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1_Introduction.pdf

Dynamics and Control

Topic 1: Introduction

Tentative Lecture Schedule

What is Control?

Think of some “control” examples:

• From your study

• From industry

• From real-life

Commonly used terms in “control”?

Open Loop Control vs. Closed Loop Control? Open Loop Control:

• There is no feedback • Calibration is the key! • Can be sensitive to disturbances

Open Loop Control:

• Objective: make bread toasted • Control action: set timer • Control result (performance)?

• How to improve the performance? • Use temp sensor • Use microprocessor and actuator to control the spring • Worth it?

Feedback

Closed Loop Control:

• There is a feedback (sensor) • Compare actual behavior with desired behavior • Make corrections based on the error • The sensor is the key element of a feedback control • Design a proper control algorithm is the focus of this

subject!

Room Temp Control: Open or Closed?

• Feedback? • Control objective? • Controller? • System/plant? • Disturbance?

Applications of Feedback Control

• Manufacturing

Applications of Feedback Control

• Robotics

UNDERWATER ROBOT

Applications of Feedback Control

• Aerospace and Astronautics

Applications of Feedback Control: BIG DOG in Action

BOSTON DYNAMICS

Sensing Actuation Control Plant ControlSystem = + + +

BIG DOG CONTROL: Block Diagram

DISTURBANCE

Environment

Foot Trajectory Planning

Desired Walking Speed

REFERENCE

Joint angles & velocities

Robot Joint torques

Robot Robot

Virtual Leg IK

Virtual Leg FK

Virtual

Leg VM

PD Servo

Virtual Leg Coords

Virtual Leg Forces

State Machine

Leg State Gait

Coordination Mechanisms

Influences to/from other legs

IK=inverse kinematics FK=forward kinematics VM=virtual model

Exercise: • Use Block Diagram to represent this driving system • Indicate: sensor, actuator, control, plant, reference &

disturbance

Brain Hand Foot

Eye

Car _

Desired Direction Desired Speed

Direction Speed

Controller Actuators Plant Error Disturbances

Sensor

Control Actuation

Sensor

Plant _

Reference Response

Error

General Block Diagram Representation

Feedback Control Concept Not limited to engineering systems only!

Human Grasping Motion

Feedback Control Concept…cont’d Not limited to engineering systems only!

e.g. Taking Dynamics & Control…

Two Main Criteria of Good Feedback Control

• Acceptable Dynamic Responses

• Stability Crash of SAAB JAS-39 due to instability

Systematic Control Design Process

Goals of This Course

To learn the basics of feedback control systems

1. Modeling as a transfer function and a block diagram • Laplace transform (Mathematics!) • Mechanical, electrical, electromechanical systems

2. Analysis • Step response, frequency response • Stability: Routh-Hurwitz criterion, (Nyquist criterion)

3. Design • Root locus technique, frequency response technique, • PID control, lead/lag compensator

4. Programming ability / Actual Implementation • Matlab • Simulink

Course Roadmap

Recap today’s topics +

Questions? Block Diagrams

Recap today’s topics +

Questions?

Homework 1 - Part A

1. Find 2 open-loop control examples and 2 closed-loop control examples in your daily life. Represent these examples using block diagram. Clearly indicate Control Reference, Controller, Plant, Actuator, Feedback Element in the block diagram.

2. Compared to open-loop control, discuss the pros and cons of feedback control.

3. What are the two main criteria in feedback control design?

4. Describe the procedures of feedback control system design.

1_MathReview.pdf

Dynamics and Control

• Math Preparation (Review of Complex Number)

• Complex Number in Matalb

Week Topics Exam Reading

1 Introduction Ch 1-2

2 Math Review Ch 1-2

3 Modelling (Frequency + Time Domain) Ch 2

4 Time Response Ch 4

5 Block Diagram M-Exam 1 Ch 5

6 Stability Ch 6

7 Midterm Review + Make up Ch 1-6

8 Experiment

9 Spring Break

10 Steady State Errors Ch 7

11 Root Locus Ch 8

12 Design Via Root Locus M-Exam 2 Ch 9

13 Design Via Root Locus Ch 9

14 Frequency Response Ch 10

15 Design Via Frequency Response Ch 11

16 Final Review + Make up Ch 7-11

17 Final Exam Final Exam 2

Tentative Lecture Schedule

Math Prerequisites

1. Complex Numbers 2. Linear Ordinary Differential Equations 3. Laplace Transform to Solve ODE’s 4. Partial Fraction Expansion 5. Linearization

WARNING!!

THIS IS A VERY MATH INTENSIVE COURSE!

It is your responsibility to review these topics. We assume that you have passed all the pre- required calculus courses and already have a GOOD command of all these prerequisites.

ADD/DROP deadline is Sept 8!

Complex Number: A Little History

Math is used to explain our universe. When a recurring phenomenon is seen and can’t be explained by our present mathematics, new systems of mathematics are derived. In the real number system, we can’t take the square root of negatives, therefore the complex number system was created.

Girolamo’s Problem





In The Great Art, published in 1545, Girolamo Cardano discusses the

following problem.

To find x and y, use substitution.

04010

40)10(

2 



Apply the Quadratic Formula.

155

)40()1(41010 2





 x

Due to the symmetry in the problem, x and y take on ± values.

5 10 15 20

No Intersection!

Bombelli And Imaginary Numbers







Rafael Bombelli in the 1560’s figured out a way to work with imaginary

numbers. Definition of Imaginary unit, i

Using these rules, Bombelli worked Cardano’s cubic solutions to arrive

at real results.

