2. 2
n
n
z
n
x
z
X
n
x
Z )
(
)
(
)}
(
{ Vs.
n
n
j
e
n
x
X
)
(
)
(
Q: Why do we need Z-transform?
A: For some signals, Fourier transform does not converge.
)
(
2
)
( n
u
n
x n
e.g.:
Complex variable
Complex function
• Relationship between and :
)
(z
X )
(
X
)
(
)
(
X
z
X jw
e
z
Z-transform evaluated at unit circle
corresponds to DTFT.
The role of Z-transform in DT is
similar to Laplace transform in CT.
3.
n
n
z
n
x
z
X
n
x
Z )
(
)
(
)}
(
{
j
re
z r
z
z
For what values of z, does the z-transform converge?
n
n
j
n
n
re
n
x
z
n
x
z
X
)
(
)
(
)
(
n
n
j
n
e
r
n
x
z
X
)
(
)
( DTFT of n
r
n
x
)
(
DTFT of n
r
n
x
)
( exists if n
r
n
x
)
( is absolutely summable.
n
n
r
n
x )
( Z-transform exists and converges.
Convergence only depends on r.
3
4.
n
n
r
n
x )
( ROC is radially symmetric.
1
1
j
re
z
If Z-transform converges for It also converges for
2
2
j
re
z
• ROC does not depend on the angle.
• Region of Convergence (ROC): ROC is a region in “z”
domain that converges.
n
n
z
n
x )
(
4
5. There are the following possibilities:
1. All Z-domain except one or two points (origin / infinity)
2. Inside a circle
3. Outside a circle
4. Between two circles (ring)
5. ROC does not exist
5
8. )
(
)
( n
u
a
n
x n
n
n
n
n
u
z
a
z
X )
(
)
(
0
n
n
n
z
a
0
1
)
(
n
n
az
1
1
az 1
1
1
)
(
az
z
X
z
a
8
9. )
1
(
)
(
n
u
a
n
x n
1
)
(
n
n
n
z
a
z
X
0
1
1
)
(
1
m
m
m
m
m
z
a
z
a
1
1
z
a 1
1
1
)
(
az
z
X
a
z
• Same Z-transform as the
previous example. However,
the ROC is different.
9
10. We have shown that X(z) needs the ROC to uniquely
specify a sequence.
)
(
)
( 1
1 n
x
ROC
z
X
)
(
)
( 2
2 n
x
ROC
z
X
)
(
)
( n
u
a
n
x n
1
1
1
)
(
az
z
X
1
1
1
)
(
az
z
X
)
1
(
)
(
n
u
a
n
x n
10
14. 14
(2) The ROC does not contain any poles (same as in LT).
(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)
15. 15
(3) If x[n] is of finite duration, then the ROC is the entire z-
plane, except possibly at z = 0 and/or z = ∞.
Why?
Examples: CT counterpart
2
1
)
(
N
N
n
n
z
n
x
z
X
s
t
z
n all
ROC
1
)
(
all
ROC
1
s
e
e
T
t
z
z
n sT
)
(
0
ROC
1 1
s
e
e
T
t
z
z
n sT
)
(
ROC
1
Book Chapter10: Section 1
16. 16
(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.
1
1
0
1
n
faster tha
converges
N
n
n
n
N
n
r
n
x
r
n
x
Book Chapter10: Section 1
17. 17
What types of signals do the following ROC correspond to?
right-sided left-sided two-sided
(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.
Book Chapter10: Section 1
18. 18
Example:
0
,
|
|
b
b
n
x n
1
n
u
b
n
u
b
n
x n
n
b
z
z
b
n
u
b
b
z
bz
n
u
b
n
n
1
,
1
1
1
,
1
1
From：
1
1
1
Book Chapter10: Section 1
19. 19
Example, continued
Clearly, ROC does not exist if b > 1 ⇒ No z-transform for b|n|.
b
z
b
z
b
bz
z
X
1
,
1
1
1
1
)
( 1
1
1
Book Chapter10: Section 1
20. 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.
