Hossein Sameti
Department of Computer Engineering
Sharif University of Technology


Z {x(n)}  X ( z )   x(n) z

n
Vs.
n  
Complex function

X ( )   x(n)e  jn
n  
Complex variable
Q: Why do we need Z-transform?
A: For some signals, Fourier transform does not converge.
e.g.:
x ( n)  2 n u ( n)
• Relationship between X (z ) and X ( ) :
X ( z)
z  e jw
 X ( )
Z-transform evaluated at unit circle
corresponds to DTFT.
 The role of Z-transform in DT is
similar to Laplace transform in CT.
2


Z {x(n)}  X ( z )   x(n) z  n
n  
For what values of z, does the z-transform converge?
z  re j

z 
z r

X ( z )   x ( n) z
n
n  

 
  x(n) re j

n  
 
X ( z )   x(n) r  n e  jn
n  
n
DTFT of x( n)r  n
DTFT of x( n)r  n exists if x( n)r  n is absolutely summable.
n
 x ( n) r  
n
Z-transform exists and converges.
Convergence only depends on r.
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
3
• Region of Convergence
(ROC): ROC is a region in “z”

domain that n x(n) z  n converges.
n
 x ( n) r  
n
ROC is radially symmetric.
If Z-transform converges for
z1  re j1
It also converges for
z2  re j 2
• ROC does not depend on the angle.
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
4

There are the following possibilities:
1.
2.
3.
4.
5.
All Z-domain except one or two points (origin / infinity)
Inside a circle
Outside a circle
Between two circles (ring)
ROC does not exist
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
5
a n
x ( n)  
 0
0  n  N 1
 
X ( z )   x(n) z n
Otherwise
n
X ( z )  1  az 1  a 2 z 2  ...  a N 1z  N 1
a a2
a N 1
 1   2  ...  N 1
z z
z
ROC: everywhere except at the origin: z  0
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
6
a n
x( n)  
0
N n0
 
X ( z )   x(n) z n
n
Otherwise
X ( z )  1  a 1z  a 2 z 2  ...  a  N z N
ROC: everywhere except at infinity: z  
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
7
x ( n)  a n u ( n )

n n

X ( z )   a z u ( n)   a z
n  
az 1  1
n n
n0
X ( z) 


1 n
 ( az )
n 0
1
1  az 1
a  z
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
8
x(n)  a nu(n  1)
1
n n
X ( z)    a z
n  
a1z  1

  a
m m
m 1
X ( z) 

z 1   (a 1 z ) m
m0
1
1  az 1
z  a
• Same Z-transform as the
previous example. However,
the ROC is different.
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
9

We have shown that X(z) needs the ROC to uniquely
specify a sequence.
x ( n)  a n u ( n )
x(n)  a nu(n  1)
X ( z) 
1
1  az 1
X ( z) 
1
1  az 1
X ( z )  ROC1 
 x1 (n)
X ( z )  ROC2 
 x2 (n)
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
10
n
n
1
 1
x ( n)    u ( n )     u ( n )
2
 3
n
1

 1 
X ( z )    z 1      z 1 
n0  2
 n 0  3 

z 
1
2

z 
n
1
3
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
11
n
n
 1
1
x(n)     u (n)    u (n  1)
 3
2
  1 1  n
X ( z)     z
n 0  3
z 
1  1 1  n
    z

n  2
1
3
z 


1
2
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
12
n
n
1
1
x(n)    u (n)    u (n  1)
2
3
n
1