Complex Numbers: Definition

A complex number consists of a real and an imaginary term:

𝑎 + 𝑏𝑖

where 𝑎 and 𝑏 are real numbers and 𝑖 is the imaginary unit.

Both real numbers and imaginary numbers have specific physical meanings in control engineering

Complex Numbers: Addition and Subtraction

Add/subtract real to real, and imaginary to imaginary

Example: (6 + 7i) + (3 - 2i)

(6 + 3) + (7i - 2i) = 9 + 5i

When subtracting, DON’T FORGET to distribute the negative sign!

Example: (3+2i) – (5 – i)

(3 – 5) + (2i – (-i)) = -2 + 3i

Complex Numbers: Multiplication

(2 3i) (3 6i)  

26 12i 9i 18i   

6 3i 18 ( 1)    

6 3i 18 24 3i    

Complex Numbers: Division/Simplification

• In order to simplify complex numbers (they must always be in the form a + bi, you must multiply by the complex conjugate:

3 8i (4 3i) (3 8i) (4 3i)

4 3i (4 3

12 41i

12 9i 32i 24i

16 9

12 41i

25 25

      

 

   



  

 

Exercise 1

Simplify:

Solution:





13 2i 3 2i 4 i 12 11i 2i

4 i 4 i 4 i

0 11 i

17 1716 i

           

     

 

Complex Numbers: Geometrical Interpretation

Jean-Robert Argand in 1806 came up with the idea of a

geometrical interpretation of complex numbers. Replace the x-

axis with the real part of complex numbers, and the y-axis with

the imaginary part.

Thus, 𝒛 = 𝒂 + 𝒃𝒊 has the graphical interpretation:

Complex Numbers: Trigonometric and Polar Forms

A complex number can be represented in the trigonometric form:

𝒂 = 𝐫𝐜𝐨𝐬⁡(𝜽)

𝒃 = 𝐫𝐬𝐢𝐧⁡(𝜽)

𝒛 = 𝒂 + 𝒃𝒊 = 𝐫𝐜𝐨𝐬 𝛉 + 𝒓𝐬𝐢𝐧 𝜽 𝒊

Recall Euler’s formula:

𝒆𝒊𝜽 = 𝐜𝐨𝐬𝜽 + 𝒊𝐬𝐢𝐧𝜽

Based on this, a complex number can also be represented in polar form:

𝒛 = 𝒓𝒆𝒊𝜽 = 𝒓∠𝜽

Complex Numbers: Angle of Quadrants Conjugate Pair

Exercise 2:

Represent these complex numbers in polar form

First key in these complex numbers in Matlab

  21505.24

88302

65 2

izzez

jzezz i



 





The variable z in function compass (z) can be a single variable, or an array. In this example, it is an array , holding 6 elements

Homework 1 - Part B

1. Practice all the exercises in this PPT.

For the following problems 2-7, you need to verify your solutions in Matlab.

2. For this complex number, 𝟒⁡ − ⁡𝟒𝒊, give its exponential form.

Solution: 4 𝟐𝒆−𝒊 𝝅

𝟒

3. For this complex number, 𝒆−𝒊 𝟐𝝅

𝟑 , give its Cartesian form.

Solution: -1/2-i 𝟑/2

4. Multiplication problems • (2 + 5i)(4 − i) [Solution 13 + 18i] • (1 − 2i)(8 − 3i) [Solution 2 − 19i]

5. Find the conjugates for the following complex numbers • 12 +7i [Solution 12 − 7i]

• 2i( 𝟏

𝟐 − i) [Solution 2 − i]

6. Division problems • (1+4i)/(3+2i) [Solution 11/13 +i*10/13] • 1/(1+i) [Solution =1/2+i*1/2]

7. Let z = 2 − 2i and w = −1 − 3i, • Carefully plot z and w in the complex plane. • Plot the complex number 𝒛 − w (𝒛 is the conjugate of z). Solution:

2_Modelling(3).pdf

2/1/2016

Dynamics and Control

Topic II: Modelling of Physical Systems

• Review of ODE • Laplace Transform • Linear Time Invariant (LTI) systems • Transfer Function • Modelling of Physical Systems • Modelling in Matlab

Math Prerequisites

1. Complex Numbers 2. Linear Ordinary Differential Equations 3. Laplace Transform to Solve ODE’s 4. Partial Fraction Expansion 5. LTI 6. Transfer Function

Schedule

Road Map

Recap today’s topics +

Questions? Block Diagrams

Why Mathematical Modelling?

Review of ODE’s

Spring-mass system f 𝒕 is an external force 𝒙 𝒕 is the displacement of the mass block

𝒙 𝒕 = 𝟎 is the equilibrium point

Free body diagram Newton’s 2nd Law: 𝑭 = 𝒎𝒂

𝚺𝑭 = 𝒇 𝒕 − 𝒌𝒙 𝒕 = 𝒎𝒙 (𝒕) 𝒇 𝒕 = 𝒎𝒙 (𝒕)+ 𝒌𝒙 𝒕

(input – output relationship)

A differential equation (or "DE") contains derivatives or differentials

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Physics is filled with all sort of differential relations

Damped Vibration Laminar Flow Electrical Circuit

Solving ODEs

Example 1: Differential equation: Rewrite this equation using differentials: Integrate both sides: The solution is:

Solving ODEs

Example 2: Differential equation: Rewrite this equation using differentials: Integrate both sides: The solution is:

)2.0sin(



 

 t

Solving ODEs

More examples:

WE DON’T LIKE integration!! We want a simpler method!!

