h[0] εmax



      



     

Hd (

Ω1 )

Hd (

Ω2 )

Hd (

ΩL+1 )

Hd (

ΩL+2 )



     

(15.52)

Once the extremal frequencies {Ω1 < Ω2 < · · · < ΩL+2} are known, Eq.

(15.52) can be used to solve for the coefficients of the FIR filter. The extremal

frequencies are computed using the Remez algorithm, which is based on Eq.

(15.52) (though it does not solve the simultaneous equations explicitly) and

consists of the following steps.

Initialization: pick {Ω1 < Ω2 < · · · < ΩL+2} ∈ S evenly over the pass and stop bands.

Given: transfer function Hd(Ω) of the ideal filter and the weighting function W (Ω).

Step 1 Solve Eq. (15.52) to calculate εmax. To compute εmax, we do not need to solve the complete set of simultaneous equations given in Eq. (15.52). Instead

the following expression, obtained from Eq. (15.52) is solved:

εmax = (−1)L+3

|� (cos (kΩ))|

∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

1 cos (Ω1) · · · cos (LΩ1) Hd (Ω1) 1 cos (Ω2) · · · cos (LΩ2) Hd (Ω2) ...

... . . .

... ...

1 cos (ΩL+1) · · · cos (LΩL+1) Hd (ΩL+1) 1 cos (ΩL+2) · · · cos (LΩL+2) Hd (ΩL+2)

∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

(15.53)

where |�(·)| denotes the determinant of the matrix �(·).

Step 2 Substituting the value of εmax determined in step 1, compute the values of G(Ωm) at discrete frequencies {Ω1 < Ω2 < · · · < ΩL+2} using Eq. (15.51).

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Step 3 Using the values of G(Ωm) computed in step 2, sketch a line plot of G(Ω) as a function ofΩby interpolating intermediate values of G(Ω). Generally, G(Ω) is interpolated over a large grid of discrete frequencies within S.

Step 4 Using the line plot of G(Ω) obtained in step 3, sketch the line plot of ε(Ω) as a function of Ω using the following expression:

ε(Ω) = W (Ω)[Hd(Ω) − G(Ω)],

derived from Eq. (15.50).

Step 5 Update the L + 2 extremal frequencies {Ω1 < Ω2 < · · · < ΩL+2} ∈ S by determining the L + 2 maxima and minima in ε(Ω) plotted in step 4.

Step 6 Check if the L + 2 maxima and minima observed in step 5 have the same value. If they do, then the alternation theorem is satisfied and the updated

frequencies {Ω1 < Ω2 < · · · < ΩL+2} can be used to solve Eq. (15.52) for the

filter coefficients. If not, then go back to step 1 and repeat steps 1–6.

The Parks–McClellan algorithm, highlighted in the aforementioned discus-

sion, designs a lowpass FIR filter. Extension to other types of FIR filters is

straightforward, provided that the required filter can be expressed in terms of

a lowpass filter. Equation (15.24) illustrates how the design of a highpass filter

can be transformed to the design of a lowpass filter. Similarly, Eqs. (15.32) and

(15.38), respectively, provide transformations for bandpass and bandstop filters.

Once the specifications of the required filter are expressed in terms of a low-

pass filter, the impulse response of the optimal lowpass FIR filter is computed

using the Parks–McClellan algorithm. The impulse response of the required FIR

filter is then calculated from the impulse response of the optimal lowpass FIR

filter.

15.6 M A T L A B examples

The design algorithms covered in this chapter are incorporated as library func-

tions in most signal processing software packages. In this section, we introduce

several M-files available in M A T L A B for the design of FIR filters. In par-

ticular, we cover rectwin, bartlett, hann, hamming, and blackman

functions, which are used to implement the elementary windows covered in

Section 15.1. In addition, we consider the fir1 function to derive the impulse

response of the FIR filter. The kaiser function used to design FIR filters

using the Kaiser window and the firpm function used to implement the Parks–

McClellan algorithm are also presented in this section. In each case, we write

the M A T L A B code for the design of the FIR filter specified in Example 15.2.

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701 15 FIR filter design

For convenience, the specifications of the lowpass filter in Example 15.2 are

given by

pass-band edge frequency (ωp) = 3π kradians/s, stop-band edge frequency (ωs) = 4π kradians/s, maximum allowable pass-band ripple − 20 log10(δp) = 25 dB,

i.e. δp = 0.0562, minimum stop-band attenuation −20 log10(δs) = 50 dB,

i.e. δs = 0.0032, sampling frequency ( f0) = 8 ksamples/s.

Example 15.9

Design the lowpass FIR filter considered in Example 15.2 using the rectangular,

Bartlett, Hanning, Hamming, and Blackman windows. Sketch and compare the

magnitude response of the resulting FIR filters.

Solution

As shown in Example 15.2, the values of the normalized cut-off frequency

and the normalized transition bandwidth for the lowpass filter are given by

Ωn = 0.4375 and �Ωn = 0.125, respectively. Since the minimum stop-band attenuation is 50 dB, only the Hamming and

Blackman windows may be used for the filter design. The value of length N of the FIR filters for the two windows is given by

Hamming window 6.6 /

N = 0.1250 ⇒ N = 6.6/0.125 = 52.8; Blackman window 11

N = 0.1250 ⇒ N = 11/0.125 = 88.

M A T L A B provides the fir1 function to derive the impulse response of the

FIR filter. The syntax for the fir1 function is given by

fir coeff. = fir1(order, norm cut off, type,window);

where the input argument order denotes the order of the FIR filter. For

an FIR filter of length N , the order is given by N – 1. The input argument norm cut off specifies the normalized cut-off frequency of the FIR filter.

Its value should lie between zero and one. The input argument type specifies

the type of the FIR filter. Two possible choices for type are ’low’ for the

lowpass FIR filter and ’high’ for the highpass FIR filter. Finally, the input

argument window accepts coefficients w[k] of the window type being used in the FIR filter design. Any of the elementary windows covered in Section

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702 Part III Discrete-time signals and systems

15.1 can be used by naming the appropriate window function. The syntaxes for

various types of length-N window functions are as follows:

>> win coeff = rectwin(N); % rectangular window >> win coeff = bartlett(N); % bartlett window >> win coeff = hann(N); % hanning window >> win coeff = hamming(N); % hamming window >> win coeff = blackman(N); % blackman window

For Example 15.2, the M A T L A B code for the design of the FIR filter using the

Hamming window is given by

% lowpass filter design using Hamming window

>> wn = 0.4375; % Normalized cut-off % frequency

>> N = 53; % Hamming Window >> h hamm = fir1 (N-1,wn, ’low’,hamming(N));

% Impulse response of

% the LPF

>> w = 0:0.001*pi:pi; % discrete frequencies % for response

>> H hamm = freqz(h hamm,1,w); % transfer function >> plot(w,20*log10(abs(H hamm))); % magnitude response

>> axis([0 pi -120 20]); % set axis

>> title(’FIR filter using Hamming window’)

>> grid on

The magnitude response of the FIR filter obtained with the Hamming window

is shown in Fig. 15.19(a). Note that the magnitude response satisfies the filter

specifications.

The M A T L A B code for the design of the FIR filter using the Blackman

window is similar, except for a few minor changes, which are shown below.

% lowpass filter design using Blackman window

>> wn = 0.4375; % Normalized cut-off % frequency

>> N = 88; % Blackman Window size >> h black = fir1(N-1,wn, ’low’,blackman(N));

% Impulse response of

% the LPF

>> w = 0:0.001*pi:pi; % discrete frequencies % for response

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0 0.5 1 1.5 2 2.5 3 −120

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20 FIR filter using Hamming window

0 0.5 1 1.5 2 2.5 3 −120

−100

−80

−60

−40

−20

20 FIR filter using Blackman window

(a) (b)

Fig. 15.19. FIR filter design for

Example 15.9 using MATLAB.

(a) Hamming window

(b) Blackman window.

>> H black = freqz(h black,1,w); % transfer function >> plot(w,20*log10(abs(H black))); % magnitude response

>> axis([0 pi -120 20]); % set axis

>> title(’FIR filter using Blackman window’);

>> grid on

The magnitude response of the FIR filter obtained with the Blackman window is

shown in Fig. 15.19(b). On comparing with Fig. 15.19(a), we note that the stop-

band attenuation in Fig. 15.19(b) is higher. The improvement in the stop-band

attenuation is the result of the shape of the Blackman window.

Although the above example uses only the Hamming and Blackman windows,

any of the elementary windows covered in Section 15.1 can be used by speci-

fying the appropriate window coefficients in the fir1 function.

Example 15.10

Design the lowpass FIR filter considered in Example 15.3 using the Kaiser

window. Sketch and compare the magnitude response of the resulting FIR filter

with those of the FIR filters obtained in Example 15.3.

Solution

As shown in Example 15.3, the normalized cut-off frequencyΩn = 0.4375 and the normalized transition bandwidth �Ωn = 0.1250. The design parameters for the Kaiser window were calculated as β = 4.5513 and N = 47.

The MATLAB code for the design of the FIR filter using the Kaiser window

is similar to the MATLAB code in Example 15.9. The major difference is in the

fir1 instruction, where the window argument is now replaced by the kaiser

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20 FIR filter using Kaiser windowFig. 15.20. FIR filter design for

Example 15.10 with the Kaiser

window using MATLAB.

function. A Kaiser window of length N and shape parameter β can be generated by the following instruction:

Win = kaiser(N, beta)

The M A T L A B code is given by

% lowpass filter design using Kaiser window

>> wn = 0.4375; % Normalized Cutoff % frequency

>> N = 47; % Kaiser Window length >> beta = 4.5513; % Kaiser Shape control

% parameter

>> h kaiser = fir1(N-1,wn, ’low’,kaiser(N,beta)); % Impulse response of

% the LPF

>> w = 0:0.001*pi:pi; % discrete frequencies % for response

>> H kaiser = freqz(h kaiser,1,w); % transfer function >> plot(w,20*log10(abs(H kaiser))); % magnitude response

>> axis([0 pi -120 20]); % set axis

>> title(’FIR filter using Kaiser window’);

>> grid on

The magnitude response of the FIR filter obtained with the Kaiser window is

shown in Fig. 15.20. Compared with Figs. 15.19(a) and (b), we note that the

minimum stop-band attenuation in Fig. 15.20 is exactly 50 dB. Being able to

provide the exact specified attenuation, the Kaiser window is able to reduce

the length of the lowpass FIR filter to 47. Among the three filters, the FIR

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705 15 FIR filter design

filter obtained from the Kaiser window is therefore the least expensive from the

implementation perspective.

For the design of optimal filters, M A T L A B has incorporated the firpm func-

tion, which has the following syntax:

fir coefficients = firpm(order,range norm cut off, f response,wmatrix);

where the input argumentorder denotes the order of the FIR filter. The second

input argument rang norm cut off is a vector that specifies the edges of

the normalized cut-off frequency of the FIR filter. All elements of this vector

should have a value between zero and one. For a lowpass filter, the elements of

the rang norm cut off vector are given by

rang norm cut off = [0, pass band cut off, stop band cut off, 1];

The third input argument f response specifies the four gains of the FIR

filter at the four frequencies specified in the rang norm cut off vector. For

a lowpass filter, the value of the f response vector is given by

f response =[1, 1, 0, 0];

Finally, the fourth input argument wmatrix specifies the weight matrix. Since

wmatrix has one entry per band, it is half the length ofrang norm cut off

and f response vectors.

Example 15.11 illustrates the design of an optimal FIR filter using thefirpm

function.

Example 15.11

Examples 15.9 and 15.10 designed an FIR filter using rectangular. Ham-

ming, and Kaiser windows with a given set of design specifications. It was

shown in Example 15.10 that an FIR filter of length 47, designed using a

Kaiser window, satisfies the design specifications. Design the optimal FIR filter

of length 47 using the Parks–McClellan algorithm and compare the magni-

tude frequency response with that of the FIR filter obtained using the Kaiser

window.

Solution

The values of the normalized pass- and stop-band edge frequencies are given

normalized pass-band edge frequency Ωp = (3π × 103)/(0.5 × 2π × 8 × 103) = 0.375;

normalized stop-band edge frequency Ωs = (4π × 103)/(0.5 × 2π × 8 × 103) = 0.5.

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20 optimal FIR filter

0 0.5 1 1.5 2 2.5 3 −120

−100

−80

−60

−40

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20 FIR filter using Kaiser window

(a) (b)

Fig. 15.21. Optimal FIR filter

designed in Example 15.11 using

M A T L A B. (a) Optimal FIR filter of

length N = 47. (b) FIR filter of length N = 47 using the Kaiser window.

The M A T L A B code for the design of the optimal FIR filter is similar to the

M A T L A B code in Example 15.8, except for the use of the firpm function,

which replaces the fir1 function:

% optimal lowpass filter design using Parks-McClellan

% algorithm

>> sz = 47; % Length of FIR filter >> range norm cut off = [0, 0.375, 0.5, 1];

% normalized cut-off

% frequencies

>> f response = [1, 1, 0, 0]; % gains at the cut-off % frequencies

>> wmatrix = [0.0032/0.0562, 1]; % weight matrix >> h optimal = firpm(sz-1, range norm cut off,f response,

wmatrix); % Impulse response of

% the optimal LPF

% FIR filter

>> w = 0:0.001*pi:pi; % discrete frequencies >> H optimal = freqz(h optimal,1,w);% transfer function >> plot(w,20*log10(abs(H optimal)));% magnitude response

>> axis([0 pi -120 20]); % set axis

>> title(’optimal FIR filter’);

>> grid on

The magnitude response of the optimal FIR filter obtained from the above code

is shown in Fig. 15.21(a). Comparing Fig. 15.21(a) with the magnitude response

of the FIR filter obtained from the Kaiser window shown in Fig. 15.21(b), the

following differences are noted.

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0 0.5 1 1.5 2 2.5 3 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

optimal FIR filter

0 0.5 1 1.5 2 2.5 3 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

(a) (b)

FIR filter using Kaiser window

Fig. 15.22. Same as Fig. 15.21

except the frequency responses

are plotted on a linear scale.

(1) The stop-band ripples in Fig. 15.21(a) have a uniform peak value of roughly

70 dB, which is about 20 dB less than the maximum stop-band ripple

value in Fig. 15.21(b). The stop-band attenuation of the optimal FIR fil-

ter is therefore higher than that for the filter obtained from the Kaiser

window.

(2) As illustrated in Fig. 15.22(a), where the magnitude response of the optimal

FIR filter is plotted on a linear scale, there are noticeable pass-band ripples

in the magnitude response of the optimal FIR filter. Figure 15.22(b) plots

the magnitude response of the FIR filter obtained from the Kaiser window,

where the pass-band ripples are negligible. The improvement in the stop-

band attenuation of the optimal FIR filter can therefore be attributed to the

pass-band ripples that the optimal filter has incorporated. The optimal FIR

filter distributes the distortion between the pass and stop bands. The FIR

filter obtained from the Kaiser window has most distortion concentrated in

the stop band, which leads to higher ripples (or lesser attenuation) within

its stop band.

(3) Finally, we observe that the transition bands in the two FIR filters are

roughly of the same width.

15.7 Summary

This chapter presented techniques for designing causal FIR filters. The ideal

frequency-selective filters presented in Chapter 14 are physically unrealizable

because of strict constraints on the pass- and stop-band gains of the filter and

also because of a sharp transition between the pass and stop bands. Practi-

cal implementations of the ideal filters are obtained by allowing acceptable

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708 Part III Discrete-time signals and systems

variations (or ripples) within the pass and stop bands. In addition, a transition

band is included between the pass and stop bands so that the filter gain can drop

off smoothly.

Section 15.1 introduced the windowing approach used to design FIR filters

from the ideal frequency-selective filters. The windowing approach truncates

the impulse response h[k] of an ideal filter, with a linear-phase component of exp(−jmΩ), to a finite length N within the range 0 ≤ k ≤ (N − 1). The value of m in the phase component is selected to be (N − 1)/2 such that the filter coefficients in the causal FIR filter are symmetrical with respect to m. Common elementary windows used to design FIR filters are the rectangular, Bartlett,

Hamming, Hanning, and Blackman windows. The selection of type of window

depends upon the maximum value of the pass- and stop-band ripples. The length

N of the window is determined from the allowable width of the transition band.

The minimum stop-band attenuation in the FIR filter obtained from the ele-

mentary windows is fixed. In most cases, the selected window surpasses the

given specification on the stop-band attenuation and the resulting FIR filter is

therefore of higher computational complexity than required. Section 15.2 intro-

duced the Kaiser window, which provides control over the stop-band attenuation

by including an additional design parameter, referred to as the shape control

parameter β. The order of the FIR filter designed by the Kaiser window is sig-

nificantly smaller than those of the FIR filters obtained using the elementary

window functions.

The FIR design techniques covered in Sections 15.1 and 15.2 are applicable to

all types of frequency-selective filters such as the lowpass, highpass, bandpass,

and bandstop filters. Common convention, however, is to express the transfer

functions of the highpass, bandpass, and bandstop filters in terms of the transfer

function of the lowpass filter. Using the resulting relationships, the design of any

type of filter can be reduced to the design of one or more lowpass filters. Section

15.3 covered design techniques for highpass FIR filters. We covered design

algorithms using the original highpass filter specifications as well as techniques

that transform the problem of designing a highpass FIR filter to designing

a lowpass FIR filter. Similarly, Section 15.4 presented design techniques for

bandpass FIR filters, while Section 15.5 designed bandstop FIR filters.

The windowing approaches produce a suboptimal design. Section 15.5 intro-

duced a computational procedure based on the Parks–McClellan algorithm that

exploits the inherent structure, expressed in Proposition 14.1 for the linear-

phase FIR filters. The Parks–McClellan algorithm computes the best FIR filter

of length N that minimizes the maximum absolute difference between the trans- fer function Hd(Ω) of the ideal filter and the transfer function H (Ω) of the cor- responding FIR filter. Mathematically, the Parks–McClellan algorithm solves

the minimax optimization problem, which finds the set of filter coefficients

that minimizes the maximum error between the desired frequency response and

the actual frequency response. According to Proposition 14.1, the frequency

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709 15 FIR filter design

response of a linear-phase filter can be expressed as a polynomial of cos(Ω).

It can also be shown that error ε(Ω) between the desired and actual frequency

response is also a polynomial of cos(Ω). The Parks–McClellan algorithm uses

the alternation theorem, which provides the following condition for the optimal

design of H (Ω). The transfer function H (Ω), which best approximates Hd(Ω) in the minimax

sense, produces an error function ε(Ω) with at least L + 2 discrete extremal frequencies {Ω1 < Ω2 < · · · < ΩL+2} ∈ S in ε(Ω) that alternate between the maximum and minimum peak values of the error, i.e. ε(Ωm+1) = −ε(Ωm) =

εmax, where εmax is the maximum value of error |ε(Ω)|.

The Parks–McClellan algorithm is available as a library function in most

signal processing packages. Section 15.7 covered the firpm function used

to design optimal FIR filters in M A T L A B using the Parks–McClellan algo-

rithm. In addition, we introduced other library functions including rectwin,

bartlett, hann, hamming, blackman, and kaiser functions used to

implement the elementary windows covered in Sections 15.1 and 15.2. The

fir1 function used to derive the impulse response of an FIR filter is also

covered.

Problems

15.1 The ideal DT differentiator is commonly used to differentiate a CT signal directly from its samples. The transfer function of a DT differentiator is

given by

Hdiff (Ω) = jΩ e−jmΩ 0 ≤ |Ω| ≤ π .

Determine the impulse response hdiff[k] of the ideal differentiator.

15.2 A system with the block schematic shown in Fig. 9.1 is used to process a CT signal with a digital filter. The A/D converter has a sampling rate of

8000 samples/s. Design the ideal digital filter if the overall transfer func-

tion of Fig. 9.1 represents an ideal lowpass filter with a cut-off frequency

of 2 kHz. Repeat for the sampling rates of 16 000 samples/s and 44 100

samples/s.

15.3 Calculate the amplitude of the 5-tap (N = 5) rectangular, Hanning, Ham- ming, and Blackman windows. Sketch the window functions.

15.4 The specifications for a lowpass filter are given as follows:

pass-band edge frequency = 0.25π ;

stop-band edge frequency = 0.55π ;

minimum stop-band attenuation = 35 dB.

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710 Part III Discrete-time signals and systems

Determine which of the elementary windows mentioned in Table 15.2

would satisfy these specifications. For the permissible choices, deter-

mine the lengths N of the windows that meet the width requirement for the transition band.

15.5 Repeat Problem 15.4 for the Kaiser window.

15.6 Determine the impulse response of an ideal discrete-time lowpass filter with a cut-off frequency ofΩc = 1 radian/s. Using a rectangular window, truncate the length N of the ideal filter to 51. Plot the impulse response and amplitude frequency characteristics of the FIR filter.

15.7 Repeat Problem 15.6 for the Hamming window and compare the result- ing FIR filter with the FIR filter obtained from the rectangular window

in that problem.

15.8 Design the digital FIR filter, shown as the central block and labeled as the DT system in Fig. 9.1, if the specifications of the overall system are

given as follows (the overall CT system is a lowpass filter):

pass-band edge frequency = 10.025 kHz; width of the transition band = 1 kHz; minimum stop-band attenuation = 45 dB; sampling rate = 44.1 ksamples/s.

(a) Determine the possible types of windows that may be used.

(b) Assuming that the Hamming window is used to design the FIR filter,

plot the impulse response h[k] of the resulting FIR filter. (c) Plot the amplitude–frequency characteristics of the FIR filter on both

absolute and logarithmic scales.

15.9 Repeat Problem 15.8 for a Kaiser window.

15.10 Using the Kaiser window, design a highpass FIR filter based on the following specifications:

pass-band edge frequency = 0.64π ; width of the transition band = 0.3π ; maximum pass-band ripple <0.002;

maximum stop-band ripple <0.005.

Use M A T L A B to confirm that the designed FIR filter satisfies the given

specifications:

15.11 Using the Kaiser window, design a bandpass FIR filter based on the following specifications:

pass-band edge frequencies = 0.4π and 0.6π ;

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711 15 FIR filter design

stop-band edge frequencies = 0.2π and 0.8π ; maximum pass-band ripple <0.02;

maximum stop-band ripple <0.009.

Use M A T L A B to confirm that the designed bandpass FIR filter satisfies

the given specifications.

15.12 Using the Kaiser window, design a bandstop FIR filter based on the following specifications:

stop-band edge frequencies = 0.3π and 0.7π ; pass-band edge frequencies = 0.4π and 0.6π ; maximum pass-band ripple <0.05;

maximum stop-band ripple <0.05.

Use M A T L A B to confirm that the designed bandstop FIR filter satisfies

the given specifications.

15.13 Equation (15.44) defines the expression for the normalized weighting function used in the design of a lowpass filter using the Parks–McClellan

algorithm Derive the expressions for the normalized weighting functions

for highpass, bandpass, and bandstop filters.

15.14 For a type I FIR filter of length N , show that the degree L of the error function ε(Ω) defined in Eq. (15.42) is given by (N − 1)/2.

15.15 Repeat Problem 15.14 for a type II FIR filter of length N by showing that the degree L of the error function ε(Ω) = Ŵ(cos(Ω)) defined in Eq. (15.42) is given by (N − 2)/2.

15.16 Repeat Problem 15.14 for a type III FIR filter of length N by showing that the degree L of the error function ε(Ω) defined in Eq. (15.42) is given by (N − 3)/2.

15.17 Repeat Problem 15.14 for a type IV FIR filter of length N by showing that the degree L of the error function ε(Ω) defined in Eq. (15.42) is given by (N − 2)/2.

15.18 Truncate the impulse response of an ideal bandstop FIR filter with edge frequencies of 0.25π and 0.75π with a 20-tap rectangular window filter.

Plot the magnitude response of the resulting FIR filter and compare the

frequency characteristics with a 40-tap FIR filter.

15.19 Using M A T L A B , determine the impulse response of the FIR filters designed in Problems 15.4 and 15.5. Sketch the magnitude response and

ensure that the FIR filters satisfy the given specifications. Comment on

the complexity and frequency characteristics of the designed filters.

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712 Part III Discrete-time signals and systems

15.20 Using M A T L A B , determine the impulse response of the optimal FIR filter for the specifications provided in Problem 15.4. You may use the

Kaiser window to determine the length of the optimal FIR filter. Sketch

the magnitude response of the optimal FIR filter and compare its fre-

quency characteristics with those of the FIR filters plotted in Problem

15.18.

15.21 Show that the alternation theorem is satisfied for the magnitude response of the optimal FIR filter designed in Problem 15.20.

15.22 Using the fir1 function in M A T L A B , design a 41-tap lowpass filter with a normalized cut-off frequency of Ωn = 0.55 using (i) rectangular; (ii) Hamming; (iii) Blackman; and (iv) Kaiser (with β = 4) windows. Plot the amplitude–frequency characteristics for the four filters. For each

plot, determine (i) the maximum pass-band ripple; (ii) the peak side lobe

gain; and (iii) the transition bandwidth. Assume that the transition band

is a band where the filter gain drops from –2 dB to –20 dB.

15.23 Using the fir1 function in M A T L A B , design a 45-tap linear-phase bandpass FIR filter with pass-band edge frequencies of 0.45π and 0.65π ,

stop-band edge frequencies of 0.15π and 0.9π , maximum pass-band

attenuation of 0.1 dB, and minimum stop-band attenuation of 40 dB.

Use the Kaiser window for your design and sketch the frequency char-

acteristics of the resulting filter.

15.24 The fir2 function in M A T L A B is used to design FIR filters with arbitrary frequency characteristics. Using fir2, design a 95-tap FIR

filter with the following frequency characteristics:

|H (Ω)| =



  

  

0.85 0 ≤ |Ωn| ≤ 0.15

0.55 0.20 ≤ |Ωn| ≤ 0.45

1 0.55 ≤ |Ωn| ≤ 0.75

0.5 0.78 ≤ |Ωn| ≤ 1,

where Ωn is the normalized DT frequency. Use M A T L A B to confirm

that the designed FIR filter satisfies the given specifications.

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C H A P T E R

16 IIR filter design

Based on the length of the impulse response h[k], Chapter 14 classified digital (or “discrete-time”) filters into two categories: finite impulse response (FIR)

filters and infinite impulse response (IIR) filters. The design techniques for

the FIR filter, with an impulse response h[k] of finite length, were covered in Chapter 15. In this chapter, we present design methodologies of the IIR fil-

ters. A common technique used to design IIR filters is based on mapping the

DT frequency specifications H (Ω) of the IIR filters in the Ω domain to the CT frequency specifications H (ω) specified in the ω domain. Based on the transformed specifications, a CT filter is designed, which is then transformed

back into the original DT frequency Ω domain to obtain the transfer func-

tion of the required IIR filter. In this chapter, we present two different DT to

CT frequency transformations. The first method is referred to as the impulse

invariance transformation, which provides a linear transformation between the

DT and CT frequency domains. At times, the impulse invariance transforma-

tion suffers from aliasing, which may lead to deviations from the original DT

specifications. An alternative to the impulse invariance transformation is the

bilinear transformation, which is a non-linear mapping between the CT and DT

frequency domains. The bilinear transformation eliminates aliasing to a large

extent.

A classical problem in the design of digital filters is the selection between

FIR and IIR filters. While both types of filters can be used to satisfy a given set

of specifications, the order N of IIR filters is in general much lower than that of FIR filters. As a consequence of the lower order N , the IIR filters have reduced implementation complexity and less propagation delay when compared with

FIR filters designed for the same specifications. However, IIR filters are imple-

mented using feedback loops, resulting in transfer functions with a significant

number of poles. IIR filters are, therefore, susceptible to instability issues when

realized on finite-precision DSP boards. In addition, IIR filters have a non-linear

phase, whereas FIR filters can be designed with a linear phase. An appropriate

digital filter type is selected based on the requirement of a given application.

713

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714 Part III Discrete-time signals and systems

The organization of this chapter is as follows. IIR filter design principles are

introduced in Section 16.1. Sections 16.2 and 16.3 present design principles

of lowpass IIR filters based on the frequency transformation methods. In Sec-

tion 16.2, we introduce the impulse invariance transformation, and in Section

16.3 we present the bilinear transformation. The analytical design procedure is

illustrated through a series of examples. We also provide the M A T L A B code,

which can also be used in the design of IIR filters. Section 16.4 covers the design

techniques for bandpass, bandstop, and highpass filters. Finally, Section 16.5

compares the frequency characteristics of IIR filters with those of FIR filters

designed for the same specifications. Section 16.6 presents a summary of the

important concepts covered in the chapter.

16.1 IIR filter design principles

As specified in Chapter 14, the transfer function of an IIR filter is given by

H(z) = b0 + b1z−1 + · · · + bM z−M

1 + a1z−1 + · · · + aN z−N , (16.1)

where br , for 0 ≤ r ≤ M , and ar , for 0 ≤ r ≤ N , are known as the filter coeffi- cients. In Eq. (16.1), we have also normalized the coefficient a0 (corresponding to r = 0) in the denominator to unity. Based on Eq. (16.1), the IIR filter can alternatively be modeled by the following linear, constant-coefficient difference

equation:

y[k] + a1 y[k − 1] + · · · + aN y[k − N ] = b0x[k] + b1x[k − 1]

+ · · · + bM x[k − M]. (16.2)

The objective of the IIR filter design is to calculate a set of filter coefficients br , for 0 ≤ r ≤ M , and ar , for 1 ≤ r ≤ N , such that the frequency characteristics of the IIR filter match the design specifications. IIR filter design can, therefore,

be viewed as a mathematical optimization problem.

A popular method used to design an IIR filter is based on converting its desired

frequency specifications H (Ω) into the CT frequency domain. Using the CT design techniques for the Butterworth, Chebyshev, or elliptic filters covered in

Chapter 6, the transfer function H (s) of the analog filter is determined. The z- transfer function H (z) of the desired IIR filter is then obtained by transforming H (s) back into the DT domain. Such transformation approaches yield closed- form transfer functions for the IIR filters.

A number of transformations have been proposed to convert the transfer

function H (s) of the CT (or analog) filter into the z-transfer function H (z) of the IIR filter such that the frequency characteristics of the CT filter in the s-plane

are preserved for the IIR filter in the z-plane. These transformations include the

following methods:

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715 16 IIR filter design

(a) finite-difference discretization of differential equations;

(b) mapping poles and zeros from the s-plane to the z-plane;

(d) bilinear transformation.

The finite-difference discretization of differential equations is a straightforward

method to derive difference equation representations for digital filters. First,

the s-transfer functions, obtained by using the CT filter design techniques, are

used to calculate the input–output relationship of the equivalent CT filter. These

relationships are generally in the form of linear, constant-coefficient differential

equations, and are discretized to obtain difference equations that represent the

input–output relationships of the designed DT filters.

In the second method, referred to as the matched z-transform technique, the

s-plane poles and zeros of a designed CT filter are mapped to the z-plane. The

s-plane poles and zeros are then used to derive the transfer function H (z) of the digital IIR filter.

The impulse invariance method samples the impulse response h(t) of an LTIC system to derive the impulse response h[k] of the corresponding LTID system. Finally, the bilinear transformation provides a one-to-one, non-linear

mapping from the s-plane to the z-plane. The impulse invariance and bilin-

ear transformations are the focus of this chapter. In Section 16.2, we cover

the impulse invariance method followed by the bilinear transformation, in

Section 16.3.

16.2 Impulse invariance

To derive the impulse invariance transformation, we approximate the impulse

response h(t) of a CT filter with its sampled representation,

h(t) ≈ ∞∑

n=−∞

h(t)δ(t − nT ) = ∞∑

n=−∞

h(nT )δ(t − nT ), (16.3)

obtained by sampling h(t) with an impulse train ∑

δ(t – nT). Clearly, the approximation in Eq. (16.3) improves as the sampling interval T → 0. The DT impulse response h[k] of the equivalent IIR filter is obtained from the samples h(kT) and is given by

h[k] = h(kT ) = ∞∑

n=−∞

h(nT )δ(k − n). (16.4)

Comparing the expressions for the Laplace transform of Eq. (16.3) given by

Laplace transform H (s) = ∞∑

n=−∞

h(nT )e−nT s (16.5)

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716 Part III Discrete-time signals and systems

(a) (b)

Re{z}

Im{z}

(1, 0) Re{s}

Im{s}

−

Fig. 16.1. Impulse invariance

transformation from the s-plane

(a) to the z-plane (b).

and the z-transform of Eq. (16.4) given by

z-transform H (z) = ∞∑

k=−∞

h(nT )z−n, (16.6)

we note that the two expressions are equal provided

z = eTs . (16.7)

In terms of real and imaginary components of s = σ + jω, Eq. (16.7) can be expressed as follows:

z = eσ T e jωT . (16.8)

Equation (16.7) provides a mapping between the DT variable z and the CT variable s. The mapping, commonly referred to as the impulse invariance trans- formation, is illustrated in Fig. 16.1, where we observe that the s-plane region

Re{s} = σ < 0 and |Im{s}| = |ω| < π/T,

shown as the shaded region, in Fig. 16.1(a) maps into the interior of the unit

circle |z| < 1 shown in Fig 16.1(b). Equations (16.7) and (16.8) can also be used to derive the following observations.

Right-half s-plane Re{s} > 0 Taking the absolute value of Eq. (16.8) yields

|z| = |eσ T | · |e jωT | = |eσ T |. (16.9)

In the right-half s-plane, Re{s} = σ > 0, resulting in |z| > 1. Therefore, the right-half s-plane is mapped to the exterior of the unit circle.

Origin s = 0 Substituting s = 0 into Eq. (16.7) yields z = 1. The origin s = 0 in the s-plane is therefore mapped to the coordinate (1, 0) in the z-plane.

Imaginary axis Re{s} = 0 Taking the absolute value of Eq. (16.8) and substi-

tuting Re{s} = σ = 0 yields |z| = 1. The imaginary axis Re{s}= 0 is therefore mapped on to the unit circle |z| = 1.

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717 16 IIR filter design

Left-half s-plane Re{s }< 0 Substituting Re{s}= σ < 0 in Eq. (16.9) yields |z| < 1. Therefore, we observe that the left-half s-plane is mapped to the interior of the unit circle. We now show that the mapping z = esT is not a unique one-to-one mapping and that different strips of width 2π/T are mapped into the same region within the unit circle |z| < 1.

Consider the set of points s = σ0 + j2kπ/T , with k = 0, ±1, ±2, . . . , in the s-plane. Substituting s = σ0 + j2kπ/T in Eq. (16.7) yields

z = eT (σ0+j2kπ/T ) = eσ0T e j2kπ = eσ0T . (16.10)

In other words, the set of points s = σ0 + j2kπ/T are all mapped to the same point z = exp(σ0T ) in the z-plane. Equation (16.8) is, therefore, not a unique, one-to-one mapping, and different strips of width 2π/T in the left-half s-plane are mapped to the same region within the interior of the unit circle.

We now illustrate the procedure used to obtain an equivalent H (z) from an impulse response h(t) through Examples 16.1 and 16.2.

Example 16.1

Use the impulse invariance method to convert the s-transfer function

H (s) = 1

s + α into the z-transfer function of an equivalent LTID system.

Solution

Calculating the inverse Laplace transform of H (s) yields

h(t) = e−αt u(t).

Using impulse train sampling with a sampling interval of T , the impulse response of the LTID system is given by

h(kT ) = e−αkT u(kT ) or h[k] = e−αkT u[k].

The z-transfer function of the equivalent LTID system is given by

H (z) = z{h[k]} = 1

1 − e−αT z−1 , ROC : |z| > e−αT .

Figure 16.2 compares the impulse response h(t) and transfer function H (s) of the LTIC system with the impulse response h[k] and transfer function H (z) of the equivalent LTID system obtained using the impulse invariance method. A sampling period of T = 0.1 s and α = 0.5 are used. Comparing the CT impulse response h(t), plotted in Fig. 16.2(a), with the DT impulse response h[k], plotted in Fig. 16.2(c), we observe that h[k] is a sampled ver- sion of h(t), and the shapes of the impulse responses are fairly similar to each other.

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718 Part III Discrete-time signals and systems

−5 0 5 10 15 20 25 30 35 40 45 50 0

0.2

0.4

0.6

0.8

−1 0 1 2 3 4 5 0

0.2

0.4

0.6

0.8

t −2p −1.5p −p −0.5p 0 0.5p p 1.5p 2p 0

0.5

1.5

−2p −1.5p −p −0.5p 0 0.5p p 1.5p 2p 0

(a) (b)

Fig. 16.2. Impulse invariance

method used for transforming

analog filters to digital filters in

Example 16.1. (a) Impulse

response h(t ) and

(b) magnitude spectrum H(ω) of

the analog filter. (c) Impulse

response h[k ] and

(d) magnitude spectrum H (Ω)

of the transformed digital filter.

Comparing the magnitude spectrum |H (ω)| of the LTIC system with the magnitude spectrum |H (Ω)| of the LTID system plotted in Figs. 16.2(b) and (d), respectively, we observe two major differences. First, the magnitude spectrum

|H (Ω)| is periodic with a period of 2π . Secondly, the magnitude spectrum |H (Ω)| is scaled by a factor of 1/T in comparison with |H (ω)|. In order to obtain a DT filter with a DC amplitude gain of the same value as that of the CT

filter, we multiply the sampled impulse response h[k] by a factor of T :

h[k] = Th(kT ) = T e−αkT u(k). (16.11)

Alternatively, the following transform pair can be used for the impulse invari-

ance transformation:

s + α impulse invariance

←− −→ T

1 − e−αT z−1 or

z − e−αT . (16.12)

Example 16.2 illustrates the application of Eq. (16.12) in transforming a But-

terworth lowpass filter into a digital lowpass filter.

Example 16.2

Consider the following Butterworth filter:

H (s) = 81.6475

s2 + 12.7786s + 81.6475 .

Use the impulse invariance transformation to derive the transfer function of the

equivalent digital filter.

