The document discusses the z-transform, which is a generalization of the Fourier transform used to analyze discrete-time signals and systems. It defines the z-transform, explains why it is used instead of the Fourier transform, and discusses its region of convergence and properties of poles and zeros. Examples are provided to illustrate key concepts such as how the region of convergence is determined by the signal type (left-sided, right-sided, two-sided). Important z-transform pairs are also summarized.

1. EC 2314 Digital Signal Processing
By
Dr. K. Udhayakumar

2. The z-Transform
Dr. K. Udhayakumar

3. Content
 Introduction
 z-Transform
 Zeros and Poles
 Region of Convergence
 Important z-Transform Pairs
 Inverse z-Transform
 z-Transform Theorems and Properties
 System Function

4. The z-Transform
Introduction

5. Why z-Transform?
 A generalization of Fourier transform
 Why generalize it?
– FT does not converge on all sequence
– Notation good for analysis
– Bring the power of complex variable theory deal with
the discrete-time signals and systems

6. The z-Transform
z-Transform

7. Definition
 The z-transform of sequence x(n) is defined by






n
n
z
n
x
z
X )
(
)
(
 Let z = ej.
( ) ( )
j j n
n
X e x n e
 



 
Fourier
Transform

8. z-Plane
Re
Im
z = ej







n
n
z
n
x
z
X )
(
)
(
( ) ( )
j j n
n
X e x n e
 



 
Fourier Transform is to evaluate z-transform
on a unit circle.

9. z-Plane
Re
Im
X(z)
Re
Im
z = ej


10. Periodic Property of FT
Re
Im
X(z)

 
X(ej)
Can you say why Fourier Transform is
a periodic function with period 2?

11. The z-Transform
Zeros and Poles

12. Definition
 Give a sequence, the set of values of z for which the
z-transform converges, i.e., |X(z)|<, is called the
region of convergence.



 









n
n
n
n
z
n
x
z
n
x
z
X |
||
)
(
|
)
(
|
)
(
|
ROC is centered on origin and
consists of a set of rings.

13. Example: Region of Convergence
Re
Im



 









n
n
n
n
z
n
x
z
n
x
z
X |
||
)
(
|
)
(
|
)
(
|
ROC is an annual ring centered
on the origin.

 
 x
x R
z
R |
|
r
}
|
{ 





 x
x
j
R
r
R
re
z
ROC

14. Stable Systems
Re
Im
1
 A stable system requires that its Fourier transform is
uniformly convergent.
 Fact: Fourier transform is to
evaluate z-transform on a unit
circle.
 A stable system requires the
ROC of z-transform to include
the unit circle.

15. Example: A right sided Sequence
)
(
)
( n
u
a
n
x n

1 2 3 4 5 6 7 8 9 10
-1
-2
-3
-4
-5
-6
-7
-8
n
x(n)
. . .

16. Example: A right sided Sequence
)
(
)
( n
u
a
n
x n

n
n
n
z
n
u
a
z
X 




 )
(
)
(





0
n
n
n
z
a





0
1
)
(
n
n
az
For convergence of X(z), we
require that






0
1
|
|
n
az 1
|
| 1


az
|
|
|
| a
z 
a
z
z
az
az
z
X
n
n




 



 1
0
1
1
1
)
(
)
(
|
|
|
| a
z 

17. a
a
Example: A right sided Sequence
ROC for x(n)=anu(n)
|
|
|
|
,
)
( a
z
a
z
z
z
X 


Re
Im
1
a
a
Re
Im
1
Which one is stable?

18. Example: A left sided Sequence
)
1
(
)
( 


 n
u
a
n
x n
1 2 3 4 5 6 7 8 9 10
-1
-2
-3
-4
-5
-6
-7
-8
n
x(n)
.
.
.