20
21. Linearity:
)
(
)
( 1
1 z
X
n
x z
)
(
)
( 2
2 z
X
n
x z
)
(
)
(
)
(
)
(
)
(
)
( 2
2
1
1
2
2
1
1 z
X
a
z
X
a
z
X
n
x
a
n
x
a
n
x z
1
ROC 2
ROC
ROC: at least the intersection of ROC1 and ROC2.
21
22. Example: ?
)
(
)
(
)
(cos
)
( 0
z
X
n
u
n
n
x
Solution: )
(
2
1
)
(
2
1
)
(
)
(cos
)
( 0
0
0 n
u
e
n
u
e
n
u
n
n
x n
j
n
j
Using linearity property: )}
(
2
1
)
(
2
1
{
)
( 0
0
n
u
e
n
u
e
z
z
X n
j
n
j
Earlier we showed that: )
(
)
( n
u
a
n
x n
1
1
1
)
(
az
z
X
If we set: 0
j
e
a
)
( z
a
)
(
)
( 0
n
u
e
n
x n
j
1
0
1
1
)
(
z
e
z
X j
1
1 0
0
1
1
2
1
1
1
2
1
)
(
z
e
z
e
z
X j
j
)
1
(
z
)
1
(
z
2
0
1
0
1
cos
2
1
cos
1
)
(
z
z
z
z
X
22
23. )
(
)
( z
X
n
x z
ROC
ROC: Similar ROC with some exceptions at 0 and infinity
)
(
)
( z
X
z
k
n
x k
z
Example: }
1
,
0
,
7
,
5
,
2
,
1
{
)
(
1
n
x ?
))
2
(
)
(
{ 1
2
n
x
n
x
Z
5
3
2
1
1 7
5
2
1
)
(
z
z
z
z
z
X
3
1
1
2
2 7
5
2
)
(
z
z
z
z
z
X
ROC?
23
24. )
(
)
( z
X
n
x z
2
1
: r
z
r
ROC
)
(
)
( 1
z
a
X
n
x
a z
n
Example:
2
1
: r
a
z
r
a
ROC
)
(
)
(cos
)
( 0 n
u
n
a
n
x n
)
(
)
(cos
)
( 0 n
u
n
n
x
2
0
1
0
1
cos
2
1
cos
1
)
(
z
z
z
z
X
)
1
(
z
Earlier:
)
(
)
(cos
)
( 0 n
u
n
a
n
x n
2
2
0
1
0
1
cos
2
1
cos
1
)
(
z
a
az
az
z
X
)
( a
z
24
25. )
(
)
( z
X
n
x z
2
1
: r
z
r
ROC
)
(
)
( 1
z
X
n
x z
Example:
1
2
1
1
:
r
z
r
ROC
)
( a
z
Earlier:
)
(
)
( n
u
n
x
?
)
(
z
X
)
(
)
( n
u
a
n
x n
1
1
1
)
(
az
z
X
)
(
)
( n
u
n
x 1
1
1
)
(
z
z
X )
1
(
z
)
(
)
( n
u
n
x
z
z
X
1
1
)
( )
1
(
z
25
26. )
(
)
( z
X
n
x z
ROC
dz
z
dX
z
n
nx z )
(
)
(
Example:
)
( a
z
Earlier:
)
(
)
( n
u
na
n
x n
?
)
(
z
X
)
(
)
(
1 n
u
a
n
x n
1
1
1
1
)
(
az
z
X
2
1
1
1
2
)
1
(
)
(
)
(
az
az
dz
z
dX
z
z
X
ROC
)
(
)
(
2 n
u
na
n
x n
)
( a
z
26
27. )
(
)
( 1
1 z
X
n
x z
)
(
)
( 2
2 z
X
n
x z
)
(
).
