SIGNALS AND SYSTEMS
For
Graduate Aptitude Test in Engineering
By
Dr. N. Balaji
Professor of ECE
JNTUK
Comparison of Laplace ad Z-Transform
• Fourier transform plays a key role in analyzing and representing
discrete-time signals and systems, but does not converge for all signals.
• Continuous systems: Laplace transform is a generalization of the
Fourier transform.
• Discrete systems : Z-transform, generalization of DTFT, converges for a
broader class of signals.
• In Laplace Transform we evaluate the complex sinusoidal
representation of a continuous signal.
• In the Z-Transform, it is on the complex sinusoidal representation of a
discrete-time signal.
2
2
Relation between DTFT and Z-Transform
• The DTFT provides a frequency-domain representation of discrete-
time signals and LTI discrete-time systems
• Because of the convergence condition, in many cases, the DTFT of a
sequence may not exist, thereby making it impossible to make use of
such frequency-domain characterization in these cases
• A generalization of the DTFT defined by X e x n e j n
j
n
leads to the Z-transform
• Z-transform may exist for many sequences for which the DTFT
does not exist.
• Use of Z-transform permits simple algebraic manipulations
3
Motivation of Z-transform
The Fourier transform does not converge for all
sequences and it is useful to have a generalization of
the Fourier transform.
In analytical problems the Z-Transform notation is
more convenient than the Fourier transform
notation.
4
Introduction to Z-Transform
5
Definition of Z-Transform
6
Example of Z-Transform
7
8
9
10
Elementary signals
Unit step function x (n) 1 n0
1 z
X ( z ) 1 z n 1 z 1 z 2 ............. 1
; z 1
n 0 1 z z 1
Power function x ( n) a n n0
1 z
X ( z ) a n z n 1 az 1 a 2 z 2 ........... 1
; z a
n 0 1 az za
Ramp function x ( n) n n0
z 1 z
X ( z ) nz n 1
0 z 2z 2 3
3 z ......... ; | z | 1
n0 (1 z 1 ) 2 ( z 1) 2
The values of z for which X(z) is finite are known as region of
convergence (ROC)
11 Try other signals: impulse function, ….
Significance of ROC
For causal sequence x(n) a n for n 0 and 0 for n0
1 z
X ( z) a z n n
( az 1 ) n 1
; z a
n 0 n 0 1 az za
For Anti causal x (n) 0 for n 0 and a n for n0
sequence 1 1
z
X ( z) a
n
n
z n
(a z )
n
1 n
za
; z a
The two sequences have same X(z) but their ROC is different. Without ROC we can
not uniquely determine the sequence x(n). Generally, for causal sequence, the ROC
is exterior of the circle having radius a and for anti causal sequence it is interior of
the circle.
Find X(z) and ROC for x(n) = αn u(n) + βn u(-n-1)
X (z) ROC : z
12
Answer z z 1
Z-transform pairs
n 1, ROC : all z
1
u n 1
, ROC : z 1
1 z
1
u n 1 1
, ROC : z 1
1 z
n m z m ,
R O C : a ll z e x c e p t 0 if m 0 o r if m 0
13
13 2020/6/12
Z-transform pairs
1
a u n
n
1
, ROC : z a
1 az
1
a u n 1
n
1
, ROC : z a
1 az
1
az
na u n
n
, ROC : z a
1 az 1 2
1
az
na n u n 1 , ROC : z a
1 az 1 2
14
14 2020/6/12
Z-transform pairs
1 cos w 0 z 1
cos w 0 n u n , ROC : z 1
1 2 cos w 0 z z
1 2
c o s w 0 n u n
1
2
e jw 0 n
e jw 0 n
u n
1 1 1
jw 0 1
jw 0 1
2 1 e z 1 e z
sin w 0 n u n
sin w 0 z 1 , ROC : z 1
1 2 cos w 0 z 1 z 2
15
15 2020/6/12
Z-transform pairs
1 r cos w 0 z 1
r n
cos w 0 n u n
1 2 r cos w 0 z r z
1 2 2
, ROC : z r
r c o s w 0 n u n
n 1
2
r e
n
jw 0 n
r n
e jw 0 n
u n
1 1 1
jw0 1
jw 0 1
2 1 re z 1 re z
r n
sin w 0 n u n
r sin w 0 z 1
, ROC : z r
16
1 2 r cos w 0 z r z
1 2 2
16 2020/6/12
Z-transform pairs
a n , 0 n N 1 1 a zN N
,
0 , otherwise 1 az 1
ROC : z 0
17 2020/6/12
17
Zero and pole
The Z-transform is most useful when the infinite sum can be expressed in closed form, usually a ratio of polynomials in z (or Z-1).