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719 16 IIR filter design

Solution

Expressing the transfer function of the CT filter as follows:

H (s) = 12.7786 6.3894

(s + 6.3893)2 + 6.38942 , (16.13)

and Calculating the inverse Laplace transform, the impulse response of the CT

filter is given by

h(t) = 12.7786e−6.3893t sin(6.3894t)u(t). (16.14)

Using Eq. (16.11) to derive the impulse response of the DT filter, we obtain

h[k] = Th(kT ) = 12.7786T e−6.3893kT sin(6.3894kT )u(kT ) (16.15)

h[k] = 12.7786T e−6.3893 kT sin(6.3894kT )u[k]. (16.16)

Calculating the z-transform of Eq. (16.16), the transfer function of the DT filter

is given by (see Problem 16.2)

H (z) = 12.7786T e−6.3893 T sin(6.3894T )z

z2 − 2z e−6.3893T cos(6.3894T )z + e−2×6.3893T . (16.17)

Alternative solution Equation (16.17) can also be derived by using the

impulse invariance transformation specified in Eq. (16.12). Using partial frac-

tion expansion, H (s) is expressed as follows:

H (s) = 12.7786

[ 1

s + 6.3893 − j6.3894 −

s + 6.3893 + j6.3894

]

(16.18)

which is then transformed using Eq. (16.12) in the z-domain:

H (z) = 12.7786

[ zT

z − e−(6.3893−j6.3894)T −

z − e−(6.3893+j6.3894)T

]

It is straightforward to show that the above expression reduces to Eq. (16.17).

Selection of the sampling interval To choose an appropriate sampling interval

T , we need to analyze the magnitude spectrum of h(t). Substituting s = jω in Eq. (16.13), we obtain

H (ω) = 12.7786 6.3894

(jω + 6.3893)2 + 6.38942 =

81.6489

(81.6489 − ω2) + j12.7788ω ,

(16.19)

which leads to the following magnitude spectrum:

|H (ω)| = 81.6489

√

(81.6489 − ω2)2 + 163.2977ω2 . (16.20)

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720 Part III Discrete-time signals and systems

Table 16.1. DT filters obtained in Example 16.2

for different values of the sampling interval T

The magnitude spectra of these transfer functions

are plotted in Figs. 16.3(b)–(e)

T H (z)

0.1 H (z) = 0.4023z

z2 − 0.8475z + 0.2786

0.0348 H (z) = 0.0785z

z2 − 1.5619z + 0.6410

0.01 H (z) = 0.0077z

z2 − 1.8724z + 0.8800

0.001 H (z) = 8.113 × 10−5z

z2 − 1.9872z + 0.9873

The peak value of the magnitude spectrum |H (ω)| occurs at ω = 0, with a value |H (0)| = 1. Also, the magnitude spectrum |H (ω)| is a monotonically decreasing function with respect to ω. Assuming that the maximum frequency

present in the function h(t) is approximated as ωmax such that |H (ω)| ≤ 0.01 for |ω| ≥ ωmax, it can be shown that ωmax = 90.4 radians/s. Using the Nyquist sampling rate, the sampling interval is therefore given by

T ≤ 1

2 fmax =

2π

2ωmax = 0.348 s.

Table 16.1 compares the transfer function of the transformed DT filters obtained

by substituting different values of the sampling intervals T into Eq. (16.17). The amplitude gain responses of the DT filters for different values of T = 0.1, 0.0348, 0.01, and 0.001 are given in Table 16.1. A comparison of the magnitude

spectra of the four transfer functions is illustrated in Fig. 16.3. We make the

following observations.

(1) Although the shapes of the magnitude spectra (Figs. 16.3(b)–(d)) of the

digital filters appear to be different, they are all valid representations of the

magnitude spectrum of the analog filter (Fig. 16.3(a)). Substituting s = jω and z = e jΩ into Eq. (16.7) yields

Ω = ωT .

The 3 dB frequency Ω0 of the digital implementations therefore

depends upon the sampling interval T . Based on the 3 dB frequency ω0 = 9.03 radians/s, the values of Ω0 are given by 0.2874π radians/s for

T = 0.1 s, by 0.1π radians/s for T = 0.0348 s, by 0.0287π radians/s for T = 0.01 s, and by 0.0029π radians/s for T = 0.001 s.

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721 16 IIR filter design

0.2

0.4

0.6

0.8

−0.6p −0.4p −0.2p 0.2p−p −0.8p 0 0.4p 0.6p 0.8p p W

−60 −40 −20 0 20 40 60 0

0.2

0.4

0.6

0.8

w 0

0.2

0.4

0.6

0.8

−0.6p −0.4p −0.2p 0.2p−p −0.8p 0 0.4p 0.6p 0.8p p W

0.2

0.4

0.6

0.8

−0.6p −0.4p −0.2p 0.2p−p −0.8p 0 0.4p 0.6p 0.8p p W 0

0.2

0.4

0.6

0.8

−0.6p −0.4p −0.2p 0.2p−p −0.8p 0 0.4p 0.6p 0.8p p W

(a) (b)

(e)

Fig. 16.3. Impulse invariance

transformation used to derive

digital representations of the

analog filter specified in Example

16.2. Magnitude spectra of

(a) the analog filter with transfer

function H(s ); (b) the digital

filter with sampling interval

T = 0.1 s; (c) the digital filter with T = 0.0348 s; (d) the digital filter with T = 0.01 s; (e) the digital filter with

T = 0.001 s.

(2) Among the digital implementations, Fig. 16.3(b) results in the highest

gain (i.e. lowest attenuation) at the stop-band frequencyΩ = ±π radians/s. Since the sampling interval (T =0.1 s) is greater than the Nyquist bound (T = 0.0348 s), Fig. 16.3(b) suffers from aliasing, which increases the gain within the pass band. In using impulse invariance transformation, it is crit-

ical that the effects of the aliasing be considered within the stop band.

16.2.1 Impulse invariance transformation using M A T L A B

M A T L A B provides a library function impinvar to transform CT transfer

functions into the DT domain using the impulse variance method. We illustrate

the application of impinvar for Example 16.2 with the sampling interval T set to 0.1 s. The M A T L A B code for the transformation is as follows:

>> num = [0 0 81.6475]; % numerator of CT filter

>> den = [1 12.7786 81.6475]; % denominator of CT filter

>> T = 0.1;

>> Fs = 1/T; % sampling rate

>> [numz,denz] = impinvar (num,den,Fs);

% numerator & denominator

% of DT filter

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722 Part III Discrete-time signals and systems

The above M A T L A B code results in the following values for the coefficients

of H (z):

numz = [0 0.4023 0] and denumz = [1 -0.8475 0.2786],

which correspond to the following transfer function:

H (z) = 0.4023z

z2 − 0.8475z + 0.2786 .

The above expression is the same as the one obtained analytically, and it is

included in row 1 of Table 16.1.

16.2.2 Look-up table

Examples 16.1 and 16.2 present direct methods to compute the impulse response

h[k], or correspondingly the transfer function H (z), of the DT filter by sampling the impulse response h(t) of an analog filter. The process can be simplified further in cases where the transfer function H (s) of the analog filter is a rational function. In such cases, the transfer function H (s) can be expressed in terms of partial fractions as follows:

H (s) = N∑

r=1

kr s + αr

, (16.21)

where kr is the coefficient of the r th partial fraction. Applying the impulse invariance transformation, Eq. (16.12), the transfer function H (z) of the digital filter is given by

H (z) = N∑

r=1

kr z

z − e−αr T . (16.22)

Table 16.2 lists a number of commonly occurring s-domain terms and the

equivalent representation in the z-domain. We now list the steps involved in

the design of digital IIR filters using the impulse invariance transformation.

16.2.3 IIR filter design using impulse invariance transformation

The steps involved in designing IIR filters using the impulse invariance trans-

formation are as follows.

Step 1 Using Ω = ωT , transform the specifications of the digital filter from the DT frequency Ω domain to the CT frequency ω domain. For convenience,

we choose T = 1.

Step 2 Using the analog filter techniques (see Chapter 7), design an analog

filter H (s) based on the transformed specifications obtained in step 1.

Step 3 Using the impulse invariance transformation specified in Eq. (16.12),

s + α impulse invariance

←− −→ T

1 − e−αT z−1 or

z − e−αT

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723 16 IIR filter design

Table 16.2. Analog to-digital transformation using impulse invariance method

CT domain DT domain

H (s) h(t) h[k] H (z)

1 δ(t) T δ[k] T

s u(t) Tu[k]

1 − z−1 =

T z

z − 1

s2 tu(t) kT2u[k]

T 2z−1

(1 − z−1)2 =

T 2z

(z − 1)2

s + α e−αt u(t) T e−αkT u[k]

(1 − e−αT z−1) =

T z

(z − e−αT )

(s + α)2 te−αt u(t) kT 2e−αkT u[k]

T 2e−αT z−1

(1 − e−αT z−1)2 =

T 2e−αT z

(z − e−αT )2

s + α (s + α)2 + β2

e−αt cos(βt)u(t) T e−αkT cos(βkT )u[k] T z[z − e−αT cos(βT )]

z2 − 2e−αT cos(βT )z + e−2αT

(s + α)2 + β2 e−αt sin(βt)u(t) T e−αkT sin(βkT )u[k]

T ze−αT sin(βT )

z2 − 2e−αT cos(βT )z + e−2αT

or the look-up table approach, derive the z-transfer function H (z) from the s-transfer function H (s).

Step 4 Confirm that the z-transfer function H (z) obtained in step 3 satisfies the design specifications by plotting the magnitude spectrum |H (Ω)|. If the design specifications are not satisfied, increase the order N of the analog filter designed in step 2 and repeat steps 2–4.

We now illustrate the application of the above algorithm in Example 16.3.

Example 16.3

Design a lowpass IIR filter with the following specifications:

pass band (0 ≤ |Ω| ≤ 0.25π radians/s) 0.8 ≤ |H (Ω)| ≤ 1;

stop band (0.75π ≤ |Ω| ≤ π radians/s) |H (Ω)| ≤ 0.20.

Solution

Choosing the sampling interval T = 1, step 1 transforms the given specifications of the DT filter into the corresponding specifications for the CT filter:

pass band (0 ≤ |ω| ≤ 0.25π radians/s) 0.8 ≤ |H (ω)| ≤ 1;

stop band (|ω| > 0.75π radians/s) |H (ω)| ≤ 0.20.

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724 Part III Discrete-time signals and systems

Step 2 designs the analog filter based on the transformed specifications. We use

the Butterworth filter, whose design procedure is outlined in Section 7.3.1.1.

Design of the analog Butterworth filter To determine the order N of the filter, we calculate the gain terms:

Gp = 1

(1 − δp)2 − 1 = 0.5625

and

Gs = 1

(δs)2 − 1 = 24.

The order N of the filter is therefore given by

N = 1

2 ×

ln(Gp/Gs)

ln(ωp/ωs) =

2 ×

ln(0.5625/24)

ln(0.25π/0.75π ) = 1.7083.

Using Table 7.2, the transfer function for the normalized Butterworth filter of

order N = 2 is given by

H (S) = 1

S2 + 1.414S + 1 .

Equation (7.32) determines the cut-off frequency ωc of the Butterworth filter

from the stop-band constraint as follows:

ωc = ωs

(Gs)0.5/N =

0.75π

240.25 = 0.3389π radians/s.

The transfer function H (s) of the required analog lowpass filter is given by

H (s) = H (S)|S=s/ωc = 1

S2 + 1.414S + 1

∣ ∣ ∣ ∣

S=s/0.3389π

= 1.1332

s2 + 1.5055s + 1.1332 ,

which can be expressed as follows:

H (s) = 1.5053 0.7528

(s + 0.7528)2 + 0.75282 .

Using Table 16.2, step 3 derives the z-transfer function as follows:

H (z) = 1.5053 ze−0.7528 sin(0.7528)

z2 − 2e−0.7528 cos(0.7528)z + e−2(0.7528) ,

which simplifies to

H (z) = 1.5053 0.3220z

z2 − 0.6875z + 0.2219 .

Step 4 computes the magnitude spectrum by substituting z = exp(jΩ). The resulting plot is shown in Fig. 16.4(a), where we observe that the magnitude

spectrum satisfies the pass-band requirements, though the dc gain of the filter

is not equal to unity. The stop band requirement is not satisfied, however, as the

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725 16 IIR filter design

0.2

0.4

0.6

0.8

−0.5p −0.25p 0.25p−p −0.75p 0 0.5p 0.75p p W0

0.2

0.4

0.6

0.8

−0.5p −0.25p 0.25p−p −0.75p 0 0.5p 0.75p p

0.2

0.4

0.6

0.8

−0.5p −0.25p 0.25p−p −0.75p 0 0.5p 0.75p p

(a) (b)

(c)

Fig. 16.4. Design of the IIR filter

specified in Example 16.3 based

on the analog Butterworth filter

of order (a) N = 2; (b) N = 3; (c) N = 4. The impulse invariance transformation is

used to convert the Butterworth

filter to a digital filter. Aliasing

introduced by the impulse

invariance transformation results

in a considerably higher order

(N = 4) Butterworth filter to meet the design specifications.

gain |H (Ω)| of the filter is greater than 0.20 at the stop-band corner frequency of 0.75π radians/s. The above procedure is repeated for a Butterworth filter of order

N = 3.

Iteration 2 for Butterworth filter of order N = 3 The transfer function for the normalized Butterworth filter of order N = 3 is obtained from Table 7.2 as follows:

H (S) = 1

(S + 1)(S2 + S + 1) .

The cut-off frequency ωc of the Butterworth filter is obtained from the stop-band

constraint:

ωc = ωs

(Gs)0.5/N =

0.75π

241/6 = 0.4416π radians/s.

The transfer function H (s) of the required analog lowpass filter is given by

H (s) = H (S)|S=s/ωc = 1

(S + 1)(S2 + S + 1)

∣ ∣ ∣ ∣

S=s/0.4416π

= 2.6702

s3 + 2.7747s2 + 3.8494s + 2.6702 .

Expanding H (s) in terms of partial fractions and using Table 16.2, we can derive the z-transfer function of the equivalent digital filter as follows:

H (z) = 0.4695z2 + 0.1907z

z3 − 0.6106z2 + 0.3398z − 0.0624 .

The above derivation is left as an exercise for the reader in Problem 16.3(a).

Figure 16.4(b) plots the magnitude spectrum |H (Ω)| of the third-order filter. We observe that the attenuation is increased at the stop-band corner frequency of

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726 Part III Discrete-time signals and systems

0.75π radians/s, but that it is still greater than the specified value. We therefore

repeat the above procedure for a Butterworth filter of order N = 4.

Iteration 3 for Butterworth filter of order N = 4 The transfer function for the normalized Butterworth filter of order N = 4 is obtained from Table 7.2 as follows:

H (S) = 1

(s2 + 0.7654s + 1)(s2 + 1.8478s + 1) .

The cut-off frequency ωc of the Butterworth filter is obtained from the stop-band

constraint:

ωc = ωs

(Gs)0.5/N =

0.75π

241/8 = 0.5041π radians/s.

The transfer function H (s) of the required analog lowpass filter is given by

H (s) = H (S)|S=s/ωc = 1

(s2 + 0.7654s + 1)(s2 + 1.8478s + 1)

∣ ∣ ∣ ∣

S=s/0.5041π ,

which reduces to

H (s) = 6.2902

s4 + 4.1383s3 + 8.5630s2 + 10.3791s + 6.2902 .

Problem 16.3(b) derives the z-transfer function of the equivalent digital filter

as follows:

H (z) = 0.3298z3 + 0.4274z2 + 0.0427z

z4 − 0.4978z3 + 0.3958z2 − 0.1197z + 0.0159 .

Figure 16.4(c) plots the magnitude spectrum |H (Ω)| of the fourth-order filter. We observe that both pass-band and stop-band requirements are satisfied by the

Butterworth filter of order N = 4.

Impulse invariance transformation using M AT L A B Starting with the ana-

log Butterworth filter, the IIR filters in Example 16.3 can also be designed

using the M A T L A B function impinvar. The syntax to call the function is

given by

[numz,denumz] = impinvar(nums,denums,fs)

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727 16 IIR filter design

where num and denum specify the coefficients of the numerator and denomi-

nator of the analog filter and fs is the sampling rate in samples/s. For Example

16.3, the M A T L A B code is given by

>> fs = 1; % fs = 1/T = 1

>> nums = [1.1332]; % numerator of CT filter

>> denums = [1 1.5055 1.1332]; % denominator of CT filter

>> [numz,denumz] = impinvar (nums,denums,fs);

% coefficients of the DT

% filter

which returns the following values:

numz = 0.4848 and denumz = [1.0000 -0.6876 0.2219].

The transfer function of the second-order IIR filter is given by

H (z) = 0.4848z

z2 − 0.6875z + 0.2219 ,

which yields the same expression as the one derived in Example 16.3.

For the third-order Butterworth filter, the M A T L A B code for the impulse

invariance transformation is given by

>> fs = 1; % fs = 1/T = 1

>> nums = [2.6702]; % numerator of the CT filter

>> denums = [1 2.7747 3.8494 2.6702];

% denominator of the CT filter

>> [numz,denumz] = impinvar (nums,denums,fs);

% coeffs of the DT filter

which returns the following values:

numz = [0 0.4695 0.1907] and denumz = [1.0000 -0.6106

0.3398 -0.0624].

The transfer function of the third-order IIR filter is given by

H (z) = 0.4695z2 + 0.1907z

z3 − 0.6106z2 + 0.3398z − 0.0624 .

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728 Part III Discrete-time signals and systems

Similarly, the M A T L A B code for transforming the fourth-order Butterworth

filter is given by

>> fs = 1; % fs = 1/T = 1

>> nums = [6.2902]; % numerator of the CT filter

>> denums = [1 4.1383 8.5603 10.3791 6.2902];

% denominator of CT filter

>> [numz,denumz] = impinvar (nums,denums,fs);

% coefficients of the DT filter

which returns the following values:

numz = [0 0.3298 0.4276 0.0428]

denumz = [1 -0.4977 0.3961 -0.1197 0.0159].

The transfer function of the fourth-order IIR filter is given by

H (z) = 0.3298z3 + 0.4276z2 + 0.0428z

z4 − 0.4977z3 + 0.3958z2 − 0.1197z + 0.0159 .

The above expression is similar to the one obtained in Example 16.3 for the

fourth-order Butterworth filter.

16.2.4 Limitations of impulse invariance method

As illustrated in Example 16.3, the impulse invariance method introduces alias-

ing while transforming an analog filter to a digital filter. Since the analog fil-

ter is not band-limited, the impulse invariance transformation would always

introduce aliasing in the digital domain. Therefore, a higher-order DT filter is

generally required to satisfy the design constraints. Section 16.3 introduces a

second transformation, known as the bilinear transformation, to eliminate the

effect of aliasing.

16.3 Bilinear transformation

The bilinear transformation provides a one-to-one mapping from the s-plane to

the z-plane. The mapping equation is given by

s = k z − 1 z + 1

, (16.23)

where k is the normalization constant given by 2/T , where T is the sampling interval. To derive the frequency characteristics of the bilinear transformation,

we substitute z = exp(jΩ) and s = jω in Eq. (16.23). The resulting expression

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729 16 IIR filter design

w 0

−p

Fig. 16.5. Bilinear

transformation between CT

frequency ω and DT

frequencyΩ.

is given by

ω = k tan Ω

2 or Ω = 2 tan−1

k , (16.24)

which is plotted in Fig. 16.5. We observe that the transformation is highly

non-linear since the positive CT frequencies within the range ω = [0, ∞] are mapped to the DT frequenciesΩ = [0, π ]. Similarly, the negative CT frequen-

cies ω = [−∞, 0] are mapped to the DT frequencies Ω = [−π, 0]. This non-

linear mapping is known as frequency warping, and is illustrated in Fig. 16.6,

where an analog lowpass filter is transformed into a digital lowpass filter using

Eq. (16.24) with k = 1. Since the CT frequency range [−∞, ∞] in Fig. 16.5 is mapped on to the DT frequency range [−π , π ], there is no overlap between

adjacent replicas constituting the magnitude response of the digital filter. Fre-

quency warping, therefore, eliminates the undesirable effects of aliasing from

the transformed digital filter. We now show how different regions of the s-plane

are mapped onto the z-plane.

p as

b an

st o p

b an

tr an

b an

0 w 0

|H(w)|

1+ dp

pass

band

stop

band

transition

band

w 0

wp ws

1− dp

1 +

d p

1 −

d p

d s|H (W

W p

W s

Fig. 16.6. Transformation

between a CT filter H(ω) and a

DT filter H (Ω) using the bilinear

transformation.

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730 Part III Discrete-time signals and systems

16.3.1 Mapping between the s-plane and the z-plane

For k = 1, Eq. (16.23) can be represented in the following form:

z = 1 + s 1 − s

. (16.25)

Substituting s = σ + jω into Eq. (16.25), we obtain

z = 1 + σ + jω 1 − σ − jω

, (16.26)

with an absolute value given by

|z| =

√

(1 + σ )2 + ω2

(1 − σ )2 + ω2 . (16.27)

By substituting different values of s = σ + jω corresponding to the right-half, left-half, and imaginary axes of the s-plane in Eq. (16.27), we derive the fol-

lowing observations.

Left-half s-plane (σ < 0) For σ < 0, we observe that the value of the

denominator (1 − σ )2 + ω2 in Eq. (16.27) exceeds the value of the numerator (1 + σ )2 + ω2, resulting in |z| < 1. In other words, the bilinear transformation maps the left-half of the s-plane to the interior of the unit circle within the

z-plane.

Right-half s-plane (Ω < 0) For σ > 0, the value of the numerator (1 + σ )2 + ω2 in Eq. (16.27) exceeds the value of the denominator (1 − σ )2 + ω2, resulting in |z| > 1. Consequently, the bilinear transformation maps the right-half of the s-plane to the exterior of the unit circle within the z-plane.

Imaginary axis (σ = 0) For σ = 0, the denominator and numerator in Eq. (16.27) are equal, resulting in |z| = 1. The bilinear transformation maps the imaginary axis of the s-plane onto the unit circle within the z-plane.

Note that the mapping in Eq. (16.25) is a one-to-one mapping, which means

that no two points in the s-plane will map to the same point in the z-plane, and

vice versa.

16.3.2 IIR filter design using bilinear transformation

The steps involved in designing IIR filters using the bilinear transformation are

as follows.

Step 1 Using Eq. (16.24), ω = k tan(Ω/2), transform the specifications of the digital filter from the DT frequency (Ω) domain to the CT frequency (ω) domain.

For convenience, we choose k = 1.

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731 16 IIR filter design

Step 2 Using the analog filter design techniques, design an analog filter H (s) based on the transformed specifications obtained in step 1.

Step 3 Using the bilinear transformation s = (z − 1)/(z + 1) (obtained by rear- ranging Eq. (16.25) to express z in terms of s), derive the z-transfer function H (z) from the s-transfer function H (s).

We now illustrate the application of the above algorithm in Example 16.4.

Example 16.4

Repeat Example 16.3 using the bilinear transformation.

Solution

Choosing k = 1 (sampling interval T = 2), step 1 transforms the pass-band and stop-band corner frequencies into the CT frequency domain:

pass-band corner frequency ωp = tan(0.5Ωp) = tan(0.5 × 0.25π ) = 0.4142 radians/s;

stop-band corner frequency ωs = tan(0.5Ωs) = tan(0.5 × 0.75π ) = 2.4142 radians/s.

The transformed specifications of the CT filter are given by

pass-band (0 ≤ |ω| ≤ 0.4142 radians/s) 0.8 ≤ |H (ω)| ≤ 1;

stop-band (|ω| > 2.4142 radians/s) |H (ω)| ≤ 0.20.

Step 2 designs the analog filter based on the transformed specifications. As in

Example 16.3, we use the Butterworth filter. The gain terms for the filter stay

the same as in Example 16.3:

Gp = 1

(1 − δp)2 − 1 = 0.5625

and

Gs = 1

(δs)2 − 1 = 24

The order N of the filter is given by

N = 1

2 ×

ln(Gp/Gs)

ln(ωp/ωs) =

2 ×

ln(0.5625/24)

ln(0.4142/2.4142) = 1.0646,

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732 Part III Discrete-time signals and systems

W −p 0 0

0.2

0.4

0.6

0.8

−0.75p −0.5p −0.25p 0.75p p0.5p0.25p

Fig. 16.7. Magnitude response

|H(Ω)| of the lowpass filter designed in Example 16.4 using

the bilinear transformation.

which is rounded up to N = 2. Using Table 7.2, the transfer function for the normalized Butterworth filter of order N = 2 is given by

H (S) = 1

S2 + 1.414S + 1 .

Using Eq. (7.31) to determine the cut-off frequency ωc of the Butterworth filter,

we obtain

ωc = ωs

(Gs)0.5/N =

2.4142

240.25 = 1.0907 radians/s.

The transfer function H (s) of the required analog lowpass filter is given by

H (s) = H (S)|S=s/ωc = 1

S2 + 1.414S + 1

∣ ∣ ∣ ∣

S=s/1.0907

= 1.1897

s2 + 1.5421s + 1.1897 .

Step 3 derives the z-transfer function of the digital filter using the bilinear

transformation:

H (z) = H (s)|s=(z−1)/(z+1)

= 1.1897(z + 1)2

(z − 1)2 + 1.5421(z − 1)(z + 1) + 1.1897(z + 1)2 ,

which simplifies to

H (z) = 0.3188z2 + 0.6375z + 0.3188

z2 + 0.1017z + 0.1734 .

Step 4 computes the magnitude spectrum by substituting z = exp(jΩ). The resulting plot is shown in Fig. 16.7, where we observe that the magnitude

spectrum satisfies the specified pass-band and stop-band requirements.

Bilinear transformation using M AT L A B The bilinear function is pro-

vided in M A T L A B to transform a CT filter to a DT filter using the bilinear

transformation. The syntax for calling the bilinear function is similar to

that of the impinvar function and is given by

[numz,denumz] = bilinear(nums,denums,fs)

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733 16 IIR filter design

wherenums anddenums specify the coefficients of the numerator and denom-

inator of the analog filter and fs is the sampling rate in samples/s. For

Example 16.4, the M A T L A B code is given by

>> fs = 0.5; % fs = 1/T = k/2 = 0.5

>> nums = [1.1897]; % numerator of the CT filter

>> denums = [1 1.5421 1.1897]; % denominator of CT filter

>> [numz,denumz] = bilinear (nums,denums,fs);

% coefficients of DT filter

which returns the values

numz = [0.3188 0.6376 0.3188];

denumz = [1.0000 0.1017 0.1735],

which are the same as the coefficients obtained in Example 16.4.

Filter design using M AT L A B Several additional functions are provided in

M A T L A B for directly determining the transfer function of the digital filters. The

buttord and butter functions, introduced in Chapter 7, can also be used to

compute IIR filters in the digital domain. The buttord function computes the

order N and cut-off frequency wn of the Butterworth filter, and the butter function computes the coefficients of the numerator and denominator of the

z-transfer function of the Butterworth filter. For lowpass filters, the calling

syntaxes for the buttord and butter functions are given by

buttord function: [N, wn] = buttord(wp, ws, rp, rs);

butter function: [numz, denumz] = butter(N, wn),

where N is the order of the lowest-order digital Butterworth filter that loses no

more than rp dB in the pass band and has at least rs dB of attenuation in the

stop band. The frequencies wp and ws are the pass-band and stop-band edge

frequencies, normalized between zero and unity, where unity corresponds to

π radians/s. Similarly, wn is the normalized cut-off frequency for the Butter-

worth filter. The matrix numz contains the coefficients of the numerator, while

matrix denumz contains the coefficients of the denominator of the transfer

function of the Butterworth filter.

For Example 16.4, the M A T L A B code is given by

>> [N,wn] = buttord(0.25,0.75,20*log10(0.8),20*log10

(0.20));

>> [numz,denumz] = butter(N,wn);

which results in the following coefficients:

numz = [0.3188 0.6376 0.3188];

denumz = [1.0000 0.1017 0.1735],

which are identical to those obtained analytically in Example 16.4.

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734 Part III Discrete-time signals and systems

16.4 Designing highpass, bandpass, and bandstop IIR filters

In the following examples, we design the highpass, bandpass, and bandstop IIR

filters.

Example 16.5

Example 15.5 designed a highpass FIR filter for the following specifications:

(i) pass-band edge frequency Ωp = 0.5π radians/s; (ii) stop-band edge frequency Ωs = 0.125π radians/s;

(iii) pass-band ripple ≤ 0.01 dB;

(iv) stop-band attenuation ≥ 60 dB.

Design an IIR filter with the same specifications.

Solution

Choosing k = 1 (sampling interval T = 2), step 1 transforms the pass-band and stop-band corner frequencies into the CT frequency domain:

pass-band corner frequency ωp = tan(0.5Ωp) = tan(0.25π ) = 1 radian/s; stop-band corner frequency ωs = tan(0.5Ωs) = tan(0.0625π ) = 0.1989 radians/s.

Step 2 designs the analog filter based on the transformed specifications. In

Chapter 7, we presented the design methodology for deriving the transfer

function of the analog highpass filter analytically. Here, we use M A T L A B

to calculate the analog elliptic filter based on the above specifications:

>> wp = 1; ws = 0.1989; Rp = 0.01; Rs = 60 ;

>> [N,wn] = ellipord (wp,ws,Rp,Rs, ’s’);

% Order and cut off frequency

% of the analog elliptic filter

>> [nums,denums]=ellip (N,Rp,Rs,wn,’high’,’s’);

% Tx function of the analog

% elliptic filter

which yields the following transfer function for the analog filter:

H (s) = 0.9988s4 + 0.0542s2 + 0.000373

s4 + 1.872s3 + 1.824s2 + 1.04s + 0.3732 .

Step 3 derives the z-transfer function of the digital filter using the bilinear trans-

formation. This is achieved by using the bilinear function in M A T L A B.

>> [numz,denumz] = bilinear(nums,denums,0.5) % DT Filter

The resulting filter is given by

H (z) = 0.1725z4 − 0.6539z3 + 0.9638z2 − 0.6539z + 0.1725

z4 − 0.6829z3 + 0.7518z2 − 0.138z + 0.0468 .

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735 16 IIR filter design

−60

−40

−20

W −p 0−0.75p −0.5p −0.25p 0.75p p0.5p0.25p

Fig. 16.8. Magnitude response

of the DT highpass filter

designed in Example 16.5.

Figure 16.8 shows the amplitude gain response of the designed filter. We observe

that the pass-band and stop-band specifications are both satisfied.

Example 16.6

Example 15.6 designed a bandpass FIR filter with the following specifications:

(i) pass-band edge frequencies, Ωp1 = 0.375π and Ωp2 = 0.5π radians/s; (ii) stop-band edge frequencies, Ωs1 = 0.25π and Ωs2 = 0.625π radians/s;

(iii) stop-band attenuations, δ s1 > 50 dB and δ s2 > 50 dB.

Design an IIR filter with the same specifications.

Solution

Choosing k = 1 (sampling interval T = 2), step 1 transforms the pass-band and stop-band corner frequencies into the CT frequency domain:

pass-band corner frequency I ωp1 = tan(0.5Ωp1) = tan(0.1875π ) = 0.6682 radians/s; pass-band corner frequency II ωp2 = tan(0.5Ωp2) = tan(0.25π ) = 1 radian/s; stop-band corner frequency I ωs1 = tan(0.5Ωs1) = tan(0.125π ) = 0.4142 radians/s; stop-band corner frequency II ωs2 = tan(0.5Ωs2) = tan(0.3125π ) = 1.4966 radians/s.

Step 2 designs an analog filter for the aforementioned specifications. We can

either use the analytical techniques developed in Chapter 7 or use the M A T L A B

program. In the following, we calculate the analog elliptic filter for the given

specifications using M A T L A B . Since the pass-band ripple is not specified, we

assume that it is given by 0.03 dB. The M A T L A B code is given by

>> wp = [0.6682 1]; ws = [0.4142 1.4966];

>> Rp = 0.03; Rs = 50;

>> [N, wn] = ellipord(wp,ws,Rp,Rs,’s’);

>> [nums,denums] = ellip(N,Rp,Rs,wn,’s’);

which results in an eighth-order elliptic filter with the following transfer

function:

H (s) = 0.001(3.164s8 + 30.27s6 + 57.02s4 + 13.51s2 + 0.6308)

s8 + 0.7555s7 + 3.07s6 + 1.634s5 + 3.229s4 + 1.092s3 + 1.371s2 + 0.2254s + 0.1994 .

Step 3 derives the z-transfer function of the digital filter using the bilinear trans-

formation. This is achieved by using the bilinear function in M A T L A B .

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736 Part III Discrete-time signals and systems

−60

−40

−20

W −p 0−0.75p −0.5p −0.25p 0.75p p0.5p0.25p

Fig. 16.9. Amplitude gain

response of the DT bandpass

filter designed in Example 16.6.

>> [numz,denumz]=bilinear(nums,denums,0.5) % DT Filter

The resulting filter is given by

H (z) = 0.001

(

8.317z8 − 6.94z7 + 4.236z6 − 5.952z5 + 13.52z4 − 5.952z3 + 4.236z2 − 6.94z + 8.317 )

z8 − 1.389z7 + 3.714z6 − 3.356z5 + 4.685z4 − 2.693z3 + 2.397z2 − 0.7107z + 0.4106 .

Figure 16.9 shows the amplitude gain response of the designed filter, which

illustrates that the pass-band and stop-band specifications are both satisfied.

Example 16.7

Example 15.7 designed a bandstop FIR filter with the following specifications:

(i) pass-band edge frequencies, Ωp1 = 0.25π and Ωp2 = 0.625π radians/s; (ii) stop-band edge frequencies, Ωs1 = 0.375π and Ωs2 = 0.5π radians/s;

(iii) stop-band attenuations, δ s1 > 50 db and δ s2 > 50 dB.

Design an IIR filter with the same specifications.

Solution

Choosing k = 1 (sampling interval T = 2), step 1 transforms the pass-band and stop-band corner frequencies into the CT frequency domain:

pass-band corner frequency I ωp1 = tan(0.5Ωp1) = tan(0.125π ) = 0.4142 radians/s; pass-band corner frequency II ωp2 = tan(0.5Ωp2) = tan(0.3125π ) = 1.4966 radians/s; stop-band corner frequency I ωs1 = tan(0.375Ωs1) = tan(0.1875π ) = 0.6682 radians/s; stop-band corner frequency ωs2 = tan(0.5Ωs2) = tan(0.25π ) = 1 radian/s.

Step 2 designs an analog filter for the aforementioned specifications. In the fol-

lowing, we use M A T L A B to derive the analog elliptic filter for the transformed

specifications and an assumed pass-band ripple of 0.03 dB:

>> wp = [0.4142 1.4966]; ws = [0.6682 1];

>> Rp = 0.03; Rs = 50;

>> [N,wn] = ellipord(wp,ws,Rp,Rs,’s’);

>> [nums,denums] = ellip(N,Rp,Rs,wn,’stop’,’s’);

The resulting elliptic filter is of the eighth order and has the following transfer

function:

H (s) = 0.9966s8 + 2.8s6 + 2.854s4 + 1.25s2 + 0.1987

s8 + 2.137s7 + 5.15s6 + 5.926s5 + 6.747s4 + 3.96s3 + 2.3s2 + 0.6377s + 0.1994 .

Step 3 derives the z-transfer function of the digital filter using the bilinear

function.

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737 16 IIR filter design

−60

−40

−20

W −p 0−0.75p −0.5p −0.25p 0.75p p0.5p0.25p

Fig. 16.10. Magnitude response

of the DT bandstop filter

designed in Example 16.7.

>> [numz,denumz]=bilinear(nums,denums,0.5); % DT Filter

The resulting DT filter is given by

H (z) = 0.2887z8 − 0.4484z7 + 1.363z6 − 1.372z5 + 2.149z4 − 1.372z3 + 1.363z2 − 0.4484z + 0.2887

z8 − 1.096z7 + 1.977z6 − 1.519z5 + 1.78z4 − 0.8638z3 + 0.6172z2 − 0.1739z + 0.09751 .

Figure 16.10 shows the magnitude response of the designed bandstop filter. We

observe that both the pass-band and stop-band specifications are satisfied by

the bandstop filter.

16.5 IIR and FIR filters

A classical problem in the design of digital filters is the selection between FIR

and IIR filters since both types of filters can be used to satisfy a given set of

specifications. In this section, we compare IIR and FIR filters with respect to

three criteria: stability, implementation complexity, and delay.

16.5.1 Stability

Stability is a major concern in the design of filters. When designing digital

filters, care must be taken to ensure that the designed filters are absolutely

BIBO stable to prevent infinite outputs. Recall that an LTID system is stable if

its poles lie inside the unit circle in the z-plane. Since the only poles in FIR filters

lie at the origin (z = 0), FIR filters are always BIBO stable. On the other hand, IIR filters have non-trivial poles because of the feedback loops and therefore

may run into stability issues.

Use of finite-precision DSP boards places a severe limitation on the type of

IIR filters that can be used. Even if the designed IIR filter is stable, quantization

of the filter coefficients can adversely affect its stability. To illustrate the effect

of quantization on the stability of the filter, consider the following four filters.

(1) Lowpass filter (arbitrary):

H (z) = 0.001(3.5747z7 − 13.649z6 + 20.9446z5 − 10.7188z4 − 10.7188z3 + 20.9446z2 − 13.649z + 3.5747)

z7 − 5.9664z6 + 15.5383z5 − 22.8594z4 + 20.49z3 − 11.1881z2 + 3.4416z − 0.46 .

(2) Highpass filter (Example 16.5):

H (z) = 0.1725z4 − 0.6539z3 + 0.9638z2 − 0.6539z + 0.1725

z4 − 0.6829z3 + 0.7518z2 − 0.138z + 0.0468 .

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738 Part III Discrete-time signals and systems

Table 16.3. Pole locations for lowpass IIR filter specified as item

(1) in the list of filters in Section 16.5.1 before and after coefficient

quantization

Before quantization After aquantization

0.906248860 + j0.374726030 1.052267965 + j0.282343949 0.906248860 − j0.374726030 1.052267965 − j0.282343949 0.868476456 + j0.325406471 0.884886889 + j0.435649276 0.868476456 − j0.325406471 0.884886889 − j0.435649276 0.816276165 + j0.206545545 0.720252455 + j0.304944386 0.816276165 − j0.206545545 0.720252455 − j0.304944386 0.784371333 0.651185382

(3) Bandpass filter (Example 16.6):

H (z) = 0.001(8.317z8 − 6.94z7 + 4.236z6 − 5.952z5 + 13.52z4 − 5.952z3 + 4.236z2 − 6.94z + 8.317)

z8 − 1.389z7 + 3.714z6 − 3.356z5 + 4.685z4 − 2.693z3 + 2.397z2 − 0.7107z + 0.4106 .