A Simplified Method: Laplace Transform

Definition:

For a function f(t), f(t) = 0 for t<0, Laplace Transform is defined as: where is a complex variable. The Inverse Laplace Transform is:

Laplace Transform Example – Step Input

, for t>0

 

s e

dte

dtetfsF

]10[ 11

)()(











 





Memorize the result!

Unit Step Input

2/1/2016

1)(

0)(

)()(























 

 

dtt

dtedtet

dtetsF

stst

 

 

  



 

1)(

0undefined,

0 0, )(

InputImpulseUnit

t t

Memorize the result!

Laplace Transform Example – Impulse Input

0 0

1111

1 ]00[

)()(

sss dte

dte s

e s

dtet

dtetfsF

stst



 

  

 

 

  

 



  

 







 





 



 



0for,)(

InputRampUnit

 tttf

Memorize the result!

Laplace Transform Example – Ramp Input

Laplace Transform Table

• We still have to do these

nasty integration?

• Things have not been simplified at all!!

• We normally use Laplace Transform tables rather than solving the preceding equations directly

• This greatly simplifies the transformation process

More Laplace Transform Table

More Laplace Transform Table…cont’d

How to Use Laplace Transform Table

( ) 3cos8f t t

2 ( ) 5/( 100)F s s 

Exercise 1:

, find the Laplace transform of it. Exercise 2:

, find the inverse transform of it. Exercise 3:

, find the inverse transform of it.

2)3/(1)(  ssF

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Laplace Transform of Differential Terms

Differentiation Theorem

( ) (0)sF s f

2 ( ) (0) (0)s F s sf f 

1 2

( 1)

( ) (0) (0)

(0)

n n n

s F s s f s f

 



 

. . . . . .

𝓛 𝒅𝒇

𝒅𝒕

𝓛 𝒅𝟐𝒇

𝒅𝒕𝟐

𝓛 𝒅𝒏𝒇

𝒅𝒕𝒏

Mass and Spring Exercise

( )mx kx f t 

0 (0)x x

Consider a vibration system with mass and spring

Solve this problem using

Laplace transform.

( )x t ( )f t ( )F s

0 0

( ) ( ) ( )

F s m sx x X s

ms k

  



Solution in Laplace domain:

( )X s

f(t) is a unit impulse function

1m  4k 

In this example, let

2 2 2

1 1 2 ( )

4 2 2 X s

s s  

 

Solution in Laplace domain:

1 ( ) sin2

2 x t t

Solution in time domain:

0)0( 0  xx 0)0( 0  xx 

We did not use integration, but solved the ODE using LT!

𝒇 𝒕 = 𝛅(𝐭) ⇒ 𝑭 𝒔 = 𝟏

Steps in Solving ODE Using Laplace Transform

1. Transform time domain function to frequency

domain function : 𝒙 𝒕 → 𝑿(𝒔)

2. Solving for solution (Algebraic problem)

3. Transform the solution in frequency domain

back to time domain:𝑿(𝒔) → 𝒙 𝒕

WE CAN USE LAPLACE TRANSFORM TABLES

FOR THESE TRANSFORMATIONS

Partial Fraction Expansion

Given the following differential equation, solve for 𝒚 𝒕 if all initial

conditions are zero:

𝒅𝟐𝒚

𝒅𝒕𝟐 + 𝟏𝟐

𝒅𝒚

𝒅𝒕 + 𝟑𝟐𝒚 = 𝟑𝟐𝒖 𝒕

1. The Laplace transform of the equation is:

𝒔𝟐𝒀 𝒔 + 𝟏𝟐𝒔𝒀 𝒔 + 𝟑𝟐𝒀 𝒔 = 𝟑𝟐

𝒔

2. Solving for the solution in s domain:

𝒀 𝒔 = 𝟑𝟐

𝒔(𝒔𝟐 + 𝟏𝟐𝒔 + 𝟑𝟐)

3. Solve for 𝒚 𝒕 using the terms in the table.

But no terms can be used directly!

We need to form the partial fraction expansion of 𝒀 𝒔 to match the

terms in the table.

teety t 8421)(  

Interesting Case: Complex Conjugate Poles

Example:

After expansion,

Complex numbers involved.

Terms not found from the table.

2 ( )

2 5

s X s

s s

 

 

Partial Fraction Expansion…Special Case

𝑿 𝒔 = 𝒔 + 𝟐

(𝒔 + 𝟏 + 𝟐𝒋)(𝒔 + 𝟏 − 𝟐𝒋)

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2 2

2 ( )

( 1) 2

s X s

 

 

2 2 5 0s s  Pole equation:

Therefore:

Solution:

2 2 2 2 5 ( 1) 2s s s    

Trick: remain in second order form

Partial Fraction Expansion…Complex Number…cont’d

tetetx tt 2sin 2

12cos)(  

Consider a vibration system with mass, damping, and spring.

The initial conditions are:

The system parameters are:

The input to the system is:

an impulse signal

Use Laplace transform to find the system’s response.

Partial Fraction Expansion Exercise: Mass, Damping, and Spring System

2c 1m 

(0) 0x  (0) 0x 

75.k

)()( ttf 

Consider a vibration system with mass, damping, and spring.

The initial conditions are:

The system parameters are:

The input to the system is:

a step signal f(t) = 1

Use Laplace transform to find the system’s response.