1 
X ( z )    z 1     z 1 
n 0  2
 n 0  3 

z 
1
2

z 
n
1
3
 Z-transform does not exist.
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
13
(1) The ROC of X(z) consists of a ring in the z-plane centered about
the origin (equivalent to a vertical strip in the s-plane)
(2) The ROC does not
contain any poles (same as in LT).
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
14
Book Chapter10: Section 1
(3) If x[n] is of finite duration, then the ROC is the entire zplane, except possibly at z = 0 and/or z = ∞.
Why?
N2
X ( z) 

xnz n
n  N1
Examples:
CT counterpart
 n  1 ROC all z  (t ) 1 ROC all s
 n  1  z 1 ROC z  0  (t  T )  e  sT es  
 n  1  z ROC z  
 (t  T )  e sT es  
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
15
Book Chapter10: Section 1
(4) If x[n] is a right-sided sequence, and if |z| = ro is in the ROC, then
all finite values of z for which |z| > ro are also in the ROC.


xn r1n
n  N1
converges faster tha n


xn r0n
n  N1
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
16
Book Chapter10: Section 1
(5) If x[n] is a left-sided sequence, and if |z| = ro is in the ROC, then
all finite values of z for which 0 < |z| < ro are also in the ROC.
(6) If x[n] is two-sided, and if |z| = ro is in the ROC, then the ROC
consists of a ring in the z-plane including the circle |z| = ro.
What types of signals do the following ROC correspond to?
right-sided
left-sided
two-sided
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
17
Book Chapter10: Section 1
Example:
xn  b|n| ,
b0
xn  b nun  b  n u n  1
From：
b n un 
1
,
z b
1
1  bz
1
1
b n u n  1 
,
z

b
1  b 1 z 1
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
18
Book Chapter10: Section 1
Example, continued
1
1
1
X ( z) 

, b z 
1
1 1
b
1  bz
1 b z
Clearly, ROC does not exist if b > 1 ⇒ No z-transform for b|n|.
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
19






ROC is only a function of r.
FLS ROC is everywhere except at the origin or
infinity.
RHS ROC is outside some circle.
LHS ROC is inside some circle.
BHS ROC is a ring or it does not exist.
FT exists if ROC includes unit circle.
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
20

Linearity:
z
x1 (n) 

X1 ( z )
ROC1
z
x2 (n) 

X 2 ( z)
ROC2
z
x(n)  a1x1 (n)  a2 x2 (n) 

X ( z )  a1 X1 ( z )  a2 X 2 ( z )
ROC: at least the intersection of ROC1 and ROC2.
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
21

Example:
Solution:
x(n)  (cos 0n) u(n)
x(n)  (cos 0 n) u (n) 
Using linearity property:
X ( z)  ?
1 j 0 n
e u ( n) 
2
1
X ( z )  z{ e j0 nu (n) 
2
Earlier we showed that: x(n)  a u (n)
n
If we set: a  e j0 x(n)  e j0 nu (n)
X ( z) 
1
1
1
1



2 1  e j0 z 1 2 1  e  j0 z 1
1  j 0 n
e
u ( n)
2
1  j 0 n
e
u (n)}
2
1
(a  z)
1  az 1
1
X ( z) 
1  e j0 z 1 ( z  1)
X ( z) 
( z  1)
1  z 1 cos 0
X ( z) 
1  2 z 1 cos 0  z  2
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
22
z
x(n) 

X ( z)
ROC
z
x( n  k ) 

z k X ( z)
ROC: Similar ROC with some exceptions at 0 and infinity
Example: x1 (n)  {1 ,2,5,7,0,1}

Z{x2 (n)  x1(n  2))  ?
X1 ( z )  1  2 z 1  5 z 2  7 z 3  z 5
X 2 ( z )  z 2  2 z1  5  7 z 1  z 3
ROC?
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
23
z
x(n) 

X ( z)
z
a n x(n) 

X (a 1z )
ROC : r1  z  r2
ROC : a r1  z  a r2
Example: x(n)  a n (cos 0 n) u (n)
Earlier:
x(n)  (cos 0n) u(n)
x(n)  a (cos 0 n) u (n)
n
1  z 1 cos 0
X ( z) 
1  2 z 1 cos 0  z  2
1  az 1 cos 0
X ( z) 
1  2az 1 cos 0  a 2 z  2
( z  1)
(z  a)
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
24
z
x(n) 