(4) Bandstop filter (Example 16.7):

H (z) = 0.2887z8 − 0.4484z7 + 1.363z6 − 1.372z5 + 2.149z4 − 1.372z3 + 1.363z2 − 0.4484z + 0.2887

z8 − 1.096z7 + 1.977z6 − 1.519z5 + 1.78z4 − 0.8638z3 + 0.6172z2 − 0.1739z + 0.09751 .

The poles and zeros of the four filters are plotted separately in Figs. 16.11(a)–

(d). Since in all cases the poles lie within the unit circle, the four filters are

absolutely BIBO stable when they are implemented with full precision.

Now, let us consider the effect of quantization on the stability of the lowpass

filter. Although most digital systems use binary arithmetic, we will use decimal

arithmetic for simplicity and assume that the coefficients of the lowpass filter

(item (1) above) are implemented up to an accuracy of three decimal places

leading to the following approximated transfer function:

Ĥ (z) = 0.001(4z7 − 14z6 + 21z5 − 11z4 − 11z3 + 21z2 − 14z + 4)

z7 − 5.966z6 + 15.538z5 − 22.859z4 + 20.494z3 − 11.188z2 + 3.442z − 0.46 .

Although the filters H (z) and Ĥ (z) look similar, they are not identical. The location of poles can be found by calculating the roots of the characteristic

equations of H (z) and Ĥ (z), and these are listed in Table 16.3. The pole–zero locations are shown in Fig. 16.12. It is observed that the two poles in H (z), which lie close to (but inside) the unit circle, moved outside the unit circle after

coefficient quantization. Therefore, although Ĥ (z) behaves as a lowpass filter after quantization, the filter is no longer absolutely BIBO stable.

Different implementations of IIR filters can be compared to determine relative

stability by observing how close the poles lie to the unit circle. The highpass

filter, the with pole–zero plot shown in Fig. 16.11(b), has four poles, which are

well inside the unit circle. The pole–zero plot of the bandpass filter is shown

in Fig. 16.11(c). Four of the eight poles in the bandpass filter are close to the

unit circle, which reduces its relative stability. The bandstop filter has eight

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739 16 IIR filter design

−1 −0.5 0 0.5 1

−1

−0.8

−0.6

−0.4

−0.2

0 0

0.2

0.4

0.6

0.8

real part

im ag

in ar

y p

ar t

−1 −0.5 0 0.5 1 real part

−1

−0.8

−0.6

−0.4

−0.2

0.2

0.4

0.6

0.8

im ag

in ar

y p

ar t

−1 −0.5 0 0.5 1 real part

−1

−0.8

−0.6

−0.4

−0.2

0.2

0.4

0.6

0.8

im ag

in ar

y p

ar t

−1 −0.5 0 0.5 1 real part

−1

−0.8

−0.6

−0.4

−0.2

0.2

0.4

0.6

0.8

im ag

in ar

y p

ar t

(a) (b)

Fig. 16.11. Locations of the

poles and zeros for IIR filters.

(a) Lowpass (specified as item 1

in Section 16.5.1); (b) highpass

(Example 16.5); (c) bandpass

(Example 16.6); (d) bandstop

(Example 16.7).

poles, which are plotted in Fig. 16.11(d)). Four of its poles are well inside the

unit circle, while the remaining four are somewhat close to the unit circle. On a

relative scale, the highpass filter provides a better resilience against quantization

among the latter three filters. The bandpass and bandstop filters are sensitive to

stability issues after quantization.

16.5.2 Implementation complexity

In this section, we compare the implementation complexity of the FIR fil-

ters designed in Examples 15.5–15.7 with that of the IIR filters designed in

Examples 16.5–16.7. Table 16.4 provides a list of the number of adders, mul-

tipliers, and unit delay elements required in each case. For IIR filters, we use

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Table 16.4. Implementation complexity of FIR and IIR filters

Note that N corresponds to the order of a DT filter

Number of two-input

adders

Number of scalar

multipliers Unit delay elements

Highpass filter FIR (N = 20) 21 10 20 (Examples 15.5/16.5) IIR (N = 4) 8 9 4

Bandpass filter FIR (N = 46) 47 24 46 (Examples 15.6/16.6) IIR (N = 8) 16 17 8

Bandstop filter FIR (N = 46) 47 24 46 (Examples 15.7/16.7) IIR (N = 8) 16 17 8

−1 −0.5 0 0.5 1 real part

−1

−0.8

−0.6

−0.4

−0.2

0.2

0.4

0.6

0.8

im ag

in ar

y p

ar t

2 0

−1 −0.5 0 0.5 1 real part

−1

−0.8

−0.6

−0.4

−0.2

0.2

0.4

0.6

0.8

im ag

in ar

y p

ar t

(a) (b)

Fig. 16.12. Locations of the

poles and zeros of the lowpass

filter specified as item 1 in

section 16.5.1 (a) Before

quantization; (b) after

quantization of coefficients.

the direct form II realizations, while the IIR filters are implemented using the

linear implementation (see Section 14.6.3).

It is observed in Table 16.4 that the complexity of IIR filters is significantly

lower than that for the corresponding FIR filters. For example, the highpass FIR

filter requires 21 additions, 10 scalar multiplications, and 20 unit delays. On the

other hand, the highpass IIR filter requires only 8 additions, 9 multiplications,

and 7 unit delays. The difference is more conspicuous for the bandpass and

bandstop filters, where the orders of the FIR filters are much larger than the

corresponding orders of the IIR filters.

In summary, for applications such as image and video processing, where

a smaller-order FIR filter can satisfy the design specifications, FIR filters are

generally chosen. In other applications, such as acoustics, a filter with a long

impulse response in the range of 2000 samples is required. In such cases, the

FIR filter provides a large implementation complexity compared with that for

an IIR filter designed with the same specifications. Between these two extremes,

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741 16 IIR filter design

there are a large number of applications where an appropriate filter (FIR or IIR)

is chosen based on implementation cost and robustness.

16.5.3 Delay

The propagation delay between the time an input signal is applied and the time

when the output appears is another important factor in filter selection. Because

of the larger number of implementation elements, the FIR filters generally have

a larger delay than the IIR filters.

16.6 Summary

This chapter presented transformation techniques, namely the impulse invari-

ance and bilinear transformations, used to design IIR filters. These transforma-

tion techniques are based on converting the frequency specifications H (Ω) of IIR filters from the DT frequency Ω domain into the CT frequency specifica-

tions H (ω). Based on the CT frequency specifications, a CT filter with transfer function H (s) is designed, which is then transformed back into the original DT frequency Ω domain to obtain the transfer function H (z) of the required IIR filter. Section 16.2 introduced the impulse invariance transformation used to

design lowpass filters. The impulse invariance method uses a linear expression,

Ω = ωT,

where T is the sampling interval, to convert DT specifications to the CT domain. Because of the sampling process, the impulse invariance method suffers from

aliasing when transforming the analog filter H (s) to the digital filter H (z). A consequence of aliasing is that the order N of the designed filter H (z) is much higher than the optimal design. To prevent aliasing, Section 16.3 presented

the bilinear transformation, which transforms the DT specifications to the CT

frequency domain using the following expression:

ω = k tan(Ω/2) or Ω = 2 tan−1(ω/k).

The transfer function H (s) of the CT filter is then transformed into the z-domain using the following transformation:

s = 1

z − 1 z + 1

in which k is generally set to unity. Section 16.4 extended the design techniques to highpass, bandpass, and bandstop filters.

A comparison of IIR and FIR filters was presented in Section 16.5. We

demonstrated that the order of the FIR filter is generally higher than that for IIR

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742 Part III Discrete-time signals and systems

filters for the same design specifications. Therefore, the implementation cost of

IIR filters is generally lower than for FIR filters. In addition, IIR filters generally

have a lower delay. However, a major limitation in the use of IIR filters is the

stability. Because IIR filters are implemented using feedback loops, they have

non-zero poles. Care should be taken in designing IIR filters by ensuring that

the poles are well inside the unit circle; this achieves good relative stability. FIR

filters have trivial poles (at z = 0) and are always stable. Another approach taken to design IIR filters is referred to as the direct design

method, which derives the filter recursively using a least-squares method. Unlike

the analog prototyping method, the direct design method is not constrained to

the standard lowpass, highpass, bandpass or bandstop configurations. Filters

with an arbitrary, perhaps multiband, frequency response are also possible. In

M A T L A B the yulewalk function designs IIR digital filters by performing a

least-squares fit in the time domain. For more details on FIR filter design using

direct design method, refer to refs. [1] and [2].†

Problems

16.1 Using the impulse invariance transformation and a sampling interval of

T = 0.1 s, convert the following analog transfer functions to their equiv- alent digital transfer functions:

(a) H (s) = s + 2

(s + 4)(s2 + 4s + 3) ;

(b) H (s) = s2 + 9s + 20

(s + 2)(s2 + 4s + 3) ;

(s2 + s + 1)(s2 + 2s + 5) .

16.2 Derive the following z-transform pair used in Example 16.2:

12.7786T e−6.3893 kT sin(6.3894kT )u[k]

Z ←→

12.7786T e−6.3893 T sin(6.3894T )z

z2 − 2ze−6.3893 T cos(6.3894T )z + e−2×6.3893 T .

16.3 (a) Use the impulse invariance method to show that the analog transfer

function given by

H (s) = 2.6702

s3 + 2.7747s2 + 3.8494s + 2.6702

† [1] B. Friedlander and B. Porat, the modified Yule–Walker method of ARMA spectral

estimation, IEEE Transactions on Aerospace Electronic Systems (1984), AES-20(2), 158–173. [2] L. B. Jackson, Digital Filters and Signal Processing, 3rd edn. Kluwer Academic Publishers (1996), Chap. 10, pp. 345–355.

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743 16 IIR filter design

results in the following z-transfer function:

H (z) = 0.4695z2 + 0.1907z

z3 − 0.6106z2 + 0.3398z − 0.0624 as stated in Example 16.3 for the third-order Butterworth filter.

(b) Use the impulse invariance method to show that the analog transfer

function given by

H (s) = 6.2902

s4 + 4.1383s3 + 8.5630s2 + 10.3791s + 6.2902 results in the following z-transfer function:

H (z) = 0.3298z3 + 0.4274z2 + 0.0427z

z4 − 0.4978z3 + 0.3958z2 − 0.1197z + 0.0159 as stated in Example 16.3 for the fourth-order Butterworth filter.

16.4 Using the impulse invariance transformation, design a lowpass IIR But-

terworth filter based on the following specifications:

pass-band edge frequency = 0.64π ; width of transition band = 0.3π ; maximum pass-band ripple <0.002;

maximum stop-band ripple <0.005.

16.5 Repeat Problem 16.4 for a highpass IIR Butterworth filter.

16.6 Figure 9.1 shows a schematic for processing CT signals using DT sys-

tems. The overall system should have the CT frequency characteristics as

follows:

overall CT system is a lowpass filter;

pass-band edge frequency = 3π kradians/s; width of the transition band = 4π kradians/s; minimum stop-band attenuation >50 dB

maximum pass-band attenuation <0.03 dB

sampling rate = 8 ksamples/s,

Design a digital IIR filter that will provide the above characteristics using

the following steps.

(a) Derive the DT specifications from the CT specifications using the

impulse invariance transformation with T = 1/8 × 10−3 s. (b) Design the digital IIR filter using a CT elliptic filter and the bilinear

transformation.

16.7 Repeat Problem 16.1 for the bilinear transformation.

16.8 Design a lowpass IIR Butterworth filter specified in Problem 16.4 using

the bilinear transformation.

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744 Part III Discrete-time signals and systems

16.9 Design a highpass IIR Butterworth filter specified in Problem 16.5 using

the bilinear transformation.

16.10 Using the bilinear transformation, design a highpass IIR filter based on

the following specifications:

pass-band edge frequency = 0.64π ; width of transition band = 0.3π ; maximum pass-band ripple <0.002;

maximum stop-band ripple <0.005.

16.11 Using the bilinear transformation, design a bandpass IIR filter based on

the following specifications.

pass-band edge frequencies = 0.4π and 0.6π ; stop-band edge frequencies = 0.2π and 0.8π ; maximum pass-band ripple <0.02;

maximum stop-band ripple <0.009.

16.12 Using the bilinear transformation, design a bandstop IIR filter based on

the following specifications:

pass-band edge frequencies = 0.3π and 0.7π ; stop-band edge frequencies = 0.4π and 0.6π ; maximum pass-band ripple <0.05;

maximum stop-band ripple <0.05.

16.13 Consider the lowpass filter design, using the bilinear transformation and

analog Butterworth filter in Example 16.4. Repeat the IIR filter design

using (i) Chebyshev Type 1 and (ii) Chebyshev Type 2 CT filters. Plot

the frequency characteristics of the designed DT filter.

16.14 Consider the highpass filter design using the bilinear transformation

and analog elliptical filter in Example 16.5. Repeat the IIR filter design

using (i) Chebyshev Type 1 and (ii) Chebyshev Type 2 CT filters. Plot

the frequency characteristics of the designed DT filter.

16.15 Consider the bandpass filter design using the bilinear transformation

and analog elliptical filter in Example 16.6. Repeat the IIR filter design

using (i) Chebyshev Type 1 and (ii) Chebyshev Type 2 CT filters. Plot

the frequency characteristics of the designed DT filter.

16.16 Consider the bandstop filter design using the bilinear transformation and

analog elliptical filter in Example 16.7. Repeat the IIR filter design using

(i) Butterworth and (ii) Chebyshev Type 2 CT filters. Plot the frequency

characteristics of the designed DT filter.

16.17 Quantize the coefficients of the bandpass filters obtained in Problem

16.15 with a resolution of three decimal points. Are the filters with

quantized coefficients stable?

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745 16 IIR filter design

16.18 Quantize the coefficients of the bandstop filters obtained in Problem

16.16 with a resolution of three decimal points. Are the filter with quan-

tized coefficients stable?

16.19 Repeat Problem 16.18 with a resolution of one decimal point.

16.20 By plotting the poles of the highpass filter obtained in Problem 16.10,

determine if the filter is absolutely stable. Quantize the coefficients of

the filter with a resolution of three decimal points. Are the filter with

quantized coefficients stable?

16.21 By plotting the poles of the bandpass filter obtained in Problem 16.11,

determine if the filter is absolutely stable. Quantize the coefficients of

the filter with three decimal points accuracy. Is the filter with quantized

coefficients stable?

16.22 By plotting the poles of the bandstop filter obtained in Problem 16.12,

determine if the filter is absolutely stable. Quantize the coefficients of

the filter with three decimal points accuracy. Is the filter with quantized

coefficients stable?

16.23 Compare the implementation complexity of the highpass FIR filter

designed in Example 15.5 and the IIR filters designed in Problem 16.14.

16.24 Compare the implementation complexity of the bandpass FIR filter

designed in Example 15.6 and the IIR filters designed in Problem 16.15.

16.25 Compare the implementation complexity of the bandstop FIR filter

designed in Example 15.7 and the IIR filters designed in Problem 16.16.

16.26 Using the M A T L A B , filter design function, confirm the transfer func-

tions derived in Problems 16.10–16.16.

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C H A P T E R

17 Applications of digital signal processing

With the increasing availability of digital computers and specialized digital

hardware, digital signal processing offers a cost-effective alternative to many

traditional analog signal processing applications. The digital approach is par-

ticularly attractive due to its adaptability and immunity to variations in the

operating conditions. Since the operation of digital systems does not depend

upon the exact value of the input signals or the constituent digital components,

digital signal processing allows precise replication where the same operation

can be repeated a large number of times, if required. In contrast, analog signal

processing suffers from deviations caused by degradation in the performance

of the analog components and changes in the operating conditions. Digital

implementations are also adaptable to changes in the specifications of the

system. By modifying the software, different specifications can be implemented

by the same digital hardware. An analog system, on the other hand, has to be

redesigned every time the specifications of the system change.

This chapter reviews elementary applications of digital signal processing

in the field of spectral estimation, audio and musical signal processing, and

image processing. Our aim is to motivate readers to explore the use of digital

signal processing in applications of interest to them. Section 17.1 introduces

spectral estimation, in which the spectral content of a non-stationary signal is

estimated from a limited number of signal realizations. Sections 17.2, 17.3, and

17.4 consider audio signal processing, including spectral estimation, filtering,

and compression of audio signals. As an example of multidimensional signal

processing, we consider digital image processing in Sections 17.5, 17.6, and

17.7. Finally, Section 17.8 concludes the chapter with a summary of important

concepts.

17.1 Spectral estimation

Estimating the frequency content of a signal, commonly referred to as spec-

tral analysis or spectral estimation, is an important step in signal processing

746

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747 17 Applications of digital signal processing

−0.04p −0.02p 0 0.02p 0.04p W0

0.2

0.4

0.6

0.8

−p −0.8p −0.6p −0.4p −0.2p 0 0.2p 0.4p 0.6p 0.8p p 0

0.2

0.4

0.6

0.8

−p −0.8p −0.6p −0.4p −0.2p 0 0.2p 0.4p 0.6p 0.8p p 0

0.02

0.04

0.06

0.08

W −0.2p −0.15p −0.1p −0.05p 0 0.05p 0.1p 0.15p 0.2p

0.02

0.04

0.06

0.08

(a) (b)

Fig. 17.1. DFT used to estimate

the frequency content of

stationary and non-stationary

signals in Example 17.1.

(a) Magnitude sepctrum of

x1[k ]. (b) Enlarged version of

part (a) in the frequency range

−0.05π ≤ Ω ≤ 0.05π . (c) Magnitude spectrum of

x2[k ]. (d) Enlarged version of

part (c) in the frequency range

−0.2π ≤ Ω ≤ 0.2π .

applications. For most signals of interest, the discrete Fourier transform (DFT)

provides a convenient approach for spectral estimation. Example 17.1 highlights

the DFT-based approach for two test signals.

Example 17.1

Using the DTFT, estimate the spectral content of the following DT signals:

(a) x1[k] = cos(0.01πk) + 2 cos(0.015πk);

(b) x2[k] = cos(0.0001πk 2),

from observations made over the interval 0 ≤ k ≤ 1000.

Solution

(a) The magnitude spectrum of x1[k] based on the DFT is plotted over the

frequency range −π ≤ Ω ≤ π in Fig. 17.1(a) with the magnified version

shown in Fig. 17.1(b), where the frequency range −0.05π ≤ Ω ≤ 0.05π

is enhanced. By looking at the peak values in Fig. 17.1(b), it is clear that

the frequencies Ω1 = 0.01π and Ω2 = 0.015π radians/s are the dominant

frequencies in the signal. On a relative scale, the frequency componentΩ2 =

0.015π has a higher strength compared with the frequency component

Ω1 = 0.01π .

(b) The magnitude spectrum of x2[k] based on the DFT over the frequency

range −π ≤ Ω ≤ π is plotted in Fig. 17.1(c), with the magnified version

shown in Fig. 17.1(d), where the frequency range −0.2π ≤ Ω ≤ 0.2π is

enhanced. From the subplots, it seems that all frequencies within the range

−0.2π ≤ Ω ≤ 0.2π are fairly significant in x2[k]. To confirm the validity

of our estimation, let us calculate the instantaneous frequency of the signal.

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748 Part III Discrete-time signals and systems

Note that the phase of x2[k] is given by θ0 = 0.0001πk2. By differentiating the phase θ0 with respect to k, the instantaneous frequency is obtained as

ω0 = 0.0002πk. The instantaneous frequency ω0 is a function of time k, and increases proportionately as k increases. However, this time-varying

nature of the frequency is not obvious from the magnitude spectrum shown

in Fig. 17.1(c). Since the DFT averages the frequency components over all

time k, the DFT provides a misleading result in this case.

Example 17.1 shows that the DFT magnitude spectrum based approach is

convenient for estimating the spectral content of a stationary signal comprising

sinusoidal components with fixed frequencies. However, it may provide mis-

leading results for non-stationary signals, where the instantaneous frequency

changes with time. In other words, it is difficult to visualize the time evolution

of frequency in the DFT magnitude spectrum. The short-time Fourier transform

is defined in Section 17.1.1 to address this limitation of DFT.

17.1.1 Short-time Fourier transform

In order to estimate the time evolution of the frequency components present in

a signal, the short-time Fourier transform (STFT) parses the signal into smaller

segments. The DFT of each segment is calculated separately and plotted as a

function of time k. The STFT is therefore a function of both frequency Ω and

time k. Mathematically, the STFT of a DT signal x[k] is defined as follows:

Xs(Ω, b) = ∞∑

k=−∞

x[k]g∗[k − b]e−jΩk, (17.1)

where the subscript s in Xs(Ω, b) denotes the STFT and b indicates the amount

of shift in the time-localized window g[k] along the time axis. Typical windows

used to calculate the STFT are rectangular, Hanning, Hamming, Blackman, and

Kaiser windows. Compared to the rectangular window, the tapered windows,

such as Hanning and Blackman, reduce the amount of ripple and are generally

preferred.

In most cases, the time shift b is selected such that successive STFTs are taken

over adjacent samples of x[k] and there is some overlap of samples between

successive STFTs. As discussed earlier, the STFT is a function of two variables:

the frequency Ω and the central location of the window. It is typically plotted

as an image plot, known as a spectrogram, with frequency Ω varying along

the y-axis and the time (i.e. the center of the window function) varying along

the x-axis. The intensity values of the image plot show the relative strength of

various frequency components in the original signal.

Example 17.2

Plot the spectrogram of the signal x2[k] = cos(0.0001πk 2) for duration

k = [0, 39 999].

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749 17 Applications of digital signal processing

Solution

In order to calculate the STFT, let us choose the Hanning window function of

length Nw = 901 samples to parse the data sequence of length Ns = 40 000. Further assume that the overlap No between two consecutive windows to be

No = 600 samples. The total number of complete windows is given by

M = ⌊

Ns − No Nw − No

⌋

= 130. (17.2)

The p = 0 window is centered at sample k = 450; the p = 1 window is centered at 450 + (901 − 600) = 751; the p = 2 window is centered at 750 + (901 − 600) = 1052. In general, a window p is centered at

k = Nw − 1

2 + p(Nw − No) = 450 + 301p (17.3)

for 0 ≤ p ≤ 129. To obtain improved resolution in the frequency domain and to

use the FFT algorithm efficiently, we zero-pad each time-windowed signal by

123 zero samples to make the total length of each segment equal 1024, which

is a power of 2.

Note that the DFT of each zero-padded time-windowed signal will have a

total of 1024 coefficients in the frequency domain. As the signal is real, the

DFT coefficients will satisfy the Hermitian symmetry property. In other words,

the amplitude spectrum is even-symmetric and we can ignore the second half

of the spectrum which corresponds to the negative frequencies. So, we choose

the first 513 coefficients out of a total of 1024 DFT coefficients corresponding

to each windowed signal. The spectrogram is therefore a 2D matrix of size 513

× 130 samples. Each of the 130 columns will represent the amplitude spectrum

of the signal at the time instant given by Eq. (17.3). Each row contains the

amplitude of the 513 DFT coefficients. Note that the first coefficient (r = 0)

represents frequency Ω = 0 and the last (r = 512) coefficient represents fre-

quency Ω = π , with the intermediate frequencies given by

Ωr = r

512 × π (17.4)

for 0 ≤ r ≤ 512. The resulting spectrogram is shown in Fig. 17.2, where the

black intensity points represent lower magnitudes and the light intensity points

represent higher magnitudes. Note that the spectrogram is wrapped around the

frequency range [0, π ]. Figure 17.2 illustrates that the frequency of the chirp

signal increases linearly with time.

In Example 17.2, we selected values for the window length and the overlap

period on an ad hoc basis. The choice of the window size is important as it

provides a trade-off between the resolution obtained in the frequency domain

and the localization in the time domain. A larger window allows us to observe

a signal for a longer period of time before we calculate the DFT. As a result, it

provides a higher frequency resolution in the spectrogram. On the other hand, a

shorter time window provides a better localization in time but a poor frequency

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0 5000 10000 15000 20000 25000 30000 35000 40000

0.1p

0.2p

0.3p

0.4p

0.5p

0.6p

0.7p

0.8p

0.9p

Fig. 17.2. Spectrogram of

the chirp signal

x2[k ] = cos(0.0001πk2) from Example 17.2.

resolution. A longer window, therefore, generates a narrow-band spectrogram

while a shorter window generates a wide-band spectrogram. Similarly, the over-

lap chosen between two consecutive windows provides continuity and reduces

sharp transitions in the spectrogram.

17.1.2 Spectrogram computation using M A T L A B

In M A T L A B , the signal processing toolbox includes the function specgram

for calculating the spectrogram of a signal. The spectrogram in Example 17.2

is computed using the following code:

>> k = [0:39999];

>> x2= cos(0.0001*pi*k.*k) ;

>> Fs = 1;

>> Nwind = 901; Nfft = 1024; Noverlap = 600;

>> [spgram, F, T] = specgram(x2, Nfft, Fs, hanning(Nwind),

Noverlap);

>> imagesc([0 length(x2)/Fs], 2*pi*F,

20*log10(abs(spgram) + eps));

>> colormap(gray)

The M A T L A B functionimagesc displays the spectrogram using a color map.

We can set the color map to gray using the last command in the code.

17.1.3 Random signals

The signals that we have studied so far are referred to as deterministic signals.

Such signals can be specified by unique mathematical expressions, allowing us

to calculate them precisely for all time. A second category consists of signals

that cannot be predicted precisely in advance, which are collectively referred

to as random or stochastic signals. Individual values of stochastic signals carry

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751 17 Applications of digital signal processing

little information, and therefore statistical averages such as mean, autocor-

relation, and power spectral density are commonly used to specify stochastic

signals. We start by defining the statistical mean and autocorrelation commonly

used to define a stochastic signal. If x[k], x[k1], x[k2] are discrete random vari-

ables taking on values from the set {xm, −∞ ≤ m ≤ ∞} at times k, k1, and k2, respectively, the mean and autocorrelation functions are defined as follows:

mean E{x[k]} =

∞ ∑

m=−∞

xm P[x[k] = xm]; (17.5)

autocorrelation Rxx [k1, k2] = E{x[k1]x[k2]}

∞ ∑

m=−∞

∞ ∑

n=−∞

xm xn P[x[k1] = xm ; x[k2] = xn].

(17.6)

In Eqs. (17.5) and (17.6), the operator E denotes the expectation and P[x[k] =

xm] is the probability that x[k] takes on the value xm . Likewise, P[x[k1] =

xm ; x[k2] = xn] refers to the joint probability for random signals x[k1] and x[k2]

observed at time instants k1 and k2. Estimating the mean and autocorrelation of

a stochastic signal is difficult in general. In many applications, random signals

satisfy the following two properties.

(1) The mean E{x[k]} is constant and independent of time.

(2) The autocorrelation E{x[k1]x[k2]} depends upon the duration between the

observation instants k1 and k2. In other words, the autocorrelation is inde-

pendent of the observation instants and is only determined by the duration

between the two observations.

Such signals are referred to as wide-sense stationary (WSS) random signals.

Sometimes, these are referred to as weak-sense stationary or second-order sta-

tionary random signals. Mathematically, the aforementioned two properties of

the WSS signals can be expressed as follows:

mean E{x[k]} = µx ; (17.7)

autocorrelation Rxx [k1, k2] = Rxx [k1 − k2] = Rxx [m]. (17.8)

The DTFT of the autocorrelation Rxx [m] of a WSS signal is referred to as the

power spectral density, which is defined as follows:

power spectral density Sxx (Ω) =

∞ ∑

m=−∞

Rxx [m]e −jΩm . (17.9)

Equations (17.8) and (17.9) are widely used to estimate the spectral content

of WSS signals, and the equations require the probability density functions

to estimate the spectral content, which is generally not known in most signal

processing applications. In the following, we present a method, based on the

periodogram, to estimate the spectral content of stochastic signals from a finite

number of observations.

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17.1.4 Periodogram

The periodogram method is similar to the spectrogram method and exploits the

STFT for spectrum estimation using a window function g[k] of length Nw and

centered at k = b. The time-windowed sequence ub[k], centered at k = b, is given by

ub[k] = x [

k + b − ⌊

⌋]

g[k], 0 ≤ k ≤ (Nw − 1) . (17.10)

The DFT of ub[k] is given by

Ub(Ω) =

Nw−1 ∑

k=0

ub[k]e −jΩk . (17.11)

The periodogram method estimates the power spectrum Pxx (Ω) using the fol-

lowing equation:

P̂xx (Ω) = 1

µ2 |Ub(Ω)|

2 , (17.12)

where µ is referred to as the norm of the window function g[k] and is calculated

as follows:

µ =

√

∑

g2[k]. (17.13)

While computing the STFT, different window functions attenuate the original

samples of the signal x[k] by different amounts. Inclusion of a scaling factor

of 1/µ2 in Eq. (17.12) reduces the bias introduced by a particular window

function.

If g[k] is a rectangular window, the estimate of the power spectrum Pxx (Ω)

computed with Eq. (17.12) is called the periodogram. For all other windows,

the estimate is referred to as the modified periodogram.

In its current form, Eq. (17.11) calculates the Nw-point DFT that produces

DTFT values for a set of equally spaced Nw frequency points within the range

Ω = [0, 2π ]. As for the spectrogram, we can zero-pad the time-windowed

sequence and increase the DFT length to obtain a denser plot in the frequency

domain.

17.1.5 Average periodogram

To estimate the power spectrum, Eq. (17.12) uses a single window with duration

of 0 ≤ k ≤ (Nw−1) within the input signal x[k]. Improved results are obtained

if several estimates from different locations of the signal are obtained and

the resulting values are averaged. Starting from duration 0 ≤ k ≤ (Nw−1),

the first iteration computes the periodogram from x[k] within the specified

duration. In the second iteration, the window is moved forward by (Nw – No)

samples such that there is an overlap of No between successive windows. The

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new location of the window is given by (Nw − No − 1) ≤ k ≤ (2Nw − No− 2) for the second iteration, which is used to compute the periodogram for the

second duration. The process is repeated until the entire signal is parsed and

the average value of the periodogram is selected as the estimate of the power

spectrum. This method, based on averaging the values of the power spectrum

obtained from different periodograms, is referred to as the Welch estimate of the

periodogram.

In the signal processing toolbox of M A T L A B , the built-in function psd

estimates the power spectrum of a signal using the periodogram approach. The

following example illustrates the use of the psd function.

Example 17.3

Estimate the power spectral density of the following signal:

x[k] = 3 cos(0.2πk) + 2 cos(0.3πk) + r [k], (17.14)

where r [k] is a white noise with Gaussian distribution with a variance of 4.

Solution

Note that the signal x[k] includes a deterministic component consisting of

the two sinusoids and a random component. The following code generates a

realization of x[k] and estimates the power spectrum:

>> k = [0:6000];

>> x = 3*cos(0.2*pi*k) + 2*cos(0.4*pi*k) +

2*randn(size(k));

>> Fs = 2 ; nwind = length(x);

>> nfft = length(x); noverlap = 0 ;

>> [PxxNoAvg, F] = psd(x, nfft, Fs, rectwin(nwind),

>> noverlap); Fs = 2; nwind=301;

>> nfft = 512; noverlap = floor(4*nwind/5) ;

>> [PxxWelch, F] = psd(x, nfft, Fs,

hanning(nwind),noverlap);

The random component r [k] is generated using the M A T L A B functionrandn.

As the variance of the random component is 4, we multiply randn by the

standard deviation, which equals 2. Figure 17.3 shows the first 201 samples of

an example of x[k]. Over different simulations, the signal x[k] may have slight

variations due to the presence of the random component.

The M A T L A B code computes the power spectrum in two ways. The first

estimate PxxNoAvg represents the power spectrum obtained by calculating

the DFT of the entire signal. Note that there is no averaging in this case. The

second estimate, PxxWelch, represents the power spectrum obtained by the

Welch method, where the signal is parsed into shorter sequences with a Hanning

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0 0.1p 0.2p 0.3p 0.4p 0.5p 0.6p 0.7p 0.8p 0.9p p 0

1 0

lo g

1 0 (P

xx (W

))

0 20 40 60 80 100 120 140 160 180 200 −10

−5

k 0 0.1p 0.2p 0.3p 0.4p 0.5p 0.6p 0.7p 0.8p 0.9p p

−50

−25

1 0

lo g

1 0 (P

xx (W

))

0x[ k]

(a) (b)

(c)

Fig. 17.3. Estimating the power

spectrum of a random signal

using the periodogram

approach. (a) Original random

signal. (b) Power spectrum

obtained from periodogram with

no averaging. (c) Power

spectrum obtained from

periodogram with overlap and

averaging based on the Welch

method. window of size 301. Two consecutive windows have an overlap of 240 samples,

resulting in a total of 94 data windows. Each of these sequences is zero-padded

with 211 zero-valued samples and the DFT is calculated. The averaged power

spectrum is then obtained by averaging all 94 power spectra.

The resulting power spectra are shown in Figs. 17.3(b) and (c). Although both

spectra exhibit peaks atΩ = 0.2π and 0.4π the estimatePxxNoAvg contains a substantial amount of noise. Since the estimate PxxWelch averages the power

spectrum, most of the noise is canceled out. However, averaging also reduces

the magnitudes of peaks at Ω = 0.2π and 0.4π in PxxWelch. In the latter case, the peaks are not as pronounced as the peaks in PxxNoAvg.

17.2 Digital audio

Since the 1980s, digital audio has become a very popular multimedia format

for several applications, including the audio CD, teleconferencing, and digital

movies. With the enormous growth of the World Wide Web (WWW), audio

processing techniques such as filtering, equalization, noise suppression, com-

pression, and synthesis are being used increasingly. In this section, we focus

on three aspects of audio processing: spectrum estimation, audio filtering, and

audio compression. We start by discussing how audio is stored in files and

played back in M A T L A B .

17.2.1 Digital audio fundamentals

Sound is a physical phenomenon induced by vibrations of physical matter, such

as the excitation of a violin string, clapping of hands, and movement of our vocal

tract. The vibrations in the matter are transferred to the surrounding air resulting

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0 5000 10000 15000 20000 25000 −1

−0.5

0.5

kau di o

x[ k]

Fig. 17.4. Waveform of a digital

audio signal stored in the

testaudio1.wav file.

in the propagation of pressure waves. The human auditory system processes

the air waves and uses the information contained in the pressure variations to

extract audio information from the wave. It is possible to process sound waves

directly, as in a microphone, which converts sound to electrical signals that

are amplified and played back using a loudspeaker. The term audio refers to

electronically recorded or reproduced sound, while digital audio is obtained by

the sampling and quantization of an analog audio signal. The waveform of an

audio signal is shown in Fig. 17.4.

An audio signal is described using two properties. The first property is pitch,

which describes the shrillness of sound. Pitch is directly related to the fre-

quency of the audio signal and the two terms are used interchangeably. The

second property is the loudness, which measures the amplitude or intensity

of the audio signal using the decibel (dB) scale. Generally, the audible inten-

sity of an audio signal varies between 0 and 140 dB, where 0 dB represents

the lower threshold of hearing, below which a human auditory system is inca-

pable of hearing any sound. Typical office environments have an ambient audio

level of about 70 dB. Audio above 120 dB is very loud and is injurious to

humans.

Sound generated from physical phenomena contains frequency in the range

0–10 GHz. Since the human auditory system is only intelligible to sound fre-

quencies between 20 Hz and 20 kHz, most audio signals record sound within

this audible range and neglect any higher-frequency components. For example,

the digital audio stored on an audio compact disc is obtained by filtering the

CT audio by a lowpass filter with a cut-off frequency of 20 kHz, and the fil-

tered signal is sampled using a sampling rate of 44.1 ksamples/s. The number

of quantization levels used to produce digital audio depends upon the appli-

cation and may vary from 4096 levels obtained with a 12-bit quantizer, to 65

536 levels with a 16-bit quantizer, to 4 million levels with a 24-bit quantizer.

Higher numbers of quantization levels result in lower distortion and more pre-

cise reproduction of the original sound.

17.2.2 Formats for storing digital audio

Digital audio is available in a wide variety of formats, such as the au, wav, and

mp3 formats. Both au and wav formats store audio in the uncompressed form,

while mp3 compresses audio using Layer 3 of the MPEG-1 audio compression

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standard. In this section, we will focus on the au and wav formats. Typically,

a digital audio file stored in the au format has an .au extension, while digital

audio stored in the wav format has a .wav extension.

M A T L A B provides a number of library functions to read and write audio

files stored in the au and wav formats. For the au format, M A T L A B provides the

auread and auwrite functions to read and write an audio file, respectively.

Likewise, the wavread and wavwrite functions are available to read and

write an audio file in the wav format. The following code reads the audio file

“testaudio1.wav” using the wavread function. There are three output

arguments to the wavread function. The first argument x is an array where

the audio signal is restored. For mono (single-channel) audio signals, x is a

1D vector. For stereo (dual-channel) signals, x is a 2D array corresponding to

the number of signals played by the two speakers. The second argument Fs

represents the sampling rate, while nbit represents the number of bits per

sample.

>> %Reading the input audio file

>> infile = ’f:\ MATLAB\signal\ % audio file >> testaudio1.wav’;

>> [x, Fs, nbit] = wavread(infile); % x = signal

% Fs = sampling rate

% nbit = number of

% bits per sample

The above M A T L A B program will produce a 1D array x with dimension

26 079 × 1. In other words, the audio signal is a mono signal and contains 26 079 samples. The sampling rate is 22.05 ksamples/s and the signal is quantized using

an 8-bit quantizer. The waveform of the audio signal stored in the testaudio1.wav

file is shown in Fig. 17.4. To play the audio signal stored inx, we use thesound

or soundsc function available in M A T L A B as follows:

>> sound(x,Fs);

The soundsc function normalizes the entries of vector x so that the sound is

played as loud as possible without clipping. The mean value is also removed.

After playing the vector x obtained from testaudio1.wav, you should

recognize that the file contains the spoken word “audio.” Relating the word

“audio” to Fig. 17.4, we observe that the waveform has three distinct segments.

The first segment represents the syllable “au,” the second segment represents

the syllable “di,” and the last segment represents “o.” Some silent intervals,

represented by near-zero-amplitude waveforms, are also observed in the plot.

17.2.3 Spectral analysis of speech signals

In Section 17.1, we presented techniques for estimating the spectral content

of a nonstationary signal. Audio signals such as speech, music, and ambient

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0 0.2 0.4 0.6 0.8 1

k (s)

w /2

p (

k H

0 0.2 0.4 0.6 0.8 1

k (s)

w /2

p (

k H

(a) (b)

Fig. 17.5. Spectrograms of the

speech signal recorded in

testaudio1.wav.

