z transform

SIGNALS & SYSTEMS

GATE CLOUD

R. K. Kanodia

Ashish Murolia

JHUNJHUNUWALA

SIGNALS & SYSTEMS

Jaipur

GATE CLOUD

GATE CLOUD Signals & Systems, 1e

R. K. Kanodia, Ashish Murolia

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CHAPTER 6THE Z TRANSFORM

EXERCISE 6.1

MCQ 6.1.1 The z -transform is used to analyze(A) discrete time signals and system (B) continuous time signals and system

(C) both (A) and (B) (D) none

MCQ 6.1.2 Which of the following expression is correct for the bilateral z -transform of [ ]x n ?

(A) [ ]x n zn

n 0

3

=/ (B) [ ]x n z n

n 0

3−

=/

(C) [ ]x n zn

n 3

3

=−/ (D) [ ]x n z n

n 3

3−

=−/

MCQ 6.1.3 The unilateral z -transform of sequence [ ]x n is defined as

(A) [ ]x n zn

n 0

3

=/ (B) [ ]x n zn

n 3

3

=−/

(C) [ ]x n z n

n 0

3−

=/ (D) [ ]x n z n

n 3

3−

=−/

MCQ 6.1.4 The z -transform of a causal signal [ ]x n is given by

(A) [ ]x n zn

n 3

3

=−/ (B) [ ]x n zn

n 0

3

=/

(C) [ ]x n z n

n 3

3−

=−/ (D) [ ]x n z n

n 0

3−

=/

MCQ 6.1.5 For a signal [ ]x n , its unilateral z -transform is equivalent to the bilateral z -transform of(A) [ ] [ ]x n r n (B) [ ] [ ]x n nδ

(C) [ ] [ ]x n u n (D) none of these

MCQ 6.1.6 The ROC of z -transform ( )X z is defined as the range of values of z for which ( )X z(A) zero (B) diverges

(C) converges (D) none

MCQ 6.1.7 In the z -plane the ROC of z -transform ( )X z consists of a(A) strip (B) parabola

(C) rectangle (D) ring

MCQ 6.1.8 If [ ]x n is a right-sided sequence, and if the circle z r0= is in the ROC, then (A) the values of z for which z r> 0 will also be in the ROC

Page 484 The Z Transform Chapter 6

(B) the values of z for which z r< 0 will also be in the ROC

(C) both (A) & (B)

(D) none of these

MCQ 6.1.9 The ROC does not contain any(A) poles (B) 1’s

(C) zeros (D) none

MCQ 6.1.10 Let [ ] ( )x n X zZ be a z -transform pair. If [ ] [ ]x n nδ= , then the ROC of ( )X z is(A) z 1< (B) z 1>

(C) entire z -plane (D) none of the above

MCQ 6.1.11 The ROC of z -transform of unit-step sequence [ ]u n , is(A) entire z -plane (B) z 1<

(C) z 1> (D) none of the above

MCQ 6.1.12 The ROC of the unilateral z -transform of nα is(A) z > α (B) z < α

(C) 1z < (D) z 1>

MCQ 6.1.13 Which of the following statement about ROC is not true ?(A) ROC never lies exactly at the boundary of a circle

(B) ROC consists of a circle in the z -plane centred at the origin

(C) ROC of a right handed finite sequence is the entire z -plane except z 0=

(D) ROC contains both poles and zeroes

MCQ 6.1.14 The z -transform of unit step sequence is (A) 1 (B) 1z

1−

(C) zz

1− (D) 0

MCQ 6.1.15 The ROC for the z -transform of the sequence [ ] [ ]x n u n= − is(A) z 0> (B) 1z <

(C) z 1> (D) does not exist

MCQ 6.1.16 Let [ ] ( )x n X zZ , then unilateral z -transform of sequence [ ] [ 1]x n x n1 = − will be(A) ( ) ( ) [0]X z z X z x1

1= +− (B) ( ) ( ) [ ]X z z X z x 111= −−

(C) ( ) ( ) [ 1]X z z X z x11= − −− (D) ( ) [ ] [ 1]X z z X z x1

1= + −−

MCQ 6.1.17 Let [ ] ( )x n X zZ , the bilateral z -transform of [ ]x n n0− is given by

(A) ( )zX z (B) ( )z X zn0

(C) ( )z X zn0− (D) ( )z X z1

Chapter 6 The Z Transform Page 485

MCQ 6.1.18 If the ROC of z -transform of [ ]x n is Rx then the ROC of z -transform of [ ]x n− is(A) Rx (B) Rx−

(C) /R1 x (D) none of these

MCQ 6.1.19 If ( ) { [ ]}X z x nZ= , then ( ) { [ ]}X z a x nZ n= − will be

(A) ( )X az (B) X az

a k

(C) X zaa k (D) X az

1b l

MCQ 6.1.20 If [ ]x n and [ ]y n are two discrete time sequences, then the z -transform of correlation of the sequences [ ] [ ]andx n y n is(A) ( ) ( )X z Y z1 1− − (B) ( ) ( )X z Y z 1−

(C) ( ) ( )X z Y z* (D) * ( ) * ( )X z Y z 1−

MCQ 6.1.21 If ( ) { [ ]}X z x nZ= , then, value of [0]x is equal to(A) ( )lim zX z

z 0" (B) ( ) ( )lim z X z1

z 1−

"

(C) ( )limX zz"3

(D) ( )limX zz 0"

MCQ 6.1.22 The choice of realization of structure depends on(A) computational complexity (B) memory requirements

(C) parallel processing and pipelining (D) all the above

MCQ 6.1.23 Which of the following schemes of system realization uses separate delays for input and output samples ?(A) parallel form (B) cascade form

(C) direct form-I (D) direct form-II

MCQ 6.1.24 The direct form-I and II structures of IIR system will be identical in(A) all pole system (B) all zero system

(C) both (A) and (B) (D) first order and second order systems

MCQ 6.1.25 The number of memory locations required to realize the system,

( )H zz zz z

1 21 3 2

2 4

2 3

=+ ++ +

− −

− − is

(A) 5 (B) 7

(C) 2 (D) 10

MCQ 6.1.26 The mapping z esT= from s -plane to z -plane, is(A) one to one (B) many to one

(C) one to many (D) many to many

***********

EXERCISE 6.2

MCQ 6.2.1 Consider a DT signal which is defined as follows

[ ]x n ,

,

n

n21 0

0 0<

n$

= b l*

The z -transform of [ ]x n will be

(A) zz

12 1

−−

(B) zz

2 12−

(C) z

121− (D) z2

1−

MCQ 6.2.2 If the z -transform of a sequence [ ] {1, 1, 1, }x n 1= − −-

is ( )X z , then what is the value of /X 1 2^ h ?(A) 9 (B) .1 125−

(C) 1.875 (D) 15

MCQ 6.2.3 The z -transform and its ROC of a discrete time sequence

[ ]x n , 0

,

n

n21

0 0

<n

$

=−b l*

will be

(A) ,zz z2 1

221>− (B) ,z

z z2 21<−

(C) ,zz z2 1

221<− (D) ,z

z z12

21>

1

−−

MCQ 6.2.4 The ROC of z -transform of the discrete time sequence [ ]x n | |n21= ^ h is

(A) 2z< <21 (B) z 2>

(C) z2 2< <− (D) z < 21

MCQ 6.2.5 Consider a discrete-time signal [ ]x n [ ] [ 1]u n u n31

21n n

= + − −b bl l . The ROC of its z-transform is

(A) z3 2< < (B) z 21<

(C) z 31> (D) z3

121< <


MCQ 6.2.6 For a signal [ ] [ ] [ ]x n u nn nα α= + − , the ROC of its z -transform would be

(A) ,minz 1> α αe o (B) z > α

(C) ,maxz 1> α αe o (D) z < α

MCQ 6.2.7 Match List I (discrete time sequence) with List II (z -transform) and choose the correct answer using the codes given below the lists:

List-I (Discrete Time Sequence) List-II (z -Transform)

P. [ 2]u n − 1.( )

, 1z z

z11 <2 1−− −

Q. [ 3]u n− − − 2. , 1z

z z1

<1

1

−−

−

−

R. [ 4]u n + 3.( )

, 1z z

z11 >4 1−− −

S. [ ]u n− 4. ,z

z z1

1>1

2

− −

−

Codes : P Q R S(A) 1 4 2 3(B) 2 4 1 3(C) 4 1 3 2(D) 4 2 3 1

MCQ 6.2.8 The z -transform of signal [ ] [ ]x n e u njn= π is

(A) , : 1ROCzz z1 >+ (B) , : 1ROCz j

z z >−

(C) , : 1ROCz

z z1

<2 + (D) , : 1ROCz z1

1 <+

MCQ 6.2.9 Consider the pole zero diagram of an LTI system shown in the figure which corresponds to transfer function ( )H z .


Match List I (The impulse response) with List II (ROC which corresponds to above

diagram) and choose the correct answer using the codes given below:

{Given that ( )H 1 1= }

List-I (Impulse Response) List-II (ROC)

P. [( 4)2 6(3) ] [ ]u nn n− + 1. does not exist

Q. ( ) [ ] ( ) [ ]u n u n4 2 6 3 1n n− + − − − 2. z 3>

R. ( ) [ ] ( ) [ ]u n u n4 2 1 6 3 1n n− − + − − − 3. z 2<

S. ( ) [ ] ( ) [ ]u n u n4 2 1 6 3n n− − + − 4. z2 3< <

Codes : P Q R S

(A) 4 1 3 2

(B) 2 1 3 4

(C) 1 4 2 3

(D) 2 4 3 1

MCQ 6.2.10 The z -transform of a discrete time signal [ ]x n is ( )X z ( )z zz

11= −

+ . What are the

values of [0], [1]x x and [ ]x 2 respectively ?

(A) 1, 2, 3 (B) 0, 1, 2

(C) 1, 1, 2 (D) 1, 0, 2−

MCQ 6.2.11 The z -transform of a signal [ ]x n is ( )X z e e /z z1= + , z 0! . [ ]x n would be

(A) [ ] !n n1δ + (B) [ ] !u n n

1+

(C) [ 1] !u n n− + (D) [ ] ( ) !n n 1δ + −

Statement For Q. 12 - 13 :

Consider a discrete time signal [ ]x n and its z -transform ( )X z given as

( )X z z z

z z2 3

52

2

=− −

+

MCQ 6.2.12 If ROC of ( )X z is z 1< , then signal [ ]x n would be

(A) [ 2(3) ( 1) ] [ ]u n 1n n− + − − − (B) [2(3) ( 1) ] [ ]u nn n− −

(C) ( ) [ ] ( ) [ ]u n u n2 3 1 1n n− − − − − (D) [ ( ) ] [ ]u n2 3 1n +

MCQ 6.2.13 If ROC of ( )X z is z 3> , then signal [ ]x n would be

(A) [ ( ) ( ) [ ]u n2 3 1n n− − (B) [ ( ) ( ) ] [ ]u n2 3 1 1n n− + − − −

(C) ( ) [ ] ( ) [ ]u n u n2 3 1 1n n− − − − − (D) [ ( ) ] [ ]u n2 3 1n +


MCQ 6.2.14 If ROC of ( )X z is z1 3< < , the signal [ ]x n would be

(A) [ ( ) ( ) ] [ ]u n2 3 1n n− − (B) [ ( ) ( ) ] [ ]u n2 3 1 1n n− + − − −

(C) ( ) [ ] ( ) [ ]u n u n2 3 1 1n n− − − − − (D) [ ( ) ( ) ] [ ]u n2 3 1 1n n+ − − −

MCQ 6.2.15 Consider a DT sequence [ ]x n [ ] [ ]x n x n1 2= + where, [ ]x n1 ( . ) [ ]u n0 7 1n= − and

[ ] ( 0.4) [ 2]x n u nn2 = − − . The region of convergence of z -transform of [ ]x n is

(A) . .z0 4 0 7< < (B) .z 0 7>

(C) .z 0 4< (D) none of these

MCQ 6.2.16 The z -transform of a DT signal [ ]x n is ( )X z z z

z8 2 12=

− −. What will be the z

-transform of [ ]x n 4− ?

(A) ( ) ( )

( )z z

z8 4 2 4 1

42+ − + −

+ (B)

z zz

8 2 12

5

− −

(C) z z

z128 8 1

42 − −

(D) z z z8 2

15 4 3− −

MCQ 6.2.17 If [ ] [ ]x n u nnα= , then the z -transform of [ ] [ ]x n u n3+ will be

(A) zz 2

α−−

(B) zz4

α−

(C) zz3α α−a k (D) z

z 3

α−−

MCQ 6.2.18 Let [ ], [ ]x n x n1 2 and [ ]x n3 be three discrete time signals and ( ), ( )X z X z1 2 and ( )X z3

are their z -transform respectively given as

( )X z1 ( )( . )z z

z1 0 5

2

= − − ,

( )X z2 ( )( . )z z

z1 0 5

= − −

and ( )X z3 ( )( . )z z1 0 5

1= − −

Then [ ], [ ]x n x n1 2 and [ ]x n3 are related as

(A) [ ] [ ] [ ]x n x n x n2 11 2 3− = − = (B) [ ] [ ] [ ]x n x n x n2 11 2 3+ = + =

(C) [ ] [ ] [ ]x n x n x n1 21 2 3= − = − (D) [ ] [ ] [ ]x n x n x n1 11 2 3+ = − =

MCQ 6.2.19 The inverse z -transform of a function ( )X z zz 9

α= −−

is

(A) [ ]u n 10n 10α −− (B) [ ]u n 10nα −

(C) [ ]u n/n 10α (D) [ ]u n 9n 9α −−


MCQ 6.2.20 Let [ ] ( )x n X zZ be a z -transform pair, where ( )X z zz

32

= −−

. The value of [ ]x 5 is

(A) 3 (B) 9

(C) 1 (D) 0

MCQ 6.2.21 The z -transform of the discrete time signal [ ]x n shown in the figure is

(A) z

z1

k

1− −

− (B)

zz

1

k

1+ −

−

(C) zz

11 k

1−−

−

− (D)

zz

11 k

1−+

−

−

MCQ 6.2.22 Consider the unilateral z -transform pair [ ] ( )x n X z zz

1Z = − . The z -transform

of [ ]x n 1− and [ ]x n 1+ are respectively

(A) zz

12

− , z 11− (B) z 1

1− , z

z1

2

−

(C) z 11− , z

z1− (D) z

z1− , z

z1

2

−

MCQ 6.2.23 A discrete time causal signal [ ]x n has the z -transform

( )X z . , : 0.4ROCzz z0 4 >= −

The ROC for z -transform of the even part of [ ]x n will be

(A) same as ROC of ( )X z (B) . .z0 4 2 5< <

(C) .z 0 2> (D) .z 0 8>

MCQ 6.2.24 The z -transform of a discrete time sequence [ ] [ 1] [ ]y n n n u n= + is

(A) ( )z

z1

23

2

− (B)

( )( )z

z z11

3−+

(C) ( )z

z1 2−

(D) ( )z 1

12−


MCQ 6.2.25 Match List I (Discrete time sequence) with List II (z -transform) and select the correct answer using the codes given below the lists.

List-I (Discrete time sequence) List-II (z -transform)

P. ( ) [ ]n u n1 n− 1.( )

, :ROCz

z z1

1>1 2

1

− −

−

Q. [ ]nu n 1− − − 2.( )

, : 1ROCz

z1

1 >1+ −

R. ( ) [ ]u n1 n− 3.( )

, : 1ROCz

z z1

<1 2

1

− −

−

S. [ ]nu n 4.( )

, : 1ROCz

z z1

>1 2

1

−+ −

−

Codes : P Q R S(A) 4 1 2 3(B) 4 3 2 1(C) 3 1 4 2(D) 2 4 1 3

MCQ 6.2.26 A signal [ ]x n has the following z -transform ( )X z (1 2 ), :log ROCz z < 21= − .

The signal [ ]x n is

(A) [ ]u n21 n

b l (B) [ ]n u n121 n

b l

(C) [ ]n u n121 1

n− −b l (D) [ ]u n2

1 1n

− −b l

MCQ 6.2.27 A discrete time sequence is defined as [ ]x n ( 2) [ 1]u nnn1= − − −− . The z -transform

of [ ]x n is

(A) , :log ROCz z21

21<+b l (B) , :log ROCz z2

121<−b l

(C) ( 2), : 2log ROCz z >− (D) ( 2), : 2log ROCz z <+

MCQ 6.2.28 Consider a z -transform pair [ ] ( )x n X zZ with ROC Rx . The z transform and its ROC for [ ] [ ]y n a x nn= will be

(A) , :ROCX az a Rxa k (B) ( ), :ROCX z a Rx+

(C) ( ), :ROCz X z Rax

− (D) ( ), :ROCX az a Rx


MCQ 6.2.29 Let ( )X z be the z -transform of a causal signal [ ] [ ]x n a u nn= with :ROC z a>. Match the discrete sequences , ,S S S1 2 3 and S4 with ROC of their z -transforms

,R R1 2 and R3.

Sequences ROC

:S1 [ ]x n 2− :R1 z a>

:S2 [ ]x n 2+ :R2 z a<

:S3 [ ]x n− :R3 z a1<

:S4 ( ) [ ]x n1 n−

(A) ( , ),( , ),( , ),( , )S R S R S R S R1 1 2 2 3 3 4 3

(B) ( , ),( , ),( , ),( , )S R S R S R S R1 1 2 1 3 3 4 1

(C) ( , ),( , ),( , ),( , )S R S R S R S R1 2 2 1 3 2 4 3

(D) ( , ),( , ),( , ),( , )S R S R S R S R1 1 2 2 3 2 4 3

MCQ 6.2.30 Consider a discrete time signal [ ] [ ]x n u nnα= and its z -transform ( )X z . Match List I (discrete signals) with List II (z -transform) and select the correct answer using the codes given below:

List-I (Discrete time signal) List-II (z -transform)

P. [ / ]x n 2 1. ( )z X z2−

Q. [ ] [ ]x n u n2 2− − 2. ( )X z2

R. [ ] [ ]x n u n2+ 3. ( / )X z 2β

S. [ ]x nn2β 4. ( )X z2α

Codes : P Q R S(A) 1 2 4 3(B) 2 4 1 3(C) 1 4 2 3(D) 2 1 4 3

MCQ 6.2.31 Let [ ] ( )x n X zZ be a z -transform pair. Consider another signal [ ]y n defined as

[ ]y n / ,

,

if is even

if is odd

x n n

n

2

0= 6 @

*

The z -transform of [ ]y n is

(A) ( )X z21 (B) ( )X z2

(C) ( )X z2 (D) ( / )X z 2


MCQ 6.2.32 The z -transform of a discrete sequence [ ]x n is ( )X z , then the z -transform of [ ]x n2 will be

(A) ( )X z2 (B) X z2a k

(C) ( ) ( )X z X z21 + −8 B (D) ( )X z

MCQ 6.2.33 Let ( )X z be z -transform of a discrete time sequence [ ] ( ) [ ]x n u n21 2= − . Consider

another signal [ ]y n and its z -transform ( )Y z given as ( )Y z ( )X z3= . What is the value of [ ]y n at n 4= ?(A) 0 (B) 2 12−

(C) 212 (D) 1

MCQ 6.2.34 Consider a signal [ ]x n and its z transform ( )X z given as

( )X z z z

z8 2 1

42=− −

The z -transform of the sequence [ ]y n [0] [1] [2] ..... [ ]x x x x n= + + + + will be

(A) ( )( )z z z

z1 8 2 1

42

2

− − − (B)

( )z zz z

8 2 14 12 − −

−

(C) ( )( )z z z

z1 8 2 1

42

2

+ − − (D)

( )z zz z

8 2 14 12 − −

+

MCQ 6.2.35 Let [ ] {1, 2, 0, 1, 1}h n = − and [ ] {1, 3, 1, 2}x n = − − be two discrete time sequences. What is the value of convolution [ ] [ ] [ ]y n h n x n= * at n 4= ?(A) 5− (B) 5

(C) 6− (D) 1−

MCQ 6.2.36 What is the convolution of two DT sequence [ ] { ,2,0,3}x n 1= −-

and [ ] {2,0, }h n 3=-

(A) { 2, 4, , 6, 9}3− −-

(B) { 2, 4, , 12, 0, 9}3− −-

(C) {9, 6, , 4, 2}3 − −-

(D) { , 6, 7, 4, 6}3−-

MCQ 6.2.37 If [ ] ( )x n X zZ be a z -transform pair, then which of the following is true?