19. Example: A left sided Sequence
)
1
(
)
( 


 n
u
a
n
x n
n
n
n
z
n
u
a
z
X 



 


 )
1
(
)
(
For convergence of X(z), we
require that






0
1
|
|
n
z
a 1
|
| 1


z
a
|
|
|
| a
z 
a
z
z
z
a
z
a
z
X
n
n






 



 1
0
1
1
1
1
)
(
1
)
(
|
|
|
| a
z 
n
n
n
z
a 






1
n
n
n
z
a






1
n
n
n
z
a






0
1

20. a
a
Example: A left sided Sequence
ROC for x(n)=anu( n1)
|
|
|
|
,
)
( a
z
a
z
z
z
X 


Re
Im
1
a
a
Re
Im
1
Which one is stable?

21. The z-Transform
Region of
Convergence

22. Represent z-transform as a
Rational Function
)
(
)
(
)
(
z
Q
z
P
z
X 
where P(z) and Q(z) are
polynomials in z.
Zeros: The values of z’s such that X(z) = 0
Poles: The values of z’s such that X(z) = 

23. Example: A right sided Sequence
)
(
)
( n
u
a
n
x n
 |
|
|
|
,
)
( a
z
a
z
z
z
X 


Re
Im
a
ROC is bounded by the
pole and is the exterior
of a circle.

24. Example: A left sided Sequence
)
1
(
)
( 


 n
u
a
n
x n
|
|
|
|
,
)
( a
z
a
z
z
z
X 


Re
Im
a
ROC is bounded by the
pole and is the interior
of a circle.

25. Example: Sum of Two Right Sided Sequences
)
(
)
(
)
(
)
(
)
( 3
1
2
1
n
u
n
u
n
x n
n



3
1
2
1
)
(




z
z
z
z
z
X
Re
Im
1/2
)
)(
(
)
(
2
3
1
2
1
12
1




z
z
z
z
1/3
1/12
ROC is bounded by poles
and is the exterior of a circle.
ROC does not include any pole.

26. Example: A Two Sided Sequence
)
1
(
)
(
)
(
)
(
)
( 2
1
3
1




 n
u
n
u
n
x n
n
2
1
3
1
)
(




z
z
z
z
z
X
Re
Im
1/2
)
)(
(
)
(
2
2
1
3
1
12
1




z
z
z
z
1/3
1/12
ROC is bounded by poles
and is a ring.
ROC does not include any pole.

27. Example: A Finite Sequence
1
0
,
)
( 


 N
n
a
n
x n
n
N
n
n
N
n
n
z
a
z
a
z
X )
(
)
( 1
1
0
1
0







 

Re
Im
ROC: 0 < z < 
ROC does not include any pole.
1
1
1
)
(
1





az
az N
a
z
a
z
z
N
N
N


 1
1
N-1 poles
N-1 zeros
Always Stable

28. Properties of ROC
 A ring or disk in the z-plane centered at the origin.
 The Fourier Transform of x(n) is converge absolutely iff the ROC
includes the unit circle.
 The ROC cannot include any poles
 Finite Duration Sequences: The ROC is the entire z-plane except
possibly z=0 or z=.
 Right sided sequences: The ROC extends outward from the outermost
finite pole in X(z) to z=.
 Left sided sequences: The ROC extends inward from the innermost
nonzero pole in X(z) to z=0.

29. More on Rational z-Transform
Re
Im
a b c
Consider the rational z-transform
with the pole pattern:
Find the possible
ROC’s

30. More on Rational z-Transform
Re
Im
a b c
Consider the rational z-transform
with the pole pattern:
Case 1: A right sided Sequence.

31. More on Rational z-Transform
Re
Im
a b c
Consider the rational z-transform
with the pole pattern:
Case 2: A left sided Sequence.

32. More on Rational z-Transform
Re
Im
a b c
Consider the rational z-transform
with the pole pattern:
Case 3: A two sided Sequence.

33. More on Rational z-Transform
Re
Im
a b c
Consider the rational z-transform
with the pole pattern:
Case 4: Another two sided Sequence.

