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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
AA1213
Information contained in this book has been obtained by author, from sources believes to be reliable.
However, neither Jhunjhunuwala nor its author guarantee the accuracy or completeness of any
information herein, and Jhunjhunuwala nor its author shall be responsible for any error, omissions,
or damages arising out of use of this information. This book is published with the understanding that
Jhunjhunuwala and its author are supplying information but are not attempting to render engineering
or other professional services.
Copyright by Jhunjhunuwala�
JHUNJHUNUWALAB-8, Dhanshree Tower Ist, Central Spine, Vidyadhar Nagar, Jaipur – 302023
Printed By: Jaipur Printing Centre, Jaipur
Ph : +91 141 01150.� ���
www.jhunjhunuwala.com
email : [email protected]
Preface to First Edition
Authors
GATE Question Cloud caters a versatile collection of Multiple Choice Questions to the students who are
preparing for GATE(Gratitude Aptitude Test in Engineering) examination. This book contains over 1500
multiple choice solved problems for the subject of Signals & Systems, which has a significant weightage in
the GATE examinations of EC, EE & IN branches.
which leads to some improvement.
Wish you all the success in conquering GATE.
The GATE examination is based on multiple choice
problems which are tricky, conceptual and tests the basic understanding of the subject. So, the problems
included in the book are designed to be as exam-like as possible. The solutions are presented using step by
step methodology which enhance your problem solving skills.
The book is categorized into eleven chapters covering all the topics of syllabus of the examination. Each
chapter contains :
Exercise 1 :
Exercise 2 :
Exercise 3 :
Exercise 4 :
Detailed Solutions to Exercise 2, 3 & 4
Summary of useful theorems
Although we have put a vigorous effort in preparing this book, some errors may have crept in. We shall
appreciate and greatly acknowledge the comments, criticism and suggestion from the users of this book
�
�
�
�
�
�
Theoretical & One line Questions
Level 1
Level 2
Mixed Questions taken form previous examinations of GATE & IES.
DIGITAL ELECTRONICSR. K . Kanodia & Ashish Murolia
GATE CLOUD
GATE CLOUD is an exclusive series of books which offers a completely solved question bank
to GATE aspirants. The book of this series are featured as
�
�
�
�
Over 1300 Multiple Choice Questions with full & detailed explanations.
Questions are graded in the order of complexity from basic to advanced level.
Contains all previous year GATE and IES exam questions from various
branches.
Each question is designed to GATE exam level.
� Circuit Analysis
Analog Circuit and Devices
(For EC, EE & IN branches)
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(For EC, EE & IN branches)
(For EC, EE & IN branches)Control Systems
Upcoming titles in this series
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< <
Chapter 6 The Z Transform Page 487
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 .
Page 488 The Z Transform Chapter 6
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 +
Chapter 6 The Z Transform Page 489
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 −−
Page 490 The Z Transform Chapter 6
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−
Chapter 6 The Z Transform Page 491
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
Page 492 The Z Transform Chapter 6
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
Chapter 6 The Z Transform Page 493
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
Page 494 The Z Transform Chapter 6
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−
Chapter 6 The Z Transform Page 495
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= −
Page 496 The Z Transform Chapter 6
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 < −
Chapter 6 The Z Transform Page 497
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<−
Chapter 6 The Z Transform Page 499
(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− − −− −
Page 500 The Z Transform Chapter 6
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−
Chapter 6 The Z Transform Page 501
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
Page 502 The Z Transform Chapter 6
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− − − −
Chapter 6 The Z Transform Page 503
(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
−
Page 504 The Z Transform Chapter 6
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
Chapter 6 The Z Transform Page 505
(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= − +
Page 506 The Z Transform Chapter 6
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
Chapter 6 The Z Transform Page 507
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
(C) A is true but R is false
(D) A is false but R is true
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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Chapter 6 The Z Transform Page 519
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
(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 6.4.59 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.
(C) A is true but R is false
(D) A is false but R is true
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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Page 520 The Z Transform Chapter 6
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
(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 6.4.63 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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Chapter 6 The Z Transform Page 521
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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Chapter 6 The Z Transform Page 523
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
(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 6.4.78 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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Page 524 The Z Transform Chapter 6
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/ /
Chapter 6 The Z Transform Page 527
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
Page 528 The Z Transform Chapter 6
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
= − =−
−−
−
SOL 6.2.8 Option (A) is correct.
( )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 "
Chapter 6 The Z Transform Page 529
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= +
SOL 6.2.12 Option (A) is correct.
( )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=−
Page 530 The Z Transform Chapter 6
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= − + − − −
SOL 6.2.13 Option (A) is correct.
( )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=− − − − −
Chapter 6 The Z Transform Page 531
SOL 6.2.15 Option (D) is correct.