(A) [ ] ( )x n X zZ −) ) (B) [ ] ( )x n X zZ −) )

(C) [ ] ( )x n X zZ) ) ) (D) [ ] ( )x n X zZ −) ) )

MCQ 6.2.38 A discrete time sequence is defined as follows

[ ]x n 1,

0,

is even

otherwise

n= )

What is the final value of [ ]x n ?(A) 1 (B) 1/2

(C) 0 (D) does not exist


MCQ 6.2.39 Let ( )X z be the z -transform of a DT signal [ ]x n given as

( )X z ( )( . )

.z z

z1 0 50 5 2

= − −

The initial and final values of [ ]x n are respectively(A) 1, 0.5 (B) 0, 1

(C) 0.5, 1 (D) 1, 0

MCQ 6.2.40 A discrete-time system with input [ ]x n and output [ ]y n is governed by following difference equation

[ ] [ ]y n y n21 1− − [ ]x n= , with initial condition [ ]y 1 3− =

The impulse response of the system

(A) 1 , 0n n25

2 $−a k (B) , 0n25

21 n

$b l

(C) , 0n25

21 n 1

$−

b l (D) , 0n25

21 n 1

$+

b l

MCQ 6.2.41 Consider a causal system with impulse response [ ] ( ) [ ]h n u n2 n= . If [ ]x n is the input and [ ]y n is the output to this system, then which of the following difference equation describes the system ?(A) [ ] [ ] [ ]y n y n x n2 1+ + = (B) [ ] [ ] [ ]y n y n x n2 1− − =

(C) [ ] [ ] [ ]y n y n x n2 1+ − = (D) [ ] [ ] [ ]y n y n x n21 1− − =

MCQ 6.2.42 The impulse response of a system is given as [ ]h n [ ] ( ) [ ]n u nn21δ= − − . For an input

[ ]x n and output [ ]y n , the difference equation that describes the system is(A) [ ] [ ] [ ]y n y n x n2 1 2+ − = (B) [ ] . [ ] . [ ]y n y n x n0 5 1 0 5 1+ − = −

(C) [ ] [ ] [ ]y n ny n x n2 1+ − = (D) [ ] 0.5 [ ] . [ ]y n y n x n1 0 5 1− − = −

MCQ 6.2.43 The input-output relationship of a system is given as [ ] 0.4 [ 1] [ ]y n y n x n− − = where, [ ]x n and [ ]y n are the input and output respectively. The zero state response of the system for an input [ ] ( . ) [ ]x n u n0 4 n= is(A) ( . ) [ ]n u n0 4 n (B) ( . ) [ ]n u n0 4 n2

(C) ( )( . ) [ ]n u n1 0 4 n+ (D) ( . ) [ ]n u n1 0 4 n

MCQ 6.2.44 A discrete time system has the following input-output relationship [ ] [ 1] [ ]y n y n x n2

1− − = . If an input [ ] [ ]x n u n= is applied to the system, then its zero state response will be

(A) ( ) [ ]u n21 2 n−: D (B) [ ]u n2 2

1 n− b l; E

(C) [ ]u n21

21 n

− b l; E (D) [ ( ) ] [ ]u n2 2 n−


MCQ 6.2.45 Consider the transfer function of a system

( )H z ( )

z zz z

4 42 12=+ +

−

For an input [ ] [ ] [ ]x n n n2 1δ δ= + + , the system output is(A) [ ] ( ) [ ]n u n2 1 6 2 nδ + + (B) [ ] ( ) [ ]n u n2 6 2 nδ − −

(C) [ ] ( ) [ ]n u n2 1 6 2 nδ + − − (D) [ ] [ ]n u n2 1 6 21 n

δ + + b l

MCQ 6.2.46 The signal [ ] ( . ) [ ]x n u n0 5 n= is when applied to a digital filter, it yields the following output [ ]y n [ ] [ ]n n2 1δ δ= − − . If impulse response of the filter is [ ]h n , then what will be the value of sample [ ]h 1 ?(A) 1 (B) .2 5−

(C) 0 (D) 0.5

MCQ 6.2.47 The transfer function of a discrete time LTI system is given as

( )H z , : 1ROCz

z z1

>2=+

Consider the following statements1. The system is causal and BIBO stable.

2. The system is causal but BIBO unstable.

3. The system is non-causal and BIBO unstable.

4. Impulse response [ ] [ ]sinh n n u n2π= a k

Which of the above statements are true ?(A) 1 and 4 (B) 2 and 4

(C) 1 only (D) 3 and 4

MCQ 6.2.48 Which of the following statement is not true?An LTI system with rational transfer function ( )H z is(A) causal if the ROC is the exterior of a circle outside the outermost pole.

(B) stable if the ROC of ( )H z includes the unit circle z 1= .

(C) causal and stable if all the poles of ( )H z lie inside unit circle.

(D) none of above

MCQ 6.2.49 If [ ]h n denotes the impulse response of a causal system, then which of the following system is not stable?

(A) [ ] [ ]h n n u n31 n

= b l (B) [ ] [ ]h n n31 δ=

(C) [ ] [ ] [ ]h n n u n31 n

δ= − −b l (D) [ ] [( ) ( ) ] [ ]h n u n2 3n n= −


MCQ 6.2.50 A causal system with input [ ]x n and output [ ]y n has the following relationship [ ] [ ] [ ]y n y n y n3 1 2 2+ − + − [ ] [ ]x n x n2 3 1= + −The system is(A) stable (B) unstable

(C) marginally stable (D) none of these

MCQ 6.2.51 A causal LTI system is described by the difference equation [ ]y n [ ] [ ]x n y n 1= + −Consider the following statement1. Impulse response of the system is [ ] [ ]h n u n=

2. The system is BIBO stable

3. For an input [ ] ( . ) [ ]x n u n0 5 n= , system output is [ ] [ ] ( . ) [ ]y n u n u n2 0 5 n= −

Which of the above statements is/are true?(A) 1 and 2 (B) 1 and 3

(C) 2 and 3 (D) 1, 2 and 3

MCQ 6.2.52 Match List I (system transfer function) with List II (property of system) and choose the correct answer using the codes given below

List-I (System transfer function) List-II (Property of system)

P. ( )( . )

, : 1.2ROCH zz

z z1 2

>3

3

=−

1. Non causal but stable

Q.. ( )( . )

, : 1.2ROCH zz

z z1 2

<3

2

=−

2. Neither causal nor stable

R. ( )( . )

, : 0.8ROCH zz

z z0 8

<3

4

=−

3. Causal but not stable

S. ( )( . )

, : 0.8ROCH zz

z z0 8

>3

3

=−

4. Both causal and stable

Codes : P Q R S(A) 4 2 1 3(B) 1 4 2 3(C) 3 1 2 4(D) 3 2 1 4

MCQ 6.2.53 The transfer function of a DT feedback system is

( )H z

.P zz

P1 0 9

=+ −a k

The range of P , for which the system is stable will be(A) 1.9 0.1P< <− − (B) 0P <

(C) 1P > − (D) 0.1P > − or 1.9P < −


MCQ 6.2.54 Consider three stable LTI systems ,S S1 2 and S3 whose transfer functions are

S1 : ( )H z 2z z

z2

21

163

21

=+ −

−

S2 : ( )H z z z z

z 132 3

21 2

34=

− − + ++

− −

S3 : ( )H z 1 11

z z zz z

131 1

21 1

21 2

34 1

=− −+ −

− − −

− −

^ ^h hWhich of the above systems is/are causal?(A) S1 only (B) S1 and S2

(C) S1 and S3 (D) ,S S1 2 and S3

MCQ 6.2.55 The transfer function for the system realization shown in the figure will be

(A) zz

42 3

−+ (B) z

z2

4 3−+

(C) zz2 3

4−

+ (D) zz

23

−+

MCQ 6.2.56 Consider a cascaded system shown in the figure

where, [ ]h n1 [ ] [ ]n n21 1δ δ= + − and [ ]h n2 [ ]u n2

1 n= b l

If an input [ ] ( )cosx n nπ= is applied, then output [ ]y n equals to

(A) ( )cos n31 π (B) ( )cos n6

5 π

(C) ( )cos n613 π (D) ( )cos nπ

MCQ 6.2.57 The block diagram of a discrete time system is shown in the figure below

The range of α for which the system is BIBO stable, will be(A) 1>α (B) 1 1< <α−

(C) 0>α (D) 0<α

***********

EXERCISE 6.3

MCQ 6.3.1 Let [ ] [ 1] [ 2]x n n nδ δ= − + + . The unilateral z -transform is

(A) z 2− (B) z2

(C) 2z 2− − (D) 2z2

MCQ 6.3.2 The unilateral z -transform of signal [ ] [ 4]x n u n= + is

(A) 1 3z z z z2 4+ + + + (B) z11−

(C) 1 z z z z1 2 3 4+ + + +− − − − (D) z1

11− −

MCQ 6.3.3 The z -transform of [ ], 0n k k >δ − is(A) , 0z z >k (B) , 0z z >k−

(C) , 0z zk ! (D) , 0z zk !−

MCQ 6.3.4 The z -transform of [ ], 0n k k >δ + is(A) , 0z zk !− (B) , 0z zk !

(C) ,z k− all z (D) zk , all z

MCQ 6.3.5 The z -transform of [ ]u n is

(A) , 1z

z1

1 >1− − (B) , 1z

z1

1 <1− −

(C) , 1z

z z1

<1− − (D) , 1z

z z1

>1− −

MCQ 6.3.6 The z -transform of ( [ ] [ 5])u n u n41 n

− −b l is

(A) ( . )

( . ), 0.25

z zz

z0 25

0 25 >4

5 5

−−

(B) ( . )

( . ), .

z zz

z0 25

0 250 5>4

5 5

−−

(C) ( . )

( . ), 0.25

z zz

z0 25

0 25 <3

5 5

−−

(D) ( . )

( . )z zz

0 250 25

4

5 5

−−

, all z

MCQ 6.3.7 The z -transform of [ ]u n41 n

−b l is

(A) ,zz z4 1

441>− (B) ,z

z z4 14

41<−


(C) ,z z1 41

41>− (D) ,z z1 4

141<−

MCQ 6.3.8 The z -transform of 3 [ 1]u nn − − is

(A) , 3zz z3 >− (B) , 3z

z z3 <−

(C) , 3z z33 >− (D) , 3z z3

3 <−

MCQ 6.3.9 The z -transform of 32 n

b l is

(A) ( )( )

,z z

z z2 3 3 2

523

32< <− −

− − − (B) ( )( )

,z z

z z2 3 3 2

532

23< <− −

−

(C) ( )( )

,z z

z z2 3 3 2

532

32< <− − (D)

( )( ),

z zz z

2 3 3 25

23

32< <− − − −

MCQ 6.3.10 The z -transform of [ ]cos n u n3π

a k is

(A) ( )(2 1)

,z zz z

2 12 − +−

0 1z< < (B) ( )( )

,z zz z

2 12 1

2 − +−

1z >

(C) ( )( )

,z zz z

2 11 2

2 − +−

0 1z< < (D) ( )( )

,z zz z

2 11 2

2 − +−

1z >

MCQ 6.3.11 The z -transform of {3, 0, 0, 0, 0, , 1, 4}6 −-

(A) 3 6 4 , 0z z z z <5 1 2 3#+ + −− − (B) 3 6 4 , 0z z z z< <5 1 2 3+ + −− −

(C) 3 6 4 , 0z z z z< <5 2 3+ + −− (D) 3 6 4 , 0z z z z <5 2 3#+ + −−

MCQ 6.3.12 The z -transform of [ ] {2, 4, , 7, 0, 1}x n 5=-

(A) 2 4 5 7 ,z z z z z2 3 3!+ + + +

(B) 2 4 5 7 ,z z z z z2 1 3 3!+ + + +− −

(C) 2 4 5 7 , 0z z z z z< <2 1 3 3+ + + +− −

(D) 2 4 5 7 , 0z z z z z< <2 1 3 3+ + + +− −

MCQ 6.3.13 The z -transform of [ ] { , 0, 1, 0, 1, 1}x n 1= − −-

is

(A) 1 2 4 5 , 0z z z z2 4 5 !+ − +− − − (B) 1 , 0z z z z2 4 5 !− + −− − −

(C) 1 2 4 5 , 0z z z z2 4 5 !− + − (D) 1 , 0z z z z2 4 5 !− + −

MCQ 6.3.14 The time signal corresponding to , 2z z

z z z1

321 < <2

23

2

+ −− is

(A) [ ] 2 [ 1]u n u n21n

n 1− − − −+ (B) [ ] 2 [ 1]u n u n21n

n 1− − ++

(C) [ ] 2 [ 1]u n u n21n

n 1+ ++ (D) [ ] 2 [ 1]u n u n21n

n 1− − −− −


MCQ 6.3.15 The time signal corresponding to , 4zz z

z16

3 >2

241

−−

is

(A) ( ) [ ]u n3249 4 32

47 4n n− +: D (B) [ ]u n3249 4 32

47 4n n+: D

(C) ( 4) [ ] 4 [ ]u n u n3249

3247n n− − + (D) 4 [ ] ( 4) [ ]u n u n32

493247n n+ − −

MCQ 6.3.16 The time signal corresponding to , 1z

z z z z1

2 2 2 >2

4 3 2

−− − is

(A) 2 [ 2] [1 ( 1) ] [ 2]n u nnδ − + − − − (B) 2 [ 2] [1 ( 1) ] [ 2]n u nnδ + + − − +

(C) 2 [ 2] [( 1) 1] [ 2]n u nnδ + + − − + (D) 2 [ 2] [( 1) 1] [ 2]n u nnδ − + − − −

MCQ 6.3.17 The time signal corresponding to 1 2 4 , 0z z z >6 8+ +− − is(A) [ ] 2 [ 6] 4 [ 8]n n nδ δ δ+ − + − (B) [ ] 2 [ 6] 4 [ 8]n n nδ δ δ+ + + +

(C) [ ] 2 [ 6] 4 [ 8]n n nδ δ δ− + − + + − + (D) [ ] 2 [ 6] 4 [ 8]n n nδ δ δ− + − − + − −

MCQ 6.3.18 The time signal corresponding to ,k z z1 0>k

k 5

10−

=/ is

(A) [ ]k n k1k 5

10

δ +=/ (B) [ ]k n k1

k 5

10

δ −=/

(C) [ ]k n k1k 5

10

δ − +=/ (D) [ ]k n k1

k 5

10

δ − −=/

MCQ 6.3.19 The time signal corresponding to (1 )z 1 3+ − , 0z > is(A) [ ] 3 [ 1] 3 [ 2] [ 3]n n n nδ δ δ δ− + − − + − − + − −

(B) [ ] 3 [ 1] 3 [ 2] [ 3]n n n nδ δ δ δ− + − + + − + + − +

(C) [ ] 3 [ 1] 3 [ 2] [ 3]n n n nδ δ δ δ+ + + + + +

(D) [ ] 3 [ 1] 3 [ 2] [ 3]n n n nδ δ δ δ+ − + − + −

MCQ 6.3.20 The time signal corresponding to 3 2 , 0z z z z z >6 2 3 4+ + + +− − is(A) [ 6] [ 2] 3 [ ] 2 [ 3] [ 4]n n n n nδ δ δ δ δ+ + + + + − + −

(B) [ 6] [ 2] 3 [ ] 2 [ 3] [ 4]n n n n nδ δ δ δ δ− + − + + + + +

(C) [ 6] [ 2] 3 [ ] 2 [ 3] [ 4]n n n n nδ δ δ δ δ− + + − + + − + − + + − +

(D) [ 6] [ 2] 3 [ ] 2 [ 3] [ 4]n n n n nδ δ δ δ δ− − + − − + − + − − + − −

MCQ 6.3.21 The time signal corresponding to ,z

z1

121>

41 2− −

(A) 2 , 0

0,

n neven and

otherwise

n $−

* (B) [ ]u n41 n2

b l

(C) 2 , , 0

0,

n n

n

odd

even

>n−

* (D) 2 [ ]u nn−


MCQ 6.3.22 The time signal corresponding to ,z

z1

121<

41 2− − is

(A) [ ( )]n k2 2 1( )k

k

2 1

0δ− − − +

3+

=/

(B) [ ( )]n k2 2 1( )k

k

2 1

0δ− − + +

3+

=/

(C) [ ( )]n k2 2 1( )k

k

2 1

0δ− + +

3+

=/

(D) [ 2( 1)]n k2 ( )k

k

2 1

0δ− − +

3+

=/

MCQ 6.3.23 The time signal corresponding to (1 ), 0ln z z >1+ − is

(A) ( )

[ ]k n k1 k 1

δ− −−

(B) ( )

[ ]k n k1 k 1

δ− +−

(C) ( )

[ ]k n k1 k

δ− − (D) ( )

[ ]k n k1 k

δ− +

MCQ 6.3.24 If z -transform is given by ( ) ( ), 0cosX z z z >3= − , the value of [12]x is

(A) 241− (B) 24

1

(C) 61− (D) 6

1

MCQ 6.3.25 [ ]X z of a system is specified by a pole zero pattern as following :

Consider three different solution of [ ]x n

[ ]x n1 [ ]u n2 31n

n= − b l; E

[ ]x n2 2 [ 1] [ ]u n u n31nn=− − −

[ ]x n3 2 [ 1] [ 1]u n u n31nn=− − + − −

Correct solution is(A) [ ]x n1 (B) [ ]x n2

(C) [ ]x n3 (D) All three


MCQ 6.3.26 Consider three different signal

[ ]x n1 [ ]u n2 21n

n= − b l; E

[ ]x n2 2 [ 1] [ 1]u n u n21nn=− − − + − −

[ ]x n3 2 [ 1] [ ]u n u n21nn=− − − −

Following figure shows the three different region. Choose the correct for the ROC of signal

R1 R2 R3

(A) [ ]x n1 [ ]x n2 [ ]x n3

(B) [ ]x n2 [ ]x n3 [ ]x n1

(C) [ ]x n1 [ ]x n3 [ ]x n2

(D) [ ]x n3 [ ]x n2 [ ]x n1

MCQ 6.3.27 Given the z -transform

( )X z z z

z

1 1

1

21 1

31 1

67 1

=− +

+− −

−

^ ^h h

For three different ROC consider there different solution of signal [ ]x n :

(a) , [ ] [ ]z x n u n21

21

31> n

n

1= − −− b l; E

(b) , [ ] [ 1]z x n u n31

21

31< n

n

1= − + − − +− b l; E

(c) , [ ] [ 1] [ ]z x n u n u n31

21

21

31< < n

n

1=− − − − −− b l

Correct solution are

(A) (a) and (b) (B) (a) and (c)

(C) (b) and (c) (D) (a), (b), (c)

MCQ 6.3.28 The ( )X z has poles at z 21= and 1z =− . If [1] 1, [ 1] 1x x= − = , and the ROC

includes the point z 43= . The time signal [ ]x n is

(A) [ ] ( 1) [ 1]u n u n21n

n1 − − − −− (B) [ ] ( 1) [ 1]u n u n

21n

n− − − −


(C) [ ] [ 1]u n u n21n 1 + − +− (D) [ ] [ 1]u n u n

21n + − +

MCQ 6.3.29 If [ ]x n is right-sided, ( )X z has a signal pole and [0] 2,x = [2] ,x 21= then [ ]x n is

(A) [ ]u n2n 1−

− (B) [ ]u n

2n 1−

(C) [ ]u n2n 1−

+ (D) [ ]

au n2n 1−

+

MCQ 6.3.30 The z -transform of [ ] [ 1]u n u n21

41n n

+ − −b bl l is

(A) 1

,z z

z1

1 141

21< <

21 1

41 1−

−−− −

(B) 1 1

,z z

z1 141

21< <

21 1

41 1−

+−− −

(C) 1 1

,z z

z1 121>

21 1

41 1−

−−− −

(D) None of the above

Statement for Q. 31-36 :

Given the z -transform pair [ ] , 4x nz

z z16

<2

2Z

−MCQ 6.3.31 The z -transform of the signal [ 2]x n − is

(A) z

z162

4

− (B)

( )( )

zz2 16

22

2

+ −+

(C) z 16

12 −

(D) ( )

( )z

z2 16

22

2

− −−

MCQ 6.3.32 The z -transform of the signal [ ] [ ]y n x n21n= is

(A) ( )

( )x

z2 16

22

2

+ −+

(B) z

z42

2

−

(C) ( )

( )z

z2 16

22

2

− −−

(D) z

z642

2

−

MCQ 6.3.33 The z -transform of the signal [ ] [ ]x n x n− * is

(A) z z

z16 257 162 4

2

− − (B)

( )zz16

162 2

2

−−

(C) z z

z257 16 162 4

2

− − (D)

( )zz16

162 2

2

−


MCQ 6.3.34 The z -transform of the signal [ ]nx n is

(A) ( )z

z16

322 2

2

− (B)

( )zz16

322 2

2

−−

(C) ( )z

z16

322 2−

(D) ( )z

z16

322 2−−

MCQ 6.3.35 The z -transform of the signal [ 1] [ 1]x n x n+ + − is

(A) ( )

( )( )

( )z

zz

z1 16

11 16

12

2

2

2

+ −+ +

− −−

(B) ( 1)z

z z162

2

−+

(C) ( )z

z z16

12

2

−− +

(D) None of the above

MCQ 6.3.36 The z -transform of the signal [ ] [ 3]x n x n −* is

(A) ( )z

z162 2

3

−

− (B)

( )zz162 2

7

−

(C) ( )z

z162 2

5

− (D)

( )zz162 2−

Statement for Q. 37-41 :

Given the z -transform pair 3 [ ] ( )n u n X zn z2

MCQ 6.3.37 The time signal corresponding to (2 )X z is

(A) 3 [2 ]n u nn2 (B) [ ]n u n23 n

2−b l

(C) [ ]n u n23 n

2b l (D) 6 [ ]n u nn 2

MCQ 6.3.38 The time signal corresponding to ( )X z 1− is(A) 3 [ ]n u nn2 −− (B) 3 [ ]n u nn2 −−

(C) 3 [ ]n

u n1 n2

1 (D) 3 [ ]

nu n1 n

2

1−

MCQ 6.3.39 The time signal corresponding to ( )dzd X z is

(A) ( 1) 3 [ 1]n u nn3 1− −− (B) 3 [ 1]n u nn3 −

(C) (1 ) 3 [ 1]n u nn3 1− −− (D) ( 1) 3 [ ]n u nn3 1− −

MCQ 6.3.40 The time signal corresponding to 2 ( )z z X z2 2− −

b l is

(A) ( [ 2] [ 2])x n x n21 + − − (B) [ 2] [ 2]x n x n+ − −

(C) [ 2] [ 2])x n x n21 − − + (D) [ 2] [ 2]x n x n− − +

MCQ 6.3.41 The time signal corresponding to { ( )}X z 2 is


(A) [ [ ]]x n 2 (B) [ ] [ ]x n x n*(C) ( ) [ ]x n x n−* (D) [ ] [ ]x n x n− −*

MCQ 6.3.42 A causal system has

Input, [ ]x n [ ] [ 1] [ 2]n n n41

81δ δ δ= + − − − and

Output, [ ]y n [ ] [ 1]n n43δ δ= − −

The impulse response of this system is

(A) [ ]u n31 5 2

1 2 41n n− −b bl l; E (B) [ ]u n3

1 5 21 2 4

1n n+ −

b bl l; E

(C) [ ]u n31 5 2

1 2 41n n

− −b bl l; E (D) [ ]u n3

1 5 21 2 4

1n n+b bl l; E

MCQ 6.3.43 A causal system has input [ ] ( 3) [ ]x n u nn= − and output [ ]y n ( ) ( ) [ ]u n4 2 n n21= −6 @ .