P z
X z
Q z
Zero: The value of z for which
X z 0
X z
Pole: The value of z for which
18
18 2020/6/12
Convolution
• Convolution: x[ n ] * h[ n ]
k
x[ k ]h[ n k ] X (z)H (z)
Proof: Z x [ n ] * h [ n ] Z x [ k ] h [ n k ] x [ k ] h [ n k ] z n
k n k
x [ k ] h [ n k ]z n
k n
Change of index on the second sum: m n k
(m i) k )
m
Z x [ n ] * h [ n ] x [ k ] h [ m ] z x [ k ]z h [ m ]z
k m k m
X (z)H (z)
The ROC is at least the intersection of the ROCs of x[n] and h[n], but can be a larger region if there is pole/zero cancellation.
• The system transfer function is completely analogous to the CT case:
h[ n ] H (z)
n
h[ n ] z n
• Causality:
h[ n ] 0 n 0
19 Implies the ROC must be the exterior of a circle and include z = .
Initial-Value and Final-Value Theorems (One-Sided ZT)
• Initial Value Theorem: x [ 0 ] lim X ( z )
z
Proof: lim X ( z ) lim
z z
n0
x [ n ] z n lim x [ 0 ] x [1 ] z 1 ... x [ 0 ]
z
• Final Value Theorem:
lim x [ n ] lim ( z 1 ) X ( z )
n z1
• Example:
3z2 2z 4 3z2 2z 4
X (z) 3
z 2 z 2 1 .5 z 0 .5 ( z 1 )( z 2 z 0 . 5 )
3z2 2z 4 5
lim x [ n ] [( z 1 ) X ( z )] 2 10
n z 1
z z 0 .5 z 1
.5
20
Table Common Z-transform Pairs
21
Properties of Z-Transform
22
Inverse Z Transform
• Recall the definition of the inverse Laplace transform via contour integration:
j
1 1
X s e st ds
2 j
x (t ) X ( s ) e st ds
j
2 j C
• The inverse Z-transform follows from this:
1
x[ n ]
2 j C
X ( z ) z n 1 dz
Evaluation of this integral is beyond the scope of this course. Instead, as with the Laplace transform, we will restrict our interest
in the inverse transform to rational forms (ratio of polynomials). We will see shortly that this is convenient since linear constant-
coefficient difference equations can be converted to polynomials using the Z-transform.
• As with the Laplace transform, there are two common approaches:
Long Division
Partial Fractions Expansion
• Expansion by long division Is also known as the power series expansion approach and can be easily demonstrated by an
example.
23
Long Division
• Consider: z2 1
X (z)
Solution: z3 2z 4
z 1
z 1 0 z 2 3 z 3 4 z 4
z 3
2z 4 z 2
1
z3 2z 4 z2 1
z 2 2 4 z 1
z 2 2 4 z 1
3 4 z 1
3 4 z 1
z 1 0 z 2 3 z 3 3 6 z 2 12 z 3
z3 2z 4 z2 1
1
4 z 1 6 z 2 12 z 3
z 2
2 4z
1
4 z 1 8 z 3 16 z 4
3 4z
3 6 z 2 12 z 3 6 z 2 20 z 3 16 z 4
4 z 1 6 z 2 12 z 3
X ( z ) z 1 0 z 2 3 z 3 4 z 4 ...
x [ n ] 0 [ n ] 1 [ n 1 ] 3 [ n 3 ] 4 [ n 4 ] ...
24
Inverse Z-Transform Using Partial Fractions
• The partial fractions approach is preferred if we want a closed-form solution rather than the numerical solution
long division provides.
z3 1
• Example: X (z) 3
z z2 z 2
In this example, the order of the numerator and denominator are the same. For this case, we can use a trick of
factoring X(z)/z:
A ( z ) z z z 2 ( z 2 )( z 0 . 5 j 0 . 866 )( z 0 . 5 j 0 . 866 )
3 2
X (z) c c1 c1 c3
0
z z z 0 . 5 j 0 . 866 z 0 . 5 j 0 . 866 z 2
X (z) 1
c0 ( z ) 0 .5
z z0 2
X (z)
c1 z 0 . 5 j 0 . 866 0 . 429 j 0 . 0825
z z 0 . 5 j 0 . 866
X (z)
c3 z 2 0 . 643
z z2
25
Inverse Z-Transform (Cont.)