( )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.
SOL 6.2.16 Option (D) is correct.
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)
Page 532 The Z Transform Chapter 6
SOL 6.2.20 Option (B) is correct.
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= −
SOL 6.2.23 Option (B) is correct.
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
−
Chapter 6 The Z Transform Page 533
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=
Page 534 The Z Transform Chapter 6
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
=− −
−
SOL 6.2.26 Option (C) is correct.
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
Chapter 6 The Z Transform Page 535
SOL 6.2.28 Option (A) is correct.
( )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
If, [ ]x n ( ), :ROCX z RxZ
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
If, [ ]x n ( ), :ROCX z RxZ
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= +
Page 536 The Z Transform Chapter 6
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 "
SOL 6.2.31 Option (B) is correct.
( )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
SOL 6.2.33 Option (A) is correct.
( )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
Chapter 6 The Z Transform Page 537
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)
Page 538 The Z Transform Chapter 6
{ [ ]}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$
Chapter 6 The Z Transform Page 539
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=
SOL 6.2.42 Option (B) is correct.
[ ]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= −
Taking inverse z -transform
[ ] [ ]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= = −
Page 540 The Z Transform Chapter 6
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=−
Chapter 6 The Z Transform Page 541
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=
Page 542 The Z Transform Chapter 6
[ ]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=
Chapter 6 The Z Transform Page 543
.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.
SOL 6.2.53 Option (D) is correct.
( )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
Page 544 The Z Transform Chapter 6
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−
***********
Chapter 6 The Z Transform Page 545
SOLUTIONS 6.3
SOL 6.3.1 Option (A) is correct.
( )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
=−/
SOL 6.3.2 Option (D) is correct.
( )X z+ [ ]x n z zz1
1n
n
n
n0 01= = =
−
3 3−
=
−
=−/ /
SOL 6.3.3 Option (D) is correct.
( )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.
SOL 6.3.4 Option (D) is correct.
( )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 .
SOL 6.3.5 Option (A) is correct.
( )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
−
=/ .
SOL 6.3.6 Option (D) is correct.
( )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
Page 546 The Z Transform Chapter 6
ROC : Summation I converges for all values of z because n has only four value.
SOL 6.3.7 Option (D) is correct.
( )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 .
SOL 6.3.8 Option (B) is correct.
( )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
SOL 6.3.9 Option (B) is correct.
( )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
SOL 6.3.10 Option (B) is correct.
[ ]x n [ ]cos n u n3π= a k [ ]e e u n2
j n j n3 3
= + −π π^ ^h h
Chapter 6 The Z Transform Page 547
[ ] [ ]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> .
SOL 6.3.11 Option (B) is correct.
[ ]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.
SOL 6.3.12 Option (D) is correct.
[ ]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.
SOL 6.3.13 Option (B) is correct.
[ ]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= .
SOL 6.3.14 Option (A) is correct.
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.
Page 548 The Z Transform Chapter 6
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
Chapter 6 The Z Transform Page 549
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 −−−
Page 550 The Z Transform Chapter 6
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= + + + + + − + −
SOL 6.3.21 Option (A) is correct.
( )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−
Taking inverse z -transform we get
[ ]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 $=−
*
SOL 6.3.22 Option (C) is correct.
( )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 + ++−
Chapter 6 The Z Transform Page 551
SOL 6.3.23 Option (A) is correct.
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 −
=
−/
Taking inverse z -transform we get
[ ]x n ( )
[ ]k n k1 k
k
1
1δ= − −
3 −
=/ [ ]z n kk Z 1
δ −−−
SOL 6.3.24 Option (B) is correct.
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 >
Taking inverse z -transform we get
[ ]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
= − =
SOL 6.3.25 Option (D) is correct.
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.
Page 552 The Z Transform Chapter 6
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
Chapter 6 The Z Transform Page 553
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
SOL 6.3.27 Option (B) is correct.
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
Page 554 The Z Transform Chapter 6
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
Chapter 6 The Z Transform Page 555
SOL 6.3.30 Option (D) is correct.
( )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.
SOL 6.3.31 Option (C) is correct.
[ ]x n Z z
z162
2
−, ROC 4z <
[ 2]x n − Z 16
zz
zz 16
12
22
2−=
−−c m (Time shifting property)
SOL 6.3.32 Option (B) is correct.
[ ]x n Z z
z162
2
−, ROC 4z <
[ ]x n21n Z
( )( )z
zz
z2 16
242
2
2
2
−=
− (Scaling in z -domain)
SOL 6.3.33 Option (C) is correct.
[ ]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
− −
SOL 6.3.34 Option (A) is correct.
[ ]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
−
SOL 6.3.35 Option (B) is correct.