(
)
(
)
(
*
)
(
)
( 2
1
2
1 z
X
z
X
z
X
n
x
n
x
n
x z
1
ROC 2
ROC
ROC: at least the intersection of ROC1 and ROC2
Example:
1
,
2
,
1
)
(
1 n
x
elsewhere
,
0
5
0
,
1
)
(
2
n
n
x ?
)
(
*
)
( 2
1
n
x
n
x
2
1
1 2
1
)
(
z
z
z
X 5
4
3
2
1
2 1
)
(
z
z
z
z
z
z
X
7
6
1
2
1 1
)
(
)
(
)
(
z
z
z
z
X
z
X
z
X
1
,
1
,
0
,
0
,
0
,
0
,
1
,
1
)
(n
x
27
29. )
(z
X
)
(n
x
n
n
z
n
x
n
x
Z )
(
)}
(
{
)
(z
X )
(n
x
?
• 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
C
n
dz
z
z
X
j
1
)
(
2
1
29
30. 4
3
20
5
8
)
(
z
z
z
X
n
n
z
x
x
z
x
z
n
x
z
X ...
)
1
(
)
0
(
)
1
(
...
)
(
)
( 1
)
4
(
20
)
(
5
)
3
(
8
)
(
n
n
n
n
x
n
n
nz
c
z
X )
(
• Basic concept: expand X(z) in terms of powers of z-1
30
31. Example: ?
)
(
:
1
log
)
( 1
n
x
a
z
ROC
az
z
X
...
4
3
2
)
1
log(
4
3
2
1
x
x
x
x
x
x
...
4
3
2
)
1
log(
4
1
3
1
2
1
1
1
1
1
az
az
az
az
az
az
...
)
3
(
3
)
2
(
2
)
1
(
)
0
(
.
0
)
(
3
2
a
a
a
n
x
)
1
(
1
)
( 1
n
u
n
a
n
x n
n
31
32. This method is applicable if we have a ratio of two polynomials in
the form of .
We would like to rewrite X(z) in the following manner:
This expansion is performed in a way that the inverse Z-transform
of can be easily found using a lookup table and
using the linearity property, we can find x(n):
)
(
...
)
(
)
(
)
( 2
2
1
1 z
X
a
z
X
a
z
X
a
z
X N
N
)
(
),...,
(
),
( 2
1 z
X
z
X
z
X N
)
(
...
)
(
)
(
)
( 2
2
1
1 n
x
a
n
x
a
n
x
a
n
x N
N
)
(
)
( n
u
a
n
x n
1
1
1
)
(
az
z
X
2
1
1
)
1
(
)
(
az
az
z
X
)
(
)
( n
u
na
n
x n
)
( a
z
Example:
N
N
M
M
z
a
z
a
a
z
b
z
b
b
z
Q
z
P
z
X
...
...
)
(
)
(
)
(
1
1
0
1
1
0
32
)
( a
z
33. Any non-constant single-variable polynomial of degree
N can be factored into N terms:
If the coefficients of the polynomial are real, then the
roots are either real or complex conjugates.
If is a complex root, then is also a root.
3
3
2
2
1
1
0
)
( z
a
z
a
z
a
a
z
P
)
)(
)(
( 3
2
1 z
z
z
z
z
z
k
0
z
*
0
z
33
34. Example:
5
.
0
5
.
1
5
.
0
5
.
1
1
1
)
( 2
2
2
1
z
z
z
z
z
z
X
)
5
.
0
)(
1
(
)
(
z
z
z
z
z
X
)
1
(
)
5
.
0
( 2
1
z
A
z
A
z
1
z 2
1
A
5
.
0
z 1
2
A
5
.
0
1
2
)
(
z
z
z
z
z
X
1
:
z
ROC )
(
]
5
.
0
2
[
)
(
)
5
.