34. 0 5 10 15 20
0
0.2
0.4
0.6
0.8
1
Bounded Signals
-5
0
5
a=0.4
-5
0
5
a=0.9
-5
0
5
a=1.2
0 5 10
-5
0
5
a=-0.4
0 5 10
-5
0
5
a=-0.9
0 5 10
-5
0
5
a=-1.2
0 10 20 30 40 50 60 70
-1
-0.5
0
0.5
1
0 2 4 6 8
-1
-0.5
0
0.5
1

35. BIBO Stability
 Bounded Input Bounded Output Stability
– If the Input is bounded, we want the Output is
bounded, too
– If the Input is unbounded, it’s okay for the Output to
be unbounded
 For some computing systems, the output is
intrinsically bounded (constrained), but limit
cycle may happen

36. The z-Transform
Important
z-Transform Pairs

37. Z-Transform Pairs
Sequence z-Transform ROC
)
(n
 1 All z
)
( m
n 
 m
z All z except 0 (if m>0)
or  (if m<0)
)
(n
u 1
1
1

 z
1
|
| 
z
)
1
( 

 n
u 1
1
1

 z
1
|
| 
z
)
(n
u
an 1
1
1

 az
|
|
|
| a
z 
)
1
( 

 n
u
an 1
1
1

 az
|
|
|
| a
z 

38. Z-Transform Pairs
Sequence z-Transform ROC
)
(
]
[cos 0 n
u
n
 2
1
0
1
0
]
cos
2
[
1
]
[cos
1








z
z
z
1
|
| 
z
)
(
]
[sin 0 n
u
n
 2
1
0
1
0
]
cos
2
[
1
]
[sin







z
z
z
1
|
| 
z
)
(
]
cos
[ 0 n
u
n
rn
 2
2
1
0
1
0
]
cos
2
[
1
]
cos
[
1








z
r
z
r
z
r
r
z 
|
|
)
(
]
sin
[ 0 n
u
n
rn
 2
2
1
0
1
0
]
cos
2
[
1
]
sin
[







z
r
z
r
z
r
r
z 
|
|


 


otherwise
0
1
0 N
n
an
1
1
1




az
z
a N
N
0
|
| 
z

39. Signal Type ROC
Finite-Duration Signals
Infinite-Duration Signals
Causal
Anticausal
Two-sided
Causal
Anticausal
Two-sided
Entire z-plane
Except z = 0
Entire z-plane
Except z = infinity
Entire z-plane
Except z = 0
And z = infinity
|z| < r1
|z| > r2
r2 < |z| < r1

40. Some Common z-Transform Pairs
Sequence Transform ROC
1. [n] 1 all z
2. u[n] z/(z-1) |z|>1
3. -u[-n-1] z/(z-1) |z|<1
4. [n-m] z-m all z except 0 if m>0 or ฅ if m<0
5. anu[n] z/(z-a) |z|>|a|
6. -anu[-n-1] z/(z-a) |z|<|a|
7. nanu[n] az/(z-a)2 |z|>|a|
8. -nanu[-n-1] az/(z-a)2 |z|<|a|
9. [cos0n]u[n] (z2-[cos0]z)/(z2-[2cos0]z+1) |z|>1
10. [sin0n]u[n] [sin0]z)/(z2-[2cos0]z+1) |z|>1
11. [rncos0n]u[n] (z2-[rcos0]z)/(z2-[2rcos0]z+r2) |z|>r
12. [rnsin0n]u[n] [rsin0]z)/(z2-[2rcos0]z+r2) |z|>r
13. anu[n] - anu[n-N] (zN-aN)/zN-1(z-a) |z|>0

41. The z-Transform
Inverse z-Transform

42. Inverse Z-Transform by Partial Fraction
Expansion
 Assume that a given z-transform can be expressed as
 Apply partial fractional expansion
 First term exist only if M>N
– Br is obtained by long division
 Second term represents all first order poles
 Third term represents an order s pole
– There will be a similar term for every high-order pole
 Each term can be inverse transformed by inspection
 