X ( z)
z
x(n) 

X ( z 1 )
Example:
x ( n)  u (  n)
Earlier:
x ( n)  a n u ( n )
x ( n)  u ( n)
x ( n)  u (  n)
ROC : r1  z  r2
ROC :
1
1
 z 
r2
r1
X ( z)  ?
1
1  az 1
1
X ( z) 
1  z 1
X ( z) 
X ( z) 
1
1 z
(z  a)
( z  1)
( z  1)
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
25
z
x(n) 

X ( z)
dX ( z )
z
nx(n) 

z
dz
Example:
x(n)  na nu (n)
Earlier:
x1 (n)  a nu (n)
x2 (n)  na nu (n)
ROC
ROC
X ( z)  ?
1
(z  a)
1
1  az
dX1 ( z )
az 1
(z  a)
X 2 ( z)   z

1 2
dz
(1  az )
X1( z) 
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
26
z
x1 (n) 

X1 ( z )
ROC1
z
x2 (n) 

X 2 ( z)
ROC2
z
x(n)  x1 (n) * x2 (n) 

X ( z )  X 1 ( z ). X 2 ( z )
ROC: at least the intersection of ROC1 and ROC2
Example:


x1 (n)  1 ,2,1


 1, 0  n  5
x2 (n)  
0, elsewhere
X 1 ( z )  1  2 z 1  z 2
x1(n) * x2 (n)  ?
X 2 ( z )  1  z 1  z 2  z 3  z 4  z 5
X ( z )  X 1 ( z ) X 2 ( z )  1  z 1  z 6  z 7


x(n)  1,1,0,0,0,0,1,1


Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
27
Inverse Z-Transform
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
28

Z {x(n)}   x(n) z  n
x (n )
n  
?
X (z )
X (z )
x (n ) 
1
n 1
 X ( z ) z dz
2j C
• Methods of calculating the inverse Z-transform:
• Cauchy theorem for calculating the inverse Z-transform
(not covered)
• Series expansion
• Partial fraction expansion and using look-up tables
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
29
• Basic concept: expand X(z) in terms of powers of z-1

X ( z )   cn z  n
n  
X ( z)  8z 3  5  20 z 4
X ( z )   x(n) z  n  ...  x(1) z  x(0)  x(1) z 1  ...
n
x(n)  8 (n  3)  5 (n)  20 (n  4)
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
30

Example:

X ( z)  log 1  az 1

ROC : z  a
x(n)  ?
x 1
x 2 x3 x 4
log(1  x)  x 


 ...
2
3
4
az 1 1
log(1  az 1 )  az 1 
az   az   az 
1 2
1 3
1 4
2
3
4
 ...
a2
a3
x(n)  0. (0)  a (1)   (2)   (3)  ...
2
3
an
 1n 1u(n  1)
x ( n) 
n
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
31

This method is applicable if we have a ratio of two polynomials in
the form of X ( z)  P( z). b0  b1z 1  ...  bM z M
Q( z )


a0  a1z 1  ...  a N z  N
We would like to rewrite X(z) in the following manner:
X ( z)  a1 X1( z)  a2 X 2 ( z)  ...  aN X N ( z)
This expansion is performed in a way that the inverse Z-transform
of X1( z), X 2 ( z),..., X N ( z) can be easily found using a lookup table and
using the linearity property, we can find x(n):
x(n)  a1x1(n)  a2 x2 (n)  ...  aN xN (n)
Example:
x ( n)  a n u ( n )
x(n)  na nu (n)
1
(z  a)
1
1  az
az 1
X ( z) 
(z  a)
1 2
(1  az )
X ( z) 
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
32

Any non-constant single-variable polynomial of degree
N can be factored into N terms:
P( z )  a0  a1z1  a2 z 2  a3 z 3  k ( z  z1 )( z  z2 )( z  z3 )