0
(
)
(
)
1
.(
2
)
( n
u
n
u
n
u
n
x n
n
n
)
5
.
0
)(
1
(
5
.
0
1
)
1
(
2
)
5
.
0
(
1
2
1
z
z
A
z
A
z
A z
A
z
1
1
5
.
0
1
1
1
2
z
z
34
35. Example: 2
1
1
)
1
)(
1
(
1
)
(
z
z
z
X
2
2
)
1
)(
1
(
)
(
z
z
z
z
z
X
)
1
(
)
1
)(
1
(
)
1
( 3
2
2
1
2
z
A
z
z
A
z
A
z
2
1
1
1
1
)
1
(
5
.
0
1
75
.
0
1
25
.
0
)
(
z
z
z
z
z
X
)
(
)
1
(
5
.
0
)
(
)
1
(
75
.
0
)
(
)
1
(
25
.
0
)
( n
u
n
n
u
n
u
n
x n
n
n
x(n) causal
• In the case of complex conjugate roots, the coefficients in the
partial fraction expansion will be complex conjugates.
2
3
2
1
1
1
1
z
A
z
A
z
A
2
3
)
1
)(
1
(
z
z
z
35
36. 36
Example:
Partial Fraction Expansion Algebra: A = 1, B = 2
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.
1
1
1
1
1
2
3
1
1
4
1
1
3
1
1
4
1
1
6
5
3
3
1
4
1
6
5
3
)
(
z
B
z
A
z
z
z
z
z
z
z
z
X
n
x
n
x
n
x
z
z
z
X
2
1
1
1
]
[
3
1
1
2
4
1
1
1
)
(
37. 37
ROC III:
ROC II:
ROC I:
n
u
n
x
n
u
n
x
z
n
n
3
1
2
4
1
singnal
sided
-
right
-
3
1
2
1
1
3
1
2
4
1
singnal
sided
-
two
-
3
1
4
1
2
1
n
u
n
x
n
u
n
x
z
n
n
1
3
1
2
1
4
1
singnal
sided
-
left
-
4
1
2
1
n
u
n
x
n
u
n
x
z
n
n
39. LTI Systems
h(n)
FIR IIR
With rational transfer
function
)
(
)
(
)
(
z
Q
z
P
z
H
No rational transfer
function
)
(
)
(
)
(
z
Q
z
P
z
H
Use difference
equations for
implementation
h(n),H (z)
x(n),X(z) y(n),Y(z) )
(
).
(
)
(
)
(
*
)
(
)
(
z
X
z
H
z
Y
n
x
n
h
n
y
Transfer
function
39
40. Cumulative Average System:
n
k
n
k
x
n
n
y
0
,...
1
,
0
),
(
1
1
)
(
)
(
)
(
)
(
)
1
(
1
0
n
x
k
x
n
y
n
n
k
)
(
)
1
( n
x
n
ny
)
(
1
1
)
1
(
1
)
( n
x
n
n
y
n
n
n
y
40
41. The general form for a linear constant-coefficient
difference equation is given by:
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.
M
k
k
N
k
k k
n
x
b
k
n
y
a
n
y
0
1
)
(
)
(
)
(
Order of the system
41
42. Question: How can an LCCDE correspond to an LTI system?
E.g.:
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 is a solution, is also a solution.
Q: How can we resolve this problem?
A: We need to add an initial condition (I.C.).