The impulse response of this system is

(A) [ ]u n7 21 10 2

1n n−b bl l; E (B) ( ) [ ]u n7 2 10 2

1nn

− b l; E

(C) ( ) [ ]u n10 21 7 2 n

2−b l; E (D) ( ) [ ]u n10 2 7 2

1nn

− b l; E

MCQ 6.3.44 A system has impulse response [ ] ( ) [ ]h n u nn21= . The output [ ]y n to the input [ ]x n

is given by [ ] 2 [ 4]y n nδ= − . The input [ ]x n is(A) 2 [ 4] [ 5]n nδ δ− − − − − (B) 2 [ 4] [ 5]n nδ δ+ − +

(C) 2 [ 4] [ 5]n nδ δ− + − − + (D) 2 [ 4] [ 5]n nδ δ− − −

MCQ 6.3.45 A system is described by the difference equation [ ]y n [ ] [ 2] [ 4] [ 6]x n x n x n x n= − − + − − −The impulse response of system is(A) [ ] 2 [ 2] 4 [ 4] 6 [ 6]n n n nδ δ δ δ− + + + − +

(B) [ ] 2 [ 2] 4 [ 4] 6 [ 6]n n n nδ δ δ δ+ − − − + −

(C) [ ] [ 2] [ 4] [ 6]n n n nδ δ δ δ− − + − − −

(D) [ ] [ 2] [ 4] [ 6]n n n nδ δ δ δ− + + + − +

MCQ 6.3.46 The impulse response of a system is given by [ ]h n [ 1]u n43n= − . The difference

equation representation for this system is(A) 4 [ ] [ 1] 3 [ 1]y n y n x n− − = − (B) 4 [ ] [ 1] 3 [ 1]y n y n x n− + = +

(C) 4 [ ] [ 1] 3 [ 1]y n y n x n+ − =− − (D) 4 [ ] [ 1] 3 [ 1]y n y n x n+ + = +

MCQ 6.3.47 The impulse response of a system is given by [ ]h n [ ] [ 5]n nδ δ= − − . The difference equation representation for this system is(A) [ ] [ ] [ 5]y n x n x n= − − (B) [ ] [ ] [ 5]y n x n x n= − +

(C) [ ] [ ] 5 [ 5]y n x n x n= + − (D) [ ] [ ] 5 [ 5]y n x n x n= − +


MCQ 6.3.48 Consider the following three systems [ ]y n1 0.2 [ 1] [ ] 0.3 [ 1] 0.02 [ 2]y n x n x n x n= − + − − + − [ ]y n2 [ ] 0.1 [ 1]x n x n= − − [ ]y n3 0.5 [ 1] 0.4 [ ] 0.3 [ 1]y n x n x n= − + − −The equivalent system are(A) [ ]y n1 and [ ]y n2 (B) [ ]y n2 and [ ]y n3

(C) [ ]y n3 and [ ]y n1 (D) all

MCQ 6.3.49 The z -transform function of a stable system is ( )( )( )

H zz z

z1 2 1

21

21 1

23 1

=− +

−− −

−

. The impulse response [ ]h n is

(A) 2 [ 1] [ ]u n u n21n

n− + − b l (B) 2 [ 1] [ ]u n u n2

1nn

− − − + −b l

(C) 2 [ 1] [ ]u n u n21n

n− − − − −

b l (D) 2 [ ] [ ]u n u n21n

n− b l

MCQ 6.3.50 The z -transform of a anti causal system is ( )X z z z

z3 7 12

12 212=

− +− . The value of [0]x is

(A) 47− (B) 0

(C) 4 (D) Does not exist

MCQ 6.3.51 The transfer function of a causal system is ( )H z z z

z6

52

2

=− −

. The impulse response is

(A) (3 ( 1) 2 ) [ ]u nn n n 1+ − + (B) (3 2( 2) ) [ ]u nn n1 + −+

(C) (3 ( 1) 2 ) [ ]u nn n n1 1+ −− + (D) (3 ( 2) ) [ ]u nn n1 1− −− +

MCQ 6.3.52 The transfer function of a system is given by ( )H z ( )

z zz z3 22

41=

− −−

. The system is

(A) causal and stable (B) causal, stable and minimum phase

(C) minimum phase (D) none of the above

MCQ 6.3.53 The z -transform of a signal [ ]x n is ( )X z z z13

310 1 2=

− +− − . If ( )X z converges on the unit circle, [ ]x n is

(A) ( )

[ ] [ 1]u n u n3 8

18

3n

n

1

3

− − − −−

+ (B)

( )[ ]

( )[ ]u n u n

3 81

83

n

n

1

3

− −−

+

(C) ( )

[ ]( )

[ ]u n u n3 8

18

3n

n

1

3

− −−

+ (D)

( )[ ]

( )[ ]u n u n

3 81

83

n

n

1

3

− − −−

+

MCQ 6.3.54 The transfer function of a system is ( )H z ,z

z z1

441>

41 1 2

1

=− −

−

^ h. The [ ]h n is

(A) stable (B) causal

(C) stable and causal (D) none of the above


MCQ 6.3.55 The transfer function of a system is given as

( )H z z z

z2

21

31

21

=− −

+

^ ^

^

h h

h

Consider the two statementsStatement (i) : System is causal and stable.Statement (ii) : Inverse system is causal and stable.The correct option is(A) (i) is true (B) (ii) is true

(C) Both (i) and (ii) are true (D) Both are false

MCQ 6.3.56 The system [ ]y n [ 1] 0.12 [ 2] [ 1] [ 2]cy n y n x n x n= − − − + − + − is stable if(A) 1.12c < (B) 1.12c >

(C) 1.12c < (D) 1.12c >

MCQ 6.3.57 The impulse response of the system shown below is

(A) 2 (1 ( 1) ) [ ] [ ]u n n21n2n

2 δ+ − +−^ h (B) (1 ( 1) ) [ ] [ ]u n n22

21n

n δ+ − +

(C) 2 (1 ( 1) ) [ ] [ ]u n n21n2n

2 δ+ − −−^ h (D) [1 ( 1) ] [ ] [ ]u n n22

21n

n δ+ − −

MCQ 6.3.58 The system diagram for the transfer function ( )H z z z

z12=

+ +. is shown below.

The system diagram is a(A) Correct solution (B) Not correct solution

(C) Correct and unique solution (D) Correct but not unique solution

***********

EXERCISE 6.4

MCQ 6.4.1 What is the z -transform of the signal [ ] [ ]x n u nnα= ?

(A) ( )X z z 11= − (B) ( )X z z1

1= −

(C) ( )X z zz

α= − (D) ( )X z z1

α= −

MCQ 6.4.2 The z -transform of the time function [ ]n kk 0

δ −3

=/ is

(A) zz 1−

(B) zz

1−

(C) ( )z

z1 2−

(D) ( )

zz 1 2−

MCQ 6.4.3 The z -transform ( )F z of the function ( )f nT anT= is

(A) z a

zT−

(B) z a

zT+

(C) z a

zT− − (D)

z az

T+ −

MCQ 6.4.4 The discrete-time signal [ ] ( )x n X z z23

nnn

02Z n

= 3+=/ , where denotes a

transform-pair relationship, is orthogonal to the signal

(A) [ ] ( )y n Y z z32 n

nn

1 1 0) = 3

=-

` j/ (B) [ ] ( ) ( )y n Y z n z5 ( )nn

n2 2 0

2 1) = −3

=- +/

(C) [ ] ( )y n Y z z2 nn

n3 3) =

3

3 -=-

-/ (D) [ ] ( )y n Y z z z2 3 14 44 2

) = + +- -

MCQ 6.4.5 Which one of the following is the region of convergence (ROC) for the sequence [ ]x n [ ] [ 1]; 1b u n b u n b <n n= + − −− ?(A) Region z 1<

(B) Annular strip in the region b z b1> >

(C) Region z 1>

(D) Annular strip in the region b z b1< <

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MCQ 6.4.6 Assertion (A) : The signals [ ]a u nn and [ 1]a u nn− − − have the same z -transform,

/( )z z a− .

Reason (R) : The Reason of Convergence (ROC) for [ ]a u nn is z a> , whereas

the ROC for [ 1]a u nn − − is z a< .

(A) Both A and R are true and R is the correct explanation of A

(B) Both A and R are true but R is NOT the correct explanation of A

(C) A is true but R is false

(D) A is false but R is true

MCQ 6.4.7 Which one of the following is the correct statement ?

The region of convergence of z -transform of [ ]x n consists of the values of z for

which [ ]x n r n− is

(A) absolutely integrable (B) absolutely summable

(C) unity (D) 1<

MCQ 6.4.8 The ROC of z -transform of the sequence [ ]x n [ ] [ 1]u n u n31

21n n

= − − −b bl l is

(A) z31

21< < (B) z

21>

(C) z31< (D) z2 3< <

MCQ 6.4.9 The region of convergence of z − transform of the sequence

[ ] [ 1]u n u n65

56n n

− − −b bl l must be

(A) z65< (B) z

65>

(C) z65

56< < (D) z

56 < < 3

MCQ 6.4.10 The region of convergence of the z -transform of the discrete-time signal [ ] 2 [ ]x n u nn=

will be

(A) 2z > (B) 2z <

(C) z 21> (D) z 2

1<

MCQ 6.4.11 The region of convergence of the z − transform of a unit step function is

(A) z 1> (B) z 1<

(C) (Real part of z ) 0> (D) (Real part of z ) 0<

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MCQ 6.4.12 Match List I (Discrete Time signal) with List II (Transform) and select the correct answer using the codes given below the lists :

List I List II

A. Unit step function 1. 1

B. Unit impulse function 2.coscos

z z Tz T2 12 ω

ω− +

−

C. , 0, , 2sin t t T Tω = 3.z

z1−

D. , 0, , 2 , .....cos t t T Tω = 4.cos

sinz z T

z T2 12 ω

ω− +

Codes : A B C D(A) 2 4 1 3(B) 3 1 4 2(C) 2 1 4 3(D) 3 4 1 2

MCQ 6.4.13 What is the inverse z -transform of ( )X z

(A) ( )j X z z dz21 n 1

π−# (B) ( )j X z z dz2 n 1π +#

(C) ( )j X z z dz21 n1

π−# (D) ( )j X z z dz2 ( )n 1π − +#

MCQ 6.4.14 Which one of the following represents the impulse response of a system defined by ( )H z z m= − ?

(A) [ ]u n m− (B) [ ]n mδ −

(C) [ ]mδ (D) [ ]m nδ −

MCQ 6.4.15 If ( )X z is z1

11− − with z 1> , then what is the corresponding [ ]x n ?

(A) e n− (B) en

(C) [ ]u n (D) ( )nδ

MCQ 6.4.16 The z −transform ( )X z of a sequence [ ]x n is given by [ ]X z .z1 2

0 51= − − . It is given that

the region of convergence of ( )X z includes the unit circle. The value of [ ]x 0 is(A) .0 5− (B) 0

(C) 0.25 (D) 05

MCQ 6.4.17 If ( )u t is the unit step and ( )tδ is the unit impulse function, the inverse z -transform of ( )F z 1z

1= + for k 0> is(A) ( ) ( )k1 k δ− (B) ( ) ( )k 1 kδ − −

(C) ( ) ( )u k1 k− (D) ( ) ( )u k 1 k− −

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MCQ 6.4.18 For a z -transform ( )X z z z

z2

21

31

65

=− −

−

^ ^

^

h h

h

Match List I (The sequences) with List II (The region of convergence ) and select the correct answer using the codes given below the lists :

List I List II

A. [(1/2) (1/3) ] [ ]u nn n+ 1. ( / ) ( / )z1 3 1 2< <

B. (1/2) [ ] (1/3) [ 1]u n u nn n− − − 2. ( / )z 1 3<

C. (1/2) [ 1] (1/3) [ ]u n u nn n− − − + 3. / & /z z1 3 1 2< >

D. [(1/2) (1/3) ] [ 1]u nn n− + − − 4. /z 1 2>

Codes : A B C D(A) 4 2 1 3(B) 1 3 4 2(C) 4 3 1 2(D) 1 2 4 3

MCQ 6.4.19 Which one of the following is the inverse z -transform of

( )X z ( )( )

, 2z z

z z2 3

<= − − ?

(A) [2 3 ] [ 1]u nn n− − − (B) [3 2 ] [ 1]u nn n− − −

(C) [2 3 ] [ 1]u nn n− + (D) [2 3 ] [ ]u nn n−

MCQ 6.4.20 Given ( )( )

X zz a

z2=

− with z a> , the residue of ( )X z zn 1− at z a= for n 0$ will

be(A) an 1− (B) an

(C) nan (D) nan 1-

MCQ 6.4.21 Given ( ) ,X zaz bz1 1121

131

=−

+−− − a and 1b < with the ROC specified as

a z b< < , then [ ]x 0 of the corresponding sequence is given by

(A) 31 (B) 6

5

(C) 21 (D) 6

1

MCQ 6.4.22 If ( )X zz zz z

1

3

=++

−

− then [ ]x n series has

(A) alternate 0’s (B) alternate 1’s

(C) alternate 2’s (D) alternate 1− ’s

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MCQ 6.4.23 Consider the z -transform ( ) 5 4 3; 0x z z z z< <2 1 3= + +− . The inverse z -

transform [ ]x n is

(A) 5 [ 2] 3 [ ] 4 [ 1]n n nδ δ δ+ + + −

(B) 5 [ 2] 3 [ ] 4 [ 1]n n nδ δ δ− + + +

(C) 5 [ 2] 3 [ ] 4 [ 1]u n u n u n+ + + −

(D) 5 [ 2] 3 [ ] 4 [ 1]u n u n u n− + + +

MCQ 6.4.24 The sequence [ ]x n whose z -transform is ( )X z e /z1= is

(A) ! [ ]n u n1 (B) ! [ ]n u n1− −

(C) ( 1) ! [ ]n u n1n− (D) ( )!

[ 1]n

u n1

1− + − −

MCQ 6.4.25 If the region of convergence of [ ] [ ]x n x n1 2+ is z31

32< < then the region of

convergence of [ ] [ ]x n x n1 2− includes

(A) z31 3< < (B) z

32 3< <

(C) z23 3< < (D) z

31

32< <

MCQ 6.4.26 Match List I with List II and select the correct answer using the codes given below

the lists :

List I List II

A. [ ]u nnα 1.( )z

z1 1 2

1

αα

− −

−, ROC : z > α

B. [ 1]u nnα− − − 2.( )z1

11α− − , ROC : z > α

C. [ 1]n u nnα− − − 3.( )z1

11α− − , ROC : |z < α

D. [ ]n u nnα 4.( )z

z1 1 2

1

αα

− −

−, ROC : |z < α

Codes :

A B C D

(A) 2 4 3 1

(B) 1 3 4 2

(C) 1 4 3 2

(D) 2 3 4 1

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MCQ 6.4.27 Match List-I ( [ ])x n with List-II ( ( ))X z and select the correct answer using the codes given below the Lists:

List-I List-II

A. [ ]a u nn 1.( )z a

az2−

B. [ ]a u n 2n 2 −− 2.ze a

zej

j

−−

−

C. e ajn n 3.z a

z−

D. [ ]na u nn 4.z az 1

−−

Codes : A B C D(A) 3 2 4 1(B) 2 3 4 1(C) 3 4 2 1(D) 1 4 2 3

MCQ 6.4.28 Algebraic expression for z -transform of [ ]x n is [ ]X z . What is the algebraic expression for z -transform of { [ ]}e x nj n0ω ?(A) ( )X z z0− (B) ( )X e zj 0ω−

(C) ( )X e zj 0ω (D) ( )X z ej z0ω

MCQ 6.4.29 Given that ( )F z and ( )G z are the one-sided z -transforms of discrete time functions ( )f nT and ( )g nT , the z -transform of ( ) ( )f kT g nT kT−/ is given by

(A) ( ) ( )f nT g nT z n−/ (B) ( ) ( )f nT z g nT zn n− −//

(C) ( ) ( )f kT g nT kT z n− −/ (D) ( ) ( )f nT kT g nT z n− −/

MCQ 6.4.30 The output [ ]y n of a discrete time LTI system is related to the input [ ]x n as given below :

[ ]y n [ ]x kk 0

=3

=/

Which one of the following correctly relates the z -transform of the input and output, denoted by ( )X z and ( )Y z , respectively ?(A) ( ) ( ) ( )Y z z X z1 1= − − (B) ( ) ( )Y z z X z1= −

(C) ( )( )

Y zz

X z1 1=

− − (D) ( )( )

Y z dzdX z=

MCQ 6.4.31 Convolution of two sequence [ ]x n1 and [ ]x n2 is represented as(A) ( ) ( )X z X z1 2* (B) ( ) ( )X z X z1 2

(C) ( ) ( )X z X z1 2+ (D) ( )/ ( )X z X z1 2

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MCQ 6.4.32 The z -transform of a signal is given by ( )C z ( )

( )z

z z4 1

1 11 2

1 4

=−

−−

− −

. Its final value is

(A) 1/4 (B) zero

(C) 1.0 (D) infinity

MCQ 6.4.33 Consider a system described by the following difference equation:( ) ( ) ( ) ( )y n y n y n y n3 6 2 11 1 6+ + + + + + ( ) ( ) ( )r n r n r n2 9 1 20= + + + +

Where y is the output and r is the input. The transfer function of the system will be

(A) 3z z z

z z2 6

2 203 2

2

+ + ++ + (B)

z z zz z

6 6 119 20

3 2

2

+ + ++ +

(C) z z

z z z9 20

6 6 112

3 2

+ ++ + + (D) none of the above

MCQ 6.4.34 If the function ( ) ( . )H z z z1 1 511 2= + −− − and ( ) .H z z z1 5 12

2= + − , then(A) the poles and zeros of the functions will be the same

(B) the poles of the functions will be identical but not zeros

(C) the zeros of the functions will be identical but not the poles

(D) neither the poles nor the zeros of the two functions will be identical

MCQ 6.4.35 The state model

[ 1]x k + [ ] [ ]x k u k0 1 0

1β α= − − +> >H H

[ ]y k [ ][ ]

x kx k0 1

1

2= 8 >B H

is represented in the difference equation as(A) [ 2] [ 1] [ ] [ ]c k c k c k u kα β+ + + + =

(B) [ 1] [ ] [ 1] [ 1]c k c k c k u kα β+ + + − = −

(C) [ 2] [ 1] [ ] [ ]c k c k c k u kα β− + − + =

(D) [ 1] [ ] [ 1] [ 1]c k c k c k u kα β− + + + = +

MCQ 6.4.36 The impulse response of a discrete system with a simple pole shown in the figure below. The pole of the system must be located on the

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(A) real axis at z 1=−

(B) real axis between z 0= and z 1=

(C) imaginary axis at z j=

(D) imaginary axis between z 0= and z j=

MCQ 6.4.37 Which one of the following digital filters does have a linear phase response ?(A) [ ] [ 1] [ ] [ 1]y n y n x n x n+ − = − −

(B) [ ] 1/6(3 [ ] 2 [ 1] [ 2])y n x n x n x n= + − + −

(C) [ ] 1/6( [ ] 2 [ 1] 3 [ 2])y n x n x n x n= + − + −

(D) [ ] 1/4( [ ] 2 [ 1] [ 2])y n x n x n x n= + − + −

MCQ 6.4.38 The poles of a digital filter with linear phase response can lie(A) only at z 0=

(B) only on the unit circle

(C) only inside the unit circle but not at z 0=

(D) on the left side of ( ) 0zReal = line

MCQ 6.4.39 The impulse response of a discrete system with a simple pole is shown in the given figure

The pole must be located(A) on the real axis at z 1= (B) on the real axis at z 1=−

(C) at the origin of the z-plane (D) at z 3=

MCQ 6.4.40 The response of a linear, time-invariant discrete-time system to a unit step input [ ]u n is the unit impulse [ ]nδ . The system response to a ramp input [ ]nu n would be

(A) [ ]u n (B) [ 1]u n −

(C) [ ]n nδ (D) [ ]k n kk 0

δ −3

=/

MCQ 6.4.41 A system can be represented in the form of state equations as [ 1]s n + [ ] [ ]As n Bx n= + [ ]y n [ ] [ ]Cs n Dx n= +

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where , ,A B C and D are matrices, [ ]s n is the state vector. [ ]x n is the input and [ ]y n is the output. The transfer function of the system ( ) ( )/ ( )H z Y z X z= is given by(A) ( )A zI B C D1− +− (B) ( )B zI C D A1− +−

(C) ( )C zI A B D1− +− (D) ( )D zI A C B1− +−

MCQ 6.4.42 What is the number of roots of the polynomial ( ) 4 2F z z z z83 2= − − + , lying outside the unit circle ?(A) 0 (B) 1

(C) 2 (D) 3

MCQ 6.4.43 [ ] [ ]y n x kk

n=

3=−/

Which one of the following systems is inverse of the system given above ?(A) [ ] [ ] [ ]x n y n y n 1= − − (B) [ ] [ ]x n y n=

(C) [ ] [ ]x n y n 4= + (D) [ ] [ ]x n ny n=

MCQ 6.4.44 For the system shown, [ ] [ ]x n k nδ= , and [ ]y n is related to [ ]x n as [ ] [ 1]y n y n21− −

[ ]x n=

What is [ ]y n equal to ?(A) k (B) ( / ) k1 2 n

(C) nk (D) 2n

MCQ 6.4.45 Unit step response of the system described by the equation [ ] [ 1] [ ]y n y n x n+ − = is

(A) ( )( )z z

z1 1

2

+ − (B) ( )( )z z

z1 1+ −

(C) zz

11

−+ (D)

( )( )z

z z11

+−

MCQ 6.4.46 Unit step response of the system described by the equation [ ] [ 1] [ ]y n y n x n+ − = is

(A) ( 1)( 1)z z

z2

+ − (B) ( )( )z z

z1 1+ −

(C) ( )( )zz

11

−+

(D) ( )( )z

z z11

+−

MCQ 6.4.47 System transformation function ( )H z for a discrete time LTI system expressed in state variable form with zero initial conditions is(A) ( )c zI A b d1− +− (B) ( )c zI A 1− −

(C) ( )zI A z1− − (D) ( )zI A 1− −

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MCQ 6.4.48 A system with transfer function ( )H z has impulse response (.)h defined as ( ) , ( )h h2 1 3 1= =− and ( )h k 0= otherwise. Consider the following statements.