 N
k
k
k
M
k
k
k
z
a
z
b
z
X
0
0
 
 


 












s
1
m
m
1
i
m
N
i
k
,
1
k
1
k
k
N
M
0
r
r
r
z
d
1
C
z
d
1
A
z
B
z
X

43. Partial Fractional Expression
 Coefficients are given as
 Easier to understand with examples
 
 


 












s
1
m
m
1
i
m
N
i
k
,
1
k
1
k
k
N
M
0
r
r
r
z
d
1
C
z
d
1
A
z
B
z
X
    k
d
z
1
k
k z
X
z
d
1
A 



   
   
  1
i
d
w
1
s
i
m
s
m
s
m
s
i
m w
X
w
d
1
dw
d
d
!
m
s
1
C

















44. Example: 2nd Order Z-Transform
– Order of nominator is smaller than denominator (in terms of z-1)
– No higher order pole
 
2
1
z
:
ROC
2
1
1
4
1
1
1
1
1


















z
z
z
X
 

















 1
2
1
1
z
2
1
1
A
z
4
1
1
A
z
X
  1
4
1
2
1
1
1
z
X
z
4
1
1
A 1
4
1
z
1
1 
























 


  2
2
1
4
1
1
1
z
X
z
2
1
1
A 1
2
1
z
1
2 























 



45. Example Continued
 ROC extends to infinity
– Indicates right sided sequence
 
2
1
z
z
2
1
1
2
z
4
1
1
1
z
X
1
1




















     
n
u
4
1
-
n
u
2
1
2
n
x
n
n














46. Example #2
 Long division to obtain Bo
   
 
1
z
z
1
z
2
1
1
z
1
z
2
1
z
2
3
1
z
z
2
1
z
X
1
1
2
1
2
1
2
1























1
z
5
2
z
3
z
2
1
z
2
z
1
z
2
3
z
2
1
1
1
2
1
2
1
2














 
 
1
1
1
z
1
z
2
1
1
z
5
1
2
z
X















  1
2
1
1
z
1
A
z
2
1
1
A
2
z
X 
 




  9
z
X
z
2
1
1
A
2
1
z
1
1 











    8
z
X
z
1
A
1
z
1
2 





47. Example #2 Continued
 ROC extends to infinity
– Indicates right-sides sequence
  1
z
z
1
8
z
2
1
1
9
2
z
X 1
1





 

       
n
8u
-
n
u
2
1
9
n
2
n
x
n










48. An Example – Complete Solution
3
8
6z
z
14
14z
3z
lim
U(z)
lim
c 2
2
z
z
0 










4
-
z
14
14z
3z
8
6z
z
14
14z
3z
2)
(z
(z)
U
2
2
2
2









2
-
z
14
14z
3z
8
6z
z
14
14z
3z
4)
(z
(z)
U
2
2
2
4









8
6z
z
14
14z
3z
U(z) 2
2





4
z
c
2
z
c
c
U(z) 2
1
0





1
4
-
2
14
2
14
2
3
(2)
U
c
2
2
1 






3
2
-
4
14
4
14
4
3
(4)
U
c
2
4
2 






4
z
3
2
z
1
3
U(z)












 

0
k
,
4
3
2
0
k
3,
u(k) 1
k
1
k

49. Inverse Z-Transform by Power Series
Expansion
 The z-transform is power series
 In expanded form
 Z-transforms of this form can generally be inversed easily
 Especially useful for finite-length series
 Example
   






n
n
z
n
x
z
X
            
 







 
 2
1
1
2
2
1
0
1
2 z
x
z
x
x
z
x
z
x
z
X
    
1
2
1
1
1
2
z
2
1
1
z
2
1
z
z
1
z
1
z
2
1
1
z
z
X


















         
1
n
2
1
n
1
n
2
1
2
n
n
x 










 



