)
(
)
1
(
)
( n
x
n
ay
n
y
)
(
1 n
y
n
ka
n
y
n
y
)
(
)
( 1
2
)
(
)
1
(
)
( n
x
n
ay
n
y
)
(
)
( n
b
n
x
0
)
1
(
:
IC y
y
42
43. )
(
)
1
(
)
( n
x
n
ay
n
y
)
(
)
( n
b
n
x
0
)
1
(
:
IC y
y
:
0
n b
ay
b
ay
y
0
)
0
(
)
1
(
)
0
(
ab
y
a
b
ay
a
ay
y
0
2
0 )
(
0
)
0
(
)
1
(
)
(
)
( 0 b
ay
a
n
y n
:
2
n )
(
)
1
(
)
( n
x
n
ay
n
y
)
(
)
(
1
)
1
( n
a
b
n
y
a
n
y
0
1
)
1
(
)
1
(
1
)
2
( y
a
a
b
y
a
y
Causal implementation
Anti-causal implementation
0
2
1
)
3
( y
a
y
0
1
)
( y
a
n
y n
:
1
n
43
44. )
1
(
)
(
)
(
)
( 0
1
0
n
u
y
a
n
u
b
ay
a
n
y n
n
)
(
)
( 0
1
n
bu
a
y
a
n
y n
n
• By Combining the two answers, we get:
• Q: Does this system correspond to a linear system?
)
(
)
1
(
)
( n
x
n
ay
n
y
)
(
)
( n
b
n
x
0
)
1
(
:
IC y
y
• A: No, zero input does not result in zero output.
• An LCCDE corresponds to a linear system, when IC=0.
44
45. )
(
)
( 0
1
n
bu
a
y
a
n
y n
n
• Q: Does this system correspond to a TI system?
)
(
)
1
(
)
( n
x
n
ay
n
y
)
(
)
( n
b
n
x
0
)
1
(
:
IC y
y
• A: Consider
• In order for an LCCDE to correspond to a TI system, IC=0.
(for 𝑛0 ≥ 0)
)
(
)
( n
b
n
x
. Now let’s shift the input: )
(
)
( 0
n
n
b
n
w
Output: )
( 0
0
1 0
n
n
u
ba
y
a n
n
n
)
(
)
( 0
0
1
0
0
0
n
n
u
ba
y
a
n
n
y n
n
n
n
Shift
variant
• In order for an LCCDE to correspond to an LTI system, IC=0.
45
46. In order an LCCDE to correspond to a causal LTI system :
◦ Initial rest condition (IRC)=0
E.g., if , for a first order LCCDE, the IRC should
be as follows:
E.g., if , for a second order LCCDE, the IRC
should be as follows:
)
(
)
( n
n
x
0
)
1
(
y
)
3
(
)
(
n
n
x
0
)
2
(
,
0
)
1
(
y
y
46
47. )
2
(
)
1
(
)
(
2
)
1
(
2
1
)
(
n
x
n
x
n
x
n
y
n
y 0
;
0
)
(
n
n
y
)
2
(
)
1
(
)
(
2
)
1
(
2
1
)
(
n
n
n
n
h
n
h
Solution1: )
2
(
)
1
(
)
(
2
)
1
(
2
1
)
(
n
n
n
n
h
n
h
0
)
1
(
h
2
0
0
2
0
0
0
)
0
(
2
)
1
(
2
1
)
0
(
h
h
1
n
0
n
1
n 1
2
2
1
0
)
0
(
0
)
0
(
2
1
)
1
(
h
h
2
n 1
1
)
2
1
(
2
)
2
1
(
1
)
1
2
2
1
(
2
1
)
0
(
0
0
)
1
(
2
1
)
2
( 2
h
h
3
n
2
1
1
)
2
1
(
2
)
2
1
(
)
1
1
)
2
1
(
2
)
2
1
((
2
1
)
2
(
2
1
)
3
( 2
3
2
h
h
3
,
1
)
2
1
(
1
)
2
1
(
2
)
2
1
(
)
1
)
2
1
(
1
)
2
1
(
2
)
2
1
((
2
1
)
1
(
2
1
)
(
2
1
3
2
1
n
n
h
n
h
n
n
n
n
n
n
47
3
n
48. )
2
(
)
1
(
)
(
2
)
1
(
2
1
)
(
n
x
n
x
n
x
n
y
n
y 0
;
0
)
(
n
n
y
)
(
)
(
)
(
2
)
(
2
1
)
( 2
1
1
z
X
z
z
X
z
z
X
z
Y
z
z
Y
1
2
1
1
1
2
1
1
2
1
1
2
1
1
2
)
(
)
(
)
(
z
z
z
z
z
z
X
z
Y
z
H
)
( a
z
)
(
)
( n
u
a
n
x n
1
1
1
)
(
az
z
X
Remember:
)
2
(
2
1
)
1
(
2
1
)
(
2
1
2
)
(
2
1
n
u
n
u
n
u
n
h
n
n
n
h(n),H (z)
x(n),X(z) y(n),Y(z)
)
(
)
(
)
(
)
(
).