S1 : ( )H z is a low-pass filter. S2 : ( )H z is an FIR filter.Which of the following is correct?(A) Only S2 is true

(B) Both S1 and S2 are false

(C) Both S1 and S2 are true, and S2 is a reason for S1

(D) Both S1 and S2 are true, but S2 is not a reason for S1

MCQ 6.4.49 The z -transform of a system is ( )H z .zz0 2= − . If the ROC is .z 0 2< , then the

impulse response of the system is(A) ( . ) [ ]u n0 2 n (B) ( . ) [ ]u n0 2 1n − −

(C) ( . ) [ ]u n0 2 n− (D) ( . ) [ ]u n0 2 1n− − −

MCQ 6.4.50 A sequence ( )x n with the z −transform ( ) 2 2 3X z z z z z4 2 4= + − + − − is applied as an input to a linear, time-invariant system with the impulse response [ ] 2 [ 3]h n nδ= − where

[ ]nδ ,

,

n1 0

0 otherwise=

=)

The output at n 4= is(A) 6− (B) zero

(C) 2 (D) 4−

MCQ 6.4.51 The z-transform of a signal [ ]x n is given by z z z z4 3 2 6 23 1 2 3+ + − +- -

It is applied to a system, with a transfer function ( )H z z3 21= −-

Let the output be [ ]y n . Which of the following is true ?(A) [ ]y n is non causal with finite support

(B) [ ]y n is causal with infinite support

(C) [ ] ;y n n0 3>=

(D) [ ( )] [ ( )]

[ ( )] [ ( )] ;

Re Re

Im Im

Y z Y z

Y z Y z <z e z e

z e z e

j j

j j #π θ π=−= −

= =

= =

i i

i i

-

-

MCQ 6.4.52 ( )H z is a transfer function of a real system. When a signal [ ] (1 )x n j n= + is the input to such a system, the output is zero. Further, the Region of convergence (ROC) of z1 2

1 1− -^ h H(z) is the entire Z-plane (except z 0= ). It can then be

inferred that ( )H z can have a minimum of(A) one pole and one zero (B) one pole and two zeros

(C) two poles and one zero (D) two poles and two zeros

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MCQ 6.4.53 ( ) 1 3 , ( ) 1 2X z z Y z z1 2= − = +− − are z -transforms of two signals [ ], [ ]x n y n respectively. A linear time invariant system has the impulse response [ ]h n defined by these two signals as [ ] [ ] * [ ]h n x n y n1= − where * denotes discrete time convolution. Then the output of the system for the input [ ]n 1δ −(A) has z -transform ( ) ( )z X z Y z1−

(B) equals [ ] [ ] [ ] [ ]n n n n2 3 3 2 4 6 5δ δ δ δ− − − + − − −

(C) has z -transform 1 3 2 6z z z1 2 3− + −− − −

(D) does not satisfy any of the above three

MCQ 6.4.54 A discrete-time signal, [ ]x n , suffered a distortion modeled by an LTI system with ( ) (1 )H z az 1= − − , a is real and 1a > . The impulse response of a stable system

that exactly compensates the magnitude of the distortion is

(A) [ ]a u n1 n

b l (B) [ 1]a u n1 n− − −b l

(C) [ ]a u nn (D) [ 1]a u nn − −

MCQ 6.4.55 Assertion (A) : A linear time-invariant discrete-time system having the system function

( )H z z

z21=

+ is a stable system.

Reason (R) : The pole of ( )H z is in the left-half plane for a stable system.(A) Both A and R are true and R is the correct explanation of A

(B) Both A and R are true but R is NOT a correct explanation of A



MCQ 6.4.56 Assertion (A) : An LTI discrete system represented by the difference equation [ 2] 5 [ 1] 6 [ ] [ ]y n y n y n x n+ − + + = is unstable.

Reason (R) : A system is unstable if the roots of the characteristic equation lie outside the unit circle.(A) Both A and R are true and R is the correct explanation of A.

(B) Both A and R are true but R is NOT the correct explanation of A.

(C) A is true but R is false.

(D) A is false but R is true.

MCQ 6.4.57 Consider the following statements regarding a linear discrete-time system

( )H z ( . )( . )z z

z0 5 0 5

12

= + −+

1. The system is stable

2. The initial value ( )h 0 of the impulse response is 4−

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3. The steady-state output is zero for a sinusoidal discrete time input of frequency equal to one-fourth the sampling frequency.

Which of these statements are correct ?(A) 1, 2 and 3 (B) 1 and 2

(C) 1 and 3 (D) 2 and 3

MCQ 6.4.58 Assertion (A) : The discrete time system described by [ ] [ ] [ ]y n x n x n2 4 1= + − is unstable, (here [ ]y n is the output and [ ]x n the input)Reason (R) : It has an impulse response with a finite number of non-zero samples.(A) Both A and R are true and R is the correct explanation of A




MCQ 6.4.59 If the impulse response of discrete - time system is [ ] [ ]h n u n5 1n=− − − , then the system function ( )H z is equal to

(A) z

z5−

− and the system is stable (B) z

z5−

and the system is stable

(C) z

z5−

− and the system is unstable (D) z

z5−

and the system is unstable

MCQ 6.4.60 ( )H z is a discrete rational transfer function. To ensure that both ( )H z and its inverse are stable its(A) poles must be inside the unit circle and zeros must be outside the unit circle.

(B) poles and zeros must be inside the unit circle.

(C) poles and zeros must be outside the unit circle

(D) poles must be outside the unit circle and zeros should be inside the unit circle

MCQ 6.4.61 Assertion (A) : The stability of the system is assured if the Region of Convergence (ROC) includes the unit circle in the z -plane.Reason (R) : For a causal stable system all the poles should be outside the unit circle in the z -plane.(A) Both A and R are true and R is the correct explanation of A

(B) Both A and R are true but R is NOT the correct explanation of A.



MCQ 6.4.62 Assertion (A) : For a rational transfer function ( )H z to be causal, stable and causally invertible, both the zeros and the poles should lie within the unit circle in the z -plane.

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Reason (R) : For a rational system, ROC bounded by poles.

(A) Both A and R are true and R is the correct explanation of A




MCQ 6.4.63 The transfer function of a discrete time LTI system is given by

( )H z 1 z z

z2

43 1

81 2

43 1

=− +

−− −

−

Consider the following statements:

S1: The system is stable and causal for ROC: /z 1 2>S2: The system is stable but not causal for ROC: 1/z 4<S3: The system is neither stable nor causal for ROC: / /z1 4 1 2< <Which one of the following statements is valid ?

(A) Both S1 and S2 are true (B) Both S2 and S3 are true

(C) Both S1 and S3 are true (D) S1, S2 and S3 are all true

MCQ 6.4.64 A causal LTI system is described by the difference equation

[ ]y n2 [ 2] 2 [ ] [ ]y n x n x n 1= α β− − + −The system is stable only if

(A) 2α = , 2<β (B) ,2 2> >α β

(C) 2<α , any value of β (D) 2<β , any value of α

MCQ 6.4.65 Two linear time-invariant discrete time systems s1 and s2 are cascaded as shown in

the figure below. Each system is modelled by a second order difference equation.

The difference equation of the overall cascaded system can be of the order of

(A) 0, 1, 2, 3 or 4 (B) either 2 or 4

(C) 2 (D) 4

MCQ 6.4.66 Consider the compound system shown in the figure below. Its output is equal to

the input with a delay of two units. If the transfer function of the first system is

given by

( )H z1 ..

zz

0 80 5= −

− ,

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then the transfer function of the second system would be

(A) ( )..H zz

z z1 0 4

0 22 1

2 3

=−−

−

− − (B) ( )

..H zz

z z1 0 5

0 82 1

2 3

=−−

−

− −

(C) ( )..H zz

z z1 0 4

0 22 1

1 3

=−−

−

− − (D) ( )

..H zz

z z1 0 5

0 82 1

2 3

=++

−

− −

MCQ 6.4.67 Two systems ( ) ( )andH z H z1 2 are connected in cascade as shown below. The overall

output [ ]y n is the same as the input [ ]x n with a one unit delay. The transfer

function of the second system ( )H z2 is

(A) ( . )

.z z

z1 0 4

1 0 61 1

1

−−

− −

− (B)

( . )( . )

zz z

1 0 41 0 6

1

1 1

−−

−

− −

(C) ( . )( . )

zz z

1 0 61 0 4

1

1 1

−−

−

− −

(D) ( . )

.z z

z1 0 6

1 0 41 1

1

−−

− −

−

MCQ 6.4.68 Two discrete time system with impulse response [ ] [ 1]h n n1 δ= − and [ ] [ 2]h n n2 δ= −

are connected in cascade. The overall impulse response of the cascaded system is

(A) [ 1] [ 2]n nδ δ− + − (B) [ 4]nδ −

(C) [ 3]nδ − (D) [ 1] [ 2]n nδ δ− −

MCQ 6.4.69 A cascade of three Linear Time Invariant systems is causal and unstable. From

this, we conclude that

(A) each system in the cascade is individually causal and unstable

(B) at least on system is unstable and at least one system is causal

(C) at least one system is causal and all systems are unstable

(D) the majority are unstable and the majority are causal

MCQ 6.4.70 The minimum number of delay elements required in realizing a digital filter with

the transfer function

( )H z cz dz ez

az bz1

11 2 3

1 2

=+ + +

+ +− − −

− −

(A) 2 (B) 3

(C) 4 (D) 5

MCQ 6.4.71 A direct form implementation of an LTI system with ( ). .

H zz z1 0 7 0 13

11 2=

− +− − is

shown in figure. The value of ,a a0 1 and a2 are respectively

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(A) 1.0, 0.7 and 0.13− (B) 0.13,0.7− and 1.0

(C) 1.0, 0.7− and 0.13 (D) 0.13, 0.7− and 1.0

MCQ 6.4.72 A digital filter having a transfer function ( )H zd z

p p z p z1 3

30 1

13

3

=+

+ +−

− −

is implemented

using Direct Form-I and Direct Form-II realizations of IIR structure. The number

of delay units required in Direct Form-I and Direct Form-II realizations are,

respectively

(A) 6 and 6 (B) 6 and 3

(C) 3 and 3 (D) 3 and 2

MCQ 6.4.73 Consider the discrete-time system shown in the figure where the impulse response

of ( )G z is (0) 0, (1) (2) 1, (3) (4) 0g g g g g g= = = = = =

This system is stable for range of values of K

(A) [ 1, ]21− (B) [ 1, 1]−

(C) [ , 1]21− (D) [ , 2]2

1−

MCQ 6.4.74 In the IIR filter shown below, a is a variable gain. For which of the following cases,

the system will transit from stable to unstable condition ?

(A) 0.1 0.5a< < (B) 0.5 1.5a< <

(C) 1.5 2.5a< < (D) 2 a< < 3

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MCQ 6.4.75 The poles of an analog system are related to the corresponding poles of the digital system by the relation z es= τ. Consider the following statements.1. Analog system poles in the left half of s -plane map onto digital system poles

inside the circle z 1= .

2. Analog system zeros in the left half of s -plane map onto digital system zeros inside the circle z 1= .

3. Analog system poles on the imaginary axis of s -plane map onto digital system zeros on the unit circle z 1= .

4. Analog system zeros on the imaginary axis of s -plane map onto digital system zeros on the unit circle z 1= .

Which of these statements are correct ?(A) 1 and 2 (B) 1 and 3

(C) 3 and 4 (D) 2 and 4

MCQ 6.4.76 Which one of the following rules determines the mapping of s -plane to z -plane ?(A) Right half of the s -plane maps into outside of the unit circle in z -plane

(B) Left half of the s -plane maps into inside of the unit circle

(C) Imaginary axis in s -plane maps into the circumference of the unit circle

(D) All of the above

MCQ 6.4.77 Assertion (A) : The z -transform of the output of an ideal sampler is given by

[ ( )]f tZ ....K zK

zK

zK

nn

01

22= + + + +

Reason (R) : The relationship is the result of application of z e sT= − , where T stands for the time gap between the samples.(A) Both A and R are true and R is the correct explanation of A




MCQ 6.4.78 z and Laplace transform are related by

(A) lns z= (B) lns Tz=

(C) s z= (D) ln zT

MCQ 6.4.79 Frequency scaling [relationship between discrete time frequency ( )Ω and continuous time frequency ( )ω ] is defined as(A) 2ω Ω= (B) 2 /TSω Ω=

(C) /T2 SωΩ = (D) TSωΩ =

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MCQ 6.4.80 A casual, analog system has a transfer function ( )H s s aa

2 2= + . Assuming a sampling time of T seconds, the poles of the transfer function ( )H z for an equivalent digital system obtained using impulse in variance method are at

(A) ( , )e eaT aT− (B) ,jTa jT

a−a k

(C) ( , )e ejaT jaT− (D) ( , )e e/ /aT aT2 2−

***********

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SOLUTIONS 6.1

Answers

1. (B) 5. (C) 9. (A) 13. (D) 17. (C) 21. (C) 25. (B)

2. (D) 6. (C) 10. (C) 14. (C) 18. (A) 22. (D) 26. (B)

3. (C) 7. (D) 11. (C) 15. (B) 19. (A) 23. (C)

4. (D) 8. (A) 12. (A) 16. (D) 20. (B) 24. (C)

SOLUTIONS 6.2

SOL 6.2.1 Option (B) is correct.z -transform is given as

( )X z z z21

21n

nn

nn 00

= =33

−

==b bl l//

z1 21

1=−

zz

2 12= −

SOL 6.2.2 Option (B) is correct.The z -transform of given sequence is ( )X z z z z z3 2 1 0= + − − z z z 13 2= + − −

Now X 21

b l 21

21

21 1

3 2= + − −b b bl l l .1 125=−

SOL 6.2.3 Option (C) is correct.The z -transform is

( )X z [ ]x n z n

n

=3

3−

=−/ z2

1 nn

n

1

= −3

−

=−

−

b l/

z z21 2

n

n

n

n

11 1

=− =−3 3

−

=−

−−

=−

−

b ^l h/ /

( )z2 m

m 1

=−3

=/ Let n m− = so,

The above series converges if 1z2 < or z 21<

( )X z zz

1 22=− − ,z

z z2 12

21<= −

SOL 6.2.4 Option (A) is correct.

We have [ ]x n 21 | |n

= b l [ ] [ ]u n u n21

21 1

n n= + − −

−

b bl l

z -transform is ( )X z [ ]x n z n

n

=3

3−

=−/

[ ] [ ]z u n z u n21 1 2

1nn

n

nn

n

= − − +3

3

3

3− −

=−

−

=−b bl l/ /

z z21

21n

n

n

n

n

1

0

= +3

3− −

=−

−

=b bl l/ /


z z21

21

II I

n

n

n

n1 0

= +3 3

= =b bl l

1 2 344 44 1 2 344 44

/ /

1 1 z

z1z21

21

21

=−

+−

z

zz

z2

21=

−− −

Series I converges, if z21 1< or z 2<

Series II converges, if z21 1< or z 2

1>

ROC is intersection of both, therefore ROC : z21 2< <

SOL 6.2.5 Option (D) is correct.

( )X z [ ]x n z n

n

=3

3−

=−/

[ ] [ ]z u n z u n21 1 3

1nn

n

nn

n

= − − +3

3

3

3−

=−

−

=−b bl l/ /

z z21

31n

n

n

n

n

1

0

= +3

3−

=−

−

=b bl l/ /

( )z z2 31n

n

n

n1 0

= +3 3

= =b l/ /

zz

z1 22

11

III

31 1= − +

− −

1 2 344 4S

Series I converges, when z2 1< or z 21<

Series II converges, when z31 1< or z 3

1>

So ROC of ( )X z is intersection of both ROC: z31

21< <

SOL 6.2.6 Option (C) is correct.z -transform of [ ]x n

( )X z [ ]x n z n

n

=3

3−

=−/

[ ] [ ]z u n z u nn n n n

nn

α α= +3

3

3

3− − −

=−=−//

( ) ( )z zn n

nn

1

00

α α= +33

− −

==//

( )z z1

11

1

I II

1 1α α=

−+

−− −

1 2 344 44 1 2 344 44

Series I converges, if z 1<1α − or z > α

Series II converges, if ( )z 1<1α − or z 1>α or z 1> αSo ROC is interaction of both

ROC : z ,max 1> α αe o


SOL 6.2.7 Option (C) is correct.( 4)P " [ ]x n1 [ 2]u n= −

( )X z1 [ ]u n z2 n

n

= −3

3−

=−/ ,z

zz z

11>n

n 21

2

= =−

3−

=−

−

/( 1)Q " [ ]x n2 [ 3]u n=− − −

[ ]X z2 [ ]u n z3 n

n

=− − −3

3−

=−/

z n

n

3

=−3

−

=−

−

/ zm

m 3

=−3

=/ Let, n m=−

( )

, 1zz

z zz1 1

1 <3

2 1= −− =

−− −

( 3)R " [ ]x n3 ( ) [ ]u n1 4n= +

( )X z3 [ ]u n z4 n

n

= +3

3−

=−/ z n

n 4

=3

−

=−/

( )

, 1z

zz z

z1 1

1 >1

4

4 1=−

=−− − −

( 2)S " [ ]x n4 ( ) [ ]u n1 n= −

[ ]u n z n

n

= −3

3−

=−/

z n

n

0

=3

−

=−/ zm

m 0

=3

=/ , 1z z

z z11

1<1

1

= − =−

−−

−


( )X z [ ]e z u njn n

n

=3

3π −

=−/ ( )e zj n

n

1

0

=3

π −

=/

e z11

j 1=− π − , z 1>

z e

zj=

− π zz

1= + e 1ja =−π

SOL 6.2.9 Option (D) is correct.We can write, transfer function

( )H z ( )( )z z

Az2 3

2

= − −

( )H 1 ( )( )

A1 2

1= − − = or A 2=

so, ( )H z ( )( )z z

z2 32 2

= − −

( )z

H z

( )( )z zz

2 32= − −

( )H z ( ) ( )z

zz

z2

43

6= −− + − From partial fraction

We can see that for : 3ROC z > , the system is causal and unstable because ROC is exterior of the circle passing through outermost pole and does not include unit circle.so, [ ]h n [( ) ( )( ) ] [ ]u n4 2 6 3n n= − + , 3z > ( 2)P "


For ROC z2 3< < , The sequence corresponding to pole at z 2= corresponds to right-sided sequence while the sequence corresponds to pole at z 3= corresponds to left sided sequence [ ]h n ( ) [ ] ( ) [ ]u n u n4 2 6 3 1n n= − + − − − ( 4)Q "

For : 2ROC z < , ROC is interior to circle passing through inner most pole, hence the system is non causal. [ ]h n ( ) [ ] ( ) [ ]u n u n4 2 1 6 3 1n n= − − + − − − ( 3)R "

For the response [ ]h n ( ) [ ] ( ) [ ]u n u n4 2 1 6 3n n= − − + −

: 2ROC z < and z 3> which does not exist ( 1)S "

SOL 6.2.10 Option (B) is correct.

( )X z ( )z zz

11= −

+

2z z z z zz1

12 1

11=− + − =− + −

−a k By partial fraction

Taking inverse z -transform [ ]x n [ ] [ ]n u n1 2 1δ=− − + − [ ]x 0 0 0 0=− + = [ ]x 1 1 2 1=− + = [ ]x 2 0 2 2=− + =

SOL 6.2.11 Option (A) is correct. ( )X z e e /z z1= +

( )X z ! ! ..... 2! .....z z zz z

1 2 3 1 1 1 12 3

2= + + + + + + + +c bm l

2! ! .... 2! ....z z z z z1 3 12 3

12

= + + + + + + + +−−

c bm l

Taking inverse z -transform

[ ]x n [ ] !n n1δ= +


( )X z z z

z z2 3

52

2

=− −

+ ( )( )

( )z z

z z3 1

5= − ++

( )z

X z

( )( )z zz3 1

5= − ++

z z32

11= − − + By partial fraction

Thus ( )X z zz

zz

32

1= − − +

Poles are at z 3= and z 1=−


ROC : z 1< , which is not exterior of circle outside the outermost pole z 3= . So,

[ ]x n is anticausal given as

[ ]x n [ ( ) ( ) ] [ ]u n2 3 1 1n n= − + − − −


( )X z zz

zz

32

1= − − +

If z 3> , ROC is exterior of a circle outside the outer most pole, [ ]x n is causal.

[ ] [2(3) ( 1) ] [ ]x n u nn n= − −

SOL 6.2.14 Option (C) is correct.

( )X z zz

zz

32

1= − − +

If ROC is z1 3< < , [ ]x n is two sided with anticausal part zz3

2− , z 3< and

causal part zz1+

− , 1z >

[ ] 2(3) [ 1] ( 1) [ ]x n u n u nn n=− − − − −



( )X z1 ( . ) [ ]z u n0 7 1n n

n

= −3

3−

=−/ ( . )z0 7 n

n

1

1

=3

−

=/

.