If the coefficients of the polynomial are real, then the
roots are either real or complex conjugates.
If z0 is a complex root, then z 0* is also a root.
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
33

Example:
1
z2
X ( z) 
 2
1
2
1  1.5 z  0.5 z
z  1.5 z  0.5
X ( z)
z
A1( z 0.5) A2 ( z 1)
A
A
1
2




z
( z  1)( z  0.5)
z  1 z  0.5 ( z  1)( z  0.5)
z  A1( z  0.5)  A2 ( z  1)
2z
z
2
1
X ( z) 




1
z  1 z  0.5 1  z
1  0.5z 1
ROC : z  1

z 1
A1  2
z  0.5
A2  1
x(n)  2.(1)n u(n)  (0.5)n u(n)  [2  0.5n ]u(n)
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
34

Example:
1
X ( z) 
(1  z 1 )(1  z 1 ) 2

z3
( z  1)( z  1) 2
A3
A1
A2
X ( z)
z2




2
2
z

1
z

1


z

1
z
( z  1)( z  1)
z 2  A1 ( z  1) 2  A2 ( z  1)( z  1)  A3 ( z  1)
0.25
0.75
0.5z 1
X ( z) 


1  z 1 1  z 1 (1  z 1 )2
x(n) causal
x(n)  0.25(1) n u (n)  0.75(1) n u (n)  0.5n(1) n u (n)
• In the case of complex conjugate roots, the coefficients in the
partial fraction expansion will be complex conjugates.
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
35
Example:
5 1
z
A
B
6
X ( z) 



1 1
1
1 
1 
1 1 
1 1 

1  z 1
 z   z   1  z 1  z  1  z
4
3
4 
3 
4
3



3z 2 
5
z
6
3
Partial Fraction Expansion Algebra:
1
1
1  z 1
4


x1 n
X ( z) 

x[n]
A = 1, B = 2
2
1
1  z 1
3


x2 n

Note, particular to z-transforms:
1) When finding poles and zeros,
express X(z) as a function of z.
2) When doing inverse z-transform
using PFE, express X(z) as a
function of z-1.
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
36
z
ROC III:
1
3
- right - sided singnal
n
1
x1 n    un
4
n
1
x2 n  2    un
 3
ROC II:
1
1
 z
4
3
- two - sided singnal
n
1
x1 n    un
4
n
1
x2 n  2    u n  1
3
ROC I:
z
1
4
- left - sided singnal
n
1
x1 n    u n  1
4
n
1
x2 n  2    u n  1
3
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
37
Linear Constant- Coefficient
Difference Equation (LCCDE)
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
38
x(n),X(z)
h(n),H (z)
y(n),Y(z)
y ( n)  h( n) * x ( n)
Y ( z )  H ( z ). X ( z )
LTI Systems
h(n)
FIR
Use difference
equations for
implementation
Transfer
function
IIR
With rational transfer
function
P( z )
H ( z) 
Q( z )
No rational transfer
function
H ( z) 
P( z )
Q( z )
39

Cumulative Average System:
1 n
y ( n) 
 x(k ), n  0,1,...
n  1 k 0
n 1
(n  1) y (n)   x(k )  x(n)
k 0
 ny ( n  1)  x( n)
y ( n) 
n
1
y (n  1) 
x ( n)
n 1
n 1
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
40

The general form for a linear constant-coefficient
difference equation is given by:
N
M
k 1
k 0
Order of the system
y (n)    ak y (n  k )   bk x(n  k )

Given the difference equation of an IIR system and the initial
condition of y(n), we can compute the output y(n) through the
difference equation more efficiently using finite number of
computations.
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
41