Partial Fraction Expansion Exercise: Mass, Damping, and Spring System

2c 1m 

(0) 0x  (0) 0x 

75.k

Linear Time Invariant (LTI) System

Consider this mass-spring system again. The input and output relation of the system can be described using the following differential equation:

𝒎𝒙 (𝒕)+ 𝒌𝒙 𝒕 = 𝒇 𝒕

Let us use a more general differential equation to describe the input and output relation of a physical system:

𝒅𝒏𝒚(𝒕)

𝒅𝒕𝒏 + 𝒂𝒏−𝟏

𝒅𝒏−𝟏𝒚(𝒕)

𝒅𝒕𝒏−𝟏 +⋯+ 𝒂𝒐𝒚 𝒕 = 𝒃𝒎

𝒅𝒎𝒖 𝒕

𝒅𝒕𝒎 +⋯+ 𝒃𝒐𝒖(𝒕)

Linear Time Invariant (LTI) System We assume that all systems studied in this subject are linear time invariant.

• That means, this equation contain no powers or other

functions or products of the independent variables or derivatives

• Nonlinear examples: 𝒙 𝒕 𝒚 𝒕 + 𝒚 𝒕 = 𝒖 𝒕 𝒙𝟐 𝒕 + 𝒙 𝒕 = 𝒖 𝒕 𝒔𝒊𝒏 𝒕 𝒙 𝒕 + 𝒙 𝒕 + 𝟐 = 𝒖(𝒕)

• All coefficients are constant coefficients • Time-variant example: a moving car consuming gas is not

strictly time-invariant – the total mass is reducing

𝒅𝒏𝒚(𝒕)

𝒅𝒕𝒏 + 𝒂𝒏−𝟏

𝒅𝒏−𝟏𝒚(𝒕)

𝒅𝒕𝒏−𝟏 +⋯+ 𝒂𝒐𝒚 𝒕 = 𝒃𝒎

𝒅𝒎𝒖 𝒕

𝒅𝒕𝒎 +⋯+ 𝒃𝒐𝒖(𝒕)

Properties of Linear Time Invariant (LTI) System

 Superposition

 Homogeneity

2/1/2016

Linear Time Invariant (LTI) System Example 1

Suppose the input to the plant consists of three impulses at times t = 0 of size 7, 8, 9. What is the value of y(t) at time t = 5? Use both superposition and homogeneity properties, we can find:

= 7ℎ 5 + 8ℎ 5 + 9ℎ(5)

Suppose the inputs are three impulses at times t = 0, 1, 2 of size 7, 8, 9. What is the value of y(t) at time t = 5?

Linear Time Invariant (LTI) System Example 2

Any signal can be thought of as the summation (integral) of many impulses at different points in time

By the principal of superposition, if we can find the response of the system to one impulse, we will be able to find the response to an arbitrary input

𝑦(𝑡)

Arbitrary input

h(t-

Convolution Integral

Impulse at time t Impulse at time τ

Arbitrary input

The convolution integral Its abbreviation y(t)=u(t)*g(t)

Cascaded LTI System

The Laplace transform of the convolution integral For this cascaded system, what is the output? Output:

y(t) = u(t)*g(t)

𝒀 𝒔 = 𝓛 𝒚 𝒕 = 𝓛 𝒖 𝒕 ∗ 𝒈(𝒕) = 𝑼 𝒔 𝑮(𝒔)

The transfer function of a system is defined as the

ratio of the Laplace transforms of the output and input

with zero initial conditions.

Example 1:

Find the transfer function for a system modelled by the

following ODE

𝒅𝒄(𝒕)

𝒅𝒕 + 𝟐𝒄 𝒕 = 𝒓(𝒕)

Transfer Function

𝑮 𝒔 = 𝒀(𝒔)

𝑼(𝒔)

Example 2:

Use the result in Example 1, to find the response to an

unit step input, assuming zero initial conditions

Transfer Function…cont’d

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Example 3:

Find the transfer function corresponding to the ODE

𝒅𝟑𝒄

𝒅𝒕𝟑 + 𝟑

𝒅𝟐𝒄

𝒅𝒕𝟐 + 𝟕

𝒅𝒄

𝒅𝒕 + 𝟓𝒄 =

𝒅𝟐𝒓

𝒅𝒕𝟐 + 𝟒

𝒅𝒓

𝒅𝒕 + 𝟑𝒓

Transfer Function…cont’d

Modelling of Physical Systems

Mechanical Systems Electrical Systems

ODE ODE Impedance

Modelling of Translational Mechanical Systems Translational Mechanical Systems…Example

Find the transfer function for this system.

Spring Damper Mass System

ODE ODE Impedance

Modelling of Rotational Mechanical Systems

Rotational Mechanical Systems…Example

Torsion Disk Control System 41

Find the transfer function.

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ODE ODE ODE

Modelling of Electrical Systems

TF TF

Modelling of Electrical Systems

Electrical Systems…Example

RLC Circuit

Find the transfer function for the system, assuming input is V(t) and output is i(t).

Commonality of Physical Systems

Compare the transfer functions of these systems. What do you find?

Recap today’s topics +

Questions?

Homework 2

All the problems are from Chapter 2 Problem 2 Problem 4 Problem 7 Problem 8 Problem 16 Problem 23

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Appendix: MATLAB EXERCISES

Ex. 1. Spring-Mass-Damper System Response When subjected to an impulse input, the output is Use y <– x, x <- time in MATLAB program

)(3)( 5.15. tt eetx  

You can save it as a file exervise1.m You can run the program using “Run” button. Or type in the file name

This the result (does it make sense, given in impulse input?) Change the increment step, and increase the response time in your program, and observe the effects.