(a) Narrow-band spectrogram;

(b) wide-band spectrogram.

sound are examples of non-stationary signals. Therefore, the techniques pre-

sented in Section 17.1 can also be used to estimate the spectral content of audio

signals.

To calculate the spectrogram of the audio signal stored in testaudio1.wav, we

use the following M A T L A B code:

>> %Reading the input audio file

>> infile = ’testaudio1.wav’; % audio file

>> [x, Fs, nbit] = wavread(infile); % x = signal

% Fs = sampling rate

% nbit = number of

% bits per sample

>> nfft = 1024; nwind = 1024; noverlap = 768;

>> [spgram,F,T] = specgram(x, nfft,Fs,hanning(nwind),

noverlap);

>> spgramdB = 20*log10 (abs (spgram) + eps);

>> imagesc([0 length (x)/Fs], 2*pi*F, spgrandB);

>> colormap(gray)

The above code calculates the spectrogram using a window size of 1024, shown

in Fig. 17.5(a). As the window size is a power of 2, we choose to calculate the

DFT without any zero padding. For the audio signal testaudio1.wav, the

sampling rate of the signal is given by 22 050 samples/s. A window size of

1024 samples therefore corresponds to a duration of 1024/22 050 = 0.0461 s. Hence, the time resolution of the spectrogram is limited to 46 ms.

The frequency resolution in the spectrogram plotted in Fig. 17.5(a) is obtained

by dividing the sampling frequency by the total number of samples in the fre-

quency domain, which gives 22 050/1024 = 21.53 Hz. During the computation of the spectrogram, it is possible to trade-off time resolution for the frequency

resolution, and vice versa. To improve the time resolution of the spectrogram

in Fig. 17.5(a), we decrease the window size to 256 with an overlap of 128

samples between two successive windows:

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>> %Reading the input audio file

>> infile = ’testaudio1.wav’; % audio file

>> [x, Fs, nbit] = wavread(infile); % x = signal

% Fs = sampling rate

% nbit = number of

% bits per sample

>> nfft = 256; nwind = 256; noverlap = 128;

>> [spgram,F,T] = specgram(x,nfft, Fs,hanning(nwind),

noverlap);

>> spgramdB = 20*log10 (abs (spgram) + eps);

>> imagesc([0 length (x)/Fs], 2*pi*F, spgrandB);

>> colormap(gray)

The resulting spectrogram is shown in Fig. 17.5(b). Choosing a window size of

256 samples improves the time resolution to 11.6 ms. However, the frequency

resolution is reduced to 22 050/256 = 86.13 Hz. Comparing the two histograms in Fig. 17.5, we observe that the time resolution of Fig. 17.5(b) is better than that

of Fig. 17.5(a). However, the improvement in the time resolution is obtained at

the cost of the frequency resolution. Clearly, Fig. 17.5(b) has a relatively lower

frequency resolution compared with that of Fig. 17.5(a). Therefore, Fig. 17.5(a),

with a better frequency resolution, is considered a narrow-band spectrogram,

whereas Fig. 17.5(b), with a lower frequency resolution, is considered a wide-

band spectrogram.

17.2.4 Power spectrum

Using the techniques discussed in Section 17.1.5, the power spectrum of the

speech signal stored in vector x obtained from the testaudio1.wav file can

be computed using the psd function available in M A T L A B as follows:

>> nwind=512; nfft = 512; noverlap = floor(3*nwind/4) ;

>> [Pxx, F] = psd(x, nfft, Fs, hanning(nwind),noverlap);

>> plot(F,10*log10(Pxx));

The resulting power spectrum is shown in Fig. 17.6, where we observe that

most of the energy of the signal is concentrated in the frequency band 0–2 kHz.

17.2.5 Spectral analysis of music signals

In this section, we analyze the spectral content of the music signal stored in

testaudio2.wav using the spectrogram and periodogram methods. The

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0 2 4 6 8 10 −60

−40

−20

f (kHz)

1 0

lo g

1 0 (P

xx (W

))

Fig. 17.6. Power spectrum of the

speech signal stored in the

testaudio1.wav file.

music signal is read using the following M A T L A B code and the resulting

time-varying waveform of the music signal is plotted in Fig. 17.7(a):

>> %Reading the input audio file

>> infile = ’testaudio2.wav’; % audio file

>> [x, Fs, nbit] = wavread(infile); % Fs = sampling rate,

% nbit = # bits/sample

>> plot(1/Fs*[0:length(x)-1],x);

>> nfft=1024; nwind=1024; noverlap=512;

>> [spgram, F, T] = specgram(x, nfft,Fs,hanning(nwind)

noverlap);

>> imagesc([0 length(x)/Fs], F/1000, 20*log10

(abs(spgram) + eps));

>> colormap(gray)

>> [Pxx, F] = psd(x,nfft,Fs, hanning(nwind),noverlap);

>> plot(F,10*log10(Pxx));

The resulting spectrogram is shown in Fig. 17.7(b), where the horizontal axis

represents time and the vertical axis represents frequency. As the speech signal

is real-valued, the spectrum is plotted for the positive frequencies only. Since

the bright intensity regions represent higher energy, it can be seen that the signal

has most energy at the lower frequencies.

The average periodogram of the music signal is plotted in Fig. 17.7(c). It is

observed that the peak power of about 6.5 dB occurs at 100 Hz and that the

power decreases as the frequency is increased.

17.3 Audio filtering

Frequency-selective filtering emphasizes certain frequency components by

attenuating the remaining frequency components present in a signal. Four types

of digital filters, namely lowpass, highpass, bandpass, and bandstop filters, were

covered in Chapters 14–16. In this section, we process audio signals using these

digital filters.

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0 10 12 14 16 18 20

−1

−0.5

0.5

k (s)

x[ k]

8642

0 5 10 15 20

6.5 0

−15

−30

−45

f (kHz)

1 0

lo g

1 0 (P

xx (W

))

0 5 10 15 20

15f (k

H z)

k (s)

(a) (b)

(c)

Fig. 17.7. Frequency analysis of

the music signal stored in the

testaudio2.wav file.

(a) Time representation;

(b) spectrogram; (c) power

spectrum of the music signal.

Example 17.4

Consider the audio signal stored in the bell.wav file, which was sampled

at a sampling rate of 22 050 samples/s and quantized using an 8-bit quantizer.

The power spectral density, shown in Fig. 17.8(b), illustrates that the signal

has frequency components across the entire 0–11 025 Hz frequency range.

We now process the audio signal with the lowpass, highpass, and bandpass

filters.

Lowpass filtering A lowpass FIR filter with a cut-off frequency of 3 kHz and

order 64 is designed using the fir1M A T L A B library function. The following

M A T L A B code designs the lowpass filter:

>> filtLow = fir1(64,3000/ % Filter: Order = 64

(Fs/2)); % cutoff = 3kHz

>> w = 0:0.001*pi:pi; % discrete frequencies for

% spectrum

>> HLpf = freqz(filtLow,1,w); % transfer function

>> plot(w*Fs/(2*pi),20*log10 % magnitude spectrum

(abs(HLpf) + eps));

By default, the fir1 function uses the Hamming window. Since the fir1

function accepts normalized frequencies, the cut-off frequency is normalized

with half the sampling frequency. The magnitude spectrum of the resulting

lowpass filter is shown in Fig. 17.9(a).

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0 0.4 0.8 1.2 1.6 2

−1

−0.5

0.5

k (s)

x[ k]

0 4 6 8 10 −40

−30

−20

−10

f (kHz)

1 0

lo g

1 0 (P

xx (W

))

(a) (b)

Fig. 17.8. Audio signal stored in

the bell.wav file. (a) Time

representation; (b) power

spectrum.

To derive the output of the lowpass filter when the audio signal stored in

bell.wav is applied at the input of the filter, the following M A T L A B code

is used:

>> xLpf = filter(filtLow,1,x); % Lowpass filtered audio

% signal

To hear the resulting audio signal and plot its power spectrum, we use the

following M A T L A B code:

>> sound(xLpf,Fs); % Play filtered sound

>> nfft=1024; nwind=1024; noverlap=512;

>> [Pxx, F] = psd(xLpf,nfft,Fs, hanning(nwind),noverlap);

>> plot(F,10*log10(Pxx));

Listening to the lowpass filtered sound, we observe that the sound is less shrill

with a lower pitch. This is also apparent from the power spectrum shown in Fig.

17.9(b), where we observe that the frequency components above 3 kHz have

a much lower magnitude than the corresponding frequency components of the

original bell sound.

0 2 4 6 8 10 −80

−60

−40

−20

f (kHz)

2 0

lo g

1 0 |H

(W )|

0 4 6 8 10 −100

−80

−60

−40

−20

f (kHz)

1 0

lo g

1 0 (P

xx (W

))

(a) (b)

Fig. 17.9. Lowpass filtering of

the audio signal stored in the

bell.wav file. (a) Frequency

characteristics of a 64-tap FIR

lowpass filter designed using a

Hamming window with a cut-off

frequency of 3000 Hz.

(b) Power spectrum of the

filtered signal.

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0 2 8 10 −80

−60

−40

−20

f (kHz)

2 0

lo g

1 0 |H

(W )|

0 2 8 10 −100

−80

−60

−40

−20

1 0

lo g

1 0 (P

xx (W

))

64 64 f (kHz)

(a) (b)

Fig. 17.10. Bandpass filtering of

the audio signal stored in the

bell.wav file. (a) Frequency

characteristics of a 64-tap FIR

bandpass filter designed using a

Hamming window with cut-off

frequencies of 2000 and

5000 Hz. (b) Power spectrum of

the filtered signal.

Bandpass filtering As was the case for the lowpass filter, we design the band-

pass filter using the fir1 command. The M A T L A B code is given below.

>> fBp = fir1(64,[2000 %Filter: order = 64

5000]/(Fs/2)); % cutoff = [2 5]kHz

>> w = 0:0.001*pi:pi; % discrete frequencies for

% spectrum

>> HBpf = freqz(fBp,1,w); % transfer function

>> plot(w*Fs/(2*pi),20*log

10(abs(HBpf) + eps)); % magnitude spectrum

The magnitude spectrum of the bandpass filter is plotted in Fig. 17.10(a), which

filters the bell sound using the following M A T L A B code:

>> xBpf = filter(fBp,1,x); % Bandpass filtered audio

% signal

>> sound(xBpf,Fs); % Play filtered sound

>> nfft=1024; nwind=1024; noverlap=512;

>> [Pxx, F] = psd(xBpf,nfft, Fs,hanning(nwind),noverlap);

>> plot(F,10*log10(Pxx + eps));

The power spectrum of the resulting bandpass signal is plotted in Fig. 17.10(b).

We see that the frequency components within the pass band of [2000 5000] Hz

are retained in the filtered signal. The remaining frequency components are

attenuated by the bandpass filter.

Highpass filtering The highpass filter with a cut-off frequency of 4 kHz is

designed using the following M A T L A B code:

>> fHp = fir1(64,4000/(Fs/2),’high’);

% Filter: order = 64

% cutoff = 4kHz

>> w = 0:0.001*pi:pi; % discrete frequencies for

% spectrum

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0 2 8 10 −80

−60

−40

−20

f (kHz)

2 0

lo g

1 0 |H

(W )|

0 4 6 8 10 −100

−80

−60

−40

−20

f (kHz)

1 0

lo g

1 0 (P

xx (W

))

64 2

(a) (b)

Fig. 17.11. Highpass filtering of

the audio signal stored in the

bell.wav file. (a) Frequency

characteristics of a 64-tap FIR

highpass filter, with cut-off

frequency of 4000 Hz, designed

using a Hamming window.

(b) Power spectrum of the

filtered signal.

>> HHpf = freqz(fHp,1,w); % transfer function

>> plot(w*Fs/(2*pi),20*log10 % magnitude spectrum

(abs(HHpf) + eps));

The magnitude spectrum of the highpass filter is plotted in Fig. 17.11(a), which

filters the bell sound using the following code:

>> xHpf = filter(fHp,1,x); % Highpass filtered audio

% signal

>> sound(xHpf,Fs) % play the sound

>> nfft=1024; nwind=1024; noverlap=512;

>> [Pxx, F] = psd(xHpf,nfft, Fs,hanning(nwind),noverlap);

>> plot(F,10*log10(Pxx + eps));

The power spectrum of the highpass filtered signal is shown in Fig. 17.11(b),

where we observe that the frequency components below 4 kHz are strongly

attenuated. The higher frequency components are left unattenuated. The obser-

vation is confirmed on playing the filtered sound, which sounds shriller, with a

higher pitch than the original bell sound.

Example 17.4 demonstrates the effects of lowpass, bandpass, and highpass

filtering on an audio signal. The following example uses a bandstop filter to

eliminate noise from a noisy signal.

Example 17.5

Consider the audio signal stored in the testaudio3.wav file with the time-

domain representation shown in Fig. 17.12(a). The audio signal is sampled at

a sampling rate of 22 050 samples/s. Using the average periodogram method

discussed in Section 17.1.5, the power spectral density of the audio signal

is estimated and plotted in Fig. 17.12(b). From the power spectral density

plot, we observe that there is a sharp peak at 8 kHz, which is identified as

noise corrupting the audio signal. The noise can be heard if we play the audio

signal.

To suppress the noise, we use a bandstop filter of order 128 with a stop band

that ranges from 7800–8200 Hz. The order of the bandstop filter is chosen

arbitrarily in this example. In more sophisticated applications, the order is

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k(s)

x[ k]

0 0.5 1 1.5 2

−1

−0.5

0.5

0 4 6 8 10 −40

−30

−20

−10

f (kHz)

1 0

lo g

1 0 (P

xx (W

)) noise impulse

(a) (b)

Fig. 17.12. Noise-corrupted

signal stored in the

testaudio3.wav file.

(a) Time representation;

(b) power spectrum.

computed from the amount of attenuation required within the stop band. Using

M A T L A B , the transfer function of the bandpass filter is computed as follows:

>> wc =[7800 8200]/11025; % Normalized cutoff

% frequency

>> fBs = fir1(128,wc,’stop’); % order-128 filter, 129 tap

>> w = 0:0.001*pi:pi; % discrete frequencies

% for spectrum

>> HBs = freqz(fBs,1,w); % transfer function

>> plot(w*Fs/(2*pi),20*log10 (abs(HBs)));

% magnitude spectrum

The magnitude spectrum of the resulting bandstop filter is plotted in Fig.

17.13(a), which shows strong attenuation at 8 kHz. The gain at the remain-

ing frequencies is close to unity. The noisy signal is filtered with the bandstop

filter and the power spectral density of the filtered signal is calculated using the

following M A T L A B code:

>> xBsf = filter(fBs,1,x); % Bandstop filtered audio

% signal

>> nfft=1024; nwind=1024; noverlap=512;

>> [Pxx, F] = psd (xBsf,nfft,Fs,hanning (nwind),noverlap);

>> plot(F,10*log10(Pxx));

The power spectral density of the filtered output is shown in Fig. 17.13(b),

which shows a strong attenuation in the noise impulse present at 8 kHz. On

playing the filtered signal, we observe that the effects of the noise have been

f (kHz)

2 0

lo g

1 0 |H

(W )|

0 4 6 8 10 −30

−20

−10

0 2 4 6 8 10

−30

−20

−10

f (kHz)

1 0

lo g

1 0 (P

xx (W

))

(a) (b)

Fig. 17.13. Bandstop filtering to

eliminate noise from the noise

corrupted signal shown in Fig.

17.12. (a) Frequency

characteristics of a 129-tap FIR

bandstop filter, with cut-off

frequencies of [7800 8200] Hz,

designed using a Hamming

windor. (b) Power spectrum of

the filtered signal.

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765 17 Applications of digital signal processing

f (kHz)

2 0

lo g

1 0 |H

(W )|

0 2 4 6 8 10 −80

−60

−40

−20

f (kHz)

1 0

lo g

1 0 (P

xx (W

))

0 2 8 10 −60

−40

−20

(a) (b)

Fig. 17.14. Bandstop filtering to

eliminate noise from the

noise-corrupted signal shown in

Fig. 17.12. (a) Frequency

characteristics of a 201-tap FIR

bandstop filter, with cut-off

frequencies of [7800 8200] Hz,

designed using a Hamming

window. (b) Power spectrum of

the filtered signal.

reduced, but not completely eliminated. Therefore, we increase the order of the

bandstop FIR filter to 200. Using the above code with the order set to 200, we

compute the impulse response of the 201-tap bandstop FIR filter. The magnitude

spectrum of the filter is plotted in Fig. 17.14(a). The power spectral density of

the filtered signal obtained from the 201-tap bandstop filter is shown in Fig.

17.14(b). On playing the filtered signal, we observe that the noise component

has been successfully suppressed. However, the suppression of noise is at the

cost of eliminating certain frequency components which neighbor the frequency

of the impulse noise.

17.4 Digital audio compression

Audio data in the raw format requires a large number of bits for repre-

sentation. For example, the CD-quality stereo audio requires a data rate of

176.4 kbytes/s for transmission or storage. This data rate is not supported by

many networks, including the internet, hence real-time audio applications can-

not be supported if the audio data are transmitted in the raw format. Similarly,

storing at a data rate of 176.4 kbytes/s requires a large storage capacity, even to

save a five-minute session. Compressing audio is therefore imperative for real-

time audio transmission or for storing an audio session of meaningful length.

Audio compression is defined as the process through which digital audio can

be represented by a lower number of bits. Most compression techniques can be

classified into two categories, namely lossy compression and lossless compres-

sion. While lossless techniques are ideal as they allow perfect reconstruction

of audio, they limit the amount of compression that can be achieved. Lossy

techniques exploit the psychoacoustic characteristics of the human auditory

system and achieve higher compression by eliminating audio components that

are not audible to humans. In this section, we present the basic principles of

audio compression. Example 17.6 emphasizes the need for audio compression.

Example 17.6

(a) A stereo (dual-channel) audio signal is to be transmitted through a 56 kbps

network in real time. If the sampling rate of the digital audio signal is

22.05 ksamples/s, what is the maximum average number of bits that can be

used to represent an audio sample?

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(b) If the quantizer uses 8 bits/sample for each channel, what is the maximum

allowable sampling rate such that the audio signal can be transmitted over

a 56 kbps network?

through a 56 kbps channel if the sampling rate is given by 22.05 ksamples/s

and the quantizer uses 8 bits/sample.

Solution

(a) Assuming that the quantizer uses n bits to represent each sample,

number of bits produced per second = n bits/sample × 22 050 samples/s × 2 channels = 44 100n bps

Equating this with the transmission rate of 56 kbps, we obtain

n = 56 000/44 100 = 1.27 bits/sample.

(b) Assuming that the sampling rate is given by fs samples/s,

number of bits produced per second = 8 bits/sample × fs samples/s × 2 channels = 16 fs bits/s.

Equating this with the transmission rate of 56 kbps, we obtain

fs = 56 000/16 = 3500 samples/s.

produced per second:

number of bits produced per second = 8 bits/sample × 22 050 samples/s × 2 channels = 352 800 samples/s.

The compression ratio is therefore given by

compression ratio = number of bits per second in the raw data

number of bits per second in the compressed data

= 352 800

5600 = 6.3.

Example 17.6 demonstrates that digital audio can be transmitted over a low-

capacity transmission channel in real time using three different approaches.

The first approach reduces the number of bits used to represent each sample.

This approach is not useful as it reduces the number of quantization levels

such that considerable distortion is introduced into the transmitted audio. The

second approach uses a low sampling rate, which is not practical as the sampling

rate is dependent on the maximum frequency present in the audio signal. The

maximum frequency of the audio signal can be reduced by lowpass filtering,

but this will again introduce distortion. The third approach compresses the raw

audio data. Compression of digital audio is achieved by eliminating redundancy

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767 17 Applications of digital signal processing

present in a signal. There are primarily three types of redundancies present in

an audio signal that may be exploited.

Statistical redundancy In most audio signals, samples with lower magnitudes

have a higher probability of occurrence than samples with higher magnitude. In

such cases, an entropy coding scheme, such as the Huffman code, can be used

to allocate fewer bits to frequently occurring values and a higher number of

bits to the other values. This reduces the bit rate for representing audio signals

when compared with a coding scheme with an equal number of bits allocated

per sample.

Temporal redundancy Neighboring audio samples typically have a strong

correlation between themselves such that the value of a sample can be predicted

with fairly high accuracy from the last few sample values. Predictive coding

schemes exploit this temporal redundancy by subtracting the predicted value

from the actual sample value. The resulting difference signal is then compressed

using an entropy based coding scheme, such as the dictionary or Huffman codes.

Psychoacoustics redundancy There are many idiosyncrasies in the human

auditory system. For example, the sensitivity of the human auditory system is

maximum for frequencies within the 2000–4000 Hz band and the sensitivity

decreases above or below this band. In addition, a strong frequency component

masks the neighboring weaker frequency components. The unequal frequency

sensitivity and masking properties are exploited to compress the audio.

In the following section, we present a simplified audio compression technique,

known as the differential pulse-code modulation (DPCM) technique. To achieve

compression, the DPCM reduces the temporal redundancy present in an audio

signal.

17.4.1 Differential pulse-code modulation

Most audio signals encoded with pulse-code modulation (PCM) exhibit a strong

correlation between neighboring samples. This is especially true if the signal is

sampled above the Nyquist sampling rate. Figure 17.15 plots 30 samples of an

audio signal stored in the chord.wav file. We observe that the neighboring

samples are correlated such that their values are fairly close to each other. In

DPCM, an audio sample s[k] is predicted from the past samples. An M-order

predictor calculates the predicted value of an audio sample at time instant k

using the following equation:

ŝ[k] = M

∑

m=1 αms[k − m], (17.15)

where s[k – m] is the value of the audio sample at time instant k − m and αm are the predictor coefficients. The DPCM encoder quantizes the prediction

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700 705 710 715 720 725 730 −0.2

−0.1

0.1

0.2

0.3

x[k] Fig. 17.15. Selected samples

(sample 700 to 730) of the

audio signal stored in the

chord.wav file. The

neighboring samples exhibit a

strong correlation between

themselves.

error as follows:

e[k] = s[k] − M

∑

m=1 αms[k − m], (17.16)

which is followed by a lossless entropy coding scheme. The DPCM decoder

takes the inverse of the above steps in the reverse order. Since the actual sample

values s[k − m] are not accessible at the decoder, the decoder uses the recon- structed values. In order to use the same prediction model at the encoder and

decoder, Eq. (17.16) is modified as follows:

e[k] = s[k] − M

∑

m=1 αms

′[k − m], (17.17)

where s ′[k − m] is the reconstructed value of the audio sample s[k − m]. The values of the predictor coefficients αm are usually estimated based on a max-

imum likelihood (ML) estimator. Alternatively, a universal prediction model

may be used where the predictor coefficients are kept constant for different audio

signals. Examples of the universal prediction models include the following:

first-order prediction model ŝ[k] = 0.97s ′[k − 1]; (17.18) second-order prediction model ŝ[k] = 1.8s ′[k − 1] − 0.84s ′[k − 2]; (17.19) third-order prediction model ŝ[k] = 1.2s ′[k − 1] + 0.5s ′[k − 2]

− 0.78s ′[k − 3]. (17.20)

Σ quantization entropy coding

prediction Σ

−

audio signal

s[k]

e[k]

s' [k]

][ˆ ke

[k]ŝ

[k]

compressed audio *

Σentropy decoding dequantization reconstructed audio[k]ê

[k]ŝ

][

compressed audio *

prediction

(a)

(b)

Fig. 17.16. Schematic of

differential pulse-code

modulator used for lossy

compression. (a) DPCM encoder

used to compress a signal;

(b) DPCM decoder used to

reconstruct a signal. The

difference e[k ] between the

original input signal s [k ] and its

predicted value ŝ [k ]is quantized

and transmitted to the receiver.

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769 17 Applications of digital signal processing

The block diagrams of DPCM encoding and decoding systems are shown in

Fig. 17.16. Example 17.7 illustrates various steps of the DPCM coding.

Example 17.7

Assume that the first four samples of a digital audio sequence are given by [70,

75, 80, 82]. The audio samples are encoded using DPCM with the first-order

predictor defined in Eq. (17.18). The error samples obtained by subtracting the

predicted sample values from the actual audio sample values are divided by a

quantization factor of 2 and then rounded to the nearest integer. Determine the

values of the reconstructed signal.

Solution

In DPCM, the first sample value is encoded independent of other samples in

the sequence. In this example, we assume that the first audio sample, at k = 0, with a value of 70 is encoded without any quantization error. In other words,

e[0] = ê[0] = 0 and the reconstructed sample value s ′[0] = 70. At k = 1, the predicted sample, the associated error, and the quantized error

are given by

predicted value ŝ[1] = 0.97 × 70 = 67.9; error e[1] = 75 − 67.9 = 7.1; quantized error ê[1] = round(7.1/2) = 4.

The reconstructed value of the sample at k = 1 is therefore given by

s ′[1] = 0.97 × 70 + 4 × 2 = 75.9.

At k = 2, the predicted sample, the associated error, and the quantized error are given by

predicted value ŝ[2] = 0.97 × 75.9 = 73.623; error e[2] = 80 − 73.623 = 6.377; quantized error ê[2] = round(6.377/2) = 3.

The reconstructed value of the sample at k = 2 is therefore given by

s ′[2] = 0.97 × 75.9 + 3 × 2 = 79.623.

At k = 3, the predicted sample, the associated error, and the quantized error are given by

predicted value ŝ[3] = 0.97 × 79.623 = 77.2343; error e[3] = 82 − 77.2343 = 4.7657; quantized error ê[3] = round(4.7657/2) = 2.

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Table 17.1. Various steps of DPCM coding for Example 17.7

Time index, k

0 1 2 3

Original signal, s[k] 70 75 80 82

Error signal, e[k] 0 75 − 67.9 = 7.1 80 − 3.6 = 6.4 82 − 7.2 = 4.8 Quantized error signal, ê[k] 0 7.1/2 = 4 6.4/2 = 3 4.8/2 = 2 Reconstructed error 0 4 ×2 = 8 3 ×2 = 6 2 ×2 = 4 Reconstructed signal, s ′[k] 70 67.9 + 8 = 75.9 73.6 + 6 = 79.6 77.2 + 4 = 81.2 Reconstruction error 0 −0.9 0.4 0.8 Predicted signal for next sample 70 × 0.97 = 67.9 75.9 × 0.97 = 73.6 79.6 × 0.97 = 77.2 81.2 × 0.97 = 78.8

The reconstructed value of the sample at k = 2 is therefore given by

s ′[3] = 77.2343 + 2 × 2 = 81.2343.

The values of the audio samples reconstructed from DPCM are given by

[70, 75.9, 79.623, 81.2343],

which implies that the following distortion is introduced by DPCM:

[0, −0.9, 0.377, 0.7657].

The above steps are summarized in Table 17.1. The third row contains the

quantized values of the error signal, which is compressed with a lossless scheme

and transmitted to the receiver.

17.4.2 Audio compression standards

The DPCM compression scheme, as described in Section 17.4.1, is a primitive

audio compression method that provides a low compression ratio. Several more

efficient compression techniques have been developed since the 1980s. In order

to achieve compatibility between the compressed bit streams, several audio

compression standards have been developed by the International Organiza-

tion for Standardization (ISO) and the International Telecommunication Union

(ITU). These audio compression standards can be broadly classified into two

categories: the low-bit-rate audio coders for telephony, such as G.711, G.722,

and G.729 developed by the ITU, and the general-purpose high-fidelity audio

coders, such as the moving pictures expert group (MPEG) audio standards,

developed by the ISO and included in MPEG-1, MPEG-2, and MPEG-4.

The ISO standards are generic audio compression standards designed for

general-purpose audio. These standards provide a trade-off between compres-

sion ratio and quality. For example, the MPEG-1 audio algorithm has three lay-

ers. Layer 1 is the simplest algorithm and provides moderate compression. Layer

2 has moderate complexity and provides a higher compression than Layer 1.

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771 17 Applications of digital signal processing

Layer 3 has the highest complexity and provides the best performance. Note

that the MPEG-1 Layer 3 standard is also referred to as the MP3 standard.

In addition to the ITU G.7xx and ISO MPEG standards, a few other standards

have been developed. For example, Dolby Laboratories have developed multi-

channel high-fidelity audio coding standards such as AC-2 and AC-3 coders.

The AC-3 standard has been adopted in the standard and high-definition digital

television standard in North America. Readers are referred to more advanced

texts for details on audio compression standards.

17.5 Digital images

Digital images have become a part of our daily lives. In this section, we present

a brief overview of digital images and the techniques used to represent them.

17.5.1 Image fundamentals

A still monochrome image is defined in terms of its intensity or brightness i

as a function of the spatial coordinates (x , y). A still image is, therefore, a 2D

function i(x , y). For analog images, coordinates (x , y) have a continuous value.

A discrete image i[m, n] is obtained by sampling the intensity i(x , y) along a

rectangular grid M = [m�x, n�y] with resolutions of �x along the horizontal axis and �y along the vertical axis. Each discrete point [m�x, n�y] along the

rectangular grid is referred to as a picture element, or pixel. A digital image

i[m, n] is an extension of the discrete image, where the intensity i is quantized

by a uniform quantizer. The number of quantization levels varies from one

application to another and depends upon the precision required. Most digital

images are quantized using an 8-bit quantizer, leading to 128 quantization levels.

Medical images require higher precision and are quantized using a 12- or 16-bit

quantizer. Color images are further extensions of discrete images, where the

intensities of the three primary colors are measured at each pixel. Color images

are therefore represented in terms of three components r [m, n], g[m, n], and

b[m, n], where intensities are denoted by r [m, n] for red, g[m, n] for green,

and b[m, n] for blue.

As an example of still images, the back cover of this book illustrates a 450 × 366 pixel test image, referred to as “train,” using three different quantization

levels. The first figure shows the train image in the black and white (BW) format,

where a single bit is used to represent each pixel. Bit 0 represents the lowest

intensity (black), while bit 1 represents the highest intensity (white). The total

number of bits used to represent the BW image is given by 1 bit/pixel × (450 × 366) pixels = 164 700 bits. To provide more details, the second figure uses 8-bit quantization for each pixel, leading to a total number of 8 bit/pixel × (450 × 366) pixels = 1 317 600 bits. The third figure shows the train image in the color format, where each pixel is represented in terms of the intensities of the three

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primary colors. The color representation of the train image requires 8 bit/color × 3 color/pixel × (450 × 366) pixels = 3 952 800 bits.

A final extension of discrete images is obtained by measuring the color

intensities r [m, n], g[m, n], and b[m, n] at discrete time k. Exploiting the

persistence of vision and showing continuously recorded images at a uniform

rate provides the impression of a video. A digital video is therefore defined

in terms of the three color components r [m, n, k], g[m, n, k], and b[m, n, k].

In this section, we limit ourselves to 8-bit, monochrome, still images i[m, n].

However, the techniques are generalizable to color images and videos.

17.5.2 Sampling of coordinates

Chapter 9 defined the Nyquist rate as the minimum sampling rate that can be

used to sample a time-varying CT signal without introducing any distortion

during reconstruction. For a baseband signal, the Nyquist rate is twice the

maximum frequency present in the signal. For analog images, the principle

can be extended to the spatial coordinates (x , y) in two dimensions to obtain

a discrete image. The minimum sampling rates are given by the Nyquist rates

and are computed from the maximum frequencies in the two directions.

17.5.3 Image formats

Like digital audio, images are available in a wide variety of formats, including

pgm, ppm, gif, jpg, and tiff. In each format, a digital image is stored as a 1D

stream of numbers. The difference in the format lies in the manner in which the

image data is compressed before storage. The portable graymap (PGM) format

is used for storage of gray-level images, where raw data is stored without

compression in the ASCII or binary representations. A few bytes of header

information included before the image data describe the format of the file, the

representation (ASCII or binary) used and the number of rows and columns

in the image. The portable pixmap (PPM) format is an extension of the PGM

format for color images, where the intensities of the three primary colors are

stored for each pixel.

The graphical interface (GIF) format uses a compression algorithm to reduce

the size of the data file. It is limited to 8-bit (256) color images and hence is

suitable for images with a few distinctive colors. It supports interlacing and

simple animation, and it can also support grayscale images using a gray palette.

The joint photograph expert group (JPEG) format uses transform-based com-

pression and provides the user with the capability of setting the desired com-

pression ratio.

The tagged image file (TIFF) format supports different types of images,

including monochrome, grayscale, 8-bit and 24-bit RGB images that are tagged.

The images can be stored with or without compression.

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773 17 Applications of digital signal processing

M A T L A B provides two library functions imread and imwrite, respec-

tively, to read and write images. These functions can read and write the image

files in several different formats. The following code shows the syntax for call-

ing these functions:

>> x = imread(’rini.jpg’); % x is a 2-D “uint” type array

>> size(x); % displays the size of the image

>> imshow(x); % displays the image

>> xd = double(x); % xd is the image array

% with double precision

>> xmax = max(max(xd))

>> x-bright=uint8(xd*2); % increases brightness of image

>> imwrite(x-bright,’rini-bright. jpg’,’jpg’,

’Quality’,80) ;

The above code loads the Rini test image from the rini.jpg file and displays

the image in Fig. 17.17(a) using the imshow function. The imread function

used in the code returns an array stored as unsigned integers with 8-bit precision.

To carry out any arithmetic operation on the image, we need to convert the data to

other data types. The instruction double changes the data type from unsigned

integer to double. The instruction max determines the maximum gray level

present in the image, and for the Rini image the value of xmax is given by 124.

As xmax has a low value, the image has low brightness, as observed in Fig.

17.17(a). A possible way to improve the brightness of the image is to increase

the intensity level of the whole image linearly. In an 8-bit image, the maximum

gray level is 255. Therefore, we scale up the gray values by a factor of 2, which

is achieved by multiplying the intensity by a factor of 2. This is followed by

the conversion of the gray values to the uint8 type. The brightened image

represented by the matrix x-bright is shown in Fig. 17.17(b). The last line

of the M A T L A B code stores the brightened image in the JPEG format with

filename rini-bright.jpg using the imwrite function. Note that the

JPEG format compresses the gray image based on the specified quality factor,

which is a number between 0 and 100. A high value for the quality factor implies

higher quality with low compression, while a low value of the quality factor

implies lower quality with high compression. Using a quality factor of 80, the

processed rini image is compressed to a file size of 27 kbytes. Compared with

the original rini.jpg file, which has a size of 180 kbytes, this implies a

compression ratio of 6.66.

17.5.4 Spectral analysis of images

Like real audio signals, natural images are non-stationary signals. The fre-

quency content of the images is estimated by extending the 1D spectral analysis

techniques, presented in Section 17.1, to two dimensions. Here, we discuss the

average periodogram approach to calculate the power spectrum of a 2D image.

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(a) (b)

Fig. 17.17. Rini test image loaded

from the rini.jpg file.

(a) Original and (b) brightened

versions.

Step 1 Parse the input image into smaller 2D segments by applying a 2D

window g[m, n]. Depending upon the application, the parsed segments may or

may not have overlapping pixels.

Step 2 Compute the 2D DFT I (Ωm ,Ωn) of each image segment i(m, n), which

is used to estimate the power spectrum based on the following expression:

P̂I (Ω) = 1

µ2 |I (Ωm,Ωn)|2 , (17.21)

where µ is the norm of the 2D window function defined as follows:

µ = √

∑

g2[m, n]. (17.22)

Step 3 The average power spectrum is obtained by averaging the waveforms

obtained in step 2.

We illustrate the steps involved in computing the power spectrum with the

following example.

Example 17.8

Consider the synthetic image, referred to as the sinusoidal grating, defined by

the following equation:

i(x, y) = 127 cos[2π (4x + 2y)]. (17.23)

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Discretizing the analog image with a sampling rate of 20 samples/spatial unit

in each direction, the DT image is given by

i[m, n] = 127 cos[2π (4m + 2n)/20] (17.24)

for 0 ≤ m, n ≤ 255. Compute the power spectrum of the DT image using the

average periodogram approach.

Solution

We plot the DT image modeled in Eq. (17.24) using the following M A T L A B

code:

>> m = [0:1:255]; % x-coordinates

>> n = [0:1:255]; % y-coordinates

>> [mgrid, ngrid] =

meshgrid(m,n); % determine the 2D meshgrid

>> I = 127*cos(2*pi*(4*mgrid + 2*ngrid)/20);

% pixel intensities

>> imagesc(I); % sketch image

>> axis image;

>> colormap (gray);

The resulting image is shown in Fig. 17.18(a). The power spectrum is calculated

using the 2D Bartlett window of size (64 × 64) pixels, an overlap of (48 × 48)

pixels between adjacent windows, and a (256 × 256)-point DFT for each parsed

subimage. The M A T L A B code used to compute the power spectrum is given

>> m = [0:1:255]; % x-coordinates

>> n = [0:1:255]; % y-coordinates

>> [mgrid, ngrid] % determine the

= meshgrid(m,n); % 2D meshgrid

>> I = 127*cos(2*pi*(4*mgrid + 2*ngrid)/20);

% pixel intensities

% 2D Bartlett window

>> x = bartlett(64);

>> for i = 1:64

zx(i,:) = x’ ;

zy(:,i) = x ;

>> end

>> bartlett2D = zx .* zy;

>> mesh(bartlett2D) % displaying 2D

% Bartlett window

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0.5

0.5p 0.2p 0.4p 0.6p 0.8p

0 p p

(a) (b)

Fig. 17.18. (a) Synthetic

sinusoidal grating. (b) Power

spectrum of the synthetic

sinusoidal grating.

% calculate power spectrum

>> P = zeros(256,256);

>> for (i = 1:64:255)

for (j = 1:64:255)

Isub = I(i:i+63,j:j+63). *bartlett2D;

P = P + fft2(Isub,256,256);

end

% mesh plot with x and y-axis scaled by pi

>> mesh([1:128]*2/256,[1:128]*2/256,

abs(P(1:128,1:128)/max(max(P))). ˆ2);

Figure 17.18(b) illustrates a sharp peak at the horizontal frequencyΩx = 0.4π and at the vertical frequency Ωy = 0.2π . This observation is consistent with the mathematical model, Eq. (17.23), used to construct the synthetic image.

Unlike the earlier power spectrum plots, we use a linear scale along the z-axis

in Fig. 17.18(b).