Question: How can an LCCDE correspond to an LTI system?
E.g.: y (n)  ay (n  1)  x(n)
It is not even a system, since a unique input does not
correspond to a unique output.
Q: How can we show this?
A: If y1 (n) is a solution, y2 (n)  y1 (n)  kan is also a solution.
Q: How can we resolve this problem?
A: We need to add an initial condition (I.C.).
y (n)  ay (n  1)  x(n)
x ( n )  b ( n )
IC : y(1)  y0
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
42
y (n)  ay (n  1)  x(n)
x ( n )  b ( n )
IC : y(1)  y0
y(0)  ay(1)  b (0)  ay0  b
n  0:
Causal implementation
y (1)  ay (0)  0  a(ay0  b)  a 2 y0  ab
y (n)  a n (ay0  b)
n  2 :
y (n)  ay (n  1)  x(n)
y (2) 
n  1 :
1
b
1
y (1)   (1)  y0
a
a
a
1
y (3)  2 y0
a
1
b
y (n  1)  y (n)   (n)
a
a
Anti-causal implementation
y (n)  a n1 y0
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
43
• By Combining the two answers, we get:
y (n)  a n (ay0  b)u (n)  a n 1 y0u (n  1)
y (n)  a n 1 y0  a nbu (n)
• Q: Does this system correspond to a linear system?
y (n)  ay (n  1)  x(n)
x ( n )  b ( n )
IC : y(1)  y0
• A: No, zero input does not result in zero output.
• An LCCDE corresponds to a linear system, when IC=0.
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
44
• Q: Does this system correspond to a TI system?
y (n)  ay (n  1)  x(n)
x ( n )  b ( n )
IC : y(1)  y0
y (n)  a n 1 y0  a nbu (n)
• A: Consider x(n)  b (n) . Now let’s shift the input: w(n)  b (n  n0 )
Output: a n 1 y0  ba n  n u (n  n0 )
0
y (n  n0 )  a n  n0 1 y0  ba n  n0 u (n  n0 )
Shift
variant
• In order for an LCCDE to correspond to a TI system, IC=0.
(for 𝑛0 ≥ 0)
• In order for an LCCDE to correspond to an LTI system, IC=0.
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
45

In order an LCCDE to correspond to a causal LTI system :
◦ Initial rest condition (IRC)=0


E.g., if x(n)   (n) , for a first order LCCDE, the IRC should
be as follows: y (1)  0
E.g., if x(n)   (n  3) , for a second order LCCDE, the IRC
should be as follows: y(1)  0, y (2)  0
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
46
1
y (n)  0; n  0
y (n)  y (n  1)  2 x(n)  x(n  1)  x(n  2)
2
1
h(n)  h(n  1)  2 (n)   (n  1)   (n  2)
2
1
h
(
n
)

h(n  1)  2 (n)   (n  1)   (n  2)
Solution1:
n  1
n0
n 1
n2
n3
n3
2
h(1)  0
1
h(0)  h(1)  2 (0)  0  0  0  2  0  0  2
2
1
1
h(1)  h(0)  0   (0)  0   2  1
12
12 1
1 2
1
h(2)  h(1)  0  0   (0)  (  2  1)  1  ( )  2  ( )  1  1
2
2 2
2
2
1
1 1
1
1
1
1
h(3)  h(2)  (( ) 2  2  ( )  1  1)  ( )3  2  ( ) 2  1 
2
2 2
2
2
2
2
1
1 1
1
1
h(n  1)  (( ) n 1  2  ( ) n  2  1  ( ) n 3  1)
2
2 2
2
2
1
1
1
 ( ) n  2  ( ) n 1  1  ( ) n  2  1, n  3
47
2
2
2
h( n) 
Y ( z )  H ( z ). X ( z )
x(n),X(z)
h(n),H (z)
Y ( z)
H ( z) 
X ( z)
y(n),Y(z)
1
y (n  1)  2 x(n)  x(n  1)  x(n  2)
2
1 1
Y ( z )  z Y ( z )  2 X ( z )  z 1 X ( z )  z  2 X ( z )
2
y ( n) 
y (n)  0; n  0
Y ( z)
2
z 1
z 2
H ( z) 