2
n
0
1
n
2
1
0
n
1
1
n
2
1
2
n
1
n
x

50. Z-Transform Properties: Linearity
 Notation
 Linearity
– Note that the ROC of combined sequence may be larger than either ROC
– This would happen if some pole/zero cancellation occurs
– Example:
 Both sequences are right-sided
 Both sequences have a pole z=a
 Both have a ROC defined as |z|>|a|
 In the combined sequence the pole at z=a cancels with a zero at z=a
 The combined ROC is the entire z plane except z=0
 We did make use of this property already, where?
    x
Z
R
ROC
z
X
n
x 

 

        2
1 x
x
2
1
Z
2
1 R
R
ROC
z
bX
z
aX
n
bx
n
ax 



 


     
N
-
n
u
a
-
n
u
a
n
x n
n


51. Z-Transform Properties: Time Shifting
 Here no is an integer
– If positive the sequence is shifted right
– If negative the sequence is shifted left
 The ROC can change the new term may
– Add or remove poles at z=0 or z=
 Example
    x
n
Z
o R
ROC
z
X
z
n
n
x o


 

 
 
4
1
z
z
4
1
1
1
z
z
X
1
1

















   
1
-
n
u
4
1
n
x
1
-
n








52. Z-Transform Properties: Multiplication by Exponential
 ROC is scaled by |zo|
 All pole/zero locations are scaled
 If zo is a positive real number: z-plane shrinks or expands
 If zo is a complex number with unit magnitude it rotates
 Example: We know the z-transform pair
 Let’s find the z-transform of
    x
o
o
Z
n
o R
z
ROC
z
z
X
n
x
z 

 /
  1
z
:
ROC
z
-
1
1
n
u 1
-
Z


 

             
n
u
re
2
1
n
u
re
2
1
n
u
n
cos
r
n
x
n
j
n
j
o
n o
o 






  r
z
z
re
1
2
/
1
z
re
1
2
/
1
z
X 1
j
1
j o
o




 





53. Z-Transform Properties: Differentiation
 Example: We want the inverse z-transform of
 Let’s differentiate to obtain rational expression
 Making use of z-transform properties and ROC
   
x
Z
R
ROC
dz
z
dX
z
n
nx 


 

    a
z
az
1
log
z
X 1


 
   
1
1
1
2
az
1
1
az
dz
z
dX
z
az
1
az
dz
z
dX











     
1
n
u
a
a
n
nx
1
n




     
1
n
u
n
a
1
n
x
n
1
n





54. Z-Transform Properties: Conjugation
 Example
    x
*
*
Z
*
R
ROC
z
X
n
x 

 

   
     
        
 
n
x
Z
z
n
x
z
n
x
z
X
z
n
x
z
n
x
z
X
z
n
x
z
X
n
n
n
n
n
n
n
n
n
n















































55. Z-Transform Properties: Time Reversal
 ROC is inverted
 Example:
 Time reversed version of
   
x
Z
R
1
ROC
z
/
1
X
n
x 

 


   
n
u
a
n
x n

 
 
n
u
an
  1
1
1
-
1
-1
a
z
z
a
-
1
z
a
-
az
1
1
z
X 







56. Z-Transform Properties: Convolution
 Convolution in time domain is multiplication in z-domain
 Example:Let’s calculate the convolution of
 Multiplications of z-transforms is
 ROC: if |a|<1 ROC is |z|>1 if |a|>1 ROC is |z|>|a|
 Partial fractional expansion of Y(z)
        2
x
1
x
2
1
Z
2
1 R
R
:
ROC
z
X
z
X
n
x
n
x 

 


       
n
u
n
x
and
n
u
a
n
x 2
n
1 

  a
z
:
ROC
az
1
1
z
X 1
1 

    1
z
:
ROC
z
1
1
z
X 1
2 

 
     
  
1
1
2
1
z
1
az
1
1
z
X
z
X
z
Y 





  1
z
:
ROC
asume
az
1
1
z
1
1
a
1
1
z
Y 1
1











 