The above M A T L A B code is modified to construct the power spectrum of a

real test image, referred to as the Lena image. The test image has dimensions

of 512 × 512 pixels and is illustrated in Fig. 17.19(a) along with its power spectrum in Fig. 17.19(b). In computing the power spectrum, a 2D Bartlett

window of dimension 128 × 128 with an overlap of 96 × 96 pixels, and a (256 × 256)-point DFT is used. The dB scale is used along the z-axis to plot the

power spectrum. Real images typically include most frequencies and hence the

power spectrum in Fig. 17.19(b) exhibits an almost uniform distribution over

all frequencies in the horizontal and vertical directions.

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200

−200

0.5p 0.5p p p

(a) (b)

Fig. 17.19. (a) Original

(512 × 512) pixel Lena image. (b) Power spectrum of the Lena

image.

17.6 Image filtering

Real images consist of a combination of smooth regions and active regions

with edges. In smooth regions, the intensity values of the pixels do not change

significantly. Therefore, the smooth regions represent lower-frequency com-

ponents in the 2D frequency space. On the other hand, the intensity values in

the active regions change significantly over edges. The active regions represent

higher-frequency components. Extracting the low- and high-frequency compo-

nents from a real image has important applications in image processing. In this

section, we introduce frequency-selective filtering in two dimensions.

The mathematical model for filtering a 2D image g[m, n] by a filter with an

impulse response h[m, n] is given by

y[m, n] = g[m, n] ∗ h[m, n] = ∞

∑

q=−∞

∞ ∑

r=−∞

g[m − q, n − r ]h[q, r ], (17.25)

where y[m, n] is the output response of the filter and ∗ denotes the convolution

operation. Alternatively, the filtering can be performed in the frequency domain

using the following equation:

Y (Ωx ,Ωy) = G(Ωx ,Ωy)H (Ωx ,Ωy), (17.26)

where G(Ωx ,Ωy) is the Fourier transform of the input image, H (Ωx ,Ωy) is the

2D transfer function of the filter, and Y (Ωx ,Ωy) is the Fourier transform of the

resulting output. Like 1D filters, 2D filters can be broadly classified into four

categories: lowpass, bandpass, highpass, and bandstop filters. Some examples

of these filters are given in the following.

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0.5

−0.5p −0.5p

0.5p

−p

(a) (b)

Fig. 17.20. (a) Ayantika image

corrupted with high-frequency

noise. The noise appears as

vertical and horizontal lines in

the image. (b) Power spectrum

of the Ayantika image.

17.6.1 Lowpass filtering

Lowpass filtering is widely used in many image processing applications. Some

applications include reducing high-frequency noise that is corrupting an image,

band-limiting the frequency component of an image prior to decimation, and

smoothing the rough edges of an image. In Example 17.9 we provide an example

of lowpass filtering in the spatial domain.

Example 17.9

Figure 17.20(a) shows a noise-corrupted image, referred to as Ayantika. Show

that:

(a) the image has high-frequency noise by plotting the power spectrum;

(b) the lowpass filter with the following impulse response:

h[m, n] = 1







1 2 3 2 1

2 3 4 3 2

3 4 5 4 3

2 3 4 3 2

1 2 3 2 1







(17.27)

eliminates the high-frequency noise from the image.

Solution

The M A T L A B code used to plot the power spectrum is given by

>> I = imread(’ayantika.tif’);

>> I = double(I);

>> I = I - mean(mean(I));

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% 2D Bartlett window

>> x = bartlett(32);

>> for i = 1:32

zx(i,:) = x’;

zy(:,i) = x;

>> end

>> bartlett2D = zx .* zy;

>> n = 0;

% calculate power spectrum

>> P = zeros(256,256);

>> for (i = 1:16:320)

for (j = 1:16:288)

Isub = I(i:i+31,j:j+31).*bartlett2D;

P = P + fftshift(fft2(Isub,256,256));

n = n + 1;

end

>> end

>> Pabs = (abs(P)/n).ˆ2;

>> mesh([-128:127]*2/256,[-128:127]*2/256,Pabs/

max(max(Pabs)));

The resulting power spectrum is shown in Fig. 17.20(b), where we see peaks at

frequencies [Ωx , Ωy] given by [0, 0], [0, ±0.5π ], and [±0.5π , 0]. The peak at [0, 0] corresponds to the dc gain, whereas the remaining peaks are because of

the additive noise that has corrupted the image. We now attempt to eliminate

the noise with a lowpass filter.

Figure 17.21(a) shows the magnitude spectrum of the filter with the impulse

response specified in Eq. (17.27). We use the following M A T L A B code to plot

the magnitude spectrum:

>> h = 1/64*[1 2 3 2 1; 2 3 4 3 2; 3 4 5 4 3; 2 3 4 3 2; 1

2 3 2 1];

>> H = fftshift(fft2(h,256,256));

% magnitude spectrum with 256-pt fft

% 2D mesh plot with frequency axis normalized to pi

>> mesh([-128: 127]*2*1/256, [-128:127]*2*1/256, abs(H));

Since the filter provides a higher gain at the lower frequencies and lower gain

at higher frequencies, it is clear that Fig. 17.21(a) corresponds to a lowpass

filter. Note that the gain at frequencies [0, ±0.5π ] and [±0.5π , 0] is zero, therefore the lowpass filter would eliminate the additive noise. The filter2

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0 0

0−0.5p −0.5p 0.5p

(a) (b)

Fig. 17.21. (a) Magnitude

spectrum of lowpass filter used

in Example 17.9. (b) Output of

the lowpass filter.

function is used to compute the output of the lowpass filter using the following

code:

>> Y = filter2(h,I);

>> imagesc(Y);

>> axis image; colormap (gray);

The resulting output is plotted in Fig. 17.21(b). It is observed that the horizontal

and vertical strips have been suppressed by the lowpass filter. However, the low-

pass filter also suppresses some high-frequency components other than noise.

Therefore, the quality of the filtered image degrades marginally, as observed

at the edges. The image in Fig. 17.20(a) has crisp edges, whereas the edges in

Fig. 17.21(b) are somewhat blurred.

17.6.2 Highpass filtering

Highpass filtering is used to detect the edges or suppress the low-frequency

noise in an image. At times, highpass filters are also used to sharpen the edges

of an image. Example 17.10 illustrates one application of highpass filtering.

Example 17.10

Consider the pepper image shown in Fig. 17.22(a). Show that the filter with the

impulse response

h[m, n] = 1





−1 −1 −1 −1 8 −1 −1 −1 −1



 (17.28)

extracts the edges of the image.

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(a)

(c)

(b)

0 0

−p

−0.5p 0.5p

Fig. 17.22. (a) Original 512 × 512 pixels peppers image.

(b) Magnitude response of the

highpass filter with impulse

response shown in Eq. (17.28).

Solution

The following M A T L A B code is used to plot the magnitude spectrum:

>> h = 1/9*[-1 -1 -1; -1 8 -1; -1 -1 -1];

% magnitude spectrum with 256-point fft

>> H = fftshift(fft2(h,256,256));

% 2D mesh plot with frequency axis normalized to pi

>> mesh([-128:127]*2*1/256, [-128:127]*2*1/256, abs(H));

The magnitude frequency response of the filter is shown in Fig. 17.22(b). Since

the gain of the filter is almost zero at low frequencies and unity at higher

frequencies, Eq. (17.28) models a highpass filter. The output of the highpass

filter is obtained using the following code:

>> I = imread(’peppers.tif’);

>> Y = filter2(h,I);

>> imagesc(Y);

>> axis image

Figure 17.22(c) shows the output of the highpass filter. From the image plot, it is

clear that the highpass filter extracts the edges, eliminating the smooth regions

(low-frequency components) of the image.

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17.7 Image compression

Raw data from digital images requires large disk space for storage. Image

compression reduces the amount of data needed to represent an image. As in

audio compression, image compression techniques are grouped into lossless and

lossy categories. With lossless compression, exact reconstruction of the original

image is possible. However, the amount of compression that can be achieved

with lossless compression is limited. Lossy compression introduces controlled

distortion to increase the compression ratio. The redundancies exploited during

image compression are classified into the following categories.

Statistical redundancy The values of pixels in natural images have a non-

uniform probability distribution of occurrences such that some values occur

more frequently than others. Some compression can be achieved by allocat-

ing fewer bits to represent pixels that occur more frequently and more bits to

represent pixels that occur less frequently.

Spatial redundancy In real images, the value of a pixel is highly correlated to

its neighboring pixels. Image compression exploits such spatial redundancy to

compress the image.

Psychovisual redundancy The human visual system (HVS) is less sensitive

to certain features within an image. For example, slight variations in the pixel

intensities within a uniform region cannot be perceived by the HVS. Image

compression exploits such psychovisual redundancy to remove features from

the image whose presence or absence is inconceivable to the HVS.

17.7.1 Predictive coding

Predictive coding exploits spatial redundancy to compress an image. Instead of

encoding the original pixels, predictive-coding schemes calculate the difference

between the actual pixel values and the estimated pixel values predicted from the

neighboring pixels. The resulting difference or error image is instead encoded.

Since the difference image has lower correlation than the original image, more

compression is achieved by encoding the difference image. Predictive coding

may use a universal model or a localized model derived from the reference

image. Examples of universal predictive models are listed below:

first-order prediction î[m, n] = i[m, n − 1]; (17.29) î[m, n] = i[m − 1, n]; (17.30)

second-order prediction î[m, n] = 0.48i[m, n − 1] + 0.48i[m − 1, n] (17.31)

third-order prediction î[m, n] = 0.33i[m, n − 1] + 0.33i[m − 1, n] + 0.33i[m − 1, n − 1]. (17.32)

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783 17 Applications of digital signal processing

Predictive compression techniques can be considered as an extension of DPCM

in two dimensions. Example 17.11 illustrates the use of the third-order predictive

model in compressing still images.

Example 17.11

Consider the Sanjukta image shown in Fig. 17.23(a). The first 4 × 4 pixels of the image are given by

i[m, n] =







156 157 154 149

156 159 159 155

153 158 160 159

149 154 157 156







. (17.33)

Using the predictors in Eqs. (17.30) and (17.31) for the first row and column,

respectively, and the predictor in Eq. (17.32) to predict the remaining values,

calculate the error in the reconstructed image. In your calculations, assume that

the quantizer divides the difference image by a quantization factor Q = 3 and rounds to the nearest integer before quantization.

Solution

Using zero boundary conditions, the predicted sample value, the prediction

error, the quantized error, and the reconstructed sample value at m = 0, n = 0 are given by

predicted value î[0, 0] = 0; error e[0, 0] = i[0, 0] − ŝ[0, 0] = 156; quantized error ê[0, 0] = round(156/3) = 52. reconstructed value i ′[0, 0] = î[0, 0] + 3 × ê[0, 0] = 0 + 3 × 52 = 156.

For spatial location m = 0, n = 1, the predicted sample value, the prediction error, the quantized error, and the reconstructed sample value are given by

predicted value î[0, 1] = i ′[0, 0] = 156; error e[0, 1] = i[0, 1] − î[0, 1] = 157 − 156 = 1; quantized error ê[0, 1] = round(1/3) = 0; reconstructed value i ′[0, 1] = î[0, 1] + 3 × ê[0, 1] = 156.

For spatial location m = 0, n = 2, the predicted sample, the prediction error, the quantized error, and the reconstructed sample value are given by

predicted value î[0, 2] = i ′[0, 1] = 156; error e[0, 2] = i[0, 2] − î[0, 2] = 154 − 156 = −2; quantized error ê[0, 2] = round(−2/3) = −1; reconstructed value i ′[0, 2] = î[0, 2] + 3 × ê[0, 2] = 156 − 3 = 153.

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For spatial location m = 0, n = 3, the predicted sample value, the prediction error, the quantized error, and the reconstructed sample value are given by

predicted value î[0, 3] = i ′[0, 2] = 153; error e[0, 3] = i[0, 3] − î[0, 3] = 149 − 153 = −4; quantized error ê[0, 3] = round(−4/3) = −1; reconstructed value i ′[0, 3] = î[0, 3] + 3 × ê[0, 3] = 153 − 3 = 150.

For spatial location m = 1, n = 0, the predicted sample value, the prediction error, the quantized error, and the reconstructed sample value are given by

predicted value î[1, 0] = i ′[0, 0] = 156; error e[1, 0] = i[1, 0] − î[1, 0] = 156 − 156 = 0; quantized error ê[1, 0] = round(0/3) = 0; reconstructed value i ′[1, 0] = î[1, 0] + 3 × ê[1, 0] = 156 + 0 = 156.

For spatial location m = 1, n = 1, the predicted sample value, the prediction error, the quantized error, and the reconstructed sample value are given by

predicted value î[1, 1] = 0.33(i ′[1, 0] + i ′[0, 1] + i ′[0, 0]) = 0.33 × 468 = 154.44;

error e[1, 1] = i[1, 1] − î[1, 1] = 159 − 154.44 = 4.56; quantized error ê[1, 1] = round(4.56/3) = 2; reconstructed value i ′[1, 1] = î[1, 1] + 3 × ê[1, 1] = 154.44 + 6 = 160.44.

For spatial location m = 1, n = 2, the predicted sample value, the prediction error, the quantized error, and the reconstructed sample value are given by

predicted value î[1, 2] = 0.33(i ′[1, 1] + i ′[0, 2] + i ′[0, 1]) = 0.33 × 469.44 = 154.92;

error e[1, 2] = i[1, 2] − î[1, 2] = 159 − 154.92 = 4.08; quantized error ê[1, 2] = round(4.08/3) = 1; reconstructed value i ′[1, 2] = î[1, 2] + 3 × ê[1, 2] = 154.92 + 3 = 157.92.

For spatial location m = 1, n = 3, the predicted sample value, the prediction error, the quantized error, and the reconstructed sample value are given by

predicted value î[1, 3] = 0.33(i ′[1, 2] + i ′[0, 3] + i ′[0, 2]) = 0.33 × 460.92 = 152.10;

error e[1, 3] = i[1, 3] − î[1, 3] = 155 − 152.10 = 2.90; quantized error ê[1, 3] = round(2.90/3) = 1; reconstructed value i ′[1, 3] = î[1, 3] + 3 × ê[1, 3] = 152.10 + 3 = 155.10.

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Similarly, the pixel values at other locations can be calculated using the above

procedure. The computed values are as follows:

i ′[m, n] =







156 156 153 150

156 160.4 157.9 155.1

153 157.9 160.2 159.2

150 155.1 156.2 156.9







Subtracting the aforementioned values from the original values given in Eq.

(17.33) gives the following values for the error image:

ê[m, n] =







0 1 1 −1 0 −1.4 1.1 −0.1 0 0.1 −0.2 −0.2

−1 −1.1 0.8 −0.9







In image compression, the mean square error (MSE) is typically used to measure

the quantitative quality of a compressed image i ′[m, n]. The MSE is defined as

follows

MSE = 1

M N

M−1 ∑

m=0

N−1 ∑

m=0 [i[m, n] − i ′[m, n]],

where i[m, n] is the pixel intensity of the original image having (M × N ) dimensions. For Example 17.11, the MSE is given by 0.6206.

In DPCM, the first pixel is referred to as the reference pixel and is typically

encoded directly with e[0, 0] = 0. The remaining pixels are encoded using the error image, which is typically divided by a quantization factor Q before

encoding. To achieve quantization, the entire dynamic range of the error image

is divided into 2B intervals and each interval is represented by B bits. Typi-

cally, B is kept small to achieve a large compression ratio. Figure 17.23 shows

two reconstructed Sanjukta test images processed at two different compres-

sion ratios. Figure 17.23(b) is compressed with a quantization factor Q = 5 and B = 4. Similarly, Fig. 17.23(c) is compressed with a quantization factor Q = 16 and B = 2. Higher compression introduces more distortion in Fig. 17.23(c), which is illustrated by the lower subjective quality of Fig. 17.23(c)

when compared with that of Fig. 17.23(b). The superior quality of Fig. 17.23(b)

can also be quantified by computing the MSE. Figure 17.23(b) has a reconstruc-

tion MSE of 6, while Fig. 17.23(c) has a MSE of 44. If required, the quantized

error values can be further encoded using a variable-length code or an entropy

code to achieve more compression.

17.7.2 Image compression standards

DPCM provides moderate compression. Several techniques, such as transform

coding, arithmetic coding, and object-based techniques, have been developed

to achieve performances superior to DPCM. In addition, several image com-

pression standards have been developed by the International Organization for

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(a) (b)

(c)

(e)

(d)

Fig. 17.23. Subjective quality of two DPCM encoded images. (a) Sanjukta image. (b) Reconstructed image

after DPCM compression with a quantization factor Q of 5 and a 4-bit quantizer. (c) Same as (b) except the

quality factor Q is set to 16 and a 2-bit quantizer is used. (d) Difference between the original image and

reconstructed image shown in (b). (e) Difference between the original image and the reconstructed image

shown in (c). The MSE associated with image (b) is 6, while the MSE associated with image (c) is 44.

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Standardization (ISO) and the International Telecommunication Union (ITU)

to ensure compatibility between different compressed bit streams. A popular

ISO image compression standard is referred to as the JPEG standard, selected

as an acronym for the Joint Photographic Experts Group, the ISO subcommittee

responsible for developing the standard.

The JPEG standard algorithm encodes both gray level and color images.

In this standard, an image is decorrelated using the discrete cosine transform

(DCT). The DCT coefficients are quantized and the quantized coefficients are

encoded using a combination of run length and Huffman coding. The size of

the compressed bit stream is varied by changing the quality factor Q, which

has a value between 1 and 100. The highest quality representation is obtained

using a quality factor of 100, and the lowest quality representation is obtained

using quality factor of 1. A high quality factor ensures superior perceived qual-

ity, but the compression is limited. Conversely, a low quality factor increases

compression, but at the expense of quality.

The image processing toolbox in M A T L A B includes a simplified version of

the JPEG encoder and decoder, which allows images to be encoded at different

quality factors Q. If x is a 2D array containing the gray values of a test image,

the following command:

>> imwrite(x,’test-70.jpg’,’jpg’,’Quality’, 70);

creates the JPEG compressed image “test-70.jpg” with a quality factor of 70.

The following example illustrates the compression performance of the JPEG

encoder and decoder.

Example 17.12

Consider the 8-bit Sanjukta image shown in Fig. 17.23(a). Using the imwrite

command, generate different compressed JPEG images with quality factors

100, 50, 25, 10, and 5. Determine the compression ratio in each case and plot

the reconstructed images.

Solution

The following M A T L A B code creates compressed images with different quality

factors:

>> x = imread(’sanjukta-gray.tif’);

>> imwrite(x,’sanjukta-100.jpg’,’jpg’,’Quality’, 100) ;

>> imwrite(x,’sanjukta-50.jpg’,’jpg’,’Quality’, 50) ;

>> imwrite(x,’sanjukta-25.jpg’,’jpg’,’Quality’, 25) ;

>> imwrite(x,’sanjukta-10.jpg’,’jpg’,’Quality’, 10) ;

>> imwrite(x,’sanjukta-5.jpg’,’jpg’,’Quality’, 5) ;

The raw image has 126 672 pixels, with each pixel represented using 8 bits.

Therefore, the uncompressed image size is 126 672 bytes or 126.7 kbytes. The

sizes of the compressed files determined from the compressed files and their

respective compression ratio are provided in Table 17.2. Table 17.2 illustrates

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(a) (b)

(e) (f)

Fig. 17.24. Subjective quality of JPEG compressed images using different quality factors. (a) Original

image; (b) quality factor 100; (c) quality factor 50; (d) quality factor 25; (e) quality factor 10; (f) quality

factor 5.

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Table 17.2. Comparison of JPEG compression performance for sanjukta gray image

Quality factor, Q Name of the compressed file Size of the file (kbytes) Compression ratio MSE

100 sanjukta-100 41.7 3 0.05

50 sanjukta-50 11.1 11 12.34

25 sanjukta-25 4.8 26 18.98

10 sanjukta-10 2.9 43 36.69

5 sanjukta-5 2.3 54 76.05

that decreasing the quality factor increases the compression ratio at the cost of

the reconstruction quality, apparent from the increase in MSE.

To provide a subjective comparison, the reconstructed images are shown in

Fig. 17.24. We observe that the perceived quality of the reconstructed images

degrades with the decrease in the quality factor. In other words, there is a

trade-off between quality and size of the compressed file.

17.8 Summary

This chapter presented applications of digital signal processing in audio and

image processing. Digital signals, including audio, images, and video, are ran-

dom in nature. Section 17.2 presented an overview of spectral analysis methods

for random signals based on the short-time Fourier transform, spectrogram, and

periodogram. Section 17.3 covered fundamentals of audio signals, their storage

format, and spectral analysis of audio signals. Filtering of audio signals was

covered in Section 17.3, and principles of audio compression were presented

in Section 17.4.

Section 17.5 extended digital signal processing to 2D signals. In particular,

we introduced digital images, their storage format, and the spectral analysis

of image signals. Section 17.6 covers 2D filtering, including the application

of lowpass filters to eliminate high-frequency noise and highpass filters for

edge detection. In each case, we presented examples of image filtering through

M A T L A B . Section 17.7 introduced principles of image compression including

the 2D differential pulse-code modulation (DPCM) and the Joint Photographic

Expert Group (JPEG) standard. Using M A T L A B , we compared the perfor-

mance of JPEG at different compression ratios.

Problems

17.1 Consider the following deterministic signal:

x1[k] = 2 sin(0.2πk) + 3 cos(0.5πk).

Using a DFT magnitude spectrum, estimate the spectral content of

x[k] for the following cases: (a) a 20-point DFT and a sample size

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790 Part III Discrete-time signals and systems

of 0 ≤ k ≤ 19; (b) a 32-point DFT and a sample size of 0 ≤ k ≤ 31; (c)

a 64-point DFT and a sample size of 0 ≤ k ≤ 31; (d) a 128-point DFT

and a sample size of 0 ≤ k ≤ 31; and (e) a 128-point DFT and a sample

size of 0 ≤ k ≤ 63. Comment on the leakage effect in each case.

17.2 Calculate and plot the amplitude spectra of the following DT signals:

(i) x1[k] = cos(0.25πk), 0 ≤ k ≤ 2000;

(ii) x2[k] = cos(2.5 × 10 −4πk2, 0 ≤ k ≤ 2000;

(iii) x3[k] = cos(2.5 × 10 −7πk3), 0 ≤ k ≤ 11000.

Comment on the spectral content of the signals.

17.3 Calculate and plot the spectrograms of the three signals considered in

Problem 17.2. Compare the results with those obtained in Problem 17.2.

17.4 Using M A T L A B , estimate the power spectral density of the following

signal:

x[k] = 2 cos(0.4πk + θ1) + 4 cos(0.8πk + θ2),

where θ1 and θ2 are independent random variables with uniform distri-

bution between [0, π ]. Use a sample realization of x[k] with 10 000

samples, the Bartlett window with length 1024, an overlap of 600 sam-

ples, and the average Welch approach.

17.5 Determine the frequency content of the audio signal “chord.wav”,

provided in the accompanying CD using (i) a spectrogram and (ii) an

average periodogram.

17.6 Consider the “testaudio4.wav” file provided in the accompanying

CD. Load the audio signal using the wavread function available in

M A T L A B .

(a) What is the sampling rate used to discretize the signal? What is the

total number of samples stored in the file?

(b) How many bits are used to represent each sample?

(d) Estimate the power spectrum of the signal

17.7 Repeat Problem 17.6 for “testaudio3.wav” provided in the accom-

panying CD.

17.8 Repeat Problem 17.6 for “bell.wav” provided in the accompanying

CD.

17.9 Repeat Problem 17.6 for “test44k.wav” provided in the accompa-

nying CD.

17.10 Repeat Example 17.7 for the following audio samples:

x1[k] = [66, 67, 68, 69] and x2[k] = [66, 72, 61, 56].

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791 17 Applications of digital signal processing

Show that the reconstruction error is greater for the second case, where

the neighboring audio samples are less correlated.

17.11 Consider the “girl.jpg” file provided in the accompanying CD. Read

the image using the imread function available in M A T L A B .

(a) What are the dimensions of the image stored in the “girl.jpg”

file?

(b) What are the maximum and minimum values of the intensity of the

pixels stored in the file?

M A T L A B .

(d) Calculate and plot the 2D power spectrum of the image to illustrate

the dominant spatial frequency components of the image.

17.12 Consider the 2D filter defined by the following impulse response:

h[m, n] = 1







1 1 1 1







(a) Show that h[m, n] is a lowpass filter by sketching its magnitude

spectrum using the mesh plot.

(b) Assume that the image stored in “girl.jpg” is applied at the

input of the filter h[m, n]. Determine and sketch the output image.

this with the result of Problem 17.11 (d), highlight how the high-

frequency components have been attenuated in the filtered image.

17.13 Repeat Problem 17.12 for the 2D filter with the following impulse

response:

h[m, n] = 1

3.2764







0 0 0.0221 0 0

0 0.1563 0.3907 0.1563 0

0.0221 0.3907 1 0.3907 0.0221

0 0.1563 0.3907 0.1563 0

0 0 0.0221 0 0







17.14 Consider the 2D filter defined by the following impulse response:

h[m, n] = 1





−1 −1 −1 −1 8 −1 −1 −1 −1



 .

(a) Show that h[m, n] is a highpass filter by sketching its magnitude

spectrum using the mesh plot.

(b) Assume that the image stored in “girl.jpg” is applied at the

input of the filter h[m, n]. Determine and sketch the output image.

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792 Part III Discrete-time signals and systems

Show that the highpass filtering leads to the detection of edges in

the image.

this with the result of Problem 17.11 (d), highlight how the low-

frequency components have been attenuated in the filtered image.

17.15 Repeat Problem 17.14 for the 2D filter with the following impulse

response:

h[m, n] = 1

6.21







0 0 −0.0442 0 0 0 −0.3126 −0.7815 −0.3126 0

−0.0442 −0.7815 4.5532 −0.7815 −0.0442 0 −0.3126 −0.7815 −0.3126 0 0 0 −0.0442 0 0







17.16 Repeat Example 17.11 for the following selections of (4 × 4) pixels:

i1[m, n] =







156 157 158 159

150 151 151 150

153 155 154 156

155 154 157 156







and

i2[m, n] =







156 177 148 189

160 171 181 150

123 125 174 196

175 164 147 156







Show that the reconstruction error is greater for the second case, where

the neighboring pixels are less correlated.

17.17 Compress the image stored in the file “lena.tif” in the accompanying

CD using the JPEG standard with quality factors set to 80, 60, 40, 20,

and 10. Determine the compression ratio for different quality factors

and show that the subjective quality deteriorates as the quality factor is

decreased. Compute the mean square error for the compressed images.

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Appendix A Mathematical preliminaries

A.1 Trigonometric identities

e±jt = cos t ± j sin t

cos t = 1

2 [ejt + e−jt ]

sin t = 1

2j [ejt − e−jt ]

cos (

t ± π

)

= ∓ sin t

sin (

t ± π

)

= ± cos t

sin 2t = 2 sin t cos t

cos2 t + sin2 t = 1

cos2 t − sin2 t = cos 2t

cos2 t = 1

2 (1 + cos 2t)

sin2 t = 1

2 (1 − cos 2t)

cos3 t = 1

4 (3 cos t + cos 3t)

sin3 t = 1

4 (3 sin t − sin 3t)

cos(t ± θ ) = cos t cos θ ∓ sin t sin θ

sin(t ± θ ) = sin t cos θ ± cos t sin θ

tan (t ± θ ) = tan t ± tan θ

1 ∓ tan t tan θ

sin t sin θ = 1

2 [cos(t − θ ) − cos(t + θ )]

cos t cos θ = 1

2 [cos(t + θ ) + cos(t − θ )]

793

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794 Appendix A

sin t cos θ = 1

2 [sin(t + θ ) + sin(t − θ )]

a cos t + b sin t = C cos(t + θ ), C = √

a2 + b2, θ = tan−1(−b/a)

a cos(mt) + b sin(mt) = √

a2 + b2 cos(mt − θ ), θ = tan−1 b

a cos(mt) + b sin(mt) = √

a2 + b2 sin(mt + φ), φ = tan−1 a

A.2 Power series

ln(1 + t) = t − t2

2 +

3 −

4 + · · ·

et = 1 + t + t2

2 ! +

3 ! +

4 ! + · · ·

sin t = t − t3

3 ! +

5 ! −

7 ! + · · ·

cos t = 1 − t2

2 ! +

4 ! −

6 ! + · · ·

tan t = t + t3

3 +

2t5

15 +

17t7

315 + · · ·

sin−1 t = t + 1

3 +

1.3

2.4

5 + · · ·

A.3 Series summation

Arithmetic series n

∑

n=1 [a + (n − 1)d] =

2 [2a + (N − 1)d]

n ∑

n=1 n = 1 + 2 + · · · + N =

N (N + 1) 2

Geometric series N

∑

n=0 arn =

a(1 − r N+1) 1 − r

N−1 ∑

n=0 exp

[

j 2πkn

]

= {

0 1 ≤ k ≤ (N − 1) N k = 0,

∞ ∑

n=0

rn = 1

1 − r , |r | < 1

∞ ∑

n=0

nrn = r

(1 − r )2 , |r | < 1

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795 A Mathematical preliminaries

The geometric progression (GP) series sum of the form

S = N

∑

n=0 arn = a + ar + ar2 + · · · + ar N

is used frequently in this text while dealing with the discrete-time signals. Note

that the factor r can be real, imaginary, or complex.

A.4 Limits and differential calculus

lim t→∞

t−α ln t = 0, Re(α) > 0

lim t→0

tα ln t = 0, Re(α) > 0

L’Hôpital’s Rule:

If lim t→a

x(t) = lim t→a

y(t) = 0 or lim t→a

x(t) = lim t→a

y(t) = ∞, and lim t→a

x ′(t)

y′(t) has a

finite value, then lim t→a

x(t)

y(t) = lim

t→a

x ′(t)

y′(t) d

{

g(t)

}

= − 1

g2(t)

dg(t)

{

h(t)

g(t)

}

= 1

g2(t)

[

g(t) dh(t)

dt − h(t)

dg(t)

]

A.5 Indefinite integrals

∫

u dv = uv − ∫

v du ∫

f (t)g(t) dt = f (t) ∫

g(t)dt − ∫ [

d f

∫

g(t)dt

]

∫

cos at dt = 1

a sin at + C, a �= 0

∫

sin at dt = − 1

a cos at + C, a �= 0

∫

cos2 at dt = t

2 +

sin 2at

4a + C, a �= 0

∫

sin2 at dt = t

2 −

sin 2at

4a + C, a �= 0

∫

t cos at dt = 1

a2 (cos at + at sin at) + C, a �= 0

∫

t sin at dt = 1

a2 (sin at − at cos at) + C, a �= 0

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796 Appendix A

∫

t2 cos at dt = 1

a3 (2at cos at − 2 sin at + a2t2 sin at) + C, a �= 0

∫

t2 sin at dt = 1

a3 (2at sin at − 2 cos at − a2t2 cos at) + C, a �= 0

∫

cos at cos bt dt = sin(a − b)t

2(a − b) +

sin(a + b)t

2(a + b) + C, a2 �= b2

∫

sin at sin bt dt = sin(a − b)t

2(a − b) −

sin(a + b)t

2(a + b) + C, a2 �= b2

∫

sin at cos bt dt = −

[

cos(a − b)t

2(a − b) +

cos(a + b)t

2(a + b)

]

+ C, a2 �= b2

∫

sin−1 at dt = t sin−1 at + 1

√

1 − a2t2 + C, a �= 0

∫

cos−1 at dt = t cos−1 at − 1

√

1 − a2t2 + C, a �= 0

∫

eat dt = 1

a eat + C, a �= 0

∫

bat dt = bat

a ln b + C, a �= 0, b > 0, b �= 1

∫

teat dt = eat

a2 (at − 1) + C, a �= 0

∫

t2eat dt = eat

a3 (a2t2 − 2at + 2) + C, a �= 0

∫

tneat dt = 1

a tneat −

∫

tn−1eat dt, a �= 0

∫

tnbat dt = tnbat

a ln b −

a ln b

∫

tn−1bat dt, a �= 0, b > 0, b �= 1

∫

eat sin bt dt = eat

a2 + b2 (a sin bt − b cos bt) + C

∫

eat cos bt dt = eat

a2 + b2 (a cos bt + b sin bt) + C

∫

tn ln at dt = tn+1

n + 1 ln at −

tn+1

(n + 1)2 + C, n �= −1

∫ 1

t2 + a2 dt =

a tan−1

(

)

+ C, a �= 0

∫ t

t2 + a2 dt =

2 ln(t2 + a2) + C

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Appendix B Introduction to the complex-number system

In this appendix, we introduce some elementary concepts that define complex

numbers. In presenting the material, it is anticipated that most readers have

some prior exposure to complex numbers, so the information presented here

serves primarily as a review. The appendix is organized as follows. In Section

B.1, we review the definition of real numbers and then survey their arithmetic

properties, including some basic operations like addition, subtraction, multipli-

cation, and division. Section B.2 extends the arithmetic operations to complex

numbers, and Section B.3 introduces its geometric structure using the 2D Carte-

sian representation. Section B.4 presents an alternative representation, referred

to as the polar representation for complex numbers. Section B.5 concludes the

appendix.

B.1 Real-number system

A real-number system ℜ is a set of all real numbers, which is defined in terms of

two basic operations: addition and multiplication. For two arbitrarily selected

real numbers a, b ∈ ℜ, these basic operations are given by

addition s1 = a + b; (B.1)

multiplication m1 = a × b, (B.2)

such that s1, m1 ∈ ℜ. The remaining arithmetic operations, for example, sub- traction and division, are expressed in terms of Eqs (B.1) and (B.2) as follows:

subtraction s2 = a − b = a + (−b); (B.3)

division m2 = a/b = a × (1/b), (B.4)

such that s2, m2 ∈ ℜ. The real number −b is referred to as the additive inverse of b since b + (−b) = 0. Likewise, the real number 1/b is referred to as the multiplicative inverse of b since b × (1/b) = 1. For ℜ to represent a complete set of real numbers, it must satisfy the following properties.

797

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798 Appendix B

� � � � �−� −� −� −�

−∞ ∞ Fig. B.1. Representation of a

real-number system using a 1D

straight line.

(i) The addition of two real numbers a, b ∈ ℜ produces a unique real number s1 ∈ ℜ.

(ii) Subtracting a real number a ∈ ℜ from another real number b ∈ ℜ produces a unique real number s2 ∈ ℜ.

(iii) Multiplication of two real numbers a, b ∈ ℜproduces a unique real number m1 ∈ ℜ.

(iv) Dividing a real number a ∈ ℜ by another real number b ∈ ℜ, b �= 0, pro- duces a unique real number m2 ∈ ℜ.

Frequently, a real-number system is modeled graphically using a 1D straight

line, as illustrated in Fig. B.1. Each point on the line represents a real number.

The 1D line is packed with real numbers such that an uncountable number of

real numbers exists between two arbitrarily selected points on the line.

B.2 Complex-number system

Let j be the root of the equation x2 + 1 = 0, such that j = √

−1. In terms of j, a complex number x is defined as

x = a + jb, such that x ∈ C, (B.5)

where a and b represent two real numbers, a, b ∈ ℜ, and C denote a set of all possible complex numbers. Equation (B.5) is referred to as the rectangular

or Cartesian representation of the complex number x . From Eq. (B.5), it is straightforward to deduce the following:

(i) The real component of the complex number x is a. This is denoted by ℜ(x) = a.

(ii) The imaginary component of the complex number x is b. This is denoted by ℑ(x) = b.

In the following, we define the basic arithmetic operations between two complex

numbers. In our definitions, we use the following two operands: x1 = a1 + jb1 and x2 = a2 + jb2, with x1, x2 ∈ C and a1, a2, b1, b2 ∈ ℜ.

B.2.1 Addition

Addition of two complex numbers is defined as follows:

x1 + x2 = (a1 + jb1) + (a2 + jb2) = (a1 + a2) + j(b1 + b2). (B.6)

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799 B Introduction to the complex-number system

In other words, when adding two complex numbers the real and imaginary

components are added separately.

B.2.2 Subtraction

The definition of subtraction follows the same lines as that for addition. Sub-

tracting a complex number x2 from x1 is defined as follows:

x1 − x2 = (a1 + jb1) − (a2 + jb2) = (a1 − a2) + j(b1 − b2).

(B.7)

As for addition, the real and imaginary components are subtracted separately.

B.2.3 Multiplication

Multiplication of two complex numbers x1 and x2 is defined as follows:

x1x2 = (a1 + jb1)(a2 + jb2) = a1a2 + jb1a2 + ja1b2 + j2b1b2 = (a1a2 − b1b2) + j(b1a2 + a1b2), (B.8)

where the final expression is obtained by noting that j2 = −1.

B.2.4 Complex conjugation

From Eq. (B.8), it is easy to deduce that

(a1 + jb1)(a1 − jb1) = (a1)2 + (b1)2. (B.9)

In other words, the imaginary component is eliminated. The complex number

x∗1 = a1 − jb1 is referred to as the complex conjugate of x1 = a1 + jb1, and vice versa. Equation (B.9) leads to the definition of the modulus or magnitude

of a complex number, which is discussed next.

B.2.5 Modulus

The modulus (or magnitude) of a complex number x1 = a1 + jb1 is defined as follows:

|x1| = √

x1x∗1 = √

(a1)2 + (b1)2. (B.10)

B.2.6 Division

Dividing two complex numbers is more complicated. To divide x1 by x2, we multiply both the numerator and denominator by the complex conjugate of x2 and expand the numerator and denominator separately using the definition of

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800 Appendix B

multiplication from Section B.2.3; i.e.

x1 x2

= a1 + jb1 a2 + jb2

= (a1 + jb1) (a2 + jb2)

· (a2 − jb2) (a2 − jb2)

= a1a2 + b1b2

a22 + b22 + j

a2b1 − a1b2 a22 + b22

, (B.11)

where the final expression is obtained by noting that j2 = −1. We illustrate these concepts with an example.

Example B.1

Two complex numbers are given by x = 5 + j7 and y = 2 − j4. Calculate (i) ℜ(x), ℑ(x), ℜ(y), ℑ(y); (ii) x + y; (iii) x − y; (iv) xy; (v) x∗, y∗; (vi) |x |, |y|; and (vii) x/y.

Solution

(1) The real and imaginary components of the complex number x are ℜ(x) = 5 and ℑ(x) = 7. Likewise, the real and imaginary components of y are ℜ(y) = 2 and ℑ(y) = −4.