.z

z1 0 7

0 71

1

=− −

−

ROC : . z0 7 1<1− or .z 0 7>

( )X z2 ( . ) [ ]z u n0 4 2n n

n

= − − −3

3−

=−/ ( . ) z0 4 n n

n

2

= −3

−

=−

−

/

( . ) z0 4 m m

m 2

= −3

−

=/ Let n m=−

[( . ) ]z0 4 m

m

1

2

= −3

−

=/

( . )( . )

zz

1 0 40 4

1

1

=+

−−

−

ROC : ( . ) z0 4 1<1− or .z 0 4<The ROC of z -transform of [ ]x n is intersection of both which does not exist.


If [ ]x n ( )X zZ

From time shifting property

[ ]x n n0− ( )z X znZ0−

So ( [ 4])z x n − ( )z X zz z z8 2

145 4 3= =− −

−

SOL 6.2.17 Option (C) is correct. [ ]x n [ ]u nnα= Let, [ ]y n [ ] [ ]x n u n3= + [ ] [ ]u n u n3n 3α= ++

[ ]u nn 3α= + [ 3] [ ] [ ]u n u n u n+ =

( )Y z [ ]y n z n

n

=3

3−

=−/ [ ]z u n zn n n n

nn

3 3

0

α α= =3

3

3+ − + −

==−//

( )z n

n

3 1

0

α α=3

−

=/

z zz

113

13α

αα α=

−= −− a k

Note : Do not apply time shifting property directly because [ ]x n is a causal signal.

SOL 6.2.18 Option (A) is correct.We can see that ( )X z1 ( ) ( )z X z z X z1

22

3= =or ( )z X z2

1− ( ) ( )z X z X z1

2 3= =−

So [ ]x n 21 − [ ] [ ]x n x n12 3= − =

SOL 6.2.19 Option (A) is correct.We know that

[ ]u nnα zzZ

α−

[ ]u n 10n 10α −− zz z10

Z

α−−

(time shifting property)



We know that [ ]a u nn z azZ

−

or [ ]u n3n zz

3Z

−

[ ]u n3 3n 3 −− z zz

33Z

−−a k

So [ ]x n [ ]u n3 3n 3= −−

[ ]x 5 [ ]u3 2 92= =

SOL 6.2.21 Option (C) is correct.[ ]x n can be written in terms of unit sequence as

[ ]x n [ ] [ ]u n u n k= − −

so ( )X z zz z z

z1 1

k= − − −−

zz

11 k

1=−−

−

−

SOL 6.2.22 Option (C) is correct.For positive shiftIf, [ ]x n ( )X zZ

then, [ ]x n n0− ( )z X znZ0− , n 00 $

So [ ]x n 1− z zz

z1 111Z

− = −−a k

For negative shift

[ ]x n n0+ ( ) [ ]z X z x n zn m

m

n

0

1Z

0

0

− −

=

−

e o/ , n 0>0

[ ]x n 1+ ( ) [ ]z X z x 0Z −^ hWe know that [ ] [ ]x n u n= so [ ]x 0 1=

and [ ]x n 1+ ( )z X z z zz1 1 1Z − = − −^ ah k z

z1= −


Even part of [ ]x n , [ ]x ne ( [ ] [ ])x n x n21= + −

z - transform of [ ]x ne , ( )X ze ( )X z X z21 1= + b l; E [ ]x n X z

1Za − b l

. / ./

zz

zz

21

0 4 21

1 0 41

I II

= − + −a ek o1 2 344 44 1 2 3444 444

Region of convergence for I series is .z 0 4> and for II series it is .z 2 5< . Therefore, ( )X ze has ROC . .z0 4 2 5< <

SOL 6.2.24 Option (A) is correct. [ ]y n [ ] [ ]n n u n1= + [ ]y n [ ] [ ]n u n nu n2= +

We know that [ ]u n zz

1Z

−


Applying the property of differentiation in z -domainIf, [ ]x n ( )X zZ

then, [ ]nx n ( )zdzd X zZ −

so, [ ]nu n zdzd

zz

1Z − −a k

or, [ ]nu n ( )z

z1 2

Z

−Again by applying the above property

( [ ])n nu n ( )

zdzd

zz1 2

Z −−: D

[ ]n u n2 ( )( )z

z z11

3Z

−+

So ( )Y z ( ) ( )

( )z

zz

z z1 1

12 3=

−+

−+

( )z

z1

23

2

=−

SOL 6.2.25 Option (B) is correct.( 4)P " [ ]y n ( ) [ ]n u n1 n= −We know that

( ) [ ]u n1 n− , 1z

z1

1 >1Z

+ −

If [ ]x n ( )X zZ

then, [ ]nx n ( )

z dzdX zZ − (z -domain differentiation)

so, ( ) [ ]n u n1 n− zdzd

z11

1Z −

+ −; E, : 1ROC z >

( )Y z ( )

, : 1ROCzz z

1>1 2

1

=+−

−

−

( 3)Q " [ ]y n [ ]nu n 1=− − −We know that,

[ ]u n 1− − ( )

, : 1ROCz

z1

1 <1Z

−−

−

Again applying z -domain differentiation property

[ ]nu n 1− − − , : 1ROCzdzd

zz

11 <1

Z

−−

−: D

( )Y z ( )

, : 1ROCz

z z1

<1 2

1

=− −

−

( 2)R " [ ]y n ( ) [ ]u n1 n= −

( )Y z ( ) [ ]z u n1 n n

n

= −3

3−

=−/

( )z n

n

1

0

= −3

−

=/

, : 1ROCz

z1

1 >1=+ −

( 1)S " [ ]y n [ ]nu n=


We know that [ ]u n , : 1ROCz

z1

1 >1Z

− −

so, [ ]nu n , : 1ROCzdzd

zz

11 >1

Z

− −b l

( )Y z ( )

, : 1ROCz

z z1

>1 2

1

=− −

−


Given that ( )X z (1 2 )log z= − , z 21<

Differentiating

( )

dzdX z

z zz

1 22

1 21 1

1

= −− =

− −

−

or, ( )

dzzdX z

1 z

121 1=

− −

From z -domain differentiation property

[ ]nx n ( )

z dzdX zZ −

so, [ ]nx n 1 z

121 1

Z

−−

−

From standard z -transform pair, we have

[ 1]u n21 n

− −b l 1 z

121 1

Z

−−

−

Thus [ ]nx n [ 1]u n21 n

= − −b l

or, [ ]x n [ ]n u n121 1

n= − −b l

SOL 6.2.27 Option (A) is correct.Its difficult to obtain z -transform of [ ]x n directly due to the term /n1 .Let [ ]y n [ ] ( ) [ ]nx n u n2 1n= = − − −−

So z -transform of [ ]y n

( )Y z , :ROCz

z z 21<

21=

+−

Since [ ]y n [ ]nx n=

so, ( )Y z ( )

z dzdX z=− (Differentiation in z -domain)

( )

z dzdX z−

zz

21=

+−

( )

dzdX z

z

121=

+

or ( )X z 1 , :log ROCz z2 21<= +b l



( )X z [ ] , :ROCx n z Rnx

n

=3

3−

=−/

( )Y z [ ] [ ]y n z a x n zn n n

nn

= =3

3

3

3− −

=−=−//

[ ]x n az n

n

=3

3−

=−a k/ , :ROCX a

z aRx= a k

SOL 6.2.29 Option (B) is correct.Using time shifting property of z -transform

If, [ ]x n ( ), :ROCX z RxZ

then, [ ]x n n0− ( )z X znZ0−

with same ROC except the possible deletion or addition of z 0= or z 3= . So, ROC for [ ]x n 2− is Rx ( , )S R1 1

Similarly for [ ]x n 2+ , ROC : Rx ( , )S R2 1

Using time-reversal property of z -transform


then, [ ]x n− , :ROCX z R1 1

x

Z

b l

For S3, [ ]x n− X z1Z

b l,

Because z is replaced by 1/z , so ROC would be z a1< ( , )S R3 3

:( ) [ ]S x n1 n4 −

Using the property of scaling in z -domain, we have


then, [ ]x nnα X zZ

αa k

z is replaced by /z α so ROC will be Rx

αHere ( ) [ ]x n1 n− , 1X z

1Z α =a k

so, :ROC z a> ( , )S R4 1

SOL 6.2.30 Option (D) is correct.Time scaling property :If, [ ]x n ( )X zZ

then, [ / ]x n 2 ( )X z2Z ( 2)P "

Time shifting property :If, [ ]x n ( )X zZ

then, [ ] [ ]x n u n2 2− − ( )z X z2Z − ( 1)Q "

For [ ] [ ]x n u n2+ we can not apply time shifting property directly.Let, [ ]y n [ ] [ ]x n u n2= + [ ] [ ]u n u n2n 2α= ++ [ ]u nn 2α= +


so, ( )Y z [ ]y n z n

n

=3

3−

=−/ zn n

n

2

0

α=3

+ −

=/

( )z n

n

2 1

0

α α=3

−

=/ ( )X z2α= ( 4)R "

Let, [ ]g n [ ]x nn2β=

( )G z [ ]g n z n

n

=3

3−

=−/ [ ]z u nn n n

n

2β α=3

3−

=−/

zn

n

n

20

αβ

=3

=b l/ X z

2β= b l ( 3)S "


( )Y z x n z2n

n=3

3

=−

−9 C/

[ ]x k z k

k

2=3

3−

=−/ Put n k2 = or n k2=

( )X z2=

SOL 6.2.32 Option (C) is correct.Let, [ ]y n [2 ]x n=

( )Y z [ ]x n z2 n

n

=3

3−

=−/

[ ]x k z /k

k

2=3

3−

=−/ Put n k2 = or n k

2= , k is even

Since k is even, so we can write

( )Y z [ ] ( ) [ ]x k x k

z21 /

kk

k

2= + −

3

3−

=−; E/

[ ] [ ] ( )x k z x k z21

21/ /k k

kk

2 1 2= + −3

3

3

3−

=−=−//

( ) ( )X z X z21= + −8 B


( )X z [ ]x n z n

n

=3

3−

=−/

( ) ( )Y z X z3= [ ] ( ) [ ]x n z x n zn n

nn

3 3= =3

3

3

3− −

=−=−//

[ / ]x k z3 k

k

=3

3−

=−/ Put n k3 = or /n k 3=

Thus [ ]y n [ / ]x n 3=

[ ]y n ( . ) , , , , ....

, otherwise

n0 5 0 3 6

0

/n 3

= − =*

Thus [ ]y 4 0=

SOL 6.2.34 Option (A) is correct.From the accumulation property we know that


If, [ ]x n ( )X zZ

then, [ ]x kk

n

3=−/

( )( )

zz X z

1Z

−

Here, [ ]y n [ ]x kk

n

0

==/

( )Y z ( )

( )z

z X z1

= − ( )( )z z z

z1 8 2 1

42

2

=− − −

SOL 6.2.35 Option (C) is correct.By taking z -transform of [ ]x n and [ ]h n ( )H z z z z1 2 1 3 4= + − +− − −

( )X z z z z1 3 21 2 3= + − −− − −

From the convolution property of z -transform ( )Y z ( ) ( )H z X z= ( )Y z z z z z z z z1 5 5 5 6 4 21 2 3 4 5 6 7= + + − − + + −− − − − − − −

Sequence is [ ]y n {1, 5, 5, 5, 6, 4, 1, 2}= − − − [ ]y 4 6=−

SOL 6.2.36 Option (B) is correct.By taking z -transform of both the sequences ( )X z ( )z z1 2 0 31 3= − + + +− −

( )H z 2 3z2= +Convolution of sequences [ ]x n and [ ]h n is given as [ ]y n [ ] [ ]x n h n*applying convolution property of z -transform, we have ( )Y z ( ) ( )X z H z= ( )( )z z z1 2 3 2 31 3 2= − + + +− − z z z z2 4 3 12 92 1 3=− + − + +− −

or, [ ]y n { 2, 4, , 12, 0, 9}3= − −-

SOL 6.2.37 Option (C) is correct.The z -transform of signal [ ]x n) is given as follows

{ [ ]}x nZ ) [ ]x n z n

n

= )

3

3−

=−/ [ ] ( )x n z n

n

= ))

3

3−

=−6 @/ ...(i)

Let z -transform of [ ]x n is ( )X z

( )X z [ ]x n z n

n

=3

3−

=−/

Taking complex conjugate on both sides of above equation

( )X z) [ [ ] ]x n z n

n

= )

3

3−

=−/

Replacing z z" ), we will get

( )X z) ) [ ] ( )x n z n

n

= ))

3

3−

=−6 @/ ...(ii)

Comparing equation (i) and (ii)


{ [ ]}x nZ ) ( )X z= ) )

SOL 6.2.38 Option (B) is correct.[ ]x n can be written as

[ ]x n [ [ ] ( ) [ ]]u n u n21 1 n= + −

z -transform of [ ]x n

( )X z z z2

11

11

11 1=

−+

+− −; E

From final value theorem ( )x 3 ( ) ( )lim z X z1

z 1= −

"

( )lim z zz

zz

21 1 1 1z 1

= − − + +"9 C

( )( )

lim zz

z z21

11

z 1= + +

−"

= G

(1)21

21= =

SOL 6.2.39 Option (C) is correct.From initial value theorem [ ]x 0 ( )limX z

z=

"3

( )( . )

.limz z

z1 0 50 5

z

2

= − −"3

..lim

z z1 1 1 0 50 5

z=

− −"3b bl l

.0 5=

From final value theorem [ ]x 3 ( ) ( )lim z X z1

z 1= −

"

( )( )( . )

.lim zz z

z11 0 50 5

z 1

2

= − − −"

..lim z

z0 5

0 5z 1

2

= −" 1=

SOL 6.2.40 Option (B) is correct.By taking z -transform on both sides of given difference equation

( ) ( ) [ ]Y z z Y z y z21 11− + −−

6 @ ( )X z=

Let impulse response is ( )H z , so the impulse input is ( )X z 1=

( ) ( )H z z H z z21 31− +−

6 @ 1=

( ) [1 ]H z z21 1− − 2

5=

( )H z /

z15 2

21 1=

− − z

z25

21=

−b l

[ ]h n 25

21 n

= b l , n 0$


SOL 6.2.41 Option (B) is correct. [ ]h n ( ) [ ]u n2 n=Taking z -transform

( )H z ( )( )

zz

X zY z

2= − =

so, ( ) ( )z Y z2− ( )zX z=or, ( ) ( )z Y z1 2 1− − ( )X z=Taking inverse z -transform [ ] [ ]y n y n2 1− − [ ]x n=


[ ]h n [ ] [ ]n u n21 n

δ= − −b l

z -transform of [ ]h n

( )H z 1z

zz2

121

21

= −+

=+

( )( )

X zY z=

( )z Y z21+b l ( )X z2

1=

( )z Y z1 21 1+ −

b l ( )z X z21 1= −


[ ] [ ]y n y n21 1+ − [ ]x n2

1 1= −

[ ] . [ ]y n y n0 5 1+ − . [ ]x n0 5 1= −

SOL 6.2.43 Option (C) is correct.We have [ ] . [ ]y n y n0 4 1− − ( . ) [ ]u n0 4 n=Zero state response refers to the response of system with zero initial condition. So, by taking z -transform

( ) . ( )Y z z Y z0 4 1− − .zz0 4= −

( )Y z ( . )z

z0 4 2

2

=−

Taking inverse z -transform [ ]y n ( 1)(0.4) [ ]n u nn= +

SOL 6.2.44 Option (B) is correct.Zero state response refers to response of the system with zero initial conditions.Taking z -transform

( ) ( )Y z z Y z21 1− − ( )X z=

( )Y z . ( )zz X z0 5= −a k

For an input [ ]x n [ ], ( )u n X z zz

1= = −


so, ( )Y z ( . )( )z

zz

z0 5 1

= − −

( )( . )z z

z1 0 5

2

= − −

( )z

Y z

( )( . )z zz

1 0 5= − −

.z z12

0 51= − − − By partial fraction

Thus ( )Y z .zz

zz

12

0 5= − − −

Taking inverse z -transform [ ]y n [ ] ( . ) [ ]u n u n2 0 5 n= −

SOL 6.2.45 Option (C) is correct.Input, [ ]x n [ ] [ ]n n2 1δ δ= + +By taking z -transform ( )X z 2 z= +

( )( )

X zY z

( )H z= , ( )Y z is z -transform of output [ ]y n

( )Y z ( ) ( )H z X z=

( )

( )( 2)

zz z

z2

2 12=

+− +

( )( )z

z zz z

z2

2 12 2

6= +− = − +

Taking inverse z -transform [ ]y n 2 [ 1] 6( 2) [ ]n u nnδ= + − −

SOL 6.2.46 Option (B) is correct.Taking z transform of input and output

( )X z .zz0 5= −

( )Y z z zz1 2 21= − = −−

Transfer function of the filter ( )H z ( )/ ( )Y z X z=

2 0.5z

zz

z= − −b bl l

.z

z z2 5 12

2

= − +

1 . z z2 5 1 2= − +− −

Taking inverse z -transform [ ]h n { , 2.5, 1}1= −

-

Therefore [ ]h 1 .2 5=−


SOL 6.2.47 Option (B) is correct.Poles of the system function are at z j!= ROC is shown in the figure.

Causality :We know that a discrete time LTI system with transfer function ( )H z is causal if and only if ROC is the exterior of a circle outside the outer most pole.For the given system ROC is exterior to the circle outside the outer most pole ( )z j!= . The system is causal.Stability :A discrete time LTI system is stable if and only if ROC of its transfer function ( )H z includes the unit circle z 1= .The given system is unstable because ROC does not include the unit circle.Impulse Response :

( )H z z

z12=

+We know that

( ) [ ]sin n u n0Ω cos

sinz z

z2 12

0

0Z

ΩΩ

− +, z 1>

Here z 12 + cosz z2 120Ω= − +

So cosz2 0Ω 0= or 20πΩ =

Taking the inverse Laplace transform of ( )H z

[ ]h n [ ]sin n u n2π= a k

SOL 6.2.48 Option (D) is correct.Statement (A), (B) and (C) are true.

SOL 6.2.49 Option (D) is correct.First we obtain transfer function (z -transform of [ ]h n ) for all the systems and then check for stability

(A) ( )H z z

z

31 2

31

=−^ h

Stable because all poles lies inside unit circle.

(B) [ ]h n [ ]n31 δ=


[ ]h n/ 31= (absolutely summable)

Thus this is also stable.

(C) [ ]h n [ ] [ ]n u n31 n

δ= − −b l

( )H z 1z

z31= −

+Pole is inside the unit circle, so the system is stable.(D) [ ]h n [( ) ( ) ] [ ]u n2 3n n= −

( )H z zz

zz

2 3= − − −

Poles are outside the unit circle, so it is unstable.

SOL 6.2.50 Option (B) is correct.By taking z -transform ( 3 2 ) ( )z z Y z1 1 2+ +− − ( ) ( )z X z2 3 1= + −

So, transfer function

( )( )( )

H zX zY z=

( )( )

z zz

1 3 22 3

1 2

1

=+ +

+− −

−

z z

z z3 2

2 32

2

=+ +

+

or ( )z

H z

z zz3 2

2 32=+ +

+ z z21

11= + + + By partial fraction

Thus ( )H z zz

zz

2 1= + + +

Both the poles lie outside the unit circle, so the system is unstable.

SOL 6.2.51 Option (B) is correct. [ ]y n [ ] [ ]x n y n 1= + −Put [ ] [ ]x n nδ= to obtain impulse response [ ]h n [ ]h n [ ] [ ]n h n 1δ= + −For n 0= , [ ]h 0 [ ] [ ]h0 1δ= + − [ ]h 0 1= ( [ ]h 1 0− = , for causal system)n 1= , [ ]h 1 [ ] [ ]h1 0δ= + [ ]h 1 1=n 2= , [ ]h 2 [ ] [ ]h2 1δ= + [ ]h 2 1=In general form [ ]h n [ ]u n= Thus, statement 1 is true.

Let [ ]x n ( )X zZ

( )X z .zz0 5= −

[ ]h n ( )H zZ

( )H z zz

1= −

Output ( )Y z ( ) ( )H z X z=


.zz

zz

1 0 5= − −a ak k .zz

zz

12

0 5= − − − By partial fractionInverse z -transform [ ]y n [ ] ( . ) [ ]u n u n2 0 5 n= − Statement 3 is also true.

( )H z zz

1= −System pole lies at unit circle z 1= , so the system is not BIBO stable.

SOL 6.2.52 Option (C) is correct.( 3)P " ROC is exterior to the circle passing through outer most pole at .2z 1= , so it is causal. ROC does not include unit circle, therefore it is unstable.( 1)Q " ROC is not exterior to the circle passing through outer most pole at

.z 1 2= , so it is non causal. But ROC includes unit circle, so it is stable.( 2)R " , ROC is not exterior to circle passing through outermost pole .z 0 8= , so it is not causal. ROC does not include the unit circle, so it is unstable also.( 4)S " , ROC contains unit circle and is exterior to circle passing through outermost pole, so it is both causal and stable.


( )H z .( . )

z PzP z

0 90 9= − +

−

( ) .

( . )P z

P z1 0 9

0 9= + −−

..

PP

z P

z1

10 90 9= + − +

−f p

Pole at z .P1

0 9= +For stability pole lies inside the unit circle, so z 1<

or .P1

0 9+ 1<

0.9 P1< + .P 0 1> − or .9P 1< −

SOL 6.2.54 Option (A) is correct.For a system to be causal and stable, ( )H z must not have any pole outside the unit circle z 1= .

S1 : ( )H z z z

zz z

z2

21

163

21

41

43

21

=+ −

− =− +

−^ ^h h

Poles are at /z 1 4= and /z 3 4=− , so it is causal.

S2 : ( )H z 1z z

z 134

21 3=

+ −+

−^ ^h h

one pole is at /z 4 3=− , which is outside the unit circle, so it is not causal.S3 : one pole is at z 3= , so it is also non-causal.