3_TimeResponse.pdf

2/3/2016

Dynamics and Control

Topic III: Time Response

• System natural response and forced response • The location of the poles and zeroes of the transfer

function in the “s-plane” • Investigate the relationship between system

components and time response • Predict the time response of a system from its

transfer function • Special facts about 1st and 2nd order systems

Week Topics Exam Reading

1 Introduction Ch 1-2

2 Math Review Ch 1-2

3 Modelling (Frequency + Time Domain) Ch 2

4 Time Response Ch 4

5 Block Diagram M-Exam 1 Ch 5

6 Stability Ch 6

7 Midterm Review + Make up Ch 1-6

8 Experiment

9 Spring Break

10 Steady State Errors Ch 7

11 Root Locus Ch 8

12 Design Via Root Locus M-Exam 2 Ch 9

13 Design Via Root Locus Ch 9

14 Frequency Response Ch 10

15 Design Via Frequency Response Ch 11

16 Final Review + Make up Ch 7-11

17 Final Exam Final Exam 2

Schedule

Course Roadmap

Block Diagrams

Why Time Response?

System Response

• The output response of a system is the sum of two related responses

I.The natural response describes dissipation, oscillation, or unstable growth from initial conditions

II.The forced response describes how the system reacts to external inputs

• Together these elements determine the overall response of the system

A Familiar Example – Nature Response

The external input to the system is f(t) =0 (Nature Response).

The system ODE: 𝒎𝒙 𝒕 + 𝑹𝒙 𝒕 + 𝒌𝒙 𝒕 = 𝒇 𝒕 = 𝟎

𝒎[𝒔𝟐𝑿 𝒔 − 𝒔𝒙 𝟎 − 𝒙 𝟎 ] + 𝑹[𝒔𝑿 𝒔 − 𝒙(𝟎)] + 𝒌𝑿 𝒔 = 𝟎

The Laplace transform:

𝑿 𝒔 = 𝒎𝒔 +𝑹 𝒙 𝟎 +𝒎𝒙 𝟎

𝒎𝒔𝟐 + 𝑹𝒔 + 𝒌

Rearrange the function:

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A Familiar Example - Natural Response

The system parameters are: Mass m = 1 kg, Damping co. R = 4 Ns/m, Spring co. k = 3N/m

The initial conditions are: 𝒙(𝟎) = .5 m, 𝒙 (0) = 0

𝑿 𝒔 = 𝒎𝒔 +𝑹 𝒙 𝟎 +𝒎𝒙 𝟎

𝒎𝒔𝟐 + 𝑹𝒔 + 𝒌

𝑿 𝒔 = . 𝟓 𝒔 + 𝟒

𝒔𝟐 + 𝟒𝒔+ 𝟑 =

. 𝟓 𝒔 + 𝟒

(𝒔 + 𝟏)(𝒔 + 𝟑) =.𝟕𝟓

𝟏

𝒔 + 𝟏 −.𝟐𝟓

𝟏

𝒔 + 𝟑

Apply inverse Laplace transform,

𝒙 𝒕 =.𝟕𝟓𝒆−𝒕−.𝟐𝟓𝒆−𝟑𝒕

A Familiar Example - Natural Response

What if we increase the spring constant? Mass m = 1 kg, Damping co. R = 4 Ns/m, Spring co. k = 20 N/m

𝑿 𝒔 = . 𝟓 𝒔 + 𝟒

𝒔𝟐 + 𝟒𝒔+ 𝟐𝟎 =

. 𝟓 𝒔 + 𝟒

(𝒔 + 𝟐)𝟐+𝟏𝟔

𝑿 𝒔 =. 𝟓 𝒔 + 𝟐

𝒔 + 𝟐 𝟐 + 𝟏𝟔 +

𝟐

𝒔 + 𝟐 𝟐 + 𝟏𝟔

𝒙 𝒕 =.𝟓𝒆−𝟐𝒕(𝒄𝒐𝒔𝟒𝒕 + 𝟎. 𝟓𝒔𝒊𝒏𝟒𝒕)

𝑿 𝒔 = 𝒎𝒔 +𝑹 𝒙 𝟎 +𝒎𝒙 𝟎

𝒎𝒔𝟐 + 𝑹𝒔 + 𝒌

A Familiar Example - Natural Response

Compare the responses in MATLAB (write the code in m file):

𝒙 𝒕 =.𝟕𝟓𝒆−𝒕−.𝟐𝟓𝒆−𝟑𝒕

𝒙 𝒕 =.𝟓𝒆−𝟐𝒕 𝒄𝒐𝒔𝟒𝒕 + 𝟎. 𝟓𝒔𝒊𝒏𝟒𝒕 Result:

When the spring constant is increased, system response is faster, but with overshoot.

MATLAB code - Natural Response % Solving for the time responses for the spring mass damper system

t=0:.1:10; % select time base

x1=.75*exp(-t)-.25*exp(-3*t); % natural response plot(t,x1,'r'); % plot in red hold on; % hold the plot

x2=.5*exp(-2*t).*(cos(4*t)+.5*sin(4*t)); % forced response % be careful with ".*" plot(t,x2,'b'); % plot in blue

grid; % add grid to the plot xlabel('Time in s'); ylabel('Response in m'); title('Comparing System Natural Responses');

pause; hold; % release the current the plot

Note! There is a dot here!!