1
1
X ( z ) 1  z 1 1  z 1 1  1 z 1
2
2
2
Remember:
x ( n)  a n u ( n )
n
1
1
h( n)  2    u ( n)   
2
2
X ( z) 
n 1
1
u (n  1)   
2
1
1  az 1
(z  a)
n2
u (n  2)
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
48

Z-transform with rational transfer function:
M
 bk z
k
B( z ) b0  b1 z 1  ...  bM z  M
X ( z) 

 k 0
N
A( z ) a  a z 1  ...  a z  N
k
0
1
N
 ak z
k 0
Zero: X(z)=0
Pole: X(z)=∞
Getting rid of the negative powers: a0  0, b0  0
B( z ) b0 z  M
X ( z) 

A( z ) a0 z  N
b 
b
z M   1  z M 1  ...  M
b0 b0  M  N ( z  z1 )( z  z2 )...( z  zM )
 b0 

z
a 
a0
( z  p1 )( z  p2 )...( z  pN )
a
z N   1  z N 1  ...  N
a0
 a0 
Poles and zeros?
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
49

A pole-zero plot can represent X(z) graphically.
Example:
x1 (n)  a nu (n)
X1( z) 
1
1  az 1
(z  a)
z
X1( z) 
za
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
50

If a polynomial has real coefficients, its roots are either
real or occur in complex conjugates.
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
51
x ( n)  a n u ( n )
X ( z) 
1
1  az 1
(z  a)
0  a 1
1  a  0
a 1
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
52
x ( n)  a n u ( n )
X ( z) 
1
1  az 1
(z  a)
a  1
a 1
a  1
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
53
az 1
X ( z) 
(1  az 1 )2
x(n)  na u (n)
n
(z  a)
0  a 1
1  a  0
a 1
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
54
az 1
X ( z) 
(1  az 1 )2
x(n)  na u (n)
n
(z  a)
a  1
a 1
a  1
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
55
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
56

For BIBO stability of an LTI system, we shall have

 h( n)  
n  

Now, let’s see what happens in the z-domain:

H ( z )   h( n) z

H ( z )   h( n) z
n
n  
n  
z 1
H ( z)
n

  h( n) z
n  
n

  h( n) z  n
n  

  h( n)
z 1
n  
•CONCLUSION: an LTI system is BIBO stable iff the ROC
of its impulse response includes the unit circle.
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
57



Causal LTI systems have h(n) = 0 for n<0.
ROC of the z-transform of a causal sequence is the
exterior of a circle.
CONCLUSION: an LTI system is causal iff the ROC of
its system function is the exterior of a circle of radius
r<∞ including the point z=∞.
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
58



Conditions for stability and causality are different and one
does not imply the other.
A causal system might be stable or unstable just as a noncausal system can be stable or unstable.
For causal systems though, the condition of stability can be
narrowed since:
◦ A causal system has z-transform with ROC outside a circle of radius r.
◦ The ROC of a stable system must contain the unit circle.
Then a causal and stable system ROC must be 𝑧 > 𝑟 with 𝑟 < 1.
This means that a causal LTI system is BIBO
stable iff all the poles of H(z) are inside the
unit circle.

 1
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
59



3  4 z 1
1
2
H ( z) 


1  3.5 z 1  1.5 z  2 1  12 z 1 1  3z 1
Example:
Specify ROC and h(n) for the following cases:
The system is stable:
ROC : 12  z  3

12 n u(n)  2(3)n u(n  1)
The system is causal:
ROC : z  3

h( n) 
h( n) 
12 n u(n)  2(3)n u(n)
The system is anti-causal:
ROC : z  0.5
h(n)  [
12 n  2(3)n ]u(n  1)
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
60



Discussed z-transform, region of convergence,
properties of z-transform and how the inverse ztransform can be calculated.
Discussed LCCDEs and the concept of the pole-zero
for their z-transform
Next: we will analyze the properties of ideal filters
Hossein Sameti, Dept. of Computer Eng.,
Sharif University of Technology
61