Homework 3

Problem 3.7

• The input to a causal, LTI system is:

• The output z-transform is:

• Determine:– (a) H(z) and ROC– (b) ROC of Y[z]– (c) y[n]

][2

1]1[][ nununx

n

)1)(2

11(

2

1

][11

1

zz

zzY

Problem 3.7

• Solve X[z]

12

1,

)2

1)(1(

2

1

2

11][

2

1,

2

1

2

11

1][

1,1

11

11][

][][][

12

011

21

zzz

z

z

z

z

zzX

zz

z

zzX

zz

z

zzzzX

zXzXzX

nn

Problem 3.7

• (a) Solve H[z]

• Causal

11

1

)2

1)(1(

2

1

)1)(2

1(

2

1

][

][][

zz

z

zz

z

zz

z

zX

zYzH

Problem 3.7

• (b) ROC of Y[z]

• Possible ROCs: , – Since one of the poles of X[z], which limited the roc OF x[Z] to be less

than 1, is cancelled by the zero of H[z], the ROC of Y[z] is the region of the z-plane that satisfies the remaining two constraints Hence Y[z] converges on .

Problem 3.7

• Solve for y[n]

Problem 3.8

• The causal system function is:

• The input is: • (a) Find h[n]• (b) Find y[n]• (c) Is H stable, absolutely summable?

1

1

4

31

1][

z

zzH

Problem 3.8

• (a) – The ROC is , since it is causal– First divide to get: H[z]=

• another way:

1

1

4

314

71

z

z

]1[4

3

4

7][][

4

3

4

71

4

3

4

71][

1

0 1

11

nunnh

zzzzH

n

nn

nn

1

1

1

4

31

4

31

1][

z

z

zzH

]1[4

3][

4

3][

1

nununhnn

Problem 3.8

• (b) Find y[n]• First solve for X[z], then Y[z] =X[z]H[z]

][4

3

13

8][

3

1

13

8][,

4

3,

4

31

13/8

3

11

13/8][

)4

31)(

3

11(3

2][][][

13

1

1

1

3

11

1][

11

11

1

11

nununyzzz

zY

zz

zzHzXzY

zzz

zX

nn

Problem 3.8

• (c) Stable and absolutely summable since ROC includes unit circle

Problem 3.17

• An LTI system with input x[n] and output y[n] satisifes the difference equation:

• Determine all possible values for the system’s impulse response h[n] at n=0

]1[][]2[]1[2

5][ nxnxnynyny

Problem 3.17

• Solve for H[z]

• 3 possible ROCs: ,

11

21

1

121

2

11

3/1

21

3/2

2

51

1][

][][][][2

5][

zzzz

zzH

zzXzXzzYzzYzY

Problem 3.17

• For

z

z

z

zzH

zzzH

21

2)3/1(

2

11

2

1

)3/2(][

2

11

3/1

21

3/2][

11

h [𝑛 ]=− 232𝑛𝑢 [−𝑛−1 ]− 1

3 ( 12 )𝑛

𝑢 [−𝑛−1 ] , h [0 ]=0

Problem 3.17

12<|𝑧|<23/1]0[

][2

1

3

1]1[2

3

2][

h

and

nununhn

n

Problem 3.17

1]0[

][2

1

3

1][2

3

2][

h

and

nununhn

n

Problem 3.32

• Determine inverse transform:

• For the 3rd term use the identity:

)2

11(

1225

58

)2

11(

35

1

)31(1225

2700

)21(1225

1568

][

)31)(21()2

11(

1][

12111

1121

zzzzzX

zzzzX

dz

zdXznnx

][][

Problem 3.32

• 3rd term continued

• Let

• Let and

• Then therefore

1

2

11

1][

zdz

dzzW

1

2

11

1][

zzX

and

][2

1][

]1[35

2][

][][

nunx

nwny

nxnnw

n

]1[

2

1)1(

35

2][

1

nunnyn

Problem 3.32

• The other terms are done by inspection and a stable sequence implies 2-sided sequence by pole observation

][)2

1(0473.0]1[)

2

1)(1(0571.0]1[)3(204.2]1[)2(28.1][ 1 nununnununx nnnn

Problem 3.32 (b)

(b)

1!0][!

1][

......!2

]2[

!1

]1[][][

......!3!2!1

1][

Re

][

321

wherenun

nx

nnnnx

zzzzX

SeriesTaylorthewrite

ezX z

Problem 3.32 (c)

• (c)

]1[)2(2]1[2]2[][

][)2(2][2]1[2]2[][

2

11

222][

2

2][

2

3

nunnnx

nunnnnx

sidedleft

zzzzX

Dividez

zzzX

n

n