(2) Adding x and y yields

x + y = (5 + j7) + (2 − j4) = (5 + 2) + j(7 − 4) = 7 + j3.

Since addition is commutative, the order of the operands does not matter,

i.e. x + y = y + x . (3) Subtracting y from x yields

x − y = (5 + j7) − (2 − j4) = (5 − 2) + j(7 − (−4)) = 3 + j11.

Subtraction is not commutative. In fact, x − y = −(y − x). (4) Multiplication of x and y is performed as follows:

xy = (5 + j7)(2 − j4) = 10 + j14 − j20 − j228 = (10 + 28) + j(14 − 20) = 38 − j6.

Multiplication is commutative, therefore xy = yx. (5) The complex conjugate of the complex number x = 5 + j7 is x∗ = 5 − j7.

Likewise, the complex conjugate of y = 2 − j4 is y∗ = 2 + j4. (6) The modulus of x = 5 + j7 is given by |x | =

√ 52 + 72 =

√ 74. Likewise,

the modulus of y = 2 − j4 is |y| = √

22 + (−4)2 = √

20.

(7) Dividing x by y yields

y = 5 + j7

2 − j4 = (5 + j7) (2 − j4) ·

(2 + j4) (2 + j4)

= (5)(2) − (7)(4) 22 + 42 + j

(7)(2) + (5)(4) 22 + 42 = −

20 + j 34

20 .

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801 B Introduction to the complex-number system

B.3 Graphical interpertation of complex numbers

Any complex number x = a + jb can be associated with an ordered pair of real numbers (a, b), i.e.

x = (a + jb) ←→ (a, b). (B.12)

The ordered pair of numbers (a, b) is represented by a point in the Cartesian coordinate system as shown in Fig. B.2(a), in which the horizontal axis rep-

resents the real component ℜ of the complex number and the vertical axis

represents the imaginary component ℑ of the complex number. Alternatively, the complex number x can be associated with a vector r originating from the coordinate (0, 0) and extending to the point (a, b). The rules for vector addi- tion and subtraction can be used to add and subtract complex numbers, and

vice versa. Since the two representations are equivalent, it is common to map a

complex number to a vector.

b (a, b) ℑ

ℜ

(a)

rx = a

ry = b ℑ

ℜθ

(b)

Fig. B.2. Graphical

representations for a complex

number x = a + jb. (a) Cartesian representation;

(b) polar representation.

Similar to the rectangular and polar representations of a vector, there are two

alternative and equivalent representations for complex numbers. The rectangu-

lar representation was introduced in Section B.2. The polar representation is

derived in Section B.4 by using Fig. B.2(b) and applying the geometric proper-

ties associated with vectors. Here, we define the notation used in the derivation

of the polar representation. The length or magnitude of the vector r , shown in Fig. B.2(b), is denoted by | r |, or simply r . The angle that the vector r makes with the positive horizontal axis is denoted by θ . The projection of the vector

r onto the horizontal axis is denoted by rx , while the projection on the vertical axis is denoted by ry . In terms of r and θ , the two projections are given by

rx = r cos θ and ry = r sin θ. (B.13) Using Pythagoras’s theorem, it is straightforward to prove that the length or

magnitude r of vector r is given by r =

√

r2x + r2y , (B.14) and the angle θ that the vector makes with the horizontal axis is given by

θ = tan−1(ry/rx ). (B.15)

B.4 Polar representation of complex numbers

To derive the polar representation of a complex number, we base our discussion

on Euler’s formula:†

ejθ = cos θ + j sin θ. (B.16) The polar representation of a complex number x = a + jb is then defined as

x = rejθ , (B.17) † Euler’s formula is named after Leonhard Euler (1707–1783), a prolific eighteenth century Swiss

mathematician and physicist.

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802 Appendix B

where r represents the magnitude or length of the vector r obtained by mapping the complex number x onto the Cartesian plane. The length r and angle θ associated with vector r are obtained from Eqs (B.14) and (B.15) with rx = a and ry = b. We demonstrate the conversion of a complex number from one representation to another with a series of examples.

Example B.2

Converting rectangular format into polar format Consider a complex num- ber x = 2 + j4. Clearly, x is represented in the rectangular format. To derive its equivalent polar format, we map the complex number into the Cartesian plane

and calculate the parameters r and θ . Using Eqs (B.14) and (B.15), we obtain

r = √

22 + 42 = √

and

θ = tan−1(4/2) = 0.35π radians. The polar representation of x = 2 + j4 is x =

√ 20e j0.35π .

Example B.3

Converting polar format into rectangular format Consider a complex num- ber in the polar format x = 4ejπ/3. The rectangular representation of x is derived using Eq. (B.13) as

a = rx = 4 cos (π

)

= 2

and

b = ry = 4 sin (π

)

= 2 √

The rectangular representation of x = 4ejπ/3 is x = 2 + j2 √

In terms of polar representations, the basic arithmetic operations between two

complex numbers x1 = r1ejθ1 and x2 = r2ejθ2 are defined as follows.

B.4.1 Addition

Addition of two complex numbers in polar format:

x1 + x2 = r1ejθ1 + r2ejθ2 = (r1 cos θ1 + jr1 sin θ1) + (r2 cos θ2 + jr2 sin θ2) = (r1 cos θ1 + r2 cos θ2) + j (r1 sin θ1 + r2 sin θ2)

= √

(r1 cos θ1 + r2 cos θ2)2 + (r1 sin θ1 + r2 sin θ2)2

× exp [

j tan−1 (

r1 sin θ1 + r2 sin θ2 r1 cos θ1 + r2 cos θ2

)]

= √

r21 + r22 + 2r1r2 cos(θ1 − θ2)

× exp [

j tan−1 (

r1 sin θ1 + r2 sin θ2 r1 cos θ1 + r2 cos θ2

)]

. (B.18)

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803 B Introduction to the complex-number system

B.4.2 Subtraction

Subtraction of two complex numbers in polar format:

x1 − x2 = r1ejθ1 − r2ejθ2

= (r1 cos θ1 − r2 cos θ2) + j (r1 sin θ1 − r2 sin θ2)

= √

(r1 cos θ1 − r2 cos θ2)2 + (r1 sin θ1 − r2 sin θ2)2

× exp [

j tan−1 (

r1 sin θ1 − r2 sin θ2 r1 cos θ1 − r2 cos θ2

)]

= √

r21 + r22 − 2r1r2 cos(θ1 − θ2)

× exp [

j tan−1 (

r1 sin θ1 − r2 sin θ2 r1 cos θ1 − r2 cos θ2

)]

. (B.19)

B.4.3 Multiplication

Multiplication of two complex numbers x1 and x2 in polar format:

x1x2 = r1ejθ1 · r2ejθ2

= r1r2ej(θ1+θ2). (B.20)

B.4.4 Complex conjugation

The complex conjugate of complex number x1 is given by

x∗1 = r1e −jθ1 . (B.21)

B.4.5 Modulus

The modulus (or magnitude) of a complex number x1 = r1ejθ1 is |x1| = r1.

B.4.6 Division

Dividing two complex numbers in polar format:

x1 x2

= r1ejθ1

r2ejθ2 =

r1 r2

ej(θ1−θ2). (B.22)

Before we end this section, we note that both rectangular and polar formats

have their advantages. It is easier to add or subtract complex numbers in the

rectangular format. Multiplication and division are, however, simpler in the

polar representation. We illustrate the concepts discussed in Section B.4 with

the following example.

Example B.4

Consider the two complex numbers

x = 5 + j7 = √

74ej0.3026π

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804 Appendix B

and

y = 2 − j4 = √

20e−j0.3524π .

Repeat Example B.1 but by selecting one of the two formats (rectangular or

polar) for which the arithmetic operation is computationally simpler.

Solution

(1) The real and imaginary components of the complex number x are obtained from the rectangular format, i.e. ℜ(x) = 5 and ℑ(x) = 7. Likewise, for y the components are ℜ(y) = 2 and ℑ(y) = −4.

(2) Addition of x and y is performed in the rectangular format as follows:

x + y = (5 + j7) + (2 − j4) = (5 + 2) + j(7 − 4) = 7 − j3.

If polar format is required, we can express the above answer for (x + y) in the polar format as x + y =

√ 58ej tan

−1(−3/7) = 7.62e−j0.13π . (3) Subtraction is also performed in the rectangular format as follows:

x − y = (5 + j7) − (2 − j4) = (5 − 2) + j(7 − (−4)) = 3 − j11.

Converting the above answer into polar form, we obtain x − y =√ 130ej tan

−1(−11/3) = 11.40e−j0.415π . (4) Multiplication of x and y is performed in the polar format as follows:

xy = √

74ej0.3026π · √

20e−j0.3524π

= √

1480e−j0.0498π .

The rectangular format is xy = √

1480(cos(0.0498π ) + j sin(−0.0498π )) = 38 − j6.

(5) In rectangular format, the complex conjugate of the complex number x = 5 + j7 is x∗ = 5 − j7. Likewise, the complex conjugate of y = 2 − j4 is y∗ = 2 + j4 in rectangular format. The complex conjugates in polar format are x∗ =

√ 74e−j0.3026π and y∗ =

√ 20ej0.3524π .

(6) The moduli of x and y are obtained directly from the polar format as |x | =

√ 74 and |y| =

√ 20.

(7) Dividing x by y is performed in polar format, yielding

y =

√ 74ej0.3026π√

20e−j0.3524π =

√ 3.7ej0.655π ,

which, in rectangular format, is √

3.7 (cos(0.655π ) + j sin(0.655π )) = −0.9 + j1.7.

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805 B Introduction to the complex-number system

B.5 Summary

Complex numbers in rectangular and polar formats were reviewed. Basic arith-

metic operations such as addition, subtraction, multiplication, division, and

complex conjugation were illustrated in both rectangular and polar domains.

Problems

B.1 Calculate the polar representations for (a) 1; (b) j; (c) − 1; (d) −j; (e) 3 + j4; (f) 8 − j6; and (g) 12 + j4.

B.2 Calculate the rectangular representations for (a) 11 exp(j2π ); (b) 125 exp(jπ/2); (c) 72 exp(−jπ ); (d) 125 exp(jπ/8); (e) 25.47 exp(−j3π/4); and (f) 0.85 exp(−jπ/4).

B.3 Consider the complex function

g(t) = 2 + j3t 1 + j2t

Plot the magnitude and phase of the function g(t) each as a function of the independent variable t .

B.4 Determine and sketch the roots of the equation ex + 10 = 0 in the Cartesian plane. [Hint: The polar representation for −10 = eln(10)+j(2m+1)π .]

B.5 Prove the following identities:

(i) cos θ = ejθ + e−jθ

2 ;

(ii) sin θ = ejθ − e−jθ

2j ;

(iii) ejmπ = (−1)m and ej(2mπ+θ ) = ejθ ;

(iv) cos θ = 1 − θ2

2 ! +

θ4

4 ! −

θ6

6 ! + · · · ;

(v) sin θ = θ − θ3

3 ! +

θ5

5 ! −

θ7

7 ! + · · · .

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Appendix C Linear constant-coefficient differential equations

It was shown in Chapters 2 and 3 that linear constant-coefficient differential

equations play an important role in LTIC systems analysis. In this appendix,

we review a direct method for solving differential equations of the form

n ∑

k=0 ak

dk y(t)

dtk =

m ∑

k=0 bk

dk x(t)

dtk , (C.1)

where the aks and bks are constants, and the derivatives

y(t), dy(t)

dt ,

d2 y(t)

dt2 , . . . ,

dn−1 y(t)

dtn−1 (C.2)

of the output signal y(t) are known at a given time instant, say t = t0. We will use the compact notation ẏ(t)to denote the first derivative of y(t) with respect to t . Therefore, ẏ(t) = dy/dt, ÿ(t) = dy2/dt2, and similarly for the higher-order derivatives. In the context of LTIC systems, the differential equation, Eq. (C.1),

provides a linear relationship between the input signal x(t) and the output y(t). The values of the derivatives of y(t), Eq. (C.2), for such LTIC systems are typically specified at t0 = 0 and are referred to as the initial conditions. The highest derivative in Eq. (C.1) denotes the order of the differential equation.

Equation (C.1) is therefore either of order n or m. The method discussed in this appendix is direct, in the sense that it solves Eq.

(C.1) in the time domain and does not require calculation of any transforms. The

direct approach expresses the output y(t) described by a differential equation as the sum of two components:

(i) zero-input response yzi(t) associated with the initial conditions; (ii) zero-state response yzs(t) associated with the applied input x(t).

The zero-input response yzi(t) is the component of the output y(t) of the sys- tem when the input is set to zero. The zero-input response describes the manner

in which the system dissipates any energy or memory of the past as specified

by the initial conditions. The zero-state response yzs(t) is the component of the output y(t) of the system with initial conditions set to zero. It describes the

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807 C Linear constant-coefficient differential equations

behavior of the system forced by the input. In the following, we outline the

procedure to evaluate the zero-input and zero-state responses.

C.1 Zero-input response

The zero-input response yzi(t) is the output of the system when the input is zero. Hence, yzi(t) is the solution to the following homogeneous differential equation:

n ∑

k=0 ak

dk y(t)

dtk = 0, (C.3)

with known initial conditions

y(t), dy(t)

dt ,

d2 y(t)

dt2 , . . . ,

dn y(t)

dtn at t = 0. (C.4)

To determine the zero-input response yzi(t), assume that the zero-input response is given by yzi(t) = Aest , substitute yzi(t) in the homogeneous dif- ferential equation, Eq. (C.3), and solve the resulting equation. We illustrate the

procedure for calculating the homogeneous solution by considering an example.

Example C.1

Consider a CT system modeled by the following differential equation:

d2 y

dt2 + 5

dt + 4y(t) = 3x(t). (C.5)

Compute the zero-input response of the system for initial conditions y(0) = 2 and ẏ(0) = −5.

Solution

Substituting yzi(t) = Aest in the homogeneous equation

d2 y

dt2 + 5

dt + 4y(t) = 0, (C.6)

obtained by setting input x(t) = 0, yields

Aest (s2 + 5s + 4) = 0. (C.7)

Ignoring the trivial solution, i.e. assuming Aest �= 0, Eq. (C.7) reduces to the

following quadratic equation, referred to as the characteristic equation, in s:

s2 + 5s + 4 = 0, (C.8)

which has two roots at s = −1, −4. The zero-input solution is given by

yzi(t) = A0e −t + A1e

−4t , (C.9)

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808 Appendix C

where A0 and A1 are constants to be determined from the given initial condi- tions. Substituting the initial conditions in Eq. (C.9) yields

A0 + A1 = 2, −A0 − 4A1 = −5, (C.10)

which has solution A0 = 1 and A1 = 1. The zero-input response for Eq. (C.5) is therefore given by

yzi(t) = e−t + e−4t . (C.11)

C.1.1 Repeated roots

The form of the zero-input response changes slightly when the characteristic

equation has repeated roots. If a root s = a is repeated J times, then we include J distinct terms in the zero-input response associated with aby using the following J functions:

eat , teat , t2eat , . . . , t J−1eat . (C.12)

The zero-input response of an LTIC system is then given by

yzi(t) = A0eat + A1teat + A2t2eat + · · · + AJ−1t J−1eat . (C.13)

The procedure for calculating the homogeneous solution for differential

equations with repeated roots is illustrated in Example C.2.

Example C.2

Consider a CT system modeled by the following differential equation:

d3 y

dt2 + 4

d2 y

dt2 + 5

dt + 2y(t) = x(t). (C.14)

Compute the zero-input response of the system for initial conditions y(0) = 4, ẏ(0) = −5, ÿ(0) = 9.

Solution

By substituting yzi(t) = Aest in the homogeneous representation for Eq. (C.14), we obtain the following characteristic equation:

s3 + 4s2 + 5s + 2 = 0, (C.15)

which has three roots at s = −1, −2, −2. The zero-input solution is therefore given by

yzi(t) = A0e−t + A1e−2t + A2te−2t , (C.16)

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809 C Linear constant-coefficient differential equations

where A0, A1, and A2 are constants determined from the given initial conditions. Substituting the initial conditions into Eq. (C.16) yields

A0 + A1 = 4, −A0 + A1 − 2A2 = −5, (C.17) A0 − 2A1 + 4A2 = 9,

which has solution A0 = 1, A1 = 2, and A2 = 3. The zero-input response for Eq. (C.14) is therefore given by

yzi(t) = e−t + 2e−2t + 3te−2t . (C.18)

C.1.2 Complex roots

Solving a characteristic equation may give rise to complex roots of the form

s = a + jb. Typically, a homogeneous differential equation, Eq. (C.3), with real coefficients, has complex roots in conjugate pairs. In other words, if s = a + jb is a root of the characteristic equation obtained from Eq. (C.3) then s = a − jb must also be a root of the characteristic equation. For such complex roots, the

zero-input response can be modified to the following form:

yzi(t) = A0eat cos(bt) + A1eat sin(bt). (C.19)

Example C.3

Compute the zero-input response of a system represented by the following

differential equation:

d4 y

dt4 + 2

d2 y

dt2 + 1 = x(t), (C.20)

with initial conditions y(0) = 2, ẏ(0) = 2, ÿ(0) = 0, ÿ(0) = −4.

Solution

Substituting yzi(t) = Aest in the homogeneous representation for Eq. (C.20) results in the following characteristic equation:

s4 + 2s2 + 1 = 0. (C.21)

The roots of the characteristic equation are given by s = j, j, −j, and −j. Note that the roots are not only complex but also repeated. The zero-input solution

is given by

yzi(t) = A0 cos(t) + A1t cos(t) + A2 sin(t) + A3t sin(t), (C.22)

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810 Appendix C

where A0, A1, A2, and A3 are constants. To calculate these constants, we sub- stitute the following initial conditions:

A0 = 2, A1 + A2 = 2,

−A0 + 2A3 = 0, −3A1 − A2 = −4,

(C.23)

which has solution A0 = 2, A1 = 1, A2 = 1, and A3 = 1. The zero-input response for the system in Eq. (C.20) is therefore given by

yzi(t) = 2 cos(t) + t cos(t) + sin(t) + t sin(t). (C.24)

C.2 Zero-state response

The zero-state response yzs(t) depends upon the input signal x(t) subject to zero initial conditions. The zero-state response consists of two components:

(i) the homogeneous component y(h)zs (t) and (ii) the particular component y (p) zs (t).

The homogeneous component is obtained by following the procedure used to

solve for the zero-input response but with zero initial conditions. The particular

component of the zero-state response is obtained from a look-up table such as

Table C.1. For example, if the input signal is x(t) = K e−at , then the partic- ular component of the zero-state response is assumed to be y(p)zs (t) = Ce−at . The constant C is determined such that yzi(t) satisfies the system’s differential equation. The procedure for computing the zero-state response is illustrated in

Example C.4.

Example C.4

Consider the system specified by the differential equation given in

Example C.1:

d2 y

dt2 + 5

dt + 4y(t) = 3x(t). (C.25)

Compute the zero-state response of the system for the input signal x(t) = cos tu(t).

Solution

The homogeneous and particular components of the zero-state response yzi(t) are solved separately in three steps as follows.

Step 1 Compute the homogeneous component y(h)zs (t) The solution for the homogeneous component is similar to the zero-input response of the system.

Using the result of Eq. (C.9), the homogeneous component of the zero-input

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811 C Linear constant-coefficient differential equations

Table C.1. Zero-state response corresponding to common input signals

Particular component of the

Input zero-state response

Impulse function, K δ(t) Cδ(t) Unit step function, Ku(t) Cu(t) Exponential, Ke−at Ce−at

Sinusoidal, A cos(ω0t + φ) C0 cos(ω0t) + C1sin(ω0t)

response is given by

y(h)zs (t) = B0e −t + B1e−4t , (C.26)

where B0 and B1 are constants.

Step 2 Determine the particular component y(p)zs (t) The particular com- ponent is obtained by consulting Table C.1. For the input signal x(t) = cos t u(t), the particular component of the zero-state response is of the form y(p)zs (t) = C0 cos t + C1 sin t for t > 0. Substituting the particular component in Eq. (C.25) yields

(−5C0 + 3C1) sin t + (3C0 + 5C1) cos t = 3 cos t. (C.27)

Equating the cosine and sine terms on the left- and right-hand sides of the

equation, we obtain the following simultaneous equations:

−5C0 + 3C1 = 0, 3C0 + 5C1 = 3, (C.28)

with solution C0 = 9/34 and C1 = 15/34. The particular component y(p)zs (t) of the zero-state response is given by

y(p)zs (t) = 9

34 cos t +

34 sin t for t > 0. (C.29)

Step 3 Determine the zero-state response from yzs(t) = y(h)zs (t) + y (p) zs (t).

The zero-state response is the sum of the homogeneous and particular com-

ponents, and is given by

yzs(t) = (B0e−t + B1e−4t ) + 9

34 cos t +

34 sin t, (C.30)

where B0 and B1 are obtained by inserting zero initial conditions, y(0) = 0 and ẏ(0) = 0. This leads to the following simultaneous equations:

B0 + B1 = − 9

34 ,

B0 + 4B1 = 15

34 ,

(C.31)

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812 Appendix C

with solution B0 = −1/2 and B1 = 4/17. The zero-state response of Eq. (C.25) is

yzs(t) = − 1

2 e−t +

17 e−4t +

34 cos t +

34 sin t. (C.32)

This approach for finding the particular component of the zero-state response

is modified when the input is of the same form as one of the terms in the homo-

geneous component of the zero-state response. We illustrate with an example

where we outline the modified procedure for calculating the particular compo-

nent of the zero-state response.

Example C.5

Repeat Example C.4 for the input signal x(t) = 2e−t .

Solution

The homogeneous component for the zero-state response is given by Eq. (C.26):

y(h)zs (t) = B0e −t + B1e−4t ,

where B0 and B1 are constants. The input signal x(t) = 2e−t . Based on Table C.1, the particular component is of the form y(p)zs (t) = Ce−t , which is similar to the first term in the homogeneous component. In such a scenario, we

assume a particular component that is different from the first term of the homo-

geneous component. To achieve this, we multiply the particular component by

the lowest power of t that will make the particular component different from the first term of the homogeneous component. The particular component, in this

example, is therefore given by y(p)zs (t) = Cte−t . In order to evaluate the value of constant C , we substitute the particular component in the system’s differential equation and solve for C ; it is found that C = 3. The overall zero-state response is therefore given by

yzs(t) = B0e−t + B1e−4t + 3te−t , (C.33)

where the values of B0 and B1 are computed using zero initial conditions. The resulting simultaneous equations are given by

B0 + B1 = 0, B0 + 4B1 = −3,

(C.34)

which has solution B0 = 1 and B1 = −1. The overall zero-state response is given by

yzs(t) = e−t − e−4t + 3te−t .

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813 C Linear constant-coefficient differential equations

C.3 Complete response

The complete response of an LTIC system is the sum of the homogeneous and

particular components. The procedure for calculating the complete response

consists of the following steps.

(1) Compute the zero-input response yzi(t) of the system using the given initial conditions.

(2) Compute the zero-state response yzs(t) of the system using zero initial conditions and the input signal. The zero-state response is obtained by

determining its homogeneous and particular components.

(3) Add the zero-input and zero-state responses of the systems to get the com-

plete response.

Example C.6

Calculate the output of an LTIC system represented by the following differential

equation:

d2 y

dt2 + 5

dt + 4y(t) = 3x(t), (C.35)

for the input signal x(t) = cos t u(t) and the initial conditions y(0) = 2 and ẏ(0) = −5.

Solution

The zero-input response was calculated in Example C.1 and is given by

Eq. (C.11), repeated below:

yzi(t) = e−t + e−4t . (C.36)

The zero-state response was calculated in Example C.4 and is given by

Eq. (C.32), repeated below:

yzs(t) = − 1

2 e−t +

17 e−4t +

34 cos t +

34 sin t. (C.37)

The complete response is the sum of Eqs (C.36) and (C.37) and is given by

y(t) = 1

2 e−t +

17 e−4t +

34 cos t +

34 sin t (C.38)

for t ≥ 0.

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Appendix D Partial fraction expansion

An alternative approach to convolution, used in calculating the output response

of a linear time-invariant (LTI) system, is to calculate the product of appropri-

ately selected transforms of the convolving signals and then evaluate the inverse

transform of the product. In most cases, the transform-based approach is more

convenient as it leads to a closed-form solution. It is therefore important to

develop methods to compute the inverse of a specified transform to determine

the output response of the LTI system in the time domain. For transforms that

can be expressed as a rational function of two polynomials, the partial fraction

expansion simplifies the evaluation of the inverse transform by expressing the

rational function as a summation of simpler terms whose inverse is obtained

from a look-up table. This appendix focuses on the partial fraction expansion

of a rational function. The partial fraction expansion techniques for the four

transforms, namely the Laplace transform, the continuous-time Fourier trans-

form (CTFT), the z-transform, and the discrete-time Fourier transform (DTFT),

covered in the text are presented separately in Sections D.1–D.4.

D.1 Laplace transform

Consider a function X (s) of the form

X (s) = N (s)

D(s) =

bmsm + bm−1sm−1 + · · · + b1s + b0 ansn + an−1sn−1 + · · · + a1s + a0

, (D.1)

where the numerator N (s) is a polynomial of degree m and the denominator D(s) is a polynomial of degree n. If m ≥ n, we can divide N (s) by D(s) and express X (s) in an alternative form as follows:

X (s) = m−n ∑

ℓ=0

αℓs ℓ+

N1(s)

D(s) . (D.2)

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815 D Partial fraction expansion

If m < n, there is no summation term in Eq. (D.2) and N1(s) = N (s). The partial fraction expansion represents the rational fraction N1(s)/D(s) as a summation of simpler terms.

The first step in obtaining the partial fraction expansion is to factorize the

denominator polynomial and express the function X (s) as follows:

N1(s)

D(s) =

N1(s)

(s − p1)(s − p2) · · · (s − pn) , (D.3)

where p1, p2, . . . , pn are the n roots of the characteristic equation,

D(s) = ansn + an−1sn−1 + · · · + a1s + a0 = 0. (D.4)

If X (s) represent the transfer function of an LTIC system, then the roots p1, p2, . . . , pn of the characteristic equation are the poles of the system. The partial fraction expansion expresses Eq. (D.3) as the following summation:

N1(s)

D(s) =

k1 s − p1

+ k2

s − p2 + · · · +

kn s − pn

, (D.5)

where kr , for 1 ≤ r ≤ n, is referred to as the coefficient (also known as the residue) of the r th partial fraction. Depending on the nature of the poles, different procedures are used to compute the partial fraction coefficients kr . We consider two cases in the following sections.

D.1.1 First-order poles The poles p1, p2, . . . , pn are of the first order if they are not repeated. In such cases, the value of the r th partial fraction coefficients kr can be calculated from the Heaviside formula:†

kr =

[

(s − pr ) N1(s)

D(s)

]

s=pr

. (D.6)

We illustrate the application of the formula with four examples.

Example D.1

For the function

X (s) = 4s2 + 20s − 2

s3 + 3s2 − 6s − 8 , (D.7)

(i) calculate the partial fraction expansion;

(ii) based on your answer to (i), calculate the inverse Laplace transform of

X (s).

† This formula is named after Oliver Heaviside (1850–1925), an English electrical engineer,

mathematician, and physicist, who developed techniques for applying the Laplace transforms to

the solution of differential equations.

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816 Appendix D

Solution

(i) The characteristic equation of X (s) is given by

s3 + 3s2 − 6s − 8 = 0,

which has roots at s = −1, 2, and −4. The partial fraction expansion of X (s) is therefore given by

X (s) = 4s2 + 20s − 2

s3 + 3s2 − 6s − 8 ≡

k1 s + 1

+ k2

s − 2 +

k3 s + 4

Using the Heaviside formula, the residues kr are given by

k1 = 4s2 + 20s − 2

(s − 2)(s + 4)

∣

s=−1

= 4 − 20 − 2

−9 = 2,

k2 = 4s2 + 20s − 2

(s + 1)(s + 4)

∣

s=2

= 16 + 40 − 2

3 × 6 = 3,

and

k3 = 4s2 + 20s − 2

(s + 1)(s − 2)

∣

s=−4

= 64 − 80 − 2

(−3) × (−6) =

−18

18 = −1.

Substituting the values of the partial fraction coefficients k1, k2, and k3, we obtain

X (s) = 2

s + 1 +

s − 2 −

s + 4 . (D.8)

(ii) Assuming the function x(t) to be causal or right-sided, we use Table 6.1 to determine the inverse Laplace transform x(t) of the X (s) as follows:

x (t) = (

2e−t + 3e2t − e−4t )

u(t) . (D.9)

Example D.2

For the function

X (s) = 6s2 + 11s + 26

s3 + 4s2 + 13s , (D.10)

(i) calculate the partial fraction expansion;

(ii) based on your answer to (i), calculate the inverse Laplace transform of

X (s).

Solution

(i) The characteristic equation of X (s) is given by

s3 + 4s2 + 13s = 0,

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817 D Partial fraction expansion

which has roots at s = 0, −2 + j3, and −2 −j3. The partial fraction expansion of X (s) is therefore given by

X (s) = 6s2 + 11s + 26 s3 + 4s2 + 13s

≡ k1 s

+ k2

s + 2 + j3 +

k3 s + 2 − j3

. (D.11)

Note that in this case, there are two complex-conjugate poles at s = −2 ±j3. Using the Heaviside formula, the residues kr are given by

k1 =

[

s 6s2 + 11s + 26

s(s + 2 + j3)(s + 2 − j3)

]

s=0

= 2,

k2 =

[

(s + 2 + j3) 6s2 + 11s + 26

s(s + 2 + j3)(s + 2 − j3)

]

s=−2−j3

= 2 − j 5

6 ,

and

k3 =

[

(s + 2 + j3) 6s2 + 11s + 26

s(s + 2 + j3)(s + 2 − j3)

]

s=−2+j3

= 2 + j 5

6 .

Substituting the values of the partial fraction coefficients k1, k2, and k3, we obtain

X (s) = 2

s +

2 − j 5 6

s + 2 + j3 +

2 + j 5 6

s + 2 − j3 . (D.12)

(ii) Assuming the function x(t) to be causal or right-sided, we use Table 6.1 to determine the inverse Laplace transform x(t) of the X (s) as follows:

x(t) =

[

2 +

(

2 − j 5

)

e−(2+j3)t +

(

2 + j 5

)

e−(2−j3)t ]

u(t)

[

2 + e−2t {(

2 − j 5

)

e−j3t +

(

2 + j 5

)

ej3t }]

u(t)

[

2 + e−2t {

2 (

e j3t + e−j3t )

+ j5

(

ej3t − e−j3t )

}]

u(t)

[

2 + e−2t {

4 cos(3t) − 5

3 sin(3t)

}]

u(t)

[

2 + 4e−2t cos(3t) − 5

3 e−2t sin(3t)

]

u(t). (D.13)

In Example D.2, the complex-valued poles of the Laplace transform X (s) occur in conjugate pairs. This is true, in general, for any polynomial with real-valued

coefficients. Although the Heaviside formula may be used to determine the

values of the partial fraction residues corresponding to the complex poles, the

procedure is often complicated due to complex algebra. Below, we present

another procedure, which expresses such complex-valued and conjugate poles

in terms of a quadratic term in the partial fraction expansion.

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818 Appendix D

Example D.3

Repeat Example D.2 by expressing the complex-valued poles as a quadratic

term.

Solution

(i) Combining the complex-valued terms in Eq. (D.11),

X (s) = 6s2 + 11s + 26 s3 + 4s2 + 13s

= k1 s

+ k2

s + 2 + j3 +

k3 s + 2 − j3

= k1 s

+ (k2 + k3) s + 2 (k2 + k3)

(s + 2)2 − ( j3)2 .

Since k2 and k3 are constants, their linear combinations can be replaced with other constants. Substituting k2 + k3 = A1 and k2 + k3 = A2, we obtain

X (s) = 6s2 + 11s + 26 s3 + 4s2 + 13s

≡ k1 s

+ A1s + A2

s2 + 4s + 13 . (D.14)

It may be noted that the above expression could have been obtained directly by

factorizing the denominator,

s3 + 4s2 + 13s = s(s2 + 4s + 13),

and writing the partial fraction expansion of X (s) in terms of two terms, one with a linear polynomial s in the denominator and the other with a quadratic polynomial (s2 + 4s + 13).

The partial fraction coefficient k1 of the linear polynomial denominator is obtained using the Heaviside formula as follows:

k1 =

[

s 6s2 + 11s + 26

s(s2 + 4s + 13)

]

s=0

= 2.

In order to calculate the remaining coefficients A1 and A2, we substitute k1 = 2 in Eq. (D.14). Cross-multiplying and equating the numerators in Eq. (D.5), we

obtain

6s2 + 11s + 26 = 2(s2 + 4s + 13) + (A1s + A2) s

(A1 + 2)s 2 + (A2 + 8)s + 26 = 6s

2 + 11s + 26.

Equating the coefficients of the polynomials of the same degree on both sides

of the above equation, we obtain:

coefficient of s2 (A1 + 2) = 6, A1 = 4;

coefficient of s (A2 + 8) = 11, A2 = 3.

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819 D Partial fraction expansion

Substituting the values of the partial fraction coefficients k1, A1, and A2 in Eq. (D.14) yields

X (s) = 2

s +

4s + 3 (s + 2)2 + 9

. (D.15)

(ii) The Laplace transform X (s) is rearranged:

X (s) = 2

s +

4(s + 2) (s + 2)2 + 9

− 5

(s + 2)2 + 9 ,

such that the second and third terms are in the same form as entries (13) and

(14) in Table 6.1. Taking the inverse transform gives the following transform

pairs:

x(t) = [

2 + 4e−2t cos(3t) − 5

3 e−2t sin(3t)

]

u(t). (D.16)

Note that the inverse Laplace transform x(t) obtained in Eq. (D.13) is identical to the answer obtained in Example D.2. The procedure followed in Example D.3

avoids complex numbers and is preferable. In cases where the roots of the char-

acteristic equations are complex-valued, we will express the Laplace transform

directly in terms of partial fraction terms with quadratic denominators.

Example D.4

For the function

H (s) = 2s3 + 10s2 + 8s − 18

s3 + 3s2 − 6s − 8 , (D.17)

(i) calculate the partial fraction expansion;

(ii) based on your answer to (i), calculate the inverse Laplace transform of

X (s).

Solution

(i) Since the degree of both the numerator and denominator polynomials is

3, we divide the numerator polynomial by the denominator polynomial and

express H (s) as follows:

H (s) = 2 + 4s2 + 20s − 2

s3 + 3s2 − 6s − 8 ︸︷︷︸

X (s)

The second term in H (s) is the same as the rational fraction X (s) specified in Example D.1. Using the results of Example D.1, the partial fraction expansion

of H (s) is given by

H (s) = 2 + 2

s + 1 +

s − 2 −

s + 4 . (D.18)

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820 Appendix D

(ii) Assuming that the inverse Laplace transform x(t) is right-sided, we use Table 6.1 to determine the inverse Laplace transform x(t) of the X (s):

h(t) = 2δ(t) + (2e−t + 3e2t − e−4t )u(t). (D.19)

D.1.2 Higher-order poles The residues in a partial fraction can be calculated using the Heaviside formula in Eq. (D.4) when the poles are not repeated.

However, when there are multiple poles at the same location, Eq. (D.4) cannot

be directly used to calculate the coefficients corresponding to the fractions at

multiple pole locations. To illustrate the partial fraction expansion for repeated

poles, consider a Laplace transform X1(s) with r− 1 unrepeated poles at s = p1, p2, . . . , pr−1 and q repeated poles at s = pr . To be consistent with the rational fraction expression in Eq. (D.1), r − 1 + q = n. The Laplace transform X1(s) can be expressed as follows:

N1(s)

D(s) =

N1(s)

(s − p1)(s − p2) · · · (s − pr−1)(s − pr )q . (D.20)

The partial fraction expansion of the above rational function is given by

N1(s)

D(s) =

k1 s − p1

+ k2

s − p2 + · · ·

kr−1 s − pr−1

+ kr,1

s − pr +

kr,2 (s − pr )2

+ · · · + kr,q

(s − pr )q . (D.21)

The coefficients k1, k2, k3, . . . and kr−1 corresponding to the unrepeated roots can be calculated using the Heaviside formula, Eq. (D.6). The last coefficient

kr,q can also be calculated using Eq. (D.6) as follows:

kr,q = [

(s − pr )q N1(s)

D(s)

]

s=pr . (D.22)

However, the coefficients kr,m for 1 ≤ m ≤ (q−1), corresponding to the repeated poles, cannot be calculated using Eq. (D.6). Instead, these coefficients

are calculated using the following formula

kr,m = 1

(q − m)!

[ dq−m

dsq−m (s − pr )

q N1(s)

D(s)

]

s=pr

for 1 ≤ m ≤ (q − 1).

(D.23)

Example D.5

For the function

X (s) = s3 + 10s2 + 27s + 20

(s + 1)(s + 2)3 , (D.24)

(i) calculate the partial fraction expansion;

(ii) based on your answer to (i), calculate the inverse Laplace transform of

X (s).

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821 D Partial fraction expansion

Solution

(i) The partial fraction expansion of Eq. (D.24) is given by

X (s) = s3 + 10s2 + 27s + 20

(s + 1)(s + 2)3 ≡

k1 s + 1

+ k2,1

s + 2 +

k2,2 (s + 2)2

+ k2,3

(s + 2)3 .

The partial fraction coefficient k1is calculated using the Heaviside formula, Eq. (D.6), as follows:

k1 = s3 + 10s2 + 27s + 20

(s + 2)3

∣

s=−1

= 2

1 = 2.

The partial fraction coefficient kr,3 is calculated using Eq. (D.22) as follows:

k2,3 = s3 + 10s2 + 27s + 20

s + 1

∣

s=−2

= −8 + 40 − 54 + 20

−1 = 2.

The remaining partial fraction coefficients are calculated using Eq. (D.22) as

follows:

k2,2 =

{ 1

(3 − 2)!