SOL 6.2.55 Option (B) is correct.Comparing the given system realization with the generic first order direct form II realization


Difference equation for above realization is [ ] [ ]y n a y n 11+ − [ ] [ ]b x n b x n 10 1= + −Here 2, ,a b b3 41 1 0=− = =So [ ] [ ]y n y n2 1− − [ ] [ ]x n x n4 3 1= + −Taking z -transform on both sides ( ) ( )Y z z Y z2 1− − ( ) ( )X z z X z4 3 1= + −

Transfer function

( )( )( )

H zX zY z=

zz

zz

1 24 3

24 3

1

1

=−+ = −

+−

−

SOL 6.2.56 Option (A) is correct.The z -transform of each system response

( )H z1 1 , ( )z H zz

z21 1

221= + =

−−

The overall system function ( )H z ( ) ( )H z H z1 2=

zz

z1 21 1

21= +

−−

b bl l zz

2121

=−+

^

^

h

h

Input, [ ]x n ( )cos nπ=

So, z 1=− and ( 1)11

H z 31

2121

=− =− −− + =

Output of system

[ ]y n ( ) [ ]H z x n1= =− cosn31 π=

SOL 6.2.57 Option (B) is correct.From the given block diagram ( )Y z ( ) ( )z X z z Y z1 1α α= +− −

( )( )Y z z1 1α− − ( )z X z1α= −

Transfer function

( )( )

X zY z

z

z1 1

1

αα=− −

−

For stability poles at z 1= must be inside the unit circle.So 1<α or 1 1< <α−

***********


SOLUTIONS 6.3


( )X z+ [ ] [ 2]x n z n z zn

n

n

n0 0

2δ= = − =3 3

−

=

−

=

−/ / [ ] [ ] [ ]f n n n f nn

0 0δ − =3

3

=−/


( )X z+ [ ]x n z zz1

1n

n

n

n0 01= = =

−

3 3−

=

−

=−/ /


( )X z [ ] [ ] , 0x n z n k z z zn

n

n

n

k !δ= = − =3

3

3

3−

=−

−

=−

−/ /

ROC : We can find that ( )X z converges for all values of z except z 0= , because at z 0= ( )X z " 3.


( )X z [ ] [ ]x n z n k z zn

n

n

n

kδ= = + =3

3

3

3−

=−

−

=−/ / , all z

ROC : We can see that above summation converges for all values of z .


( )X z [ ] [ ]x n z u n zn

n

n

n= =

3

3

3

3−

=−

−

=−/ /

zz1

1n

n 01

I

= =−

3−

=−

S

/

ROC : Summation I converges if z 1> , because when z 1< , then z n

n 0

" 33

−

=/ .


( )X z [ ]x n z n

n

=3

3−

=−/

( [ ] [ 5])u n u n z41 n

n

n

= − −3

3−

=−b l/

z41 n

n

1

0

4

I

= −

=b l

1 2 344 44

/ [ ] [ 5] 1, 0 4u n u n nfor # #− − =

( . )

( . )zz

z zz

11

0 50 25

41 141 1 5

4

5 5

=−−

=−

−−

−

^

^

h

h, all z


ROC : Summation I converges for all values of z because n has only four value.


( )X z [ ]x n z n

n

=3

3−

=−/ [ ]u n z4

1 nn

n

= −3

3−

=−b l/

z z41 4

n

n

n

n

10 0

= =3 3

−

=−

−

=−b ^l h/ /

( ) ,z z z4 1 41

41<m

m 0

I

= = −3

=1 2 344 44

/ Taking n m"− ,

ROC : Summation I converges if 1z z4 or< < 41 .


( )X z [ ]x n z n

n

=3

3−

=−/ 3 [ 1]u n zn n

n

= − −3

3−

=−/

( )z z3 31n

n n

n1

1

1

I

= =3

3−

=−

−

=b l

1 2 344 44

/ / [ 1] 1, 1u n n #− − = −

z

z1 3

131

=−

zz

3= − , 3z <

ROC : Summation I converges when 1 3z zor< <31


( )X z [ ]x n z n

n

=3

3−

=−/ z3

2 nn

n

=3

3−

=−b l/

z z32

32n

nn

nn

n

0

1

= +3

3

− −

==−

−−

b bl l//

In first summation taking n m=− ,

( )X z z z32

32m

mn

n

nm 01= +

33−

==b bl l//

z z32

32m

m

n

n1

1

0

I II

= +3 3

=

−

=b bl l

1 2 344 44 1 2 344 44

/ /

z

zz1 1

132

32

32 1=

−+

− −^ ^h h

z z11

11

23 1

32 1=

−− +

−− −^ ^h h

ROC : Summation I converges if 1z zor< <32

23 and summation II converges if

1z zor< >32 1

32− . ROC of ( )X z would be intersection of both, that is z< <3

223


[ ]x n [ ]cos n u n3π= a k [ ]e e u n2

j n j n3 3

= + −π π^ ^h h


[ ] [ ]e u n e u n21

21j n j n3 3= + −π π

^ ^h h

[ ]X z e z e z2

11

11

11 1

I II

j j3 3

=−

+−− − −π π

> H1 2 344 44 1 2 344 44

[ ] ,a u naz

z a1

1 >n1

Z

− −

z e e z

z e e21

1

21 2

1

j j

j j

3 3

3 3

=− + +

− +− − −

− −

π π

π π

^ h

6>

@H, z 1>

( )

( ), 1z

z zz

z2 12 1 >2=− +

−

ROC : First term in ( )X z converges for z e z 1> >3j

&π

. Similarly II term also

converges for 1z e z> >j3 &− π

, so ROC would be simply z 1> .


[ ]x n 3 [ 5] 6 [ ] [ 1] 4 [ 2]n n n nδ δ δ δ= + + + − − −

[ ]X z 3 6 4 , 0z z z z< <5 1 2 3= + + −− − [ ]n n z n0

Z0!δ !

ROC : ( )X z is finite over entire z plane except z 0= and z 3= because when

z 0= negative power of z becomes infinite and when z " 3 the positive powers of

z tends to becomes infinite.


[ ]x n 2 [ 2] 4 [ 1] 5 [ ] 7 [ 1] [ 3]n n n n nδ δ δ δ δ= + + + + + − + −

( )X z 2 4 5 7 , 0z z z z z< <2 1 3 3= + + + +− − [ ]n n z n0

Z0!δ !

ROC is same as explained in previous question.


[ ]x n [ ] [ 2] [ 4] [ 5]n n n nδ δ δ δ= − − + − − −

( )X z 1 , 0z z z z2 4 5 != − + −− − − [ ]n n z n0

Z0!δ !

ROC : ( )X z has only negative powers of z , therefore transform ( )X z does not

converges for z 0= .


Using partial fraction expansion, ( )X z can be simplified as

( )X z z z

z zz z

z1

31

1 32

23

2

23 1 2

1

=+ −

− =+ −

−− −

−

(1 2 )(1 )

(1 3 )z z

z1

21 1

1

+ −−

− −

−

z z1 2

21

11

21 1

I II

=+

−−− −

1 2 344 4S

Poles are at z 2=− and z 21= . We obtain the inverse z -transform using relationship

between the location of poles and region of convergence as shown in the figure.


ROC : 2z< <21 has a radius less than the pole at z 2=− therefore the I term of

( )X z corresponds to a left sided signal

z1 2

21+ − 2(2) [ 1]u nnZ 1

− − −−

(left-sided signal)

While, the ROC has a greater radius than the pole at z 21= , so the second term of

( )X z corresponds to a right sided sequence.

z1

121 1− − [ ]u n

21n

Z 1−

(right-sided signal)

So, the inverse z -transform of ( )X z is

[ ]x n 2(2) [ 1] [ ]u n u n21nn=− − − −

SOL 6.3.15 Option (A) is correct.Using partial fraction expansion ( )X z can be simplified as follows

( )X z zz z

163

2

241

=−−

zz

1 163

141 1

=−−

−

−

z z1 4 1 41

3249

13247

=+

+−− −

Poles are at z 4=− and z 4= . Location of poles and ROC is shown in the figure below


ROC : 4z > has a radius greater that the pole at 4z =− and z 4= , therefore both the terms of ( )X z corresponds to right sided sequences. Taking inverse z - transform we have

[ ]x n ( ) [ ]u n3249 4 32

47 4n n= − +: D

SOL 6.3.16 Option (C) is correct.Using partial fraction expansion ( )X z can be simplified as

( )X z z

z z z1

2 2 22

4 3 2

=−

− −

z

z z z1

2 2 22

1 22=

−− −

−

− −

; E

2z z

z1

11

11 1

2= ++

+−−

− −; E

Poles are at z 1=− and z 1= . Location of poles and ROC is shown in the following figure

ROC : z 1> has radius grater than both the poles at z 1=− and z 1= , therefore both the terms in ( )X z corresponds to right sided sequences.

z1

11+ − ( 1) [ ]u nnZ 1

−−

(right-sided)

z1

11− − [ ]u nZ 1−

(right-sided)

Now, using time shifting property the complete inverse z -transform of ( )X z is [ ]x n 2 [ 2] (( 1) 1) [ 2]n u nnδ= + + − − +

SOL 6.3.17 Option (A) is correct.We have, ( )X z 1 2 4 , 0z z z >6 8= + +− −

Taking inverse z -transform we get

[ ]x n [ ] 2 [ 6] 4 [ 8]n n nδ δ δ= + − + − [ ]z n nn0

Z0

1

δ −−−

SOL 6.3.18 Option (B) is correct.Since [ ]x n is right sided,

[ ] [ ]x n k n k1k 5

10

δ= −=/ [ ]z n nn

0Z

01

δ −−−


SOL 6.3.19 Option (D) is correct.We have, ( )X z (1 )z 1 3= + −

1 3 3z z z1 2 3= + + +− − − , 0z >Since [ ]x n is right sided signal, taking inverse z -transform we have

[ ]x n [ ] 3 [ 1] 3 [ 2] [ 3]n n n nδ δ δ δ= + − + − + − [ ]z n nn0

Z0

1

δ −−−

SOL 6.3.20 Option (A) is correct.We have, ( )X z 3 2 , 0z z z z z >6 2 3 4= + + + +− −

Taking inverse z transform we get [ ]x n [ 6] [ 2] 3 [ ] 2 [ 3] [ 4]n n n n nδ δ δ δ δ= + + + + + − + −


( )X z ,z

z1

121>

41 2=

− −

The power series expansion of ( )X z with z > 21 or z 1<4

1 2− is written as

( )X z 1 .......z z z4 4 4

2 2 2 2 3

= + + + +− − −

b bl l

z z41

41k

k

k

k

k2

0 0

2= =3 3

−

= =

−b bl l/ / , Series converges for 1z zor< >4

1 221−


[ ]x n [ 2 ]n k41 k

k 0δ= −

3

=b l/ [ ]z n k2k2 Z 1

δ −−−

,

,

n n

n41 0

0

even and

odd

n2

$= b l*

,

,

n n

n

2 0

0

even and

odd

n $=−

*


( )X z ,z

z1

121<

41 2=

− −

Since ROC is left sided so power series expansion of ( )X z will have positive powers of z , we can simplify above expression for positive powers of z as

( )X z ( )

,z

z z1 2

421<2

2

=−−

The power series expansion of ( )X z with z < 21 or z4 1<2 is written as

( )X z 4 [1 (2 ) (2 ) (2 ) .......]z z z z2 2 4 6=− + + + +

( )X z 4z z2 k

k

2 2

0

=−3

=^ h/ z2 ( ) ( )k k

k

2 1 2 1

0=−

3+ +

=/

Taking inverse z -transform, we get

[ ]x n [ ( )]n k2 2 1( )k

k

2 1

0δ=− + +

3+

=/ [ ( )]z n k2 1( )k2 1 Z 1

δ + ++−



Using Taylor’s series expansion for a right-sided signal, we have

(1 )ln α+ .......2 3 42 3 4

α α α α= − + − +

( )

( )k1 k

k

k

1

1α= −3 −

=/

( )X z (1 )ln z 1= + −

( )

( )k z1 k

k

k1

1

1= −3 −

=

−/


[ ]x n ( )

[ ]k n k1 k

k

1

1δ= − −

3 −

=/ [ ]z n kk Z 1

δ −−−


We know that,

cosα 1 ! ! ! ! ..........2 4 6 8

2 4 6 8α α α α= − + − + −

( ) !( )

k21 k

k

k

0

2α= −3

=/

Thus, ( )X z ( )( ) !( )

( )cos zk

z2

1 k

k

k3

0

3 2= = −3−

=

−/ , 0z >


[ ]x n ( ) !( )

[ 6 ]k

n k2

1 k

k 0δ= − −

3

=/

Now for 12n = we get, 12 6 0 2k k&− = =

Thus, [12]x !( )

41

2412

= − =


From the given pole-zero pattern

( )X z z z

Az23

1=− −^ ^h h

, A " Some constant

Using partial fraction expansion, we write

( )z

X z

z z 231

α β=−

+ − , α and β are constants.

( )X z 1 1 2z z

I II

31 1 1

α β=−

+−− −

^ ^h h1 2 344 44 1 2 344 44

...(i)

Poles are at z 31= and 2z = . We obtain the inverse z -transform using relationship

between the location of poles and region of convergence as shown in following

figures.


ROC : 2z >

ROC is exterior to the circle passing through right most pole so both the term in

equation (i) corresponds to right sided sequences

[ ]x n1 [ ] ( ) [ ]u n u n31 2

nnα β= +b l

ROC : 2z31 < <

Since ROC has greater radius than the pole at z 31= , so first term in equation (i)

corresponds to right-sided sequence

z1 3

1 1α

− −^ h

[ ]u n31z n1

α−

b l (right-sided)

ROC z 2< has radius less than the pole at z 2= , so the second term in equation

(i) corresponds to left sided sequence.

( )z1 2 1

β− − ( ) [ ]u n2 1z n

1

β −−

(left-sided)

So, [ ]x n2 [ ] ( ) [ ]u n u n31 2 1

nnα β= + −b l


ROC : z 31<

ROC is left side to both the poles of ( )X z , so they corresponds to left sided signals.

[ ]x n3 [ ] ( ) [ ]u n u n31 1 2 1

nnα β= − − + − −b l

All gives the same z -transform with different ROC so, all are the solution.

SOL 6.3.26 Option (C) is correct.The z -transform of all the signal is same given as

( )X z z z1 2

11

11

21 1=

−−

−− −

Poles are at 2z = and z 21= . Now consider the following relationship between ROC

and location of poles.1. Since [ ]x n1 is right-sided signal, so ROC is region in z -plane having radius

greater than the magnitude of largest pole. So, 2z > and z > 21 gives :R1

2z >

2. Since [ ]x n2 is left-sided signal, so ROC is the region inside a circle having radius equal to magnitude of smallest pole. So, 2z < and z < 2

1 gives :R2 z < 21

3. Since [ ]x n3 is double sided signal, So ROC is the region in z -plane such asz > 2

1 and 2z < which gives R3 : 2z< <21


We have ( )X z z z

z

1 1

1

21 1

31 1

67 1

=− +

+− −

−

^ ^h h

z z12

11

21 1

31 1=

−+

+−

− −

( )X z has poles at z 21= and z 3

1−= , we consider the different ROC’s and location of poles to obtain the inverse z -transform.1. ROC z > 2

1 is exterior to the cicle which passes through outtermost pole, so both the terms in equation (i) contributes to right sided sequences.

[ ]x n [ ] [ ]u n u n22

31

n

n= − −

b l


2. ROC z < 31 is interior to the circle passing through left most poles, so both

the terms in equation (i) corresponds to left sided sequences.

[ ]x n [ 1]u n22

31

n

n= − + − − −b l; E

3. ROC z< <31

21 is interior to the circle passing through pole at z 2

1= so the first term in equation (i) corresponds to a right sided sequence, while the ROC is exterior to the circle passing through pole at z 3

1−= , so the second term corresponds to a left sided sequence. Therefore, inverse z -transform is

[ ]x n [ 1] [ ]u n u n22

31

n

n=− − − − −

b l

SOL 6.3.28 Option (A) is correct.The location of poles and the ROC is shown in the figure. Since the ROC includes the point z 4

3= , ROC is 1z< <21

( )X z z

Az

B1 12

1 1 1=−

++− −

ROC is exterior to the pole at z 21= , so this term corresponds to a right-sided

sequence, while ROC is interior to the pole at z 1=− so the second term corresponds to a left sided sequence. Taking inverse z -transform we get

[ ]x n [ ] ( 1) [ 1]A u n B u n2n

n= + − − −

For n 1= , [ ]x 1 (1) 0A B2 1#= + = A2 1& = & 2A =

For 1n =− , [ 1]x − ( )A B B0 1 1 1&#= + − = =−

So, [ ]x n [ ] ( 1) [ 1]u n u n21n

n1= − − − −−

SOL 6.3.29 Option (B) is correct.Let, [ ]x n [ ]Cp u nn= (right-sided sequence having a single pole) [0]x 2 C= =

[2]x 2p p21

212 &= = = ,

So, [ ]x n 2 ( )u n21 n

= b l



( )X z z z21

41n n

nn

1 11

0= +

3

3− −

=−

−

=b bl l//

( )z z21 4

I II

n

n

nm

m

1

0 1= +

3−

= =b l

1 2 344 44 1 2 344 44

/ / z z1 2

11

1 411

1 1=

−−

−− −

ROC : Summation I converges if 1 orz z< >21 1

21− and summation II converges

if 1 orz z4 < < 41 . ROC would be intersection of both which does not exist.


[ ]x n Z z

z162

2

−, ROC 4z <

[ 2]x n − Z 16

zz

zz 16

12

22

2−=

−−c m (Time shifting property)


[ ]x n Z z

z162

2

−, ROC 4z <

[ ]x n21n Z

( )( )z

zz

z2 16

242

2

2

2

−=

− (Scaling in z -domain)


[ ]x n Z z

z162

2

−, ROC 4z <

[ ]x n− Z ( ) 16

( )z

z1 2

1 2

− (Time reversal property)

[ ] [ ]x n x n− * Z ( ) 16

( )16z

z1 2

1 2

2

2

z

z

− −= ;G E (Time convolution property)

z z

z257 16 162 4

2Z

− −


[ ]x n Z z

z162

2

−, ROC 4z <

[ ]nx n Z zdzd

zz

162

2

−−

(Differentiation in z -domain)

Z ( )z

z16

322 2

2

−


[ ]x n Z z

z162

2

−, ROC 4z <

[ 1]x n + Z ( )zX z (Time shifting)

[ 1]x n − Z ( )z X z1− (Time shifting)

[ 1] [ 1]x n x n+ + − Z ( ) ( )z z X z1+ − (Linearity)


Z ( )

zz z z z

zz z

16 1612 2

2

1

2

2

−+

=−

+−^ ^h h


[ ]x n Z z

z162

2

−, ROC 4z <

[ ]x n 3− Z 16

zz

zz

z162

23

2

1

−=

−−

−

c m (Time shifting property)

[ ] [ 3]x n x n −* Z 16 16z

zz

z2

2

2

1

− −

−

c cm m (Time convolution property)

Z ( )z

z162 2−


We have, ( )X z Z 1−

3 [ ]n u nn 2

( )X z2 Z 1−

3 [ ]n u n21 2n

n" , (Scaling in z -domain)


( )X z Z 1−

3 [ ]n u nn 2

X z1

b l Z 1−

3 ( ) [ ]n u n( )n 2− −− (Time reversal)

Z 1−

3 [ ]n u nn 2 −−


( )X z Z 1−

3 [ ]n u nn 2

( )zdzd X z− Z 1−

[ ]nx n (Differentiation in z -domain)

( )z zdzd X z1 −−

: D Z 1−

( ) [ ]n x n1 1− − (Time shifting)

So, ( )

dzdX z

Z 1−

( 1) [ 1]n x n− − − ( )z zdzd X z1− −−

: D ( )

dzdX z=

Z 1−

( 1)3 ( 1) [ 1]n n u nn 1 2− − − −−

Z 1−

( 1) 3 [ 1]n u nn3 1− − −−


( )z X z21 2 Z 1−

[ 2]x n21 + (Time shifting)

( )z X z21 2− Z 1−

[ 2]x n21 − (Time shifting)

( )z z X z22 2− −

Z 1−

( [ 2] [ 2])x n x n21 + − − (Linearity)


( )X z Z 1−

3 [ ]n u nn 2

( ) ( )X z X z Z 1−

[ ] [ ]x n x n* (Time convolution)



( )X z 1 , ( ) 1z z Y z z4 8 4

31 2 1

= + − = −− − −

( )H z ( )( )

X zY z

z z1 121 135

41 132

= =+

+−−

− −

For a causal system impulse response is obtained by taking right-sided inverse z-transform of transfer function ( )H z . Therefore,

[ ]h n [ ]u n31 5 2

1 2 41n n

= − −b bl l; E

SOL 6.3.43 Option (D) is correct.We have [ ]x n ( 3) [ ]u nn= −

and [ ]y n ( ) [ ]u n4 2 21n

n= − b l; E

Taking z transform of above we get

( )X z z1 3

11=

+ −

and ( )Y z z z1 2

41

11

21 1=

−−

−− − ( )( )z z1 2 1

31

21 1=

− −− −

Thus transfer function is

( )H z ( )( )

X zY z

z z1 210

17

121 1= =

−+

−−

− −

For a causal system impulse response is obtained by taking right-sided inverse z-transform of transfer function ( )H z . Therefore,

[ ]h n ( ) [ ]u n10 2 7 21n

n= − b l; E

SOL 6.3.44 Option (D) is correct.We have [ ]h n ( ) [ ]u nn

21=

and [ ]y n 2 [ 4]nδ= −Taking z -transform of above we get

( )H z z1

121 1=

− −

and ( )Y z 2z 4= −

Now ( )X z ( )( )

2H zY z

z z4 5= = −− −

Taking inverse z -transform we have [ ]x n 2 [ 4] [ 5]n nδ δ= − − −

SOL 6.3.45 Option (C) is correct.We have, [ ]y n [ ] [ 2] [ 4] [ 6]x n x n x n x n= − − + − − −Taking z -transform we get ( )Y z ( ) ( ) ( ) ( )X z z X z z X z z X z2 4 6= − + −− − −

or ( )H z ( )( )

(1 )X zY z

z z z2 4 6= = − + −− − −


Taking inverse z -transform we have [ ]h n [ ] [ 2] [ 4] [ 6]n n n nδ δ δ δ= − − + − − −


We have [ ]h n [ 1]u n43

41 n 1

= −−

b l

Taking z -transform we get

( )H z ( )( )

X zY z

zz

1 41 1

43 1

= =− −

−

[ ]1

1u n zz

141 1

n41 1 1Z−

−− −

−^ ch mo

or, ( ) ( )Y z z Y z41 1− − ( )z X z4

3 1= −

Taking inverse z -transform we have

[ ] [ 1]y n y n41− − [ 1]x n4

3= −

SOL 6.3.47 Option (A) is correct.We have, [ ]h n [ ] [ 5]n nδ δ= − −Taking z -transform we get

( )H z ( )( )

1X zY z

z 5= = − −

or ( )Y z ( ) ( )X z z X z5= − −

Taking inverse z -transform we get [ ]y n [ ] [ 5]x n x n= − −

SOL 6.3.48 Option (A) is correct.Taking z transform of all system we get ( )Y z1 . ( ) ( ) . ( ) . ( )z Y z X z z X z z X z0 2 0 3 0 021 1 2= + − +− − −

( )H z1 ( )( )

.. .