(
)
(
z
X
z
Y
z
H
z
X
z
H
z
Y
48
49. Z-transform with rational transfer function:
N
k
k
k
M
k
k
k
N
N
M
M
z
a
z
b
z
a
z
a
a
z
b
z
b
b
z
A
z
B
z
X
0
0
1
1
0
1
1
0
...
...
)
(
)
(
)
(
Zero: X(z)=0 Pole: X(z)=∞
0
,
0 0
0
b
a
Getting rid of the negative powers:
0
1
0
1
0
1
0
1
0
0
...
...
)
(
)
(
)
(
a
a
z
a
a
z
b
b
z
b
b
z
z
a
z
b
z
A
z
B
z
X
N
N
N
M
M
M
N
M
)
)...(
)(
(
)
)...(
)(
(
2
1
2
1
0
0
N
M
N
M
p
z
p
z
p
z
z
z
z
z
z
z
z
a
b
Poles and zeros?
49
50. A pole-zero plot can represent X(z) graphically.
)
( a
z
Example: )
(
)
(
1 n
u
a
n
x n
1
1
1
1
)
(
az
z
X
a
z
z
z
X
)
(
1
50
51. If a polynomial has real coefficients, its roots are either
real or occur in complex conjugates.
51
52. )
( a
z
)
(
)
( n
u
a
n
x n
1
1
1
)
(
az
z
X
1
0
a
0
1
a
1
a
52
53. )
( a
z
)
(
)
( n
u
a
n
x n
1
1
1
)
(
az
z
X
1
a
1
a
1
a
53
54. )
( a
z
)
(
)
( n
u
na
n
x n
2
1
1
)
1
(
)
(
az
az
z
X
1
0
a
0
1
a
1
a
54
55. )
( a
z
)
(
)
( n
u
na
n
x n
2
1
1
)
1
(
)
(
az
az
z
X
1
a
1
a
1
a
55
57. For BIBO stability of an LTI system, we shall have
Now, let’s see what happens in the z-domain:
n
n
h )
(
n
n
z
n
h
z
H )
(
)
(
n
n
z
n
h
z
H )
(
)
(
n
n
n
n
z
n
h
z
n
h )
(
)
(
1
z
n
z
n
h
z
H )
(
)
( 1
•CONCLUSION: an LTI system is BIBO stable iff the ROC
of its impulse response includes the unit circle.
57
58. 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=∞.
58
59. 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 non-
causal 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
59
60. Example:
Specify ROC and h(n) for the following cases:
The system is stable:
The system is causal:
The system is anti-causal:
1
1
2
1
2
1
1
3
1
2
1
1
5
.
1
5
.
3
1
4
3
)
(
z
z
z
z
z
z
H
3
: 2
1
z
ROC )
1
(
)
3
(
2
)
(
)
( 2
1
n
u
n
u
n
h n
n
3
:
z
ROC )
(
)
3
(
2
)
(
)
( 2
1
n
u
n
u
n
h n
n
5
.
0
:
z
ROC )
1
(
]
)
3
(
2
[
)
( 2
1
n
u
n
h n
n
60
61. Discussed z-transform, region of convergence,
properties of z-transform and how the inverse z-
transform 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
61