We can compute the inverse using our table of common transforms:
c1 z c1 z c3 z
X (z) c0
z 0 . 5 j 0 . 866 z 0 . 5 j 0 . 866 z 2
c1 c1 c3
c0 1
1
1 0 . 5 j 0 . 866 z 1 0 . 5 j 0 . 866 z 1 2 z 1
x [ n ] c 0 [ n ] c 1 ( 0 . 5 j 0 . 866 ) n u [ n ] c 1 ( 0 . 5 j 0 . 866 ) n u [ n ] c 3 2 n u [ n ]
The exponential terms can be converted to a single cosine using a magnitude/phase conversion:
p1 ( 0 . 5 ) 2 ( 0 . 866 ) 2 1
0 . 866 4
p 1 tan 1
rad
0 .5 3
c1 ( 0 . 429 ) 2 ( 0 . 0825 ) 2 0 . 437
1 0 . 0825
c 1 tan 0 . 19 rad (10.89 )
0 . 429
x [ n ] c 0 [ n ] c 1 ( 0 . 5 j 0 . 866 ) n u [ n ] c 1 ( 0 . 5 j 0 . 866 ) n u [ n ] c 3 2 n u [ n ]
c 0 [ n ] 2 c 1 p 1 cos( p 1 n c 1 ) c 3 ( 2 ) n u [ n ]
4
0 . 5 [ n ] 0 . 874 cos( n 0 . 19 ) 0 . 643 ( 2 ) n u [ n ]
26 3
Transfer Functions
• In addition to our normal transfer function components, x[ n ] y [ n ] x [ n 1]
D
such as summation and multiplication, we use one
important additional component: delay. x[ n ] y [ n ] x [ n 1]
Z-1
• This is often denoted by its Z-transform equivalent. Y ( z ) z 1 X ( z )
• We can illustrate this with an example (assume initial conditions are zero):
27
Transfer Function Example
• Redraw using Z-transform:
• Write equations for the behavior at
each of the summation nodes:
zQ 1 ( z ) Q 2 ( z ) X ( z )
zQ 2 ( z ) Q 1 ( z ) 3 Y ( z )
Y ( z ) 2Q1 ( z ) Q 2 ( z )
• Three equations and three unknowns: solve the first for Q1(z) and substitute into the other two equations.
Q 1 ( z ) z 1Q 2 ( z ) z 1 X ( z )
zQ 2 ( z ) z 1 Q 2 ( z ) z 1 X ( z ) 3 Y ( z )
Q 2 ( z ) z 2 Q 2 ( z ) z 2 X ( z ) 3 z 1Y ( z )
Q 2 (z)
1 z
1
2
z 2 X ( z ) 3 z 1Y ( z )
Y ( z ) 2 z 1
1
2
z 2 X ( z ) 3 z 1Y ( z ) 2 z 1 X ( z )
1
2
z 2 X ( z ) 3 z 1Y ( z )
1 z 1 z
Simplify.. .
Y (z) 2z 1
H (z) 2
28 X (z) z 3z 5
The Properties of ROC.
Figure : The relationship between the ROC and the time extent of a signal.
(a) A right-sided signal has an ROC of the form |z| > r+.
(b) A left-sided signal has an ROC of the form |z| < r–.
29 29 (c) A two-sided signal has an ROC of the form r+ < |z| < r–.
Contd…
Poles: -1/2,1/4
Figure : ROCs .
(a) Two-sided signal x[n] has ROC in between the poles.
(b) Right-sided signal y[n] has ROC outside of the circle containing the pole of largest magnitude.
(c) Left-sided signal w[n] has ROC inside the circle containing the pole of smallest magnitude.
30 30
Contd…
Taking a path analogous to that used the development of the
Laplace transform, the z transform of the causal DT signal is
A n
u n , 0
n
Xz A n unz n A n z n A
n n0
n0 z
and the series converges if |z| > |α|. This
defines the ROC as the exterior of a circle in
the z plane centered at the origin, of radius|α|.