X zY z

zz z

1 0 21 0 3 0 021

1

1 2

= =−

− +−

− −

( . )

( . )( . )( . )

zz z

z1 0 2

1 0 2 1 0 11 0 11

1 11=

−− − = −−

− −−

( )Y z2 ( ) . ( )X z z X z0 1 1= − −

( )H z2 ( )( )

(1 0.1 )X zY z

z2 1= = − −

( )Y z3 . ( ) . ( ) . ( )z Y z X z z X z0 5 0 4 0 31 1= + −− −

( )H z3 ( )( )

.. .

X zY z

zz

1 0 50 4 0 33

1

1

= =−−

−

−

( ) ( )H z H z1 2= , so y1 and y2 are equivalent.


We have ( )H z z z1 2

11

11

21 1=

−+

+− −

Poles of ( )H z are at z 2= and z 21−= . Since [ ]h n is stable, so ROC includes unit

circle 1z = and for the given function it must be 2z< <21 . The location of

poles and ROC is shown in the figure below


Consider the following two cases :1. ROC is interior to the circle passing through pole at z 2= , so this term

corresponds to a left-sided signal.

z1 2

11− − (2) [ 1]u nnZ 1

− − −−

(left-sided)

2. ROC is exterior to the circle passing through pole at z 21=− , so this term

corresponds to a right-sided signal.

z1

121 1+ − [ ]u n2

1 nZ 1 −−

b l (right-sided)

Impulse response,

[ ]h n (2) [ 1] [ ]u n u n21n

n=− − − + −

b l

SOL 6.3.50 Option (C) is correct.For anti causal signal initial value theorem is given as,

[0]x ( ) 4lim limX zz z

z3 7 12

12 21312

z z 2= =− +

− = =" "3 3


We have ( )H z ( )( )z z

zz z

z6

53 25

2

2 2

=− −

= − +

( )( )z z z z1 3 1 2

51 3

31 2

21 1 1 1=

− +=

−+

+− − − −

Since [ ]h n is causal, therefore impulse response is obtained by taking right-sided inverse z -transform of the transfer function ( )X z [ ]h n [3 2( 2) ] [ ]u nn n1= + −+


Zero at : 0z = , 32 , poles at z 2

1 2!=

(i) For a causal system all the poles of transfer function lies inside the unit circle 1z = . But, for the given system one of the pole does not lie inside the unit

circle, so the system is not causal and stable.

(ii) Not all poles and zero are inside unit circle 1z = , the system is not minimum


phase.


( )X z z z1 1 33

1 183

1827

=−−

+−− −

Poles are at z 31= and z 3= . Since ( )X z converges on 1z = , so ROC must include

this circle. Thus for the given signal ROC : 3z< <31

ROC is exterior to the circle passing through the pole at z 31= so this term will

have a right sided inverse z -transform. On the other hand ROC is interior to the circle passing through the pole at z 3= so this term will have a left sided inverse z -transform.

[ ]x n [ ] [ 1]u n u n3 8

18

3n

n

1

3

=− − − −−

+


Since ROC includes the unit circle 1z = , therefore the system is both stable and causal.

SOL 6.3.55 Option (C) is correct.(i) Pole of system ,z 2

131−= lies inside the unit circle z 1= , so the system is

causal and stable.

(ii) Zero of system ( )H z is z 21−= , therefore pole of the inverse system is at z 2

1−= which lies inside the unit circle, therefore the inverse system is also causal and stable.

SOL 6.3.56 Option (C) is correct.Taking z -transform on both sides ( )Y z ( ) . ( ) ( ) ( )cz Y z z Y z z X z z X z0 121 2 1 2= − + +− − − −

Transfer function,

( )( )( )

H zX zY z=

.cz zz z

1 0 121 2

1 2

=− +

+− −

− −

.z czz

0 121

2=− +

+


Poles of the system are .z c c2

0 482!= − .

For stability poles should lie inside the unit circle, so z 1<

.c c2

0 482! − 1<

Solving this inequality, we get 1.12c < .

SOL 6.3.57 Option (C) is correct.Writing the equation from given block diagram we have [2 ( ) ( )]Y z X z z 2+ − ( )Y z=

or ( )H z z

zz z1 2 2

11 2 1 22

2

141

141

=−

=− +−

++−

−

− −

Taking inverse laplace transform we have

[ ]h n [ ] {( ) ( ) } [ ]n u n21

41 2 2n nδ=− + + −

SOL 6.3.58 Option (D) is correct. ( )Y z ( ) { ( ) ( ) }X z z Y z z Y z z1 1 2= − +− − −

( )( )

X zY z

z zz

z zz

1 11 2

1

2=+ +

=+ +− −

−

So this is a solution but not unique. Many other correct diagrams can be drawn.

***********

SOLUTIONS 6.4

SOL 6.4.1 Option (C) is correct.z -transform of [ ]x n

( )X z [ ]x n z n

n=

3

3−

=−/ [ ]u n zn n

nα=3

3−

=−/

( )z n

n

1

0α=

3−

=/

z zz

11

1α α=−

= −−


We have [ ]x n [ ]n kk 0

δ= −3

=/

( )X z [ ]x n z n

k 0=

3−

=/ [ ]n k z n

kn 0δ= −

3

3

3−

==−; E//

Since [ ]n kδ − defined only for n k= so

( )X z z k

k 0=

3−

=/

( / )z1 11= −

( )z

z1

= −

SOL 6.4.3 Option (A) is correct.We have ( )f nT anT=Taking z -transform we get

( )F z a znT n

n=

3

3−

=−/ ( )a zT n n

n=

3

3−

=−/

zaT n

n 0=

3

=b l/

z az

T=−

SOL 6.4.4 Option ( ) is correct.

SOL 6.4.5 Option (A) is correct. [ ]x n [ ] [ ]b u n b u n 1n n= + − −−

z -transform of [ ]x n is given as

( )X z [ ]x n z n

n=

3

3−

=−/

[ ] [ ]b u n z b u n z1n n n n

nn= + − −

3

3

3

3− − −

=−=−//

b z b zn n n n

nn

1

0= +

3

3− − −

=−

−

=//

In second summation, Let n m=−

( )X z b z b zn n m m

mn 10= +

33−

==//


( ) ( )bz bz

I II

n

n

m

m

1

0 1= +

3 3−

= =1 2 344 44 1 2 344 4

/ /

Summation I converges, if bz 1<1− or z b>

Summation II converges, if bz 1< or z b1<

since b 1< so from the above two conditions ROC : z 1< .

SOL 6.4.6 Option (B) is correct.z -transform of signal [ ]a u nn is

( )X z [ ]a u n zn n

n=

3

3−

=−/ ( )az

I

n

n

1

0=

3−

=1 2 344 44

/ [ ] 1, 0u n n $=

az z a

z1

11=

−= −−

Similarly, z -transform of signal [ ]a u n 1n − − is

( )X z [ ]a u n z1n n

n= − − −

3

3−

=−/

a zn n

n

1

=−3

−

=−

−

/ a [ ]u n 1 1− − = , n 1#−Let n m=− , then

( )X z a zm m

m 1=−

3−

=/ ( )a z

II

m

m

1

1= −

3−

=1 2 3444 444

/

a z

a z1 1

1

=−− −

− z a

z= −

z -transform of both the signal is same.(A) is trueROC : To obtain ROC we find the condition for convergences of ( )X z for both the transform.Summation I converges if a z 1<1− or z a> , so ROC for [ ]a u nn is z a>Summation II converges if a z 1<1− or z a< , so ROC for [ ]a u n 1n− − − is z a< .

(R) is true, but (R) is NOT the correct explanation of (A).

SOL 6.4.7 Option (B) is correct.z - transform of [ ]x n is defined as

( )X z [ ]x n z n

n=

3

3−

=−/ [ ]x n r en j n

n=

3

3Ω− −

=−/ Putting z rej= Ω

z -transform exists if ( )X z < 3

[ ]x n r en j n

n 3

3Ω− −

=−/ < 3

or [ ]x n r n

n 3

3−

=−/ < 3

Thus, z -transform exists if [ ]x n r n− is absolutely summable.



[ ]x n [ ] [ 1]u n u n31

21n n

= − − −b bl l

Taking z transform we have

( )X z z z31

21

n

n nn

n

n

nn

0

1

= −3

3=

=−

=−

=−−

b bl l/ / z z31

21n

n

n n

n

n1

0

11

= −3

3

−

=

=−

=−

=−

b bl l/ /

First term gives

z31 1− z1

31< <"

Second term gives

z21 1− z1

21> >"

Thus its ROC is the common ROC of both terms. that is

z31

21< <


Here [ ]x n1 [ ]u n65 n

= ` j

( )X z1 z1

165 1=

− −^ h

ROC : R z65>1 "

[ ]x n2 [ 1]u n56 n

=− − −` j

( )X z1 1z1

156 1= −

− −^ h

ROC : R z56<2 "

Thus ROC of [ ] [ ]x n x n1 2+ is R R1 2+ which is z65

56< <

SOL 6.4.10 Option (A) is correct. [ ]x n [ ]u n2n=z -transform of [ ]x n

( )X z [ ]x n z n

n=

3

3−

=−/ [ ]u n z2n n

n=

3

3−

=−/

( )z2 n

n

1

0=

3−

=/ ( ) ...z z1 2 21 1 2= + + +− −

z1 2

11=

− −

the above series converges if z2 1<1− or z 2>

SOL 6.4.11 Option (A) is correct.We have [ ]h n [ ]u n=

( )H z [ ]x n z n

n

=3

3−

=−/ ( )z z1 n

n

n

n0

1

0

= =3 3

−

=

−

=/ /

( )H z is convergent if

( )z <n

n

1

03

3−

=/

and this is possible when 1z <1− . Thus ROC is 1z <1− or 1z >


SOL 6.4.12 Option (B) is correct.(Please refer to table 6.1 of the book Gate Guide signals & Systems by same authors)

(A) [ ]u n zz

1Z

− ( 3)A "

(B) [ ]nδ 1Z ( 1)B "

(C) sin tt nT

ω=

cos

sinz z T

z T2 12

Z

ωω

− + ( 4)C "

(D) cos tt nT

ω=

coscos

z z tz T2 12

Z

ωω

− +− ( 2)D "

SOL 6.4.13 Option (A) is correct.Inverse z -transform of ( )X z is given as

[ ]x n ( )j X z z dz21 n 1

π= −#

SOL 6.4.14 Option (B) is correct. ( )H z z m= −

so [ ]h n [ ]n mδ= −

SOL 6.4.15 Option (C) is correct.We know that

[ ]u nnα z1

11

Z

α− −

For 1α = , [ ]u n z1

11

Z

− −


( )X z .z1 2

0 51=

− −

Since ROC includes unit circle, it is left handed system [ ]x n (0.5)(2) [ 1]u nn=− − −−

( )x 0 0=If we apply initial value theorem

( )x 0 ( )limX zz

="3

. 0.5limz1 2

0 5z 1=

−=

"3 −

That is wrong because here initial value theorem is not applicable because signal [ ]x n is defined for n 0< .


z -transform ( )F z z 1

1=+

1z

z1

= −+

1z1

11= −

+ −

so, ( )f k ( ) ( 1)k kδ= − −

Thus ( 1)k− z1

11

Z

+ −



( )X z z zz z2

21

31

65

=− −

−

^ ^

^

h h

h

( )z

X z

z zz2

21

31

65

=− −

−

^ ^

^

h h

h

z z1 1

21

31=

−+

−^ ^h h Using partial fraction

or ( )X z z

zz

z

term I term II

21

31=

−+

−^ ^h hS S

...(i)

Poles of ( )X z are at z 21= and z 3

1=

ROC : z > 21 Since ROC is outside to the outer most pole so both the terms in

equation (i) corresponds to right sided sequence.

So, [ ]x n [ ] [ ]u n u n21

31n n

= +b bl l ( 4)A "

ROC : z < 31 :Since ROC is inside to the innermost pole so both the terms in

equation (i) corresponds to left sided signals.

So, [ ]x n [ ] [ ]u n u n21 1 3

1 1n n

=− − − − − −b bl l ( 2)D "


ROC : z< <31

21 : ROC is outside to the pole z 3

1= , so the second term of equation (i) corresponds to a causal signal. ROC is inside to the pole at z 2

1= , so First term of equation (i) corresponds to anticausal signal.

So, [ ]x n [ ] [ ]u n u n21 1 3

1n n=− − − +b bl l ( 1)C "

ROC : &z z< >31

21 : ROC : z < 3

1 is inside the pole at z 31= so second term

of equation (i) corresponds to anticausal signal. On the other hand, ROC : z > 21

is outside to the pole at z 21= , so the first term in equation (i) corresponds to a

causal signal.

So, [ ]x n [ ] [ ]u n u n21

31 1

n n= − − −b bl l ( 3)B "


Given, ( )X z ( )( )z z

z2 3

= − − , z 2<

( )z

X z

( )( )z z2 31= − − z z3

12

1= − − − By partial fraction

or, ( )X z zz

zz

3 2= − − − ...(i)


Poles of ( )X z are 2z = and z 3=ROC : z 2<

Since ROC is inside the innermost pole of ( )X z , both the terms in equation (i) corresponds to anticausal signals. [ ]x n 3 [ 1] 2 [ ]u n u n nn n=− − − + − − ( ) [ ]u n2 3 1n n= − − −


Given that ( )X z ( )z a

z2=

−, z a>

Residue of ( )X z zn 1− at z a= is

( ) ( )dzd z a X z zn

z a2 1= − −

=

( )( )dz

d z az a

z zn

z a

22

1= −−

−

=

dzd zn

z a=

= nzn

z a1= −

= nan 1= −


( )X z az bz1 1121

131

=−

+−− − , ROC : a z b< <

Poles of the system are z a= , z b=ROC : a z b< <


Since ROC is outside to the pole at z a= , therefore the first term in ( )X z corresponds to a causal signal.

az1 121

− − ( ) [ ]a u n21 nZ 1−

ROC is inside to the pole at z b= , so the second term in ( )X z corresponds to a anticausal signal.

bz1 131

− − ( ) [ ]b u n31 1nZ 1

− − −−

[ ]x n ( ) [ ] ( ) [ ]a u n b u n21

31 1n n= − − −

[ ]x 0 [ ] 3 [ ]u u21 0 1 1= − − 2

1=


( )X z ( )( )z zz z

1

3

=++

−

−

( )( )

z zz z

11

2

4

=++

−

−

( )( )z z1 14 2 1= + +− − −

Writing binomial expansion of ( )z1 2 1+ − − , we have ( )X z ( )( ....)z z z z1 14 2 4 6= + − + − +− − − −

1 2 2 ...z z z2 4 6= − + − +− − −

For a sequence [ ]x n , its z -transform is

( )X z [ ]x n z n

n=

3

3−

=−/

Comparing above two [ ]x n [ ] [ ] [ ] [ ] ...n n n n2 2 4 2 6δ δ δ δ= − − + − − − + , , , , , , , ....1 0 1 0 2 0 2= − −" ,[ ]x n has alternate zeros.


We know that Z aα ! [ ]n aZ 1

!αδ−

Given that ( )X z z z5 4 32 1= + +−

Inverse z -transform [ ]x n [ ] [ ] [ ]n n n5 2 4 1 3δ δ δ= + + − +

SOL 6.4.24 Option (A) is correct. ( )X z e /z1=

( )X z e /z1= 1 ....z z z1

21

31

2 3= + + + +

z -transform of [ ]x n is given by

( )X z [ ]x n z n

n 0=

3−

=/ [0]

[ ] [ ] [ ]....x z

xz

xz

x1 2 32 3= + + + +

Comparing above two

[ ], [ ], [ ], [ ] ....x x x x0 1 2 3" , , , , , ....111

21

31= ' 1

[ ]x n [ ]n

u n1=


SOL 6.4.25 Option (D) is correct.The ROC of addition or subtraction of two functions [ ]x n1 and [ ]x n2 is R R1 2+ . We have been given ROC of addition of two function and has been asked ROC of subtraction of two function. It will be same.

SOL 6.4.26 Option (D) is correct.(A) [ ]x n [ ]u nnα=

( )X z [ ]x n z n

n=

3

3−

=−/ zn n

n 0α=

3−

=/ [ ] 1, 0u n n $=

( )z n

n

1

0α=

3−

=/

z1

11α

=− − , z 1<1α − or z > α ( 2)A "

(B) [ ]x n [ ]u n 1nα=− − −

( )X z [ ]u n z1n n

nα=− − −3

3−

=−/

zn n

n

1

α=−3

−

=−

−

/ [ 1] 1u n− − = , 1n #−

Let n m=− , ( )X z zm m

m 1α=−

3−

=/ ( )z m

m

1

1α=−

3−

=/

,z

z1 1

1

αα=

−−

−

− z 1<1α− or z < α

( )z1

11α

=− − , z < α ( 3)B "

(C) [ ]x n [ ]n u n 1nα=− − −We have,

[ ]u n 1nα− − − ( )

,z

z1

1 <1Z

αα

− −

From the property of differentiation in z -domain

[ ]n u n 1nα− − − zdzd

z11

1Z

α−

− −: D, z < α

( )

,z

z1 1 2

1Z

αα

− −

− z < α ( 4)C "

(D) [ ]x n [ ]n u nnα=

We have, [ ]u nnα ( )z1

11

Z

α− − , z > α


[ ]n u nnα ( )

zdzd

z11

1Z

α−

− −; E, z > α

( )z

z1 1 2

1Z

αα

− −

−, z > α ( 1)D "


SOL 6.4.27 Option (C) is correct.We know that,

[ ]a u nn z azZ

− ( 3)A "

From time shifting property

[ ]a u n 2n 2 −− z z az2Z

−− ( 4)B "

From the property of scaling in z -domain

If, [ ]x n ( )X zZ

then, [ ]x nnα X zZ

αa k

so ( )e aj n n

ez a

ez

ze aze

j

j

j

jZ

−=

−−

−

a

a

k

k ( 2)C "


If, [ ]a u nn z azZ

−

then, [ ]na u nn ( )dz

dz a

zz a

az2

Z

− =−a k ( 1)D "

SOL 6.4.28 Given that, z transform of [ ]x n is

( )X z [ ]x n z n

n=

3

3−

=−/

z -transform of { [ ] }x n ej n0ω

( )Y z [ ]x n e zj n n

n

0=3

3ω −

=−/ [ ] ( )x n ze j n

n

0=3

3ω− −

=−/

( )X zz ze j 0

== ω−

ll

so, ( )Y z ( )X ze j 0= ω−


SOL 6.4.30 Option (C) is correct.The convolution of a signal [ ]x n with unit step function [ ]u n is given by

[ ]y n [ ] [ ] [ ]x n u n x kk 0

= =3

=* /

Taking z -transform

( )Y z ( )X zz1

11=

− −

SOL 6.4.31 Option (B) is correct.From the property of z -transform.

[ ] [ ]x n x n1 2* ( ) ( )X z X z1 2Z


SOL 6.4.32 Option (C) is correct.Given z transform

( )C z ( )( )

zz z4 1

11 2

1 4

=−

−−

− −

Applying final value theorem ( )lim f n

n"3 ( ) ( )lim z f z1

z 1= −

"

( ) ( )lim z F z1z 1

−"

( )( )( )

lim zz

z z1

4 11

z 1 1 2

1 4

= −−

−"

−

− −

( )

( )( )lim

zz z z

4 11 1

z 1 1 2

1 4

=−

− −"

−

− −

( )

( )( )lim

z zz z z z

4 11 1

z 1 2 2

1 4 4

=−

− −"

−

− −

( )

( )( )( )( )lim z

zz z z z

4 11 1 1 1

z 1

3

2

2

=−

− + + −"

−

( 1)( 1)lim z z z4 1z 1

32= + + =

"

−

SOL 6.4.33 Option (C) is correct. ( )H z1 . z z1 1 5 1 2= + −− −

z z1 2

3 12= + −

zz z

22 3 2

2

2

= + −

Poles z 02 = z 0& =

zeros ( )z z2 3 22 + − ( )z z0 21 2 0&= − + =b l ,z z2

1 2& = =−

zeros of the two systems are identical.

