Given the answers to the sample problems i need the questions.

1

Fall 2006

Solution to ENEE 3530 Exam #1

1.  

  

 π 2

sin 2 e = 2e =7.39 [Note that other than at 6=τ , the function is zero everywhere]

2. (a) It is linear because for any input x1 that yields the output y1 and any input x2 that

yields the output y2, we have input (a1x1+a2x2) → output (a1y1+a2y2).

(b) It is time invariant because y(t−t0)= x(t−t0−5) for any t0.

3.

- x(t) R

C

L + y

L = C = 1 and R = 2

(a) Input-Output Relation

Dyx)1D(Dyx C

1 LDyx

CD

1 LDx

1

2

1

2

11 =+⇒=



  

 +⇒=+

2

y x

R

y xx

1 −=−=

Dy) 2

y x)(1D(

2 =−+

⇒ x)2D2(y)1D2D( 22 +=++

(b) State and Output Equations

yxxL =+ 21

& ⇒ yxx =+ 21

&

1212 xxxxC =⇒= &&

)(2)( 11 xxyxx

R

y −=⇒−=

We can write the above equations in the standard matrix form as below.

State equation: x 0

2

x

x

01

12

x

x

2

1

2

1

 

  

 + 

  

  

  

 −− =



  

 &

&

Output equation: [ ] x2 x

x 02y

2

1 + 

  

 −=

(c) Unit Impulse Response

Characteristic equation: 0122 =++ λλ 1,1 −−=λ

( ) ( ) tetccty −+= 210

x1

x2

Here, for impulse response

( ) 000 =y and ( ) 100 =y&

This part can be represented in many

other alternate forms.

2

( ) ( ) tetcccty −−+−= 2210& ( ) 00 10 == cy ( ) 110 2210 =⇒=+−= cccy&

Hence, ( ) ttety −=0 Unit impulse response: ( ) ( ) ( ) ( ) ( )tutyDtth ]22[2 02 ++= δ

( ) ( ) ( ) ( )tuettth t−−+= 142δ

4.

(a) Let [ ]ky be the balance just after the kth montly payment is made (n is replaced by k) and [ ]kf be the input. Here, [ ] xkf = for k=1,…,n.

[ ] 80000 −=y [ ] [ ] [ ]11 +=−+ kfkyky γ 100/1%1 and 1 ==+= rrγ or ( ) [ ] [ ]kEfkyE =−γ (b) y[2] = 1.01�[1.01�(−8000)+x]+x = −8160.8+2.01x

The impulse response of this system is: [ ] [ ]kukh kγ= The zero input response is: [ ] [ ]kucky ko γ= Setting 0=k and substituting [ ] 80000 −=y , we find 8000−=c [ ] [ ]kuky ko γ8000−= The zero-state response [ ]ky is: [ ] [ ] [ ]kfkhky *= The first monthly payment x is made one month after receiving the loan (at k=1).

Hence, the input is: [ ] [ ]1−= kxukf

[ ] [ ] [ ]1* −= kxukhky [ ] [ ] [ ]1* −= kukuxky kγ

Here we have [ ] [ ] [ ] [ ]kukukukz k

k

1

1 *

1

− −

== +

γ γ

γ

Using the shift property of convolution [ ] [ ] [ ] [ ]1ku 1

1 x1kz1ku*kux

k k −

−γ −γ

=−=−γ

The total balance is [ ] [ ]kyky +0 [ ] [ ] [ ] [ ]1 1

1 80000 −

− −

+−=+ kuxkukyky k

k

γ γ

γ

For [ ] [ ] 11 ,1 =−=> kukuk

Therefore, the loan balance is 1

1 8000

− −

+− γ γ

γ k

k x , k>1.

Here, 2=nb

Only the boxed portion above with

explanation of the terms is

necessary for the exam problem.

The rest part is one possible way to

solve for x (20% bonus).

3

At 484*12 ==k the loan balance becomes zero, i.e., 1

1 80000

48 48

− −

+−= γ γ

γ x .

7.210 1

1 8000

1 48

48 =  

   



−γ −γ

γ=

−

x

5. Convolutions.

(a) ( ) ( ) ( )tuetuetc tt 2* −=

∫ τ= τ−−τ

t

t dee

0

)(2

t

t ee

0

32

3

1 

 

 = τ− ( )tuee tt ][

3

1 2−−=

( )tueetftftc tt ][ 3

1 )(*)()(

2

21

−−==

(b) [ ] [ ] ( ) [ ] ][ 25.11

25.11 4.05.2][4.05.05.24.0*5.0

1

0

1 nununununc

n n

n

i

ininn

  

   



− −

⋅=  





 



 ⋅==

+

=

−− ∑

( ) ][4.0105.05.12][*][][ 43 nunfnfnc nn ⋅−⋅==