A Familiar Example – Forced Response

The external input f(t) ≠ 𝟎. We will assume initial conditions are 0 when considering forced response.

The system ODE: 𝒎𝒙 𝒕 + 𝑹𝒙 𝒕 + 𝒌𝒙 𝒕 = 𝒇(𝒕)

𝑿 𝒔

𝑭(𝒔) =

𝟏

𝒎𝒔𝟐 + 𝑹𝒔 + 𝒌

𝒎𝒔𝟐𝑿 𝒔 +𝑹𝒔𝑿 𝒔 + 𝒌𝑿 𝒔 = 𝑭(𝒔)

The Laplace transform:

The transfer function:

A Familiar Example – Forced Response

Assume f(t) = 1. The system parameters are: Mass m = 1 kg, Damping co. R = 4 Ns/m, Spring co. k = 3 N/m

𝑿 𝒔 = 𝟏

𝒎𝒔𝟐 + 𝑹𝒔 + 𝒌 𝑭(𝒔) =

𝟏

𝒔

𝟏

𝒔𝟐 + 𝟒𝒔 + 𝟑

The transfer function is:

When the spring constant is increased to k = 20 N/m,

the transfer function is :

𝑿 𝒔 = 𝟏

𝒎𝒔𝟐 + 𝑹𝒔 + 𝒌 𝑭(𝒔) =

𝟏

𝒔

𝟏

𝒔𝟐 + 𝟒𝒔 + 𝟐𝟎

Compare the responses in MATLAB (Write the code in m file).

𝒙 𝒕 = 𝟏

𝟑 − 𝟏

𝟐 𝒆−𝒕 +

𝟏

𝟔 𝒆−𝟑𝒕

𝒙 𝒕 = 𝟏

𝟐𝟎 −

𝟏

𝟐𝟎 𝒆−𝟐𝒕𝐜𝐨𝐬𝟒𝒕 +

𝟑

𝟒 𝒆−𝟐𝒕𝐬𝐢𝐧𝟒𝒕

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MATLAB code – Forced Response

% Forced Response for Step Input % 1st system: Y(s)/F(s) = 1/(s^2 + 4s + 3) sys = tf(1,[1,4,3]); % tf = transfer function step(sys); % response to a step input pause; % 2nd system: Y(s)/F(s) = 1/(s^2 + 4s + 20) sys = tf(1,[1,4,20]); % tf = transfer function step(sys); % response to a step input

Ist click

2nd click

% Another way to program a transfer function

t=0:.1:7;

% 1st system: Y(s)/F(s) = 1/(s^2 + 4s + 3)

numt1=[1]; % define numerator of T1

dent1=[1 4 3]; % define denominator of T1

T1=tf(numt1, dent1); % create transfer f. for T1

[y1,t1]=step(T1,t); % run step r. and save points

% 2nd system: Y(s)/F(s) = 1/(s^2 + 4s + 20)

numt2=[1]; % define numerator of T2

dent2=[1 4 20]; % define denominator of T2

T2=tf(numt2, dent2); % create transfer f. for T2

[y2,t2]=step(T2,t); % run step r. and save points

plot(t1,y1,'-.', t2,y2,'-'); % chose plot pattern

xlabel('Time in s');

ylabel('Response in m');

title('Response Comparison');

MATLAB code – Forced Response

Natural Response and Forced Response

Natural Response Forced Response

K=3Nm

K=20Km

K=3Nm

Poles, Zeroes, and Responses

Transfer function

• The poles of a transfer function are the values of s that cause the transfer function to become infinite

• The zeroes of a transfer function are the values of s that cause the transfer function to go to zero

• A qualitative understanding of the effect of poles and zeroes can help us to quickly estimate performance

𝑮 𝒔 = 𝒀(𝒔)

𝑼(𝒔)

The s-plane

In the forced response example, when k = 3N/m, the output is Apply inverse Laplace transform, we can find

𝒙 𝒕 = 𝟏

𝟑 −

𝟏

𝟐 𝒆−𝒕 +

𝟏

𝟔 𝒆−𝟑𝒕

𝑿 𝒔 = 𝟏

𝒎𝒔𝟐 + 𝑹𝒔 + 𝒌 𝑭 𝒔 =

𝟏

𝒔𝟐 + 𝟒𝒔 + 𝟑

𝟏

𝒔 =

𝟏

𝒔(𝒔 + 𝟏)(𝒔 + 𝟑)

Pole zero

The s-plane In the forced response example, when k = 20 N/m, the output is

Apply inverse Laplace transform, we can find

𝒙 𝒕 = 𝟏

𝟐𝟎 −

𝟏

𝟐𝟎 𝒆−𝟐𝒕(𝐜𝐨𝐬𝟒𝒕 + 𝟒𝐬𝐢𝐧𝟒𝒕)

𝑿 𝒔 = 𝟏

𝒎𝒔𝟐 + 𝑹𝒔 + 𝒌 𝑭 𝒔 =

𝟏

𝒔𝟐 + 𝟒𝒔 + 𝟐𝟎

𝟏

𝒔

= 𝟏

𝒔(𝒔 + 𝟐 + 𝟒𝒋)(𝒔 + 𝟐 − 𝟒𝒋)

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Qualitative Effect of Poles

Assume a system has the following transfer function:

If the input to the system is a unit impulse, we can find the time response is:

𝒚 𝒕 = 𝒆−𝝈𝒕 Sketch system response: when 𝝈 > 𝟎, pole is located on the left hand side of the s-plane, system stable when 𝝈 < 𝟎, pole is located on the right hand side of the s-plane, system unstable

𝑮 𝒔 = 𝒀(𝒔)

𝑼(𝒔) =

𝟏

𝒔 + 𝝈

Qualitative Effect of Poles

• So we can conclude that in general, if the poles of the system are on the left hand side of the s-plane, the system is stable

• Poles in the right hand plane will

introduce components that grow without bound

• Poles on real axis generate exponential response

• Poles as conjugate pairs lead to oscillation

First Order Systems

Consider this RL circuit. I(s) is the output, and V(s) is the input to the circuit. The transfer function of this system is:

𝑰(𝒔)

𝑽(𝒔) =

𝟏

𝑳𝒔 + 𝑹 =

𝟏/𝑳

𝒔 + 𝑹/𝑳 =

𝒂

𝒔 + 𝝈

In general, we represent a first order system like this:

𝑮 𝒔 = 𝑪(𝒔)

𝑹(𝒔) =

𝒂

𝒔 + 𝝈

First Order Systems

If the input R(s) is a unit step, the system response is:

𝑪(𝒔) =

𝟏

𝒔

𝒂

𝒔 + 𝝈

The system time response is:

𝒄 𝒕 = 𝒂

𝝈 (𝟏 − 𝒆−σ𝒕)

Natural response Forced response

The system time response is: 23 𝒄 𝒕 = 𝒂

𝝈 (𝟏 − 𝒆−σ𝒕)

Response Characteristics

𝑎/𝜎

• Time constant 1/σ

𝒕 = 𝟏/σ, 𝒄 𝒕 =.𝟔𝟑 𝒂

𝝈

• Rising time Tr Time for the response to go from 10% to 90% of the final response

• Settling time Ts Time to reach 98% of the final response

1/σ 2/σ 3/σ 4/σ 5/σ

• The dynamic model of a system is not known.

• The unit step response of the system is given on the left.

• Estimate the dynamic model of the system.

First Order Systems Example

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• For a first order system we have a single, real pole • Many systems of interest are of higher order • Second order systems are quite common. The

general dynamic model is:

• In control engineering, we normally use the following standard form for a second order system:

Second Order Systems

cbss

a sG

 

2 )(

ss CsG



  

Second Order Systems – Some Examples

The transfer functions for both systems: or

cbss

a sG

 

2 )(

ss CsG



  

Damping Ratio & Natural Frequency

• 𝝎𝒏 is the natural frequency of the system, i.e., the frequency of a oscillation of the system without damping

• 𝝇 is the damping ratio of the system

No damper

Location of Poles

• Consider a general second order system:

• The poles are the roots of the denominator: recall:

2 )(

ss CsG



  

02 22  nnss

nn j 21roots

Location of Poles

• When there are two identical roots:

Laplace transform term

1 nroots

1

nn j 21roots

Location of Poles

• When , there are two real poles:

1

nn  1roots 2

nn j 21roots

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Location of Poles

• When , there are two complex poles:

1

nn j 21roots

Location of Poles

• When , there are two imaginary poles:

0

njroots

nn j 21roots

33 Second Order System Behaviors

Different Damping Cases

• Find damping ratio for each system, report the expected response and sketch the response.

Second Order Systems Example

Underdamped Second Order System

• A common model for physical systems • Damping ratio 𝝇 < 𝟏 • Step response for a 2nd-order underdamped system

𝑪 𝒔 = 𝝎𝒏 𝟐

𝒔(𝒔𝟐 + 𝟐𝝇𝝎𝒏𝒔 +𝝎𝒏 𝟐)

𝒄 𝒕 = 𝟏 − 𝟏

𝟏−𝝇𝟐 𝒆−𝝇𝝎𝒏𝒕𝐜𝐨𝐬(𝝎𝒏 𝟏 − 𝝇𝟐𝒕 − 𝝓)

where 𝝓 = tan−𝟏(𝝇/ 𝟏 − 𝝇𝟐)

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2nd-Order System Responses for damping ratios

• Tr, rise time. Time required for the response to go from .1 to .9 of the final value

• Tp, peak time. Time required to reach the 1st maximum peak

• Ts, settling time. Time required to reach and stay within 2% of the steady-state value

• %OS, percent overshoot

• 𝝎𝒏 natural frequency

• 𝝇 damping ratio

Response Characteristics



• Tp, peak time. 𝑻𝒑 = 𝝅

𝝎𝒏 𝟏−𝝇𝟐

• %OS, overshoot. %𝐎𝐒 = 𝒆

−( 𝝇𝝅

𝟏−𝝇𝟐 )

∗ 𝟏𝟎𝟎

• Ts, settling time. 𝑻𝒔= 𝟒

𝝇𝝎𝒏

• Tr, rise time. No analytical form. Can be found from plot. 𝐓𝐫 ≈ 𝟏. 𝟖/𝝎𝒏 (empirical formula)

Evaluation of Response Characteristics

Second-Order System Example

Finding Tp, %𝐎𝐒, Ts and Tr for the following system when subjected to a step input:

𝑮 𝑺 = 𝟏𝟎𝟎

𝒔𝟐 + 𝟏𝟓𝒔 + 𝟏𝟎𝟎

Relating Poles to Response Characteristics

ddnn jj  21roots

Pole Location and Response Characteristics

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Pole Location and Response Characteristics

Example 1: Given pole location, predicting response Matlab Exercise 1 (code given)