     
 
n
u
a
n
u
a
1
1
n
y 1
n




57. The z-Transform
z-Transform Theorems
and Properties

58. Linearity
x
R
z
z
X
n
x 
 ),
(
)]
(
[
Z
y
R
z
z
Y
n
y 
 ),
(
)]
(
[
Z
y
x R
R
z
z
bY
z
aX
n
by
n
ax 



 ),
(
)
(
)]
(
)
(
[
Z
Overlay of
the above two
ROC’s

59. Shift
x
R
z
z
X
n
x 
 ),
(
)]
(
[
Z
x
n
R
z
z
X
z
n
n
x 

 )
(
)]
(
[ 0
0
Z

60. Multiplication by an Exponential Sequence



 x
x- R
z
R
z
X
n
x |
|
),
(
)]
(
[
Z
x
n
R
a
z
z
a
X
n
x
a 

 
|
|
)
(
)]
(
[ 1
Z

61. Differentiation of X(z)
x
R
z
z
X
n
x 
 ),
(
)]
(
[
Z
x
R
z
dz
z
dX
z
n
nx 


)
(
)]
(
[
Z

62. Conjugation
x
R
z
z
X
n
x 
 ),
(
)]
(
[
Z
x
R
z
z
X
n
x 
 *)
(
*
)]
(
*
[
Z

63. Reversal
x
R
z
z
X
n
x 
 ),
(
)]
(
[
Z
x
R
z
z
X
n
x /
1
)
(
)]
(
[ 1


 
Z

64. Real and Imaginary Parts
x
R
z
z
X
n
x 
 ),
(
)]
(
[
Z
x
R
z
z
X
z
X
n
x
e 

 *)]
(
*
)
(
[
)]
(
[ 2
1
R
x
j R
z
z
X
z
X
n
x 

 *)]
(
*
)
(
[
)]
(
[ 2
1
Im

65. Initial Value Theorem
0
for
,
0
)
( 
 n
n
x
)
(
lim
)
0
( z
X
x
z 



66. Convolution of Sequences
x
R
z
z
X
n
x 
 ),
(
)]
(
[
Z
y
R
z
z
Y
n
y 
 ),
(
)]
(
[
Z
y
x R
R
z
z
Y
z
X
n
y
n
x 

 )
(
)
(
)]
(
*
)
(
[
Z

67. Convolution of Sequences






k
k
n
y
k
x
n
y
n
x )
(
)
(
)
(
*
)
(
 















n
n
k
z
k
n
y
k
x
n
y
n
x )
(
)
(
)]
(
*
)
(
[
Z
 









k
n
n
z
k
n
y
k
x )
(
)
(  









k
n
n
k
z
n
y
z
k
x )
(
)
(
)
(
)
( z
Y
z
X


68. The z-Transform
System Function

69. Signal Characteristics from Z-
Transform
 If U(z) is a rational function, and
 Then Y(z) is a rational function, too
 Poles are more important – determine key
characteristics of y(k)
m)
u(k
b
...
1)
u(k
b
n)
y(k
a
...
1)
y(k
a
y(k) m
1
n
1 
















 m
1
j
j
n
1
i
i
)
p
(z
)
z
(z
D(z)
N(z)
Y(z)
zeros
poles

70. Why are poles important?













m
1
j j
j
0
m
1
j
j
n
1
i
i
p
z
c
c
)
p
(z
)
z
(z
D(z)
N(z)
Y(z)






m
1
j
1
-
k
j
j
impulse
0 p
c
(k)
u
c
Y(k)
Z-
1
Z domain
Time domain
poles
componen
ts

71. Various pole values (1)
-1 0 1 2 3 4 5 6 7 8 9
0
0.5
1
1.5
2
2.5
-1 0 1 2 3 4 5 6 7 8 9
0
0.2
0.4
0.6
0.8
1
-1 0 1 2 3 4 5 6 7 8 9
0
0.2
0.4
0.6
0.8
1
-1 0 1 2 3 4 5 6 7 8 9
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
-1 0 1 2 3 4 5 6 7 8 9
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
-1 0 1 2 3 4 5 6 7 8 9
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
p=1.1
p=1
p=0.9
p=-1.1
p=-1
p=-0.9