The z transform is
z
Xz A , z
z Causal
31 31
Contd…
By similar reasoning, the z transform and region of convergence of the anti-causal signal
below, are
A n
u n , 0
A Az 1 1
Xz 1 , z
1 z z
Anti-Causal
32 32
LTI System Stability
LTI DT causal system is BIBO stable provided that all poles of
the system transfer function lie inside the unit circle in Z-plane
33
BIBO Stable
34
BIBO Stable
35
Causality and Stability
The impulse response of a causal system is zero for n<0.
The impulse response of a casual LTI system is determined from the transfer function by using right-
sided inverse transform.
The poles inside the unit circle, contributes an exponentially decaying term to the impulse response.
The poles outside the unit circle, contributes an exponentially increasing term.
Fig.
Pole and impulse response characteristic of a causal
system.
(a) A pole inside the unit circle contributes an
exponentially decaying term to the impulse response.
(b) A pole outside the unit circle contributes an
exponentially increasing term to the impulse response.
36 36
Contd…
Stable system the impulse response is absolute summable and the DTFT of impulse
response exist.
The impulse response of a casual LTI system is determined from the transfer
function by using right-sided inverse transform.
The poles inside the unit circle, contributes a right-sided decaying exponential term
to the impulse response.
The poles outside the unit circle, contributes a left-sided decaying exponentially term
to the impulse response.
Refer to Figure.
Figure : Pole and impulse response characteristics for a stable
system.
(a) A pole inside the unit circle contributes a right-sided term to
the impulse response. (b) A pole outside the unit circle
contributes a left-sided term to the impulse response.
37 37
Contd… Stable/Causal
From the ROC below the system is stable, because all the poles within the unit circle
and causal because the right-sided decaying exponential in terms of impulse response.
Figure 7.16: A system that is both stable and causal must have all its poles inside the unit circle in the Z-plane, as
illustrated here. 38
38
Z-Plane and Stability
39
Implementing Discrete-Time LTI System.
The system is represented by the differential equation.
y[ n ] a1 y[ n 1] a 2 y[ n 2 ] b0 x[ n ] b1 x[ n 1] b2 x[ n 2 ]
Taking the Z-transform of difference equation gives,
(1 a1 z 1 a 2 z 2 )Y ( z ) ( b 0 b1 z 1 b 2 z 2 ) X ( z )
The transfer function of the system,
Y (z)
H (z)
X (z)
b0 b1 z 1 b2 z 2
H (z)
40
1 a1 z401 a 2 z 2
Example: Causality and Stability.
Can this system be both stable and causal?
Solution: Step 1: Find the characteristic equation of the system.
From the plot, the system is unstable
because the pole at z = 1.2 is outside
the unit circle.
41 . 41
Contd…
(1 a1 z 1 a 2 z 2 )Y ( z ) ( b0 b1 z 1 b2 z 2 ) X ( z )
Figure : Block diagram of the transfer function.
42 42
Contd…
Figure : Development of the direct form II representation of an LTI system. (a) Representation of the transfer
function H(z) as H2(z)H1(z).
(b) Direct form II implementation of the transfer function H(z) obtained from (a) by collapsing the two sets of z–1
43 43 blocks.
44
45
Problem
46
Solution
47
Problem
48
Solution
49
Problem
50
Solution
51
Problem
52
Solution:
53
Problem
54
Solution:
55
Problem
56
Solution:
57
Problem:
58
Solution:
59
Solution
60
Problem
61
Solution
62
Problem and Solution
63
Problem
64
Solution
65
Problem
66
Solution
67
Problem
68
Solution:
69
Problem
70
Solution
71
Problem and Solution
72
Problem
73
Solution
74
Sources, References and Acknowledgement
i) Lecture slides of Michael D. Adams
ii) Lecture slides of Prof. Paul Cuff
iii) A.V. Oppenheim, A.S. Willsky and S. H. Nawab, “Signals and Systems”, PHI,
2nd Edition, 1997.
iv) GATE Previous Questions with Solutions, Nodia Publications
v) MIT Open Courseware http://ocw.mit.edu
vi) https://www.aceenggacademy.com/
vii) https://www.madeeasy.in/
Disclaimer: The material presented in this presentation is taken from various standard Textbooks
and Internet Resources and the presenter is acknowledging all the authors.
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