[ ]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)
Page 556 The Z Transform Chapter 6
Z ( )
zz z z z
zz z
16 1612 2
2
1
2
2
−+
=−
+−^ ^h h
SOL 6.3.36 Option (D) is correct.
[ ]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−
SOL 6.3.37 Option (C) is correct.
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)
SOL 6.3.38 Option (B) is correct.
( )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 −−
SOL 6.3.39 Option (C) is correct.
( )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− − −−
SOL 6.3.40 Option (A) is correct.
( )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)
SOL 6.3.41 Option (B) is correct.
( )X z Z 1−
3 [ ]n u nn 2
( ) ( )X z X z Z 1−
[ ] [ ]x n x n* (Time convolution)
Chapter 6 The Z Transform Page 557
SOL 6.3.42 Option (A) is correct.
( )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= = − + −− − −
Page 558 The Z Transform Chapter 6
Taking inverse z -transform we have [ ]h n [ ] [ 2] [ 4] [ 6]n n n nδ δ δ δ= − − + − − −
SOL 6.3.46 Option (A) is correct.
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.
SOL 6.3.49 Option (B) is correct.
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
Chapter 6 The Z Transform Page 559
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
SOL 6.3.51 Option (B) is correct.
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= + −+
SOL 6.3.52 Option (D) is correct.
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
Page 560 The Z Transform Chapter 6
phase.
SOL 6.3.53 Option (A) is correct.
( )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
=− − − −−
+
SOL 6.3.54 Option (C) is correct.
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=− +
+
Chapter 6 The Z Transform Page 561
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=−
= −−
SOL 6.4.2 Option (B) is correct.
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−
==//
Chapter 6 The Z Transform Page 563
( ) ( )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.
Page 564 The Z Transform Chapter 6
SOL 6.4.8 Option (A) is correct.
[ ]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< <
SOL 6.4.9 Option (C) is correct.
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 >
Chapter 6 The Z Transform Page 565
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
− −
SOL 6.4.16 Option (B) is correct.
( )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< .
SOL 6.4.17 Option (B) is correct.
z -transform ( )F z z 1
1=+
1z
z1
= −+
1z1
11= −
+ −
so, ( )f k ( ) ( 1)k kδ= − −
Thus ( 1)k− z1
11
Z
+ −
Page 566 The Z Transform Chapter 6
SOL 6.4.18 Option (C) is correct.
( )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 "
Chapter 6 The Z Transform Page 567
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 "
SOL 6.4.19 Option (A) is correct.
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)
Page 568 The Z Transform Chapter 6
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= − − −
SOL 6.4.20 Option (D) is correct.
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= −
SOL 6.4.21 Option (C) is correct.
( )X z az bz1 1121
131
=−
+−− − , ROC : a z b< <
Poles of the system are z a= , z b=ROC : a z b< <
Chapter 6 The Z Transform Page 569
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=
SOL 6.4.22 Option (A) is correct.
( )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.
SOL 6.4.23 Option (A) is correct.
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=
Page 570 The Z Transform Chapter 6
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
From the property of differentiation in z -domain
[ ]n u nnα ( )
zdzd
z11
1Z
α−
− −; E, z > α
( )z
z1 1 2
1Z
αα
− −
−, z > α ( 1)D "
Chapter 6 The Z Transform Page 571
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 "
From the property of differentiation in z -domain
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 6.4.29 Option (A) is correct.
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
Page 572 The Z Transform Chapter 6
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=
Chapter 6 The Z Transform Page 573
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= −
Page 574 The Z Transform Chapter 6
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=
− −
Chapter 6 The Z Transform Page 575
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=
SOL 6.4.44 Option (B) is correct.
[ ] [ ]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=
Taking inverse z -transform
[ ]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
= + −
Page 576 The Z Transform Chapter 6
SOL 6.4.46 Option (A) is correct.
SOL 6.4.47 Option (A) is correct.
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.
SOL 6.4.49 Option (D) is correct.
( )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= − + − − + −− − −
Chapter 6 The Z Transform Page 577
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=− − −
SOL 6.4.55 Option (C) is correct.
( )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=
Page 578 The Z Transform Chapter 6
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.
SOL 6.4.57 Option (C) is correct.
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<
Chapter 6 The Z Transform Page 579
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.
Page 580 The Z Transform Chapter 6
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).
SOL 6.4.63 Option (C) is correct.
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
Chapter 6 The Z Transform Page 581
( )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
Page 582 The Z Transform Chapter 6
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.
SOL 6.4.70 Option (B) is correct.
( )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.
Chapter 6 The Z Transform Page 583
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.
Page 584 The Z Transform Chapter 6
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=
SOL 6.4.79 Option (D) is correct.
SOL 6.4.80 Option (C) is correct.
( )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 =
***********