[ s3 + 10s2 + 27s + 20

s + 1

]}

s=−2

{ 1

(s + 1)2

[

(s + 1) d

ds (s3 + 10s2 + 27s + 20)

−(s3 + 10s2 + 27s + 20) d

ds (s + 1)

]}

s=−2

{ 1

(s + 1)2 [(s + 1)(3s2 + 20s + 27) − (s3 + 10s2 + 27s + 20)]

}

s=−2

{ 1

(s + 1)2 [2s3 + 13s2 + 20s + 7]

}

s=−2

= 3

and

k2,1 =

{ 1

(3 − 1)!

ds2

[ s3 + 10s2 + 27s + 20

s + 1

]}

s=−2

{ 1

[ 2s3 + 13s2 + 20s + 7

(s + 1)2

]}

s=−2

= 1

{ 1

(s + 1)4

[

(s + 1)2 d

ds (2s3 + 13s2 + 20s + 7)

−(2s3 + 13s2 + 20s + 7) d

ds (s + 1)2

]}

s=−2

= 1



  

  

(s + 1)4 ︸︷︷︸



(s + 1)2 ︸︷︷︸

(6s2 + 26s + 20) ︸︷︷︸

=−8

− (2s3 + 13s2 + 20s + 7) ︸︷︷︸

(2s + 2) ︸︷︷︸

=−2











s=−2

= 1

2 {−8 + 6} = −1.

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822 Appendix D

Therefore, the partial fraction expansion for X (s) is given by

X (s) = 2

s + 1 −

s + 2 +

(s + 2)2 +

(s + 2)3 . (D.25)

(ii) Assuming that the inverse Laplace transform x(t) is right-sided, we use Table 6.1 to determine the inverse Laplace transform x(t) of the X (s):

x(t) = (2e−t − e−2t + 3te−2t + t2e−2t )u(t) = [2e−t + (t2 + 3t − 1)e−2t ]u(t). (D.26)

D.2 Continuous-time Fourier transform

The partial fraction expansion method, described above, may also be applied

to decompose the CTFT functions to a summation of simpler terms. Consider

the following rational function for CTFT:

X (ω) = N (ω)

D(ω) =

bm( jω) m + bm−1( jω)m−1 + · · · + b1( jω) + b0

an( jω) n + an−1( jω)n−1 + · · · + a1( jω) + a0

, (D.27)

where the numerator N (ω) is a polynomial of degree m and the denominator D(ω) is a polynomial of degree n. If m ≥ n, we can divide N (ω) by D(ω) and express X (ω) as follows:

X (ω) = m−n ∑

ℓ=0

αℓ( jω) −ℓ +

N1(ω)

D(ω) ︸︷︷︸

X1(ω)

. (D.28)

The procedure for decomposing X1(ω) in simpler terms remains the same as that discussed for the Laplace transform, except that the expansion is now made

with respect to (jω). For example, if the denominator polynomial D(ω) has n first-order, non-repeated roots, p1, p2, . . . , pn , such that

X1(ω) = N1(ω)

D(ω) =

N1(ω)

( jω − p1)( jω − p2) · · · ( jω − pn) , (D.29)

the function X1(ω) may be decomposed as follows:

N1(ω)

D(ω) =

k1 jω − p1

+ k2

jω − p2 + · · · +

kn jω − pn

, (D.30)

where the partial fraction coefficients kr are calculated using the Heaviside formula:

kr =

[

( jω − pr ) N1(ω)

D(ω)

]

jω=pr

. (D.31)

Using the CTFT pair

e−at u(t) CTFT ←→

a + jω ,

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823 D Partial fraction expansion

the inverse CTFT of Eq. (D.30) is given by

x1(t) = (k1ep1t + k2ep2t + · · · + knepn t )u(t). (D.32)

Similarly, the complex roots and repeated roots may be expanded in partial

fractions by following the procedure outlined for the Laplace transform.

Example D.6

Using the partial fraction method, calculate the inverse CTFT of the following

function:

X (ω) = 2( jω) + 7

( jω)3 + 10( jω)2 + 31( jω) + 30 . (D.33)

Solution

The characteristic equation of X (ω) is given by

(jω)3 + 10(jω)2 + 31(jω) + 30 = 0,

which has roots at jω = −2, −3, and −5. The partial fraction expansion of X (ω) is therefore given by

X (ω) = 2( jω) + 7

( jω + 2)( jω + 3)( jω + 5) ≡

k1 jω + 2

+ k2

jω + 3 +

k3 jω + 5

The partial fraction coefficients are calculated using the Heaviside formula:

k1 =

[

( jω + 2) 2( jω) + 7

( jω + 2)( jω + 3)( jω + 5)

]

jω=−2

= 1,

k2 =

[

( jω + 3) 2( jω) + 7

( jω + 2)( jω + 3)( jω + 5)

]

jω=−3

= − 1

2 ,

and

k3 =

[

( jω + 5) 2( jω) + 7

( jω + 2)( jω + 3)( jω + 5)

]

jω=−5

= − 1

2 .

Therefore, the partial fraction expansion of X (ω) is given by

X (ω) = 1

jω + 2 −

( jω + 3) −

( jω + 5) . (D.34)

Using Table 5.2, the inverse DTFT x(t) of X (ω) is given by

x(t) =

[

e−2t − 1

2 e−3t −

2 e−5t

]

u(t). (D.35)

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824 Appendix D

Example D.7

Using the partial fraction method, calculate the inverse CTFT of the following

function:

X (ω) = 4( jω)2 + 20( jω) + 19

( jω)3 + 5( jω)2 + 8( jω) + 4 . (D.36)

Solution

The characteristic equation of X (ω) is given by

( jω)3 + 5( jω)2 + 8( jω) + 4 = 0,

which has roots at jω = −1, −2, and −2. The partial fraction expansion of X (ω) is therefore given by

X (ω) = 4( jω)2 + 20( jω) + 19

( jω)3 + 5( jω)2 + 8( jω) + 4 ≡

k1 ( jω + 1)

+ k2,1

( jω + 2) +

k2,2 ( jω + 2)2

The partial fraction coefficients k1 and k2,2 are calculated using the Heaviside formula:

k1 =

[

( jω + 1) 4( jω)2 + 20( jω) + 19

( jω + 1)( jω + 2)2

]

jω=−1

= 3

and

k2,2 =

[

( jω + 2)2 4( jω)2 + 20( jω) + 19

( jω + 1) ( jω + 2)2

]

jω=−2

= 5.

The remaining partial fraction coefficient is calculated using Eq. (D.23):

k2,1 = 1

(2 − 1)!

[ d

d( jω)

4( jω)2 + 20( jω) + 19

( jω + 1)

]

jω=−2

, (D.37)

where the differentiation is with respect to jω. To simplify the notation for

differentiation, we substitute s = jω in Eq. (D.37) to obtain:

k2,1 = 1

(2 − 1)!

[ d

4s2 + 20s + 19

(s + 1)

]

s=−2

[ (s + 1)(8s + 20) − (4s2 + 20s + 19)

(s + 1)2

]

s=−2

= 1.

The partial fraction expansion of X (ω) is therefore given by

X (ω) = 4( jω)2 + 20( jω) + 19

( jω)3 + 5( jω)2 + 8( jω) + 4 =

( jω + 1) +

( jω + 2) +

( jω + 2)2 .

(D.38)

Using Table 5.2, the inverse CTFT x(t) of X (ω) is given by

x(t) = [3e−t + e−2t + 5te−2t ]u(t). (D.39)

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825 D Partial fraction expansion

D.3 Discrete-time Fourier transform

To illustrate the partial fraction expansion of the DTFT, consider the following

rational function:

X (Ω) = N (Ω)

D (Ω) =

bmejmΩ + bm−1ej(m−1)Ω + · · · + b1ejΩ + b0 anejnΩ + an−1ej(n−1)Ω + · · · + a1ejΩ + a0

, (D.40)

where the numerator N (Ω) is a polynomial of degree m and the denominator D(Ω) is a polynomial of degree n. An alternative representation for Eq. (D.40) is obtained by dividing both the numerator and the denominator by ejnΩ as

follows:

X (Ω) = N (Ω)

D(Ω) = ej(m−n)Ω ·

bm + bm−1e−jΩ + · · · + b1e−j(m−1)Ω + b0e−jmΩ

an + an−1e−jΩ + · · · + a1e−j(n−1)Ω + a0e−jnΩ ︸︷︷︸

X ′ (ω)

(D.41)

We need to express Eq. (D.41) in simpler terms using the partial fraction expan-

sion with respect to e−jΩ. To simplify the factorization process, we substitute

z = ejΩ:

X (z) = z(m−n) · bm + bm−1z−1 + · · · + b1z−(m−1) + b0z−m

an + an−1z−1 + · · · + a1z−(n−1) + a0z−n . (D.42)

The process for the partial fraction expansion of Eq. (D.41) is the same as for

the CTFT and Laplace transform, except that the expansion is performed with

respect to z−1. Below we illustrate the process with an example.

Example D.8

Using the partial fraction method, calculate the inverse CTFT of the following

function:

X (Ω) = N (Ω)

D(Ω) =

2ej2Ω − 5ejΩ

ej2Ω − (4/9)ejΩ + (1/27) . (D.43)

Solution

Dividing both the numerator and the denominator of Eq. (D.43) by ej2Ω yields

X (Ω) = 2 − 5e−jΩ

1 − (4/9)e−jΩ + (1/27)e−2jΩ .

Substitute z = ejΩ in the above equation to obtain

X (z) = 2 − 5z−1

1 − (4/9)z−1 + (1/27)z−2 ,

with the characteristic equation

1 − 4

9 z−1 +

27 z−2 = 0 or z2 −

9 z +

27 = 0,

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826 Appendix D

which has two poles at z = 1/3 and 1/9. The partial fraction expansion of X (z) is therefore given by

X (z) = 2 − 5z−1

(1 − (1/3)z−1)(1 − (1/9)z−1) ≡

k1 1 − (1/3)z−1

+ k2

1 − (1/9)z−1 .

Using the Heaviside formula, the partial fraction coefficients are given by

k1 =

[

(1 − (1/3)z−1) 2 − 5z−1

(1 − (1/3)z−1)(1 − (1/9)z−1)

]

z−1=3

= − 39

and

k2 =

[

(1 − (1/9)z−1) 2 − 5z−1

(1 − (1/3)z−1)(1 − (1/9)z−1)

]

z−1=9

= 43

2 .

The partial fraction expansion of Eq. (D.43) is given by

X (z) = − 39

1 − (1/3)z−1 +

1 − (1/9)z−1 .

We substitute z = ejΩ = z to express the above equation in terms of the discrete frequency Ω as follows:

X (Ω) = − 39

1 − (1/3)e−jΩ +

1 − (1/9)e−jΩ .

Using Table 11.2, the inverse DTFT x[k] of X (ejΩ) is given by

x(t) =

[

− 39

( 1

+ 43

( 1

)k ]

u[k]. (D.44)

D.4 The z-transform

The partial fraction expansion method can also be applied to evaluate the inverse

transform of the z functions. Consider a z function of the following form:

X (z) = N (z)

D(z) =

bm zm + bm−1zm−1 + · · · + b1z + b0 anzn + an−1zn−1 + · · · + a1z + a0

(D.45)

X (z) = N (z)

D(z) = zm−n

bm + bm−1z−1 + · · · + b1z−(m−1) + b0z−m

an + an−1z−1 + · · · + a1z−(n−1) + a0z−n . (D.46)

Either of the two forms, Eq. (D.45) or Eq. (D.46), may be used to calculate

the partial fraction expansion and eventually the inverse z-transform. If we use

the format specified in Eq. (D.45), the partial fraction of the function X (z)/z is performed with respect to z. As illustrated in Example D.9, the partial fraction of X (z)/z leads to expansion terms for which the inverse z-transform is readily available in Table 13.1. If instead Eq. (D.46) is used, the partial fraction of the

function X (z) is performed with respect to z−1. We illustrate the procedure for both formats in Examples D.9 and D.10.

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827 D Partial fraction expansion

Example D.9

Using Eq. (D.45) for the partial fraction expansion, calculate the inverse

z-transform of the following function:

X (z) = z2 − 3z

z3 − z2 + 0.17z + 0.028 . (D.47)

Solution

The transform X (z) is expressed in the following form:

X (z)

z =

z − 3 z3 − z2 + 0.17z + 0.028

, (D.48)

which has poles at z = −0.1, 0.4, and 0.7. The partial fraction expansion of Eq. (D.48) is given by

X (z)

z =

z − 3 z3 − z2 + 0.17z + 0.028

≡ k1

z + 0.1 +

k2 z − 0.4

+ k3

z − 0.7 .

The partial fraction coefficients are calculated using the Heaviside formula:

k1 =

[

(z + 0.1) z − 3

(z + 0.1)(z − 0.4)(z − 0.7)

]

z=−0.1

= − 31

4 ,

k2 =

[

(z − 0.4) z − 3

(z + 0.1)(z − 0.4)(z − 0.7)

]

z=0.4

= 52

3 ,

and

k3 =

[

(z − 0.7) z − 3

(z + 0.1)(z − 0.4)(z − 0.7)

]

z=0.7

= − 115

12 .

The partial fraction expansion is given by

X (z)

z = −

(z + 0.1) +

(z − 0.4) −

115

(z − 0.7)

X (z) = − 31

(z + 0.1) +

(z − 0.4) −

115

(z − 0.7) .

Assuming a right-sided sequence, the inverse z-transform x[k] of the X (z) is given by

x[k] =

[

− 31

4 (−0.1)k +

3 (0.4)k −

115

12 (0.7)k

]

u [k] .

Example D.10

Using Eq. (D.46) for the partial fraction expansion, calculate the inverse z-

transform of the following function:

X (z) = z2 − 3z

z3 − z2 + 0.17z + 0.028 .

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828 Appendix D

Solution

The transform X (z) is expressed in the following form:

X (z) = z−1 − 3z−2

1 − z−1 + 0.17z−2 + 0.028z−3 , (D.49)

which has poles at z = −0.1, 0.4, and 0.7. The partial fraction expansion of Eq. (D.49) is given by

X (z) = z−1 − 3z−2

1 − z−1 + 0.17z−2 + 0.028z−3

≡ k1

1 + 0.1z−1 +

k2 1 − 0.4z−1

+ k3

1 − 0.7z−1 .

The partial fraction coefficients are calculated using the Heaviside formula:

k1 =

[

(1 + 0.1z−1) z−1 − 3z−2

(1 + 0.1z−1)(1 − 0.4z−1)(1 − 0.7z−1)

]

z−1=−10

= − 31

4 ,

k2 =

[

(1 − 0.4z−1) z−1 − 3z−2

(1 + 0.1z−1)(1 − 0.4z−1)(1 − 0.7z−1)

]

z−1=10/4

= 52

3 ,

and

k3 =

[

(1 − 0.7z−1) z−1 − 3z−2

(1 + 0.1z−1)(1 − 0.4z−1)(1 − 0.7z−1)

]

z−1=10/7

= − 115

12 .

The partial fraction expansion is given by

X (z) = − 31

k1 (1 + 0.1z−1)

+ 52

k2 (1 − 0.4z−1)

− 115

k3 (1 − 0.7z−1)

Assuming a right-sided sequence, the inverse z-transform x[k] of the X (z) is given by

x [k] =

[

− 31

4 (−0.1)k +

3 (0.4)k −

115

12 (0.7)k

]

u[k] .

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Appendix E Introduction to M A T L A B

E.1 Introduction

M A T L A B , an abbreviation for the term “MATrix LABoratory,” is a powerful

computing environment for numerical calculations and multidimensional visu-

alization. It has become a de facto industry standard for developing engineering applications for several reasons. First, M A T L A B reduces programming to data

processing abstraction. Instead of becoming bogged down with the intrinsic

details of programming, as required with other high-level languages, it allows

the user to focus on the theoretical concepts. Developing code in M A T L A B

takes a fraction of the time necessary with other programming languages. Sec-

ondly, it provides a rich collection of library functions, referred to as toolboxes,

in virtually every field of engineering. The user can access the library functions

to build the required application. Thirdly, it supports multidimensional visual-

ization that allows experimental data to be rendered graphically in a compre-

hensible format.

In this appendix we provide a brief introduction to M A T L A B . Our intention

is to introduce the basic capabilities of M A T L A B so that the reader can start

working on the problems contained in this text. In the following discussion,

M A T L A B commands and results are shown in “Courier” font with the com-

mands preceded by the >> prompt. Results returned by M A T L A B in response

to the typed commands are also shown in the “Courier” font but are not preceded by the >> prompt.

Starting a MA T L A B session

M A T L A B is available on a variety of computing platforms. On an IBM com-

patible PC, a M A T L A B session can be initiated by selecting the M A T L A B

program or double clicking on its icon. In an X-window system, M A T L A B

is invoked by typing the complete path to the executable file of M A T L A B at

the shell prompt. Before using M A T L A B , it is recommended that you cre-

ate a subdirectory named 〈matlab〉 (all lower case letters for case-sensitive

829

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830 Appendix E

operating systems) in your home directory. Any file placed in this subdirectory

can be accessed from within the M A T L A B environment without specifying the

complete path of the file.

M A T L A B includes a comprehensive combination of demos to illustrate the

offered features and capabilities to its users. In order to explore the demo, just

type demo at the command line of the M A T L A B environment indicated by the

>> prompt:

>> demo

This will open the M A T L A B demo window. Follow the interactive options by

clicking on the features that interest you. In most cases, the M A T L A B code

used to generate the demo is also included for illustration.

Help in M A T L A B

M A T L A B provides a useful built-in help facility. You can access help either

from the command line or by clicking on the graphical “Help” menu. On the

command line, the format for obtaining help on a particular M A T L A B function

is to type help followed by the name of the function. For example, to learn

more about the plot function, type the following instruction in the M A T L A B

command window:

>> help plot

If the name of the function is not known beforehand, you can use the lookfor

command followed by a keyword that identifies the function being searched,

to enlist the available M A T L A B functions with the specified keyword. For

example, all M A T L A B functions with the keyword “Fourier” can be listed by

typing the following command:

>> lookfor Fourier

On execution of the above command, M A T L A B returns the following list,

specifying the names of the functions and a brief comment on their capabilities:

FFT Discrete Fourier transform.

FFT2 Two-dimensional discrete Fourier Transform.

FFTN N-dimensional discrete Fourier Transform.

IFFT Inverse discrete Fourier transform.

IFFT2 Two-dimensional inverse discrete Fourier trans-

form.

IFFTN N-dimensional inverse discrete Fourier transform.

XFOURIER Graphics demo of Fourier series expansion.

DFTMTX Discrete Fourier transform matrix.

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831 E Introduction to M A T L A B

E.2 Entering data into M A T L A B

Data can be entered in the M A T L A B as a scalar quantity, a row or column vector,

and a multidimensional array. In each case, both real and complex numbers can

be entered. As required in other high-level languages, there is no need to declare

the type of a variable before assigning data to it. For example, variable a can

be assigned the value (6 + j8) by typing the following command:

>> a = 6 + j*8

On the execution of the above command, M A T L A B returns the following

answer:

a = 6.0000 + 8.0000i

In the above command, we did not allocate any value to j, yet M A T L A B

recognized it as a complex operator with value j2 = 1. There is a whole range of special words that are used by M A T L A B either as the name of functions or

variables. These include pi, i, j, Inf, NaN, sin, cos, tan, exp, and rem.

Type help elfun to list the names that are used by M A T L A B to specify the

built-in functions and variables. The value of any of these special words can be

changed by assigning a new value to it. For example,

>> sin = 1

allocates the value of 1 to the variable sin. The M A T L A B definition of

the trigonometric sine is overwritten by our command. To check the cur-

rent status of the runtime environment of M A T L A B , type whos at the

prompt:

>> whos

M A T L A B returns the following answer:

Name Size Bytes Class

a 1x1 16 double array (complex)

sin 1x1 8 double array

Grand total is 2 elements using 24 bytes

Alternatively, the command who can also be used to list the name of defined

variables in the M A T L A B runtime environment. The command who does not

provide additional details such as the size and class of each variable. In the

preceding discussions, we overwrote the sin function and allocated a value of

1 to it. Consequently, we cannot access the M A T L A B built-in function sin

to evaluate the sine of an angle. To clear our definition of sin, we can use the

following command:

>> clear sin

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832 Appendix E

The original definition ofsin is restored in the M A T L A B environment. Typing

>> sin(pi/6)

calls up the built-in sin function with π /6 as the input argument. Recall

that the variable pi is a built-in variable that has been assigned the value

of 3.141 596 25. M A T L A B returns

ans =

0.5000

after execution of the sin command. For additional information on the sin

function, type help sin. To allocate the returned value of sin(pi/6)to

variable x, for example, type

>> x = sin(pi/6)

which returns

x =

0.5000

In the above examples, M A T L A B displays the result of each instruction. The

display can be suppressed by inserting a semicolon at the end of each instruction.

For example, the command

>> x = sin(pi/6);

initializes x = 0.5000 without displaying the end result.

Most common arithmetic operations are available in M A T L A B . These

include + (add), − (subtract), ∗ (multiply), / (divide), ∧ (power), .∗ (array multiplication), and ./ (array division). For complex numbers, in addition to

the aforementioned operators, M A T L A B provides a collection of library func-

tions that can be used to perform more complex operations. These are illustrated

through the following example, where a brief explanation of each instruction is

included as a comment. In M A T L A B , the segment of line after the % sign on the

same line are treated as comments and ignored during execution. The returned

value is enclosed in parentheses and is also included with the explanation.

>> x = 2.3 - 4.7*i; % Initializes x as a complex

% variable.

>> x magn = abs(x); % magnitude of x, (5.2326)

>> x phas = angle(x); % phase of x in radians/s, (1.1157)

>> x real = real(x); % Real component of x, (2.3)

>> x imag = imag(x); % Imaginary component of x,(-4.7)

>> x conj = conj(x); % Complex conjugate of x,

% (2.3 + 4.7i)

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833 E Introduction to M A T L A B

M A T L A B also provides a set of functions for decimal numbers. If applied to

integers, these functions do not make any changes. On the other hand, if these

functions are applied to complex numbers, each operation is performed individ-

ually on the real and imaginary component. Below we provide a selected list.

>> x = 2.3 - 4.7*i; % Initializes x as a complex

% variable

>> x round = round(x); % rounds to nearest

% integer, (2 – 5i)

>> x fix = fix(x) % rounds to nearest integer

% towards zero, (2 – 4i)

>> x floor = floor(x) % rounds down (towards negative

% infinity), (2 – 5i)

>> x ceil = ceil(x) % rounds up (towards positive

% infinity) (3 – 4i)

We now consider initialization of multidimensional arrays through a series of

examples.

Example E.1

Consider the two row vectors

f = [

1, 4, −2, (3 − 2i) ]

and

g = [

−3, (5 + 7i), 6, 2 ]

Perform the following mathematical operations in M A T L A B on vectors f and g:

(i) addition, r1 = f + g; (ii) dot product, r2 = f · g;

(iii) mean, r3 = 1

4 ∑

k=1 f (k);

(iv) average energy, r4 = 1

4 ∑

k=1 | f (k)|2;

(v) variance, r5 = 1

4 ∑

k=1 | f (k) − r3|2, where r3 is defined in (iii).

Solution

The M A T L A B code to solve part (i) is given below with comments following

the % sign:

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834 Appendix E

>> f = [1 4 -2 3-2*i]; % initialize f

>> g = [-3 5+7*i 6 2]; % initialize g

>> r1 = f + g % Calculate the sum of f and g

% The result is displayed due

% to the absence of a

% semicolon at the end of the

% instruction

results in the following value for r1:

r1 =

-2.0000 9.0000+7.0000i 4.0000 5.0000-2.0000i

which can be confirmed by direct addition of vectors f and g. (ii) To compute part (ii), we use the M A T L A B function dot as follows:

>> r2 = dot(f,g) % dot returns dot product btw f and g

which returns

r2 =

11.0000+32.0000i

An alternative approach to compute the dot product is to multiply the row vector

f by the conjugate transpose of g. The transpose is needed to make the two

vectors conformable for multiplication. You may verify that the instruction

>> r2 = g*f’; % alternative expression for calculating

% the dot product. Operator ’ denotes

% complex-conjugate transpose

returns the same value as above.

(iii) The instruction for part (iii) is as follows:

>> r3 = sum(f)/length(f) % sum(f) adds all row entries

% of vector f length(f)

% returns no. of entries in f

which returns

r3 =

1.5000 − 0.5000i

(iv) The instruction for part (iv) is as follows:

>> r4 = sum(f.*conj(f))/length(f)

% Operation f.*g does an element by

% element multiplication of vectors f

% and g. Operation conj(f) takes complex

% conjugate of each entry in f

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835 E Introduction to M A T L A B

which returns

r4 =

8.5000

(v) To compute part (v), we can modify the code in part (iv) by preceding it

with the following instruction:

>> f zero mean = f – mean(f);% mean(f) computes the

% average value of f

>> r5 = sum(f zero mean.*conj(f zero mean))/length

(f zero mean)

which returns

r5 =

As a final note to our introduction on vectors, the second element of vector f

can be accessed by the instruction

>> f(2)

which returns

ans =

A range of elements within a vector can be accessed by specifying the integer

index numbers of the elements. To access elements 1 and 2 of row vector f, for

example, we can type the instruction

>> f(1:1:2);

Similarly, the odd number elements in f can be accessed by the instruction

>> x = f(1:2:length(f));

where we have assigned the returned value to a new variable x. Code

1:2:length(f) is referred to as a range-generating statement that generates

a row vector. The first element of the row vector is specified by the left-most

number (1 in our example). The next element in the row vector is obtained by

adding the middle element (2 in our example) to the first element and proceed-

ing all the way till the limit (length(f)) is reached. The middle element (2 in

our example) specifies the increment, while the third element (length(f))

is the ending index. If the increment is missing, M A T L A B assigns a default

value of 1 to it. As another example, the range-generating statement 1:11 pro-

duces the row vector [1 2 3 4 5 6 7 8 9 10 11]. Further, the start-

ing index, increment, or ending index can also be real-valued numbers. The

range-generating statement [0.1:0.1:0.9] produces the row vector [0.1

0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9].

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836 Appendix E

Example E.2

Initialize the following matrix:

A = [

2 4 −1 0 5 2 3 9

]

and take the pseudo-inverse of A, defined as A+ = (AT A)−1 AT with T denoting the conjugate transpose operation.

Solution

The following M A T L A B code initializes matrix A:

>> A = [2 4 -1 0;5 2 3 9]; % The semicolon inside square

% parenthesis separates

% adjacent rows of a matrix

An alternative but longer set of instructions for the initialization of A is as follows:

>> A(1,1)=2; A(1,2)=4; A(1,3)=-1; A(1,4)=0;

>> A(2,1)=5; A(2,2)=2; A(2,3)=-3; A(2,4)=9;

To calculate the pseudo-inverse of A, the following instruction may be used:

>> Ainverse = inv(A’*A)*A’; % Function inv calculates

% inverse of a matrix

% while ’ denotes conjugate

% transpose

which returns a warning that the matrix is singular. From linear algebra, we

know that the inverse of a matrix only exits if it is non-singular, hence the

pseudo-inverse does not exist for the above choice of A.

Example E.3

Initialize the following discrete-time function:

f [k] = 2∗ cos (

15 ∗k

)

for 0 ≤ k ≤ 30.

Solution

As in other high-level languages, we can use a for statement to initialize the

function f . The code is given by

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837 E Introduction to M A T L A B

for k = 0:1:30,

f(k+1) = 2*cos(pi/15*k); % In MATLAB, the index of a

% vector or a matrix must

end % not be zero.

In M A T L A B , the index of a vector or matrix cannot be zero. Therefore, we use

two row vectors k and f to store the DT function. The row vector k specifies

the time indices at which function f is evaluated, while f contains the value

of the DT function at the corresponding time index stored in k. The above

initialization can also be performed in M A T L A B more quickly and in a much

more compact way.

clear % user-defined variables are cleared

k = 0:30; % k is a row vector of dimensions 1x30

f = 2*cos(5*k) % f has the same dimensions as k

returns the following answer:

f =

Columns 1 through 7

2.0000 0.5673 -1.6781 -1.5194 0.8162 1.9824 0.3085

Columns 8 through 14

-1.8074 -1.3339 1.0506 1.9299 0.0443 -1.9048 -1.1249

Columns 15 through 21

1.2666 1.8435 -0.2208 -1.9688 -0.8961 1.4603 1.7246

Columns 22 through 28

-0.4819 -1.9980 -0.6516 1.6284 1.5754 -0.7346 -1.9922

Columns 29 through 31

-0.3956 1.7677 1.3985

In terms of execution time, implementation 2 is more efficient than the first

implementation. Since M A T L A B is an interpretive language, loops take a long

time to be executed. An efficient M A T L A B code avoids loops and, if possible,

replaces them with matrix or vector multiplications.

Example E.4

Initialize the following DT function:

g[k] = f [k] for 0 ≤ k ≤ 6.

Solution

In the above example, it has been assumed that the matrix f has been initialized as per Example E.3. The following M A T L A B code will initialize row vector

>> g = f(1:7);

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838 Appendix E

If missing, a default value of 1 is assumed as the increment in the range-

generating statement(1:7). Therefore,g(1:7) is equivalent tog(1:1:7).

E.3 Control statements

M A T L A B supports several other loop statements (while, switch, etc.) as

well as theif-else statement. In functionality, these statements are similar to

their counterparts in C but the syntax is slightly different. In the following, we

provide examples for some of the loop and conditional statements by providing

analogy with the C code. Readers who are unfamiliar with C can skip the C

instructions and study the explanatory comments that follow.

Example E.5

Consider the following set of instructions in C:

int X[2][2] ={ {2, 5},{4,6} }; /* initialize matrix X */

int Y[2][2] ={ {1, 5},{6,-2} }; /* initialize matrix Y */

int Z[2][2]; /* declare Z */

for (m = 1; m <= 2; m++) {

Z[m][n] = X[m][n] + Y[m][n];

/* Z = X + Y */

}

Write down the equivalent M A T L A B code for the above instructions. Can the

M A T L A B code be simplified?

Solution

Implementation 1 Following a step-by-step conversion of the C code into M A T L A B yields

>> X = [2 5; 4 6] % X is initialized

>> Y = [1 5; 6 -2] % Y is initialized

>> for m = 1:2,

for n = 1:2,

Z(m,n) = X(m,n)+Y(m,n);

end

Implementation 2 Thefor loops in M A T L A B can be replaced by thewhile statement as follows:

>> X = [2 5; 4 6] % X is initialized

>> Y = [1 5; 6 -2] % Y is initialized

>> m = 1;

>> while (m < 3),

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839 E Introduction to M A T L A B

n = 1;

while (n < 3),

Z(m,n) = X(m,n)+Y(m,n);

n = n + 1;

end

m = m + 1;

end

Implementation 3 We can avoid the twofor orwhile loops by performing a direct sum of matrices X and Y as follows:

>> X = [2 5; 4 6] % X is initialized

>> Y = [1 5; 6 -2] % Y is initialized

>> Z = X + Y;

Compared with the first two implementations, the third implementation is

cleaner and faster.

Example E.6

Consider the following set of instructions in C:

int a = 15; /* initialize scalar a */

int x; /* declare x */

if (a > 0)

x = 5; /* initialize x to 5 if a > 0*/

else

x = 100; /* initialize x to 5 if a <= 0 */

Write down the equivalent M A T L A B code.

Solution

Following a step-by-step conversion, we obtain the following equivalent set of

instructions in M A T L A B :

>> a = 15;

>> if a > 0,

x = 5;

else,

x = 100

end

While using the conditional statements, relational operators such as equal to,

not equal to, or less than are generally required in the code. M A T L A B provides

six basic relational operators which are defined in Table E.1.

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840 Appendix E

Table E.1. Relational operations available in

M A T L A B

Relational operator Definition

< less than

> greater than

== equal to ∼= not equal to

<= less than or equal to

>= greater than or equal to

E.4 Elementary matrix operations

M A T L A B provides several built-in functions to manipulate matrices. In the

following, we provide a brief description of some of the important matrix oper-

ations. Consider the instruction

>> f = exp(0.05*[1:30]); % Initialize row vector f

which initializes the row vector f according to the following definition:

f [k] = e0.05k for 1 ≤ k ≤ 30.

The following M A T L A B instructions provide examples of basic arithmetic

operations performed on a row or column vector. Comments against each

instruction provide a brief description of the instruction, with the value returned

by M A T L A B enclosed in parenthesis:

>> f max = max(f); % Maximum value in f (4.4817)

>> f min = min(f); % Minimum value in f (1.0513)

>> f sum = sum(f); % Sum of all entries in f (71.3891)

>> f prod = prod(f); % Product of entries in

f (1.2513e+10)

>> f mean = mean(f); % Mean of entries in f (2.3796)

>> f var = var(f); % Variance of entries in f (1.0578)

>> f size = size(f); % Dimensions of f ([1 30])

>> f length = length(f); % Length of f (30)

>> fprintf(‘\nThe min value of all matrix elements =

%f\n’, f min); % Prints the variable f min

The fprintf instruction at the end of the code is used to print the value of

the variable f min onto the screen. It returns

The min value of all matrix elements = 1.051300

The aforementioned instructions can alternatively be used for matrices and

higher dimensional arrays. The syntax stays the same, but the result may be

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841 E Introduction to M A T L A B

different. For matrices, for example, the specified operation is performed on

each column of the matrix and a row vector is returned as the answer. For

example, consider the matrix F initialized by the following instruction:

>> F = magic(5); % magic(N) returns an (N x N) matrix

% with entries between 1 through

% N∧2 having equal row, column, and

% diagonal sums

For matrix F, the values indicated in the comments are returned:

>> F max = max(F); % Maximum value along each column

% [23 24 25 21 22]

>> F min = min(F); % Minimum value along each column

% [ 4 5 1 2 3]

>> F sum = sum(F); % Sum of entries along each column

% [65 65 65 65 65]

>> F prod = prod(F); % Product of entries along each

% column

% [172040 155520 43225 94080 142560]

>> F mean = mean(F); % Mean of entries along each column

% [13 13 13 13 13]

>> F var = var(F); % Variance of entries along each

% column

% [52.5 65.0 90.0 65.0 52.5]

>> F size = size(F); % Dimensions of F; [5 5]

>> F length

= length(F); % Returns number of rows in F (5)

For completeness, we also include a list of some basic matrix operations, some

of which were introduced in Section E.1:

>> X = [2 5; 4 6]; % Initailize (2 x 2) matrix X

>> Y = [1 5; 6 -2]; % Initailize (2 x 2) matrix Y

>> Zsum = X + Y; % Adds matrices of equal dimensions.

% Returns [3 10; 10 4]

>> Zdif = X - Y; % Subtracts matrices of equal

% dimensions;

% Returns [1 0; -2 8]

>> Zprod = X*Y; % Multiplies matrices conformable for

% multiplication; Returns

% [32 0; 40 8].

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>> Ztran = X’; % Calculates transpose of X

% Returns [2 4; 5 6]

>> Zinv = inv(X); % Inverts X

% Returns [-0.75 0.62; 0.50 -0.25]

>> Zarraymul = X.*Y; % Element by element multiplication

% Returns [2 25; 24 -12]

>> Zarraydiv = X./Y; % Element by element division

% Returns [2 1; 0.6667 -3]

>> Zpower1 = X.∧2; % Each element is raised to power

% by 2

% Returns [4 25; 26 36]

>> Zpower2 = X.∧Y; % Each element in X is raised to

% power by its corresponding

% element in Y

% Returns [2 3125; 4096 0.028]

E.5 Plotting functions

M A T L A B supports multidimensional visualization that allows experimental

data to be rendered graphically in a comprehensible format. In this section, we

will focus on 2D plots for continuous-time and discrete-time variables. Readers

should check the demo for more advanced graphics including 3D plots.

Example E.7

Plot the following function:

f [k] = 2 cos(0.5k)

as a function of k for the range −20 ≤ k ≤ 20.

Solution

The following set of M A T L A B instructions will generate and plot the function:

>> k = -20:20; % Initializes k as a (1 x 41)

% row vector

>> f = 2*cos(0.5*k); % Initializes f as cos(0.5k)

>> figure(1); % selects figure 1 where plot

% is drawn

>> plot(k,f); grid on; % CT plot of f (ordinate)

% versus k (abscissa)

% Grid is turned on

>> xlabel(‘k’); % Sets label of X-axis to k

>> ylabel(‘f[k]’); % Sets label of Y-axis to f[k]

>> axis([-25 25 -3 3]) % Plot is viewed in the range

% given by

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843 E Introduction to M A T L A B

−20 −3

−2

−1

−10 0 10 20

f [k

]

−20 −3

−2

−1

−10 0 10 20

f [k

]

(a) (b)

Fig. E.1. Plots of

f [k ] = 2 cos(0.5k) versus k in the range −20 ≤ k ≤ 20. (a) CT plot; (b) stem DT plot.

% [x-min x-max y-min y-max]

>> print -dtiff plot.tiff % Saves figure in the file

% “plot.tiff” in

% the TIFF format

These instructions produce a continuous plot cosine wave, as shown in

Fig. E.1. It is also possible to construct a discrete-time plot using the stem

function:

>> figure(2)

>> stem(k,f,‘filled’); % DT plot; option ‘filled’

% fills the circles at the

% top of vertical bars

>> xlabel(‘k’); % Sets label of X-axis to k

>> ylabel(‘f[k]’); % Sets label of Y-axis to f[k]

>> axis([-25 25 -3 3])

>> print -dtiff plot2.tiff

Both plot and stem functions have a variety of options available, which may

be selected to change the appearance of the figures. The reader is encouraged

to explore these options by seeking help on these functions in M A T L A B . In

addition, there are several other 2D graphical functions in M A T L A B . These

include semilogx, semilogy, loglog, bar, hist, polar, stairs,

rose, errorbar, compass, and pie.

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3 5

0 2 4 6 8 100

0 0 2 4 6 8 10

150

100

0.5

f 1 [k

]

−0.5

−10 −5 0 5 10−5 0 k k

k k

5 −1

f 3 [k

]

f 2 [k

] f 4 [k

]

(a) (b)

Fig. E.2. Multiple plots sketched

in the same window for Example

E.8.

Plotting multiple graphs in one figure

M A T L A B provides the function subplot to sketch multiple graphs in one

figure. We demonstrate the application of the subplot function through an exam-

ple.

Example E.8

Plot the following functions over the specified range in one figure:

(a) f1[k] = sin(0.1πk) for −5 ≤ k ≤ 5; (b) f2[k] = 2−k for −7 ≤ k ≤ 7;

{

1 (0 ≤ k ≤ 4) 3 (5 ≤ k ≤ 9) ;

(d) f4[k] =

{

k (0 ≤ k ≤ 5) 0 (6 ≤ k ≤ 9).

Solution

The following set of M A T L A B instructions plots the four functions illustrated

in Fig. E.2.