P1=[1 3+7i]; P2=[1 3-7i]; deng=conv(P1, P2); omegan=sqrt(deng(3)/deng(1)) zeta=(deng(2)/deng(1))/(2*omegan) Ts=4/(zeta*omegan) Tp=pi/(omegan*sqrt(1-zeta^2)) Tr=omegan/1.8 Os=100*exp(-zeta*pi/sqrt(1-zeta^2))

Matlab Exercise 2 (write your code)

• Plot response subject to unit step input • Move poles around (in 3 directions), observe the changes • Prepare a one-page report, discussing how location of

poles affects the system response (in terms of time response characteristics), attach your code

Example 2: Find Allowable Region for Poles

Find allowable region in the s-plane for the poles of a transfer function to meet the requirements: Tr < 0.6 sec Os% < 10% Ts < 3 sec

Example 3: Transient Response Through Component Design

Given the system shown above, find J and D to yield 20% overshoot and a settling time of 2 seconds for a step input torque T(t)

Additional Poles

• The preceding discussed are second order systems (two poles)

• What happens to the system response if there are additional poles (higher order system)?

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Additional Poles

• The preceding discussed are second order systems (two poles)

• Consider a third order system with an additional pole at α

• By partial fraction expansion, we can obtain the step response

• The response of the system consists two parts:  contribution by the original

second system  contribution by the

additional pole

)(sG

Additional Poles

Infinity Far Near

α=∞, effect of Case III ≈ 0

Additional Poles

C3(t), additional pole is at infinity

Additional Zeros

Consider a second order system: An addition zero is added to the system: The response of the system consists two parts: • the derivative of the original system • a scaled version of the original system

Additional Zeros

In Class Exercise 1. Find the Transfer Function of the System

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In Class Exercise 2. Find the Transfer Function of the System

Homework 3

All the problems are from Chapter 4 Problems 1 – 8. Problems 12 -14. Problem 29.

4_Block Diagrams.pdf

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Dynamics and Control

Topic IV: Block Diagrams

Week Day Topics Exam Reading

1 Aug 25 No Class

2 Sept 1 Labor Day

3 Sept 8 Introduction/Math Review Ch. 1-2

4 Sept 15 Modelling (Frequency + Time Domain) Ch 2

5 Sept 22 Time Response Ch 4

6 Sept 29 Block Diagram M-Exam 1 Ch 5

7 Oct 6 Stability Ch 6

8 Oct 13 Fall Break

9 Oct 20 Midterm Review + Make up Ch 1-6

10 Oct 27 Steady State Errors Ch 7

11 Nov 3 Root Locus Ch 8

12 Nov 10 Design Via Root Locus M-Exam 2 Ch 9

13 Nov 17 Design Via Root Locus Ch 9

14 Nov 24 Frequency Response Ch 10

15 Dec. 1 Design Via Frequency Response Ch 11

16 Dec. 8 Final Review + Make up Ch 7-11

16 Dec. 11 Final Exam 10:30 -12:30 Final Exam

Schedule

Course Roadmap

Block Diagrams

Why Block Diagrams?

G(s)

We have been working with individual subsystems represented by a block with its input and output.

Sensing Actuation Control PlantControlSystem = + + +

Remember… BIG DOG CONTROL: Block Diagram Representation

DISTURBANCE

Environment

Foot Trajectory Planning

Desired Walking Speed

REFERENCE

Joint angles & velocities

Robot Joint torques

RobotRobot

Virtual Leg IK

Virtual Leg FK

Virtual

Leg VM

PD Servo

Virtual Leg Coords

Virtual Leg Forces

State Machine

Leg State Gait

Coordination Mechanisms

Influences to/from other legs

IK=inverse kinematics FK=forward kinematics VM=virtual model

10/18/2014

• Like the BIG DOG, in reality, many systems are composed of multiple subsystems

• Block Diagrams Method is for combining subsystems and simplifying system representation

• Schematic elements used in block diagrams

Schematic Elements

Brain Hand Foot

Eye

Car_

Desired Direction

Desired Speed

Direction

Speed Hands + Feet

Cascaded System

Parallel System

Brain Hand

Foot

Eye

Car_

Desired Direction

Desired Speed

Direction

Speed

Feedback System

General form

Block Diagram Reduction – Example 1

• Search for the general form • Apply the general formal • Reduce to a single transfer

function

Find Transient Response – Example 2

For the system shown above, find Tp, OS, Ts

numg = [25]; Ts = 4/(z*wn) deng = poly([0 -5]); Tp = pi/(wn*sqrt(1-z^2)) G = tf (numg, deng) OS=exp(-z*pi/sart(1-z^2))*100 T = feedback (G, 1) step(T) [numt, dent] = tfdata(T, ‘v’); pause wn = sqrt(dent(3)) z=dent(2)/(2*wn)

Unity feedback

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Gain Design for Transient Response – Example 3

• Find the value of K so that the system will respond with OS=10%

• Can we select Ts as a designed criterion?

Shuttle Pitch Control - Example 4

• A real system that incorporates feedback to control the

pitch of the space shuttle (amongst other things).

• We manipulate block diagrams for control design and analysis.

• The control

mechanisms include

the body flap,

elevons and engines

• Measurements are

made by the

vehicle’s inertial

unit, gyros and accelerometers

A simplified model of the pitch controller for the space shuttle

Transfer Function?

Transfer Function:

Homework 4

Reduce the block diagram to a single transfer function. Use Matlab to verify your result.

Problems 15-17.