SOL 6.4.34 Option (D) is correct.Taking z -transform on both sides of given equation. ( ) ( ) ( ) ( )z Y z z Y z zY z Y z6 11 63 2+ + + ( ) ( ) ( )z R z zR z R z9 202= + +Transfer function

( )( )

R zY z

z z z

z z6 11 6

9 203 2

2

=+ + +

+ +

SOL 6.4.35 Option (A) is correct.Characteristic equation of the system zI A− 0=

zI A− z

zz

z00 0 1 1

β α β α= − − =−+> > >H H H

zI A− ( )z z 0α β= + + = z z2 α β+ + 0=In the given options, only option (A) satisfies this characteristic equation. [ 2] [ 1] [ ]c k c k c kα β+ + + + [ ]u k= z z2 α β+ + 0=


SOL 6.4.36 Option (B) is correct.We can see that the given impulse response is decaying exponential, i.e. [ ]h n [ ]a u nn= , a0 1< <z -transform of [ ]h n

( )H z z az= −

Pole of the transfer function is at z a= , which is on real axis between a0 1< < .

SOL 6.4.37 Option (A) is correct. [ ] [ ]y n y n 1+ − [ ] [ ]x n x n 1= − −Taking z -transform ( ) ( )Y z z Y z1+ − ( ) ( )X z z X z1= − −

( )Y z ( )( )

zz

11

1

1

=+−

−

−

which has a linear phase response.

SOL 6.4.38 Option (A) is correct.For the linear phase response output is the delayed version of input multiplied by a constant. [ ]y n [ ]kx n n0= −

( )Y z ( )( )

kz X zz

kX znn

0

0= =−

Pole lies at z 0=

SOL 6.4.39 Option (B) is correct.Given impulse response can be expressed in mathematical form as [ ]h n [ ] [ ] [ ] [ ] ....n n n n1 2 3δ δ δ δ= − − + − − − +By taking z -transform ( )H z ....z z z z z1 1 2 3 4 5= − + − + − +− − − − −

( ....) ( ....)z z z z z1 2 4 1 3 5= + + + − + + +− − − − −

z z

z1

112 2

1

=−

−−− −

−

zz

zz

1 12

2

2=−

−−

( )( )

( )( )( )

zz z

z zz z

1 1 11

2

2

=−− = − +

− z

z1= +

Pole at z 1=−

SOL 6.4.40 Option (B) is correct.Let Impulse response of system [ ] ( )h n H zZ

First consider the case when input is unit step.

Input, [ ]x n1 [ ]u n= or ( )( )

X zz

z11 = −

Output, [ ]y n1 [ ]nδ= or ( )Y z 11 =so, ( )Y z1 ( ) ( )X z H z1=

1 ( )

( )z

z H z1

= −

Transfer function, ( )H z ( )

zz 1= −


Now input is ramp function [ ]x n2 [ ]nu n=

( )X z2 ( )z

z1 2=

−Output, ( )Y z2 ( ) ( )X z H z2=

( 1) ( )

( 1)z

zz

z2=

−−

: =D G ( )z 11= −

( )Y z2 [ ]y n2Z 1−

( )z 1

1− [ ]u n 1Z 1

−−

SOL 6.4.41 Option (C) is correct.Given state equations [ ]s n 1+ [ ] [ ]As n Bx n= + ...(i) [ ]y n [ ] [ ]Cs n Dx n= + ...(ii)Taking z -transform of equation (i) ( )zS z ( ) ( )AS z BX z= + ( )S z zI A−6 @ ( )BX z= unit matrixI " ( )S z ( ) ( )zI A BX z1= − − ...(iii)Now, taking z -transform of equation (ii) ( )Y z ( ) ( )CS z DX z= +Substituting ( )S z from equation (iii), we get ( )Y z ( ) ( ) ( )C zI A BX z DX z1= − +−

Transfer function

( )( )( )

H zX zY z= ( )C zI A B D1= − +−

SOL 6.4.42 Option (B) is correct. ( )F z z z z4 8 23 2= − − + ( )F z ( ) ( )z z z z4 2 22= − − − ( )( )z z4 2 22= − − 4 2z2 − 0= and ( )z 2 0− =

z 21

!= and z 2=

Only one root lies outside the unit circle.

SOL 6.4.43 Option (A) is correct.We know that convolution of [ ]x n with unit step function [ ]u n is given by

[ ] [ ]x n u n* [ ]x kk

n

=3=−

/

so [ ]y n [ ] * [ ]x n u n=Taking z -transform on both sides

( )Y z ( )( )

X zz

z1

= − ( )( )

X zz1

11=

− −


Transfer function,

( )H z ( )( )

( )X zY z

z11

1= =− −

Now consider the inverse system of ( )H z , let impulse response of the inverse system is given by ( )H z1 , then we can write ( ) ( )H z H z1 1=

( )H z1 ( )( )

1Y zX z

z 1= = − −

(1 ) ( )z Y z1− − ( )X z= ( ) ( )Y z z Y z1− − ( )X z=Taking inverse z -transform [ ] [ 1]y n y n− − [ ]x n=


[ ] [ ]y n y n21 1− − [ ]x n=

Taking z -transform on both sides

( ) ( )Y z z Y z21 1− − ( )X z=

Transfer function

( )( )( )

H zX zY z=

z1121 1=

− −

Now, for input [ ] [ ]x n k nδ= Output is ( )Y z ( ) ( )H z X z=

z

k1 2

1 1=− −

^ h ( )X z k=


[ ]y n [ ]k u n21 n

= b l k 21 n

= b l , n 0$

SOL 6.4.45 Option (A) is correct. [ ] [ ]y n y n 1+ − [ ]x n=For unit step response, [ ] [ ]x n u n= [ ] [ ]y n y n 1+ − [ ]u n=Taking z -transform

( ) ( )Y z z Y z1+ − zz

1= −

(1 ) ( )z Y z1+ − ( )z

z1

= −

( )

( )zz

Y z1 +

( )z

z1

= −

( )Y z ( )( )z z

z1 1

2

= + −




SOL 6.4.48 Option (A) is correct.We have (2) 1, (3) 1h h= =− otherwise [ ] 0h k = . The diagram of response is as follows :

It has the finite magnitude values. So it is a finite impulse response filter. Thus S2 is true but it is not a low pass filter. So S1 is false.


( )H z .z

z0 2

=−

.z 0 2<We know that

[ 1]a u nn− − − az11

1*− − z a<

Thus [ ]h n (0.2) [ 1]u nn=− − −

SOL 6.4.50 Option (B) is correct.We have [ ]h n 3 [ 3]nδ= −or ( )H z 2z 3= − Taking z transform ( )X z 2 2 3z z z z4 2 4= + − + − −

Now ( )Y z ( ) ( )H z X z= 2 ( 2 2 3 )z z z z z3 4 2 4= + − + −− −

2( 2 2 3 )z z z z z1 2 3 7= + − + −− − − −

Taking inverse z transform we have [ ]y n 2[ [ 1] [ 1] 2 [ 2] 2 [ 3] 3 [ 7]]n n n n nδ δ δ δ δ= + + − − − + − − −At n 4= , [4]y 0=

SOL 6.4.51 Option (A) is correct.z -transform of [ ]x n is ( )X z z z z z4 3 2 6 23 1 2 3= + + − +- -

Transfer function of the system ( )H z z3 21= −-

Output ( )Y z ( ) ( )H z X z= (3 2)(4 3 2 6 2 )z z z z z1 3 1 2 3= − + + − +− − −

12 9 6 18 6 8 6 4 12 4z z z z z z z z z4 2 1 2 3 1 2 3= + + − + − − − + −− − − − −

12 8 9 4 18 18 4z z z z z z4 3 2 2 3= − + − − + −− − −


Or sequence [ ]y n is[ ] 12 [ 4] 8 [ 3] 9 [ 2] 4 [ ] 18 [ 1] 18 [ 2] 4 [ 3]y n n n n n n n nδ δ δ δ δ δ δ= − − − + − − − + + + − +

[ ]y n 0=Y , 0n <So [ ]y n is non-causal with finite support.

SOL 6.4.52 Option ( ) is correct.

SOL 6.4.53 Option (C) is correct.Impulse response of given LTI system. [ ]h n [ ] [ ]x n y n1 )= −Taking z -transform on both sides.

( )H z ( ) ( )z X z Y z1= − [ 1] ( )x n z x z1Z− −

We have ( ) 1 3 ( ) 1 2andX z z Y z z1 2= − = +− −

So ( )H z (1 3 )(1 2 )z z z1 1 2= − +− − −

Output of the system for input [ ] [ 1] isu n nδ= − ,

( )y z ( ) ( )H z U z= [ ] ( )U n U z z 1Z = −

So ( )Y z (1 3 )(1 2 )z z z z1 1 2 1= − +− − − −

(1 3 2 6 )z z z z2 1 2 3= − + −− − − −

3 2 6z z z z2 3 4 5= − + −− − − −

Taking inverse z-transform on both sides we have output. [ ]y n [ ] [ ] [ ] [ ]n n n n2 3 3 2 4 6 5δ δ δ δ= − − − + − − −

SOL 6.4.54 Option (D) is correct. ( )H z ( )az1 1= − −

We have to obtain inverse system of ( )H z . Let inverse system has response ( )H z1 . ( ) ( )H z H z1 1=

( )H z1 ( )H z1=

az11

1=− −

For stability ( ) ( ),H z az z a1 >1= − − but in the inverse system z a< , for stability of ( )H z1 .so [ ]h n1 [ ]a u n 1n=− − −


( )H z z

z21=

+

Pole, z 21=−

The system is stable if pole lies inside the unit circle. Thus (A) is true, (R) is false.

SOL 6.4.56 Option (A) is correct.Difference equation of the system. [ ] [ ] [ ]y n y n y n2 5 1 6+ − + + [ ]x n=Taking z -transform on both sides of above equation. ( ) 5 ( ) 6 ( )z Y z zY z Y z2 − + ( )X z= ( 5 6) ( )z z y z2 − + ( )X z=


Transfer function,

( )( )( )

H zX zY z=

( )z z5 61

2=− +

( 3)( )z z 2

1= − −Roots of the characteristic equation are 2z = and 3z =We know that an LTI system is unstable if poles of its transfer function (roots of characteristic equation) lies outside the unit circle. Since, for the given system the roots of characteristic equation lies outside the unit circle ( , )z z2 3= = so the system is unstable.


System function, ( )H z ( . )( . )z z

z0 5 0 5

12

= + −+

Poles of the system lies at 0.5, 0.5z z= =− . Since, poles are within the unit circle, therefore the system is stable.From the initial value theorem

[ ]h 0 ( )limH zz

="3

( . )( . )

( )lim

z zz

0 5 0 51

z

2

= + −+

"3

. .lim

z z

z

1 0 5 1 0 5

1 1

z

2

=+ −

+

"3b b

b

l l

l 1=

SOL 6.4.58 Option (D) is correct. [ ]y n [ ] [ ]x n x n2 4 1= + −Taking z -transform on both sides ( )Y z ( ) ( )X z z X z2 4 1= + −

Transfer Function,

( )H z ( )( )

X zY z

z zz2 4 2 41= = + = +−

Pole of ( )H z , z 0=Since Pole of ( )H z lies inside the unit circle so the system is stable.(A) is not True. ( )H z z2 4 1= + −

Taking inverse z -transform [ ]h n [ ] [ ]n n2 4 1δ δ= + − ,2 4= " ,Impulse response has finite number of non-zero samples.(R) is true.

SOL 6.4.59 Option (B) is correct.For left sided sequence we have

[ 1]a u nn− − − az11

1Z

− − where z a<

Thus 5 [ 1]u nn− − − z1 5

11

Z

− − where z 5<

or 5 [ 1]u nn− − − z

z5

Z

− where z 5<


Since ROC is z 5< and it include unit circle, system is stable.Alternative : [ ]h n 5 [ 1]u nn=− − −

( )H z [ ]h n z n

n

=3

3−

=−/ z5n n

n

1

= −3

−

=−

−

/ ( )z5 n

n

11

=−3

−

=−

−

/

Let ,n m=− then

( )H z ( )z5 m

n

1

1

=−3

− −

=−

−

/ 1 ( )z5 m

m

1

0

= −3

− −

=/

1 ,z1 5

11= −

− − 1z5 <1− or z 5<

z z

z15

55

= −−

=−

SOL 6.4.60 Option (B) is correct.For a system to be stable poles of its transfer function ( )H z must lie inside the unit circle. In inverse system poles will appear as zeros, so zeros must be inside the unit circle.

SOL 6.4.61 Option (C) is correct.(Please refer to the section 6.7, page 439 of the book GATE GUIDE Signals & Systems by the same authors.)An LTI discrete system is said to be BIBO stable if its impulse response [ ]h n is summable, that is

[ ]h nn 3

3

=−/ < 3

z -transform of [ ]h n is given as

( )H z [ ]h n z n

n=

3

3−

=−/

Let z ej= Ω (which describes a unit circle in the z -plane), then

( )H z [ ]h n e j n

n=

3

3Ω−

=−/

[ ]h n e j n

n=

3

3Ω−

=−/

[ ]h n <n

3=3

3

=−/

which is the condition of stability. So LTI system is stable if ROC of its system function includes the unit circle z 1= .(A) is true.We know that for a causal system, the ROC is outside the outermost pole. For the system to be stable ROC should include the unit circle z 1= . Thus, for a system to be causal & stable these two conditions are satisfied if all the poles are within the unit circle in z -plane.(R) is false.


SOL 6.4.62 Option (B) is correct.We know that for a causal system, the ROC is outside the outermost pole. For the system to be stable ROC should include the unit circle z 1= . Thus, for a system to be causal & stable these two conditions are satisfied if all the poles are within the unit circle in z -plane.(A) is true.If the z -transform ( )X z of [ ]x n is rational then its ROC is bounded by poles because at poles ( )X z tends to infinity.(R) is true but (R) is not correct explanation of (A).


We have ( )H z 1

2z z

z

43 1

81 2

43 1

=− +

−− −

−

1 1z z

1 121 1

41 1=

−+

−− −^ ^h h

By partial fraction

For ROC : 1/2z >

[ ] [ ] [ ], 0 [ ],h n u n u n nz

a u n z a21

41

11> >

n nn

1= +−

=−b bl l

Thus system is causal. Since ROC of ( )H z includes unit circle, so it is stable also. Hence S1 is True

For ROC : z 41<

[ ] [ 1] [ ], ,h n u n u n z z21

41

41

21> <

n n=− − − +b bl l

System is not causal. ROC of ( )H z does not include unity circle, so it is not stable and S3 is True.

SOL 6.4.64 Option (C) is correct.We have 2 [ ]y n [ 2] 2 [ ] [ 1]y n x n x nα β= − − + −Taking z transform we get ( )Y z2 ( ) 2 ( ) ( )Y z z X z X z z2 1α β= − +− −

or ( )( )

X zY z

z

z2

22

1

αβ=

−−

−

−

c m ...(i)

or ( )H z ( )( )z

z z2

2

2=−−

α

β

It has poles at /2! α and zero at 0 and /2β . For a stable system poles must lie inside the unit circle of z plane. Thus

2α 1<

or α 2<But zero can lie anywhere in plane. Thus, β can be of any value.

SOL 6.4.65 Option (D) is correct.Let ( )H z1 and ( )H z2 are the transfer functions of systems s1 and s2 respectively. For the second order system, transfer function has the following form


( )H z1 az bz c2 1= + +− −

( )H z2 pz qz r2 1= + +− −

Transfer function of the cascaded system ( )H z ( ) ( )H z H z1 2= ( )( )az bz c pz qz r2 1 2 1= + + + +− − − −

( ) ( ) ( )apz aq bp z ar cp z br qc z cr4 3 2 1= + + + + + + +− − − −

So, impulse response [ ]h n will be of order 4.

SOL 6.4.66 Option (B) is correct.Output is equal to input with a delay of two units, that is ( )y t ( )x t 2= − ( )Y z ( )z X z2= −

Transfer function,

( )( )( )

H zX zY z= z 2= −

For the cascaded system, transfer function ( )H z ( ) ( )H z H z1 2=

z 2− ( . )( . )

( )zz

H z0 80 5

2= −−

( )H z2 ..

zz z

0 50 81 2

= −−− −

..z

z z1 0 5

0 81

2 3

=−−

−

− −

SOL 6.4.67 Option (B) is correct. [ ]y n [ 1]x n= −or ( )Y z ( )z X z1= −

or ( )( )

( )X zY z

H z= z 1= −

Now ( ) ( )H z H z1 2 z 1= −

.. ( )

zz H z

1 0 61 0 4

1

1

2−−

−

−

c m z 1= −

( )H z2 ( . )( . )

zz z

1 0 41 0 6

1

1 1

=−

−−

− −

SOL 6.4.68 Option (C) is correct.We have [ ]h n1 [ ] [ ]n or H Z Z1 1

1δ= − = −

and [ ]h n2 [ ] ( )n or H Z Z2 22δ= − = −

Response of cascaded system ( )H z ( ) ( )H z H z1 2:= z z z1 2 3

:= =− − −

or, [ ]h n [ ]n 3δ= −

SOL 6.4.69 Option (B) is correct.Let three LTI systems having response ( ), ( )H z H z1 2 and ( )H z3 areCascaded as showing below


Assume ( )H z1 1z z2 1= + + (non-causal) ( )H z2 1z z3 2= + + (non-causal)Overall response of the system ( )H z ( ) ( ) ( )H z H z H z1 2 3= ( )( ) ( )z z z z H z1 12 1 3 2

3= + + + +To make ( )H z causal we have to take ( )H z3 also causal.Let ( )H z3 1z z6 4= + +− −

( )H z ( 1)( 1)( 1)z z z z z z2 1 3 2 6 4= + + + + + +− −

( )H z causal"

Similarly to make ( )H z unstable atleast one of the system should be unstable.


( )H z cz dz ez

az bz1

11 2 3

1 2

=+ + +

+ +− − −

− −

We know that number of minimum delay elements is equal to the highest power of z 1− present in the denominator of ( )H z . No. of delay elements 3=

SOL 6.4.71 Option (A) is correct.From the given system realization, we can write ( ) ( ) ( )X z Y z z a Y z a z a2

2 11

0#+ +− −^ h ( )Y z=

System Function

( )( )( )

H zX zY z=

a z a za

1 11

22

0=− −− −

a aa z a

a z11

0 0

1 1

0

2 2=

− −− −

Comparing with given ( )H z

a10 1 1a0&= =

aa

0

1− . .a0 7 0 71&=− =

aa

0

2− 0.13 0.1a 32&= =−

SOL 6.4.72 Option (B) is correct.Let, M " highest power of z 1− in numerator. N " highest power of z 1− in denominatorNumber of delay elements in direct form-I realization equals to M N+Number of delay elements in direct form-II realization equal to N .Here, M 3= , N 3= so delay element in direct form-I realization will be 6 and in direct form realization will be 3.


SOL 6.4.73 Option (A) is correct.System response is given as

( )H z ( )

( )KG z

G z1

=−

[ ]g n [ ] [ ]n n1 2δ δ= − + − ( )G z z z1 2= +− −

So ( )H z ( )

( )K z zz z

1 1 2

1 2

=− +

+− −

− −

z Kz K

z 12=− −

+

For system to be stable poles should lie inside unit circle. z 1#

z 1K K K2

42! #= +

2K K K42! #+ 4K K2 + 2 K# − 4K K2 + 4 4K K2# − + 8K 4#

K 1/2#

SOL 6.4.74 Option (B) is correct.Input-output relationship of the system [ ]y n [ ] [ ]x n ay n 1= + −Taking z -transform ( )Y z ( ) ( )X z z aY z1= + −

Transform Function,

( )( )

X zY z

z a11

1=− −

Pole of the system ( )z a z a1 01 &− = =−

For stability poles should lie inside the unit circle z 1< so a 1< .

SOL 6.4.75 Option (D) is correct.The relation ship between Laplace transform and z -transform is given as ( )X s ( )X z

z est==

z esT= ...(i)We know that z rej= Ω ...(ii)and s jσ ω= + (iii)From equation (i), (ii) and (iii), we can write z re e( )j j T= = σ ωΩ + e eT j T= σ ω

From above relation we can find that z e T= σ and TωΩ = . It is concluded that,• If 0σ = then z 1= , the jω-axis of s -plane maps into unit circle.• If 0<σ , z 1< , it implies that left half of s -plane maps into inside of unit

circle z 1<_ i.• Similarly, if 0>σ , z 1> which implies that right half of s -plane maps into

outside of unit circle.


SOL 6.4.76 Option (D) is correct.The relation ship between Laplace transform and z -transform is given as ( )X s ( )X z

z est==

z esT= ...(i)We know that z rej= Ω ...(ii)and s jσ ω= + (iii)From equation (i), (ii) and (iii), we can write z re e( )j j T= = σ ωΩ +

e eT j T= σ ω

z re e ej T j T= = σ ωΩ

From above relation we can find that z e T= σ and TωΩ = . It is concluded that,• If 0σ = then z 1= , the jω-axis of s -plane maps into unit circle.• If 0<σ , z 1< , it implies that left half of s -plane maps into inside of unit

circle z 1<_ i.• Similarly, if 0>σ , z 1> which implies that right half of s -plane maps into

outside of unit circle.

SOL 6.4.77 Option (C) is correct.Ideal sampler output is given by

( )f t [ ]K t nTn sn 0

δ= −3

=/

where Ts " sampling period n " integer ( )f t [ ] [ 1] [ 2] ...K n K n K n0 1 2δ δ δ= + − + − + [ ( )]f tZ ...K K z K z K zn

n0 1

12

2= + + + +− − −

SOL 6.4.78 Option (B) is correct.We know that ( )X s ( )X z

z esT=

=so, z esT= ln z sT=

s lnT

z=



( )H s s a

a2 2=+

Poles in s -domain are at s ja!= . In z -domain poles will be at z esT= , so

z1 e jaT= − and z ejaT2 =

***********

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