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>> % Part (a)

>> figure(5) % Select figure 5 for plots

>> clf % Clear figure 5

>> k = [-5:5]; % k = [-5 -4 ...0 ...4 5]

>> f1 = sin(0.1*pi*k); % Calculate function f1

>> subplot(2,2,1); % Divides fig 5 into (m = 2)

% vertical and (n = 2)

% horizontal sub-figures.

% The last argument (p = 1)

% accesses sub-figures

% (1 <= p <= m*n).

>> stem(k,f1,‘filled’);

grid on; % DT plot of f1 versus k

>> xlabel(‘k’) ; % Label of X-axis

>> ylabel(‘f1[k]’) % Label of Y-axis

>> % Part (b)

>> k = [-7:7]; % k overwritten to

% [-7 -6 ...0 ...6 7]

>> f2 = 2. ∧ (-k) ; % Calculate function f2

>> subplot(2,2,2); % Select p = 2 sub-figure

>> stem(k,f2,‘filled’);

grid on; % DT plot of f2 versus k

>> xlabel(‘k’); % Label of X-axis

>> ylabel(‘f2[k]’); % Label of Y-axis

>> % Part (c)

>> k = [0:9]; % k overwritten to

[0 1 ...8 9]

>> f3 = [1 1 1 1 1 3 3 3 3 3]; % Calculate function f3

>> subplot(2,2,3); % Select p = 3 sub-figure

>> stem(k,f3,’filled’);

grid on; % DT plot of f3 versus k

>> xlabel(‘k’); % Label of X-axis

>> ylabel(‘f3[k]’); % Label of Y-axis

>> % Part (d)

>> k = [0:9];

>> f4 = [0 1 2 3 4 5 0 0 0 0]; % Calculate function f4

>> subplot(2,2,4); % Select p = 4 sub-figure

>> stem(k,f4, ‘filled’);

grid; % DT plot of f2 versus k

>> xlabel(‘k’); % Label of X-axis

>> ylabel(‘f4[k]’); % Label of Y-axis

>> print -dtiff plot.tiff; % Save the figure as a

% TIFF file

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E.6 Creating M A T L A B functions

In the preceding examples, we have used M A T L A B in an interactive mode

with each instruction individually typed at the command prompt. M A T L A B

allows for the creation of M-files, where instructions can be stored in a file.

An M-file can be of two types, called scripts and functions. A script is a list of M A T L A B instructions that are saved in file with a . m extension. The script

file can access the variables defined in the M A T L A B workspace. Likewise,

all variables declared in the script are accessible to the workspace. For exam-

ple, the instructions to solve part (a) of Example E.8 can be stored in a file

myfirstplot.m as follows:

% Content of script myfirstplot.m

% Part (a)

figure(5) % Select figure 5 for plots

clf % Clear figure 5

k = [-5:5]; % k = [-5 -4 ...0 ...4 5]

f1 = sin(0.1*pi*k); % Calculate function f1

subplot(2,2,1); % Divides fig 5 into (m = 2) vertical

% and (n = 2) horizontal sub-figures

% The last argument (p = 1) accesses

% sub-figures (1 <= p <= m*n)

stem(k,f1,‘filled’);

grid on; % DT plot of f1 versus k

xlabel(‘k’); % Label of X-axis

ylabel(‘f1[k]’) % Label of Y-axis

To executemyfirstplot.m, simply type the name of the M-file (myfirst-

plot in this case) at the command prompt. By executing the function whos,

you can determine that all variables defined in myfirstplot.m are part of

the M A T L A B workspace.

A function in M A T L A B is a special type of script file that can accept input

arguments and return output arguments. Variables declared within a function are

local to the function. Likewise, none of the variables defined in the M A T L A B

working environment are accessible by the function unless these variables are

explicitly passed as an input argument to the function. A function file must

follow a specific format. The first line defines the function by specifying a

name for the function and indicates the number of input and output arguments.

Immediately following the definition, lines that begin with a comment symbol

(%) are printed when help is requested on the function. As an example, we

modify script myfirstplot.m into a function in the following.

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function [f1] = myfirstplot(k)

% USAGE: [f1] = myfirstplot(k)

% Plots f1 = sin(0.1*pi*k) as a function of k in subplot

% (2,2,1) where k = row vector containing the indices

% where f1 is to be defined f1 is the output row vector

figure(5) % Select figure 5 for plots

clf % Clear figure 5

f1 = sin(0.1*pi*k); % Calculate function f1

subplot(2,2,1); % Divides fig 5 into (m = 2) vertical

% and (n = 2) horizontal sub-figures

% The last argument (p = 1) accesses

% sub-figures. (1 <= p <= m*n)

stem(k,f1,‘filled’);

grid on; % DT plot of f1 versus k

xlabel(‘k’) ; % Label of X-axis

ylabel(‘f1[k]’) % Label of Y-axis

end

Once a function has been created, it must be saved in a file whose name is

same as the defined name of the function. In our example, the aforementioned

function must be saved in a file myfirstplot.m. The calling format for a

function is the same as one would use to access a M A T L A B built-in function.

To access myfirstplot, the following instructions must be typed at the

M A T L A B prompt:

>> m = [-5:5]; % Define the input argument

>> [y] = myfirstplot(m); % Output value is returned to y

% with subplot plotted in

% figure 5

E.7 Summary

In this appendix, a working introduction to M A T L A B is provided. The intent

is to introduce the basic capabilities of M A T L A B to the reader. M A T L A B

supports hundreds of built-in functions from linear algebra, numerical analysis,

polynomial algebra, and numerical optimization. These built-in functions are

supported in both the student and full version of M A T L A B , and do not require

any toolboxes. A list of built-in functions is available on the Mathworks website

(www.mathworks.com). Readers are encouraged to visit the website and explore

M A T L A B in more detail.

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Appendix F About the CD

This book is accompanied by a CD that includes material for supplementary

reading, M A T L A B code used in the text, and data used in different simulations.

The organization of the CD is shown in Table F.1.

In Table F.1, we have assumed that the CD drive is mapped to the shortcut

“CD.” Check the appropriate shortcut to the CD drive on your computer. For

example, if the CD drive is mapped to the shortcut “F,” replace “CD” in the

aforementioned paths to the folders with “F” such that the path to the interactive

programs is specified by F:\InteractEnv. The other two folders can be found in a similar way. In the following we provide additional information on each folder.

F.1 Interactive environment

The “InteractEnv” folder contains three interactive learning objects used to

explain the operations of convolution integral, convolution sum, and digital

filtering. While the first two learning objects developed to explain convolution

integral and sum are based on Macromedia Flash, the third learning object uses

a graphical interface environment based on M A T L A B .

F.1.1 Convolution

Convolution is an important signal processing operation, which is extensively

used to compute the output of linear time-invariant systems. The graphical

approach to solve the convolution integral in the CT domain was presented in

Section 3.5. Likewise, the steps involved in computing the convolution sum in

the DT domain were explained in Section 10.5. To help understand the two

convolution operations, the CD includes two Shockwave Flash animations, one

each for the convolution integral and convolution sum.

The learning object for the convolution integral convolves the following CT

signal:

x(t) = u(t + 0.5) − u(t − 1) with h(t) = u(t + 0.5) − u(t + 1)

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849 F About the CD

Table F.1. Organization of the CD

Folder Comments

CD:\ InteractEnv contains interactive programs explaining important concepts such as the convolution integral, the

convolution sum, and digital filtering

CD:\Data contains selected audio clips and images used in M A T L A B simulations

CD:\ M A T L A B Codes contains M A T L A B functions used in the text

Table F.2. Values of the sequence y [k ]

k −5 −4 −3 −2 −1 0 1 2 y[k] 8 12 14 15 15 7 3 1

and describes the graphical approach to derive the output of the LTIC system. By

analytical computation, it is straightforward to derive the following expression

for the output:

y(t) =







t + 1 −1 ≤ t < 0.5 −t + 2 0.5 ≤ t < 2

0 otherwise.

The learning object for the convolution sum uses the following DT sequences:

x[k] = u[k + 2] − u[k − 3] with h[k] = 2−k(u[k + 3] − u[k − 1])

and describes the graphical approach to derive the output of the LTID system.

All non-zero values of the output sequence y[k] are specified in Table F.2. In order to run the two animations, you should open a web browser, such as

Netscape or Internet Explorer (IE), with the Flash Player incorporated within

the browser. If the Flash Player is not incorporated, it can be downloaded and

installed from http:/www.macro.media.com, which is the official website of

Macromedia. In the following, we highlight the procedure for the convolution

integral through a series of steps.

Step 1 Open the internet browser (Netscape or IE) by selecting the program

from the task bar. Within the browser, select the “File” option from the extreme

top left menu and click on the “Open” option. This opens a dialog box, where

you can provide the complete path to the convolution integral animation and

choose a file. Browse to the convolution animation and select it. In our case,

the path to the animation for the convolution integral is given by

CD:\InteractEnv\convolution\ConvolutionIntegral.swf

where CD specifies the drive name to the CD-ROM. After the execution of step

1, a frame similar to that in Fig. F.1 would be displayed on the computer screen.

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Fig. F.1. Initial Flash window for

convolution integral.

Step 2 The frame displayed in step 1 has three subwindows. The top subwindow

on the left-hand side plots the figures graphically, while the top subwindow on

the right displays different steps involved in computing the convolution integral.

The step being executed is highlighted, with the explanation included in the

bottom subwindow. To interact with the animation, three options are available.

Clicking on the “previous step” option moves the animation back by one frame,

showing the result of the previous step. Clicking on the “next step” moves the

animation forward by one frame, while clicking on the “reset” option initializes

the animation to the start.

Step 3 Play the animation according to your speed and try to understand all

operations performed to compute the result of the convolution integral. Once

the animation has been completely played, a frame similar to that in Fig. F.2

would appear on the computer screen.

The procedure for running the convolution sum animation is identical to that

of the convolution integral. Once this animation has been completely played, a

frame similar to that in Fig. F.3 would appear on the computer screen.

F.1.2 Digital audio filtering

To explain digital filtering, the CD provides a set of M A T L A B programs used to

create a digital audio filtering interactive environment (DAFIE). The programs

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851 F About the CD

Fig. F.2. Final frame for the

learning object explaining the

convolution integral operation.

are available in the following folder:

CD:\InteractEnv\filter

where CD specifies the drive name to the CD-ROM. DAFIE is a graphical user

interface (GUI), which may be used to select an audio file, read the signal,

and manipulate the signal in different ways. The following four functions are

primarily used to create the interactive environment:

dafie.m % main program for generating DAFIE

localbutton.m % function that selects the operation

% using local buttons

designfilter.m % designs filters based on the specs

% provided by the user

openfile.m % opens a dialog box to select an input

% audio file

The main program dafie uses the built-in M A T L A B function uicontrol to create the user interface. When the main program dafie is run, an inter- active window is created. A snapshot of the window is shown in Fig. F.4. The

interactive window consists of three subwindows: Command, Comments, and

Graphics. The Command subwindow controls the environment through a series

of buttons. A brief description on the functionality of each button is as follows.

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852 Appendix F

Fig. F.3. Final frame for the

learning object explaining the

convolution sum operation.

Read File: Loads the input signal from a sound file stored in the wav

format.

Plot Signal: Plots the loaded signal in the Graphics window.

Play Signal: Plays the audio signal. The user must have a sound card and

speakers to hear the audio.

Signal Spectrum: Computes the power spectrum of the audio signal and

displays it in the Graphics subwindow. The power spectrum is calculated

by parsing the audio signal in segments. Each segment has a length of

1024 samples with an overlap of 512 in between the neighboring seg-

ments. Section 17.2.3 explains the steps involved in computing the power

spectrum of a signal.

Design Filter: Designs a DT filter and displays the coefficients of the fil-

ter. If the selected filter is of the FIR type, then the impulse response

h[k] of the filter is plotted in the Graphics window. If the selected filter is of the IIR type, then the coefficients of the numerator and denomi-

nator of the transfer function of the filter are displayed using the stem

plot. DAFIE provides the option of selecting one of the Bartlett, Ham-

ming, Hanning, Blackman, or Kaiser windows in designing the FIR fil-

ter with the number of taps limited to 201. For IIR filters, the choices

are limited to the Butterworth or Chebyshev type II filter with a stop-

band attenuation of at least 50 dB and pass-band ripples limited to a

maximum level of 2 dB. The number of taps is ignored for the IIR

filters.

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853 F About the CD

Fig. F.4. The DAFIE environment

for digital audio filtering.

Freq Response: Calculates and plots the magnitude spectrum of the

designed filter. The magnitude spectrum is displayed in the Graphics win-

dow.

Apply Filter: Filters the input signal and plots the resulting output signal as

a function of time.

Play Filt Signal: Plays the output (filtered) signal as audio.

Filt Sig Spectrum: Computes the power spectrum of the output signal and

displays it in the Graphics window.

Save Output: Stores the output signal as audio in the file 〈output.wav〉 in the working directory. If you choose this command, ensure that you

have write permission to the current working directory.

Exit Dafie: Exits the DAFIE, ending the program.

F.2 Data

The Data folder in the CD contains two subfolders. These subfolders contain

different audio clips and images used in the text. The audio clips are stored

in the wav format with the .wav extension. The images are stored in the TIFF

(also referred to as the TIF) format, where the image data is stored without any

distortions. A list of the audio clips and images included in the CD is provided

in the following.

Audio clips (CD:\Data\audio)

bell.wav % Audio sampled at 22.05 kHz and

% quantized to 8-bits

test44k.wav % Audio sampled at 44.1 kHz and

% quantized to 8-bits

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854 Appendix F

noisy audio1.wav % Audio signal corrupted with

% narrowband noise

noisy audio2.wav % Audio signal corrupted with

% wideband noise

Gray images (CD:\Data\image)

{ayantika.tif, lena.tif, % Images used in this book

rini.jpg, sanjukta.tif,

train.jpg}

{castle.jpg, eiffel.jpg, % Other images given for

girl.jpg, sounio.jpg} % solving problems

Note that images with tif/tiff extension include no distortion. On the other hand,

images with jpg extension are compressed using JPEG codec.

Color images (CD:\Data\image\color)

{castle, eiffel, gardern, girl, % Selected color images

lena, sanjukta, sounio, % in JPG/TIFF format

stadium, train}

F.3 M A T L A B codes

The CD includes the M A T L A B codes used in various examples in the text. In

the following, we provide a listing of the names of the functions arranged in

terms of their inclusion in different chapters.

Chapter 1 Example 01 23.m % plots several CT functions using

% subplot and plot

Example 01 24.m % plots several DT sequences using

% subplot and stem

Chapter 3 Example 03 12.m % solves first order differential

% equation

Example 03 13.m % solves second order differential

% equation

myfunc1.m % defines a first order differential

% equation

myfunc2.m % computes vector of derivatives

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855 F About the CD

Chapter 5 bodeplot.m % plots BODE plot of a transfer

% function in section 5.10.2

myctft.m % calculates CTFT of a function in

% section 5.10.1

myinvctft.m % calculates inverse CTFT of a function

% in section 5.10.1

section 5 10 1.m % calculates CTFT of a function in

% section 5.10.1

Chapter 7 Example 07 5.m % calculates and plots frequency

% response of Butterworth filter

Example 03 7.m % calculates and plots frequency

% response of Chebyshev I filter

Example 07 8.m % calculates and plots frequency

% response of Chebyshev II filter

Example 03 9.m % calculates and plots frequency

% response of elliptic filter

Example 03 10.m % designs highpass filter and plots

% frequency response

Example 07 11.m % designs bandpass filter and plots

% frequency response

Example 03 12.m % designs bandstop filter and plots

% frequency response

Chapter 8 ImmuneSystem1.mdl % Simulink model for stable immune

% system

ImmuneSystem2.mdl % Simulink model for unstable immune

% system

Chapter 10

Example 10 17.m % calculates system output using direct

% method in Example 10.17

Example 10 18.m % calculates system output using direct

% method in Example 10.18

Example 10 19.m % calculates system output using conv

% function in Example 10.19

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Chapter 12

Example 12 6.m % calculates freq. charac. of decaying

% exponential function using dft

Example 12 7.m % calculates freq. charac. of two

% complex exponential functions

% using dft

Example 12 8.m % calculates frequency characteristics

% using N=32

Example 12 8 N64.m % calculates frequency characteristics

% using N=64

Example 12 9.m % calculates dft of a decaying

% exponential function

Example 12 11.m % calculates DTFT of an aperiodic

% sequence

mydft.m % calculates dft using direct

% calculation

myfft.m % calculates dft using radix-2 fft

% method

tfft.m % test program to compare mydft and

% myfft functions

Chapter 13

Example 13 20.m % calculates partial fraction

% coeffs of H(z)=B(z)/A(z)

Example 13 21.m % calculates poles and zeros and plots

% them in the z-plane

Example 13 22.m % calculates transfer function of a

% system from its poles and zeros

Chapter 14

section14 9 1.m % calculates the partial fraction

% coefficients in section 14.9.1

section14 9 2.m % calculates the zeros and poles of a

% transfer function in section 14.9.2

Chapter 15

Example 15 9.m % designs lowpass FIR filter using

% Hamming/Blackman windows

Example 15 10.m % designs lowpass FIR filter using

% Kaiser window

Example 15 11.m % designs lowpass FIR filter using

% Parks-McClellan algorithm

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Chapter 16 Example 16 2.m % converts a CT filter to DT using

% impulse invariance method

Example 16 3.m % converts a CT filter to DT using

% impulse invariance method

Example 16 4.m % converts a CT filter to DT using

% bilinear transformation

Example 16 5.m % designs highpass IIR filter using

% CT elliptic filter and bilinear

% transform

Example 16 6.m % designs bandpass IIR filter using

% CT elliptic filter and bilinear

% transform

Example 16 7.m % designs bandstop IIR filter using

% CT elliptic filter and bilinear

% transform

Chapter 17 Example 17 2.m % calculates spectrogram of a DT signal

Example 17 3.m % calculates power spectral density

% using Welch method

Example 17 4.m % filters (lowpass, bandpass,highpass)

% an audio signal

Example 17 5.m % bandstop filters an audio signal

Example 17 8.m % calculates 2-D spectrum of a grating

% image

Example 17 9.m % spectral analysis and lowpass

% filtering of an image

Example 17 10.m % highpass filters an image

Example 17 11.m % predictive coding of an image

Example 17 12.m % JPEG compression of an image with

% different quality factors

Section 17 2 3.m % calculates power spectral density of

% an audio signal

Section 17 2 5.m % calculates power spectral density of

% a music signal

Section 17 5 3.m % reads and manipulates an image

Appendix E

Example E 7.m % plots a CT and a DT function

Example E 8.m % plots several functions in one figure

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Bibliography

In the following, we have included selected textbooks and reference books on subjects

related to signals and systems.

Signals and systems

S. R. Devasahayam, Signals and Systems in Biomedical Engineering: Signal Processing

and Physiological Systems Modeling. Kluwer Academic/Plenum Publishers (2000).

S. Haykin and B. V. Veen, Signals and Systems. 2nd edn. Wiley (2002).

H. Hsu, Schaum’s Outline of Signals and Systems. McGraw-Hill (1995).

B. P. Lathi, Signal Processing and Linear Systems. Oxford University Press (2000).

A. V. Oppenheim, A. S. Willsky, and S. Hamid, Signals and Systems, 2nd edn. Prentice

Hall (1996).

R. E. Ziemer, Signals and Systems: Continuous and Discrete, 4th edn. Prentice Hall

(1998).

Digital signal processing and filtering

A. Antoniou, Digital Filters: Analysis, Design and Applications, 2nd edn. McGraw-Hill

(2001).

L. B. Jackson, Digital Filters and Signal Processing, 3rd edn. Kluwer Academic Pub-

lishers (1996).

S. K. Mitra, Digital Signal Processing: A Computer-Based Approach, 2nd edn. McGraw-

Hill (2001).

A. V. Oppenheim, R. W. Schafer, and J. R. Buck, Discrete-Time Signal Processing, 2nd

edn. Prentice Hall (1999).

T. W. Parks and C. S. Burrus, Digital Filter Design. Wiley-Interscience (1987).

J. G. Proakis and D. K. Manolakis, Digital Signal Processing: Principles, Algorithms

and Applications, 3rd edn. Prentice Hall (1995).

Electrical circuits

R. L. Boylestad, Introductory Circuit Analysis, 10th edn. Prentice Hall (2002).

A. M. Davis, Linear Circuit Analysis. Thomson Engineering (1998).

J. O. Malley, Schaum’s Outline of Basic Circuit Analysis, 2nd edn. McGraw-Hill (1992).

W. D. Stanley, Network Analysis with Applications, 4th edn. Prentice Hall (2002).

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Communications

A. B. Carlson, P. B. Crilly, and J. Rutledge, Communication Systems, 4th edn. McGraw-

Hill (2001).

S. Haykin, Communications Systems, 4th edn. Wiley (2000).

M. Schwartz, Information Transmission, Modulation and Noise. McGraw Hill (1980).

Multimedia

B. Furht, S. W. Smoliar, and H. Zhang, Video and Image Processing in Multimedia

Systems. Kluwer Academic Publishers (1995).

R. C. Gonzalez and R. E. Woods, Digital Image Processing, 2nd edn. Prentice Hall

(2002).

M. K. Mandal, Multimedia Signals and Systems. Kluwer Academic Publishers (2003).

John Watkinson, The MPEG Handbook. Focal Press (2001).

U. Zölzer, Digital Audio Signal Processing. John Wiley & Sons (1997).

Systems and control

D. Basmadjian, Mathematical Modeling of Physical Systems: An Introduction. Oxford

University Press (2002).

R. C. Dorf and R. H. Bishop, Modern Control Systems, 10th edn. Prentice Hall (2004).

B. C. Kuo and F. Golnaraghi, Automatic Control Systems, 8th edn. Wiley (2002).

N. S. Nise, Control Systems Engineering, 4th edn. Wiley (2003).

Mathematics

E. O. Brigham, The Fast Fourier Transform and its Applications. Prentice Hall (1988).

G. A. Korn and T. M. Korn, Mathematical Handbook for Scientists and Engineers:

Definitions, Theorems, and Formulas for Reference and Review, 2nd edn. Dover Pub-

lications (2000).

E. Kreyszig, Advanced Engineering Mathematics, 8th edn. Wiley (1998).

K. A. Stroud and D. J. Booth, Engineering Mathematics, 5th edn. Industrial Press (2001).

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CUUK852-Mandal & Asif May 28, 2007 14:21

Index

Adder, 571

Additivity property, 73, 76

Aliasing, 402

Alternation theorem, 619

Amplitude modulation, 66–67

Amplitude response, 167–169, 170–171, 172

Amplitude spectrum, 167–169, 170–171, 172

Analog signals, 6

Analog to digital (A/D) conversion, 649, 393,

526

Aperiodic signals, 9, 193, 475

Approximate bandwidth, 322

Arithmetic overflow, 586

Audio, 756

formats, 757

spectral analysis, 758

filtering, 761

compression, 767, 772

Autocorrelation function, 750, 753

Bandpass filter, 557, 612, 737

Bandstop filter, 558, 615, 738

Bandwidth, 322, 413, 539

approximate, 418

transition, 601

Baseband signal, 393

Bilateral Laplace transform, 261, 262–266

Bilateral z-transform, 567

Bilinear transformation, 730–735

Binary code, 412

Bit, 9, 71, 415

Block diagram, 62–63, 76, 307–311, 382, 385

Block diagram representation, 63, 307–311

Bode plots, 245–246, 250–251, 568

Bounded-input bounded-output (BIBO)

stability, 88–90, 128–130, 298–305,

452, 601–606, 739–741

Break frequency, see corner frequency Butterworth filter, 321, 328–338, 364, 720

Butterfly computation, 556

Carrier, 67, 369

Causal signal, 31, 266

Causal system, 84–85, 93, 127, 136, 204, 591

CCD camera, 415

Characteristic equation, 107, 110, 273, 294,

346, 597

Characteristic roots, 294–295

Charge coupled device (CCD), 3–5

Circular reflection, 442

Compact disc (CD), 413

Complex frequency, 28–31, 261, 306

Complex frequency plane, 250, 271

Complex numbers, 3–5, 799–807

arithmetical operations of, 800

graphical interpretation, 803

polar representation, 803

set of, 218

Continuous-time filter, 320–364

Continuous-time FT to DTFT, 526

Continuous-time system, 6–8, 84

forced response of, 107

frequency response of, 203, 351

Laplace-transform analysis of, 285–286

natural response of, 106

realization of, 307

stability of, 88, 128, 298–305

time-domain analysis of, 116–124

transfer function of, 181, 229, 237–239,

285

zero-input response of, 106–112

zero-state response of, 106–112

Control system, 306, 368

stability considerations in, 458

Convolution, 116–127, 430–451

circular (or periodic), 431, 439, 500

graphical, 118–125, 850

properties of, 125–127, 448

Convolution property

of DTFT, 498, 502

of DFT, 549

860

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861 Index

of DTFS, 504

of z-transform, 589

Convolution sum

graphical procedure of, 432

properties of, 448

sliding tape method of, 436

Corner (break) frequency, 332

Cut-off frequency, 321, 556

normalized, 600

Damping ratio, 69, 377

Decibel, 246, 595

Decimation (downsampling), 41, 584

Decimation-in-time algorithm, 553

Decomposition property, 182, 193

Delay element, 14–571

Demodulation of AM, 371–374

DFT, see discrete Fourier transform Difference equation, 63, 70–72, 423, 455

iterative solution of, 423

linear, 431

z-transform solution of, 594–595

Differential equation, 63, 64–67

time-domain solution of, 106–111, 131–135

classical solution of, 108, 808

Laplace transform solution of, 288–293

Digital audio, 756

filtering, 761, 852

Digital communication, 20, 70

Digital filters, 555–560

advantages of, 555

nonrecursive, 559, 591–630

recursive, 559, 715–744

Digital signals, 8

Digital to analog (D/A) conversion, 393

Dirac, Paul Adrien, 32

Dirichlet conditions, 178

Discrete Fourier transform (DFT), 525–560,

531

properties of, 547–551

Discrete Fourier transform

as matrix multiplication, 535

basis functions of, 537

spectrum analysis using, 538

computational complexity of, 551

Discrete-time Fourier series (DTFS), 465–475

spectrum, 483

Discrete-time Fourier transform, 475–482

existence of, 484

of periodic functions, 485

equations, 477

existence of, 482

properties of, 491–505

spectrum, 483

Table, 481

Discrete-time processing, 393

Discrete-time signals, 6, 30, 34

Discrete-time sinusoid, 27

Discrete-time systems, 62–63, 69–72, 393

forced response of, 424

frequency response of, 506

natural response of, 424

realization of, 570–584

stability of, 601–605

time-domain analysis of, 422–460

transfer function of, 499, 596

z-transform analysis of, 594–609

zero-input response of, 424

zero-state response of, 424

Distortionless transmission, 560

Downsampling (decimation), 41

DPCM, 769

DTFT, see Discrete-time Fourier transform Dual tone multifrequency (DTMF), 555

Duality property, 226

Dynamic systems, 83

Energy signals, 17–20

Envelop detector, 374

Euler formula, 11, 803

Even function, 21–24

Everlasting exponential, 28–31

Exponential Fourier series, 163–179

Fast Fourier transform (FFT), 553–558

radix-2 algorithm, 553–556

bit-reversal for, 558

Feedback systems, 308

Fidelity, 412

Filter realization

direct form, 572

cascaded form, 572

linear phase form, 573

parallel form, 581

transposed form, 573

Filters, 322–367, 555–744

allpass, 260, 304–305

analog, see continuous-time filter bandpass, 322, 357–361, 557, 612–615, 737

bandstop, 322–323, 361–364, 558,

615–617, 738

butterworth, 321, 328–338, 351, 720–730,

733–735

causal, 565, 592

chebyshev, 321, 338–349, 351

digital, 555

elliptic, 321, 349–352, 716, 736–738

FIR, 559

frequency transformation in, 352–364

group delay of, 561

highpass, 321–322, 353–357, 556, 609–612,

736

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862 Index

Filters (cont.) ideal, 321–324, 556–558

IIR, 559

linear phase, 562, 591

lowpass, 321, 327–352, 556, 591–608

non-ideal, 565

phase delay of , 561

realization, 570

recursive, 86, 338

passband of, 321, 511, 556–558

stopband of, 321, 511, 556–558,

567

Final value theorem, 287–288, 593

Finite impulse response, 559

Finite precision representation, 585

FIR, see finite impulse response FIR filters

linear phase, 562

Type 1–4, 562

optimal, 618

Forced response, 107, 424

Fourier, Jean-Baptiste-Joseph, 152

Fourier integral, 196

Fourier series, 141–182

dirichlet conditions for, 178

exponential, 163–176

Symmetry conditions in, 156–158

trigonometric, 153–163

Fourier spectra

of CTFT, 197, 205–208

of discrete-time Fourier series, 471

of exponential CTFS, 167–169

of DTFT, 478–479

Fourier transform

continuous-time, 193–251

discrete-time, 475–517

duality property of, 226–227

existence of, 231–233, 482

frequency-convolution property of, 227–230

frequency-shifting property of, 222–223

linearity, 216–219, 492

numerical computation of, 247–250

properties of, 491–505

scaling property of, 219–220, 493

short-time, 750

table of, 217, 481

time convolution property of, 227–230,

498

time differentiation property of, 224–225

time integration property of, 225

time shifting property of, 221–222, 493

Frequency division multiplexing (FDM),

369

Frequency-differentiation property

of DTFT, 497

of z-transform, 588

Frequency-domain analysis

of continuous-time systems, 227–230,

237–246

of discrete-time systems, 498–502,

506–514

Frequency resolution, 542, 751, 759

Frequency response, 245, 506, 606, 629

Frequency sampling, 529

Frequency shifting property

of CTFT, 222–223

of DTFT, 495

of DTFS, 504

Frequency spectrum, 245, 471, 483

Fundamental frequency, 10

Gate function, 25, 208

Generalized function, 255

Gibbs phenomenon, 158, 593

Hamming window, 594

Hanning (Von Hann) window, 594

Harmonic frequency, 13

Heaviside, Oliver, 817

Heaviside formula, 210, 817

Hermitian Symmetry Property

of DFT, 548

of DTFS, 504

of DTFT, 491

Highpass filter, 556, 782

design methods, 609, 706

Homogeneity property, 73

Ideal filter, 321–324, 556–559

IIR, see infinite impulse response Image, 773

formats, 774

spectral analysis, 775

filtering, 779

compression, 784

Impulse function, 32–34, 426

Impulse invariance method, 717–730

Impulse response, 98, 103, 113–116, 427,

556

of ideal filters, 559, 565, 597, 600

Infinite impulse response, 559

Initial conditions, 64, 423, 594

Initial value theorem, 287–288, 593

Instantaneous frequency, 750

Instantaneous (memoriless) systems, 83–84,

127

Integration table, 797–798

Interpolation, 41–43, 399, 584

zero-order hold, 407

Inverse discrete Fourier transform, 531

Inverse Fourier transform, 209–210, 477

Inverse Laplace transform, 273–276

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863 Index

Inverse z-transform, 574

partial fraction method, 575

power series method, 580

JPEG format, 421

Laplace, Pierre-Simon, 262

Laplace transform, 261–311

bilateral, 262–266

existence, 271

frequency-convolution property of,

284–287

frequency-shifting property of,

280–281

inverse, 273–276

linearity property of, 276–278

region of convergence, 271, 295–298

scaling property of, 278–279

table of, 270

time convolution property of, 284–287

time differentiation property of,

281–282

time integration property of, 282–284

time shifting property of, 279–280

unilateral, 266–269

Leakage, 543

Left half plane (LHP), 301

Legendre polynomials, 185

L’Hopital’s rule, 797

Linear phase, 562, 591

Linear system, 73–79

Linear time-invariant system, 103–137,

423

Linearity property

of DTFT, 492, 505

of DTFS, 492, 504

of DFT, 549

of z-transform, 582

Lower sidebands, 371

Lowpass filter, 556, 780

designh methods, 599, 605

Magnitude response, 508, 557

Magnitude spectrum, 471, 483

Main lobe, 595

Marginally stable system, 302–303,

604

MATLAB, 831

control statements, 840

elementary operations, 842

plotting functions, 844

user interface, 853

Maximally flat response, 324

Mean, 753

Mean square error, 788

Memoryless system, 83–84, 127, 452

Minmax optimization, 620

Modulation, 66–67, 70–72, 369–373

MP3 player, 421

Multiplexing, frequency-division, 369

Multipliers , 14–571

Natural frequencies, 95, 344

Natural response, 107, 424

Noncausal signals, 31

Noncausal system, 84–85, 93, 127, 136

Nyquist sampling rate, 247, 397

Odd function, 21–24

Operators, differential, 106

Orthogonal signal set, 142–149

Orthogonal vector set, 142

Orthogonality

in complex signals, 143

property, 465

Orthonormal set, 144, 465

Parks-McClellan algorithm, 621

Parseval’s theorem

for discrete Fourier transform, 550

for Fourier series, 170–171, 184

for Fourier transform, 230–231, 253

for discrete-time Fourier series, 504

for discrete-time Fourier transform, 503

Partial fraction expansion

for CTFT, 209–211, 824

for DTFT, 500, 816, 827

for Laplace transform, 273, 816–824

for z-transform, 575, 828–830

Passband of a filter, 320, 321–323, 556

Period of a CT signal, 9–15

Period of a DT signal, 9–15

Periodic reflection, 442

Periodic signal, continuous-time, 9–15

Periodic signal, discrete-time, 9–15

Periodicity property

of DTFT, DTFS, 491

of DFT, 548

Periodogram, 754

Phase response, 245–246, 351, 508

Phase spectrum, 245–246, 351, 471, 483

Picket fence effect, 543

Picture element (pixel), 415

Polar plot, 803

Power spectral density, 753

Poles, 294–295, 597, 612

first-order, 817

higher-order, 822

Power Series, 796

Power signals, 17–20

Probabilistic signals, 20–21

Pulse code modulation (PCM), 412

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864 Index

Quantization, 410–413

uniform/nonuniform, 410

error, 411

Random signals, 20–21, 752

spectral analysis, 758, 775

Rectangular window, 593

Recursive filter, 657

Region of convergence, 263–266, 295–298,

573

Right half plane (PHP), 301

Ripple Control parameter, 604

Ripples, 593

Round-off errors, 586

Sampling rate (frequency), 247

Sampling, 393

interval, 395

rate, 395, 397

impulse-train, 395

pulse-train, 405

bandpass, 418

sawtooth wave, 419

theorem, 247, 397

Scaling property, 126, 173–174, 219–220,

278–279, 493

Series summation, 796

Shape control parameter, 604

Shift operator, 571

Sidebands, 371

Sidelobes, 177, 595

Short-time FT, 750

Signals, 3

analog, 8–9

aperiodic, 9–15

continuous-time, 6–8

digital, 8–9

discrete-time, 6–8

energy, 16–20

energy of, 16

essential bandwidth of, 322

periodic, 9–15

power, 16–20

power of, 16

orthogonal representation of, 142–149

Signum (sign) function, sgn(t), 25–27

Sinc function, 27–28

Sinusoids, continuous-time, 10–11, 27,

29–30

Sinusoids, discrete-time, 11–12, 27

Spectral estimation, 748

Spectral folding, see aliasing Spectrogram, 752

Spectrum

magnitude, 167–169, 170–171

phase, 245–246, 351

Stability, 88–90

analysis, 601, 453, 739

bounded-input, bounded-output (BIBO),

88–90, 128–130, 298–305, 601

marginal, 302–304, 604

Steady-state response, 107, 109, 137, 608

Stopband, 320, 321–323, 556

attenuation, 601

Superposition principle, 73, 113

Symmetry conditions in Fourier series,

169–170

Systems

block-diagram of, 63, 307–311

causal, 84–85, 93, 127, 452

characteristic equation of, 107, 273, 294,

346, 575

classification of, 72–90

continuous-time, 73

control, 306, 368

discrete-time, 73, 422

dynamic, 83–84

feedback, 308

finite memory, 84

frequency response of, 245, 506, 606, 629

invertible, 130–131, 454

linear, 73–79

LTIC, 103

LTID, 422

marginally stable, 302–303, 604

memoryless, 83–84, 127, 452

overdamped, 376

realization of

response to sinusoid input, 150–152,

239–240

stability, 88–90, 128, 298–305, 452

time-domain analysis of, 103–137

time-invariance, 79–83

transform analysis of, 180–182, 237–246,

305–307

underdamped, 376

unstable, 88–90, 128, 298–305

Time-differencing property

of DTFT, 496

of DTFS, 504

of z-transform, 587

Time-differentiation property

of CTFS, 174

of CTFT, 224–225

of Laplace transform, 281–282

Time-domain analysis

of continuous-time systems, 118–126

of discrete-time systems, 422–460

Time integration property, 174, 225, 282–284

Time-invariant system, 79–83

Time inversion, 172

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CUUK852-Mandal & Asif May 28, 2007 14:21

865 Index

Time reversal property, 172

Time scaling property

of CTFS, 173

of CTFT, 219–220

of Laplace transform, 278–279

of DTFT, 493

of DTFS, 504

of z-transform, 584

Time shifting, 35–39

Time shifting property

of CTFS, 171

of CTFT, 221–222

of DFT, 549

of DTFS, 504

of DTFT, 493

of Laplace transform, 279–280

of z-transform, 585

Time-summation property

of DTFT, 498

of DTFS, 504

of z-transform, 592

Transfer function, 181, 237–239, 285, 556, 596

Transition bandwidth, 567, 595, 601

normalized, 601

Trigonometric Fourier series, 153–162

Trigonometric identities, 795

Underdamped system, 376

Unilateral Laplace transform, 266–272

Unilateral z-transform, 579

Unit impulse function, 32–35

Unit impulse response of a system, 113–116

Unstable system, 88–90, 128–130, 298–305,

453, 602

Upsampling (interpolation), 41–44, 493

Vectors, 142

Width property of convolution, 126, 449

Window function

Bartlett/triangular, 594

Blackman, 594

Hamming, 594

Hanning, 594

Kaiser, 595, 603

rectangular, 593

Zero-input response, 106–111, 424,

809

Zero-order hold, 407

Zero-padding, 546

Zero-state response, 106–111, 424, 812

Zeros, 294–295, 597

z-transform, 565

bilateral, 567

convolution property, 589

inverse, 574

shifting property of, 585

linearity property of, 582

region of convergence of, 573

Table of, 572

Unilateral, 569

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