72. Various pole values (2)
-1 0 1 2 3 4 5 6 7 8 9
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
-1 0 1 2 3 4 5 6 7 8 9
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
-1 0 1 2 3 4 5 6 7 8 9
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
p=0.9
p=0.6
p=0.3
-1 0 1 2 3 4 5 6 7 8 9
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
-1 0 1 2 3 4 5 6 7 8 9
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
-1 0 1 2 3 4 5 6 7 8 9
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
p=-0.9
p=-0.6
p=-0.3

73. Conclusion for Real Poles
 If and only if all poles’ absolute values are
smaller than 1, y(k) converges to 0
 The smaller the poles are, the faster the
corresponding component in y(k) converges
 A negative pole’s corresponding component is
oscillating, while a positive pole’s
corresponding component is monotonous

74. How fast does it converge?
 U(k)=ak, consider u(k)≈0 when the absolute
value of u(k) is smaller than or equal to 2% of
u(0)’s absolute value
|
a
|
ln
4
k
3.912
ln0.02
|
a
|
kln
0.02
|
a
| k






11
0.36
4
|
0.7
|
ln
4
k
0.7
a







Rememb
er
This!
0 2 4 6 8 10 12
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
y(k)=0.7k
y(11)=0.0198

75. When There Are Complex Poles
…
U(z)
z
a
...
z
a
1
z
b
...
z
b
Y(z) n
n
1
1
m
m
1
1










c)...
bz
(az2


2a
4ac
b
b
z
2




0,
4ac
b2


)
2a
4ac
b
b
)(z
2a
4ac
b
b
a(z
c
bz
az
2
2
2 










0,
4ac
b2

 )
2a
b
4ac
i
b
)(z
2a
b
4ac
i
b
a(z
c
bz
az
2
2
2 










If
If
Or in polar coordinates,
)
ir
r
)(z
ir
r
a(z
c
bz
az2
θ
θ
θ
θ sin
cos
sin
cos 







76. What If Poles Are Complex
 If Y(z)=N(z)/D(z), and coefficients of both D(z) and N(z) are all real
numbers, if p is a pole, then p’s complex conjugate must also be a
pole
– Complex poles appear in pairs





















l
1
j
2
2
j
j
0
l
1
j j
j
0
r
)z
(2r
z
)
r
dz(z
bzr
p
z
c
c
ir
r
z
c'
ir
r
z
c
p
z
c
c
Y(z)
θ
θ
θ
θ
θ
θ
θ
cos
cos
sin
sin
cos
sin
cos
coskθ
dr
sinkθ
br
p
c
(k)
u
c
y(k) k
k
m
1
j
1
-
k
j
j
impulse
0 




 

Z-
1
Time domain

77. An Example
0 2 4 6 8 10 12 14 16 18 20
-1
-0.5
0
0.5
1
1.5
2
)
3
kπ
cos(
0.8
)
3
kπ
sin(
0.8
2
y(k)
0.64
0.8z
z
z
z
Y(z)
k
k
2
2









Z-Domain: Complex Poles
Time-Domain:
Exponentially Modulated Sin/C

78. Poles Everywhere

79. Observations
 Using poles to characterize a signal
– The smaller is |r|, the faster converges the signal
 |r| < 1, converge
 |r| > 1, does not converge, unbounded
 |r|=1?
– When the angle increase from 0 to pi, the frequency of oscillation
increases
 Extremes – 0, does not oscillate, pi, oscillate at the maximum frequency

80. Change Angles
0.9
-0.9 Re
Im
0 5 10 15
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0 5 10 15
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0 5 10 15
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0 5 10 15
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0 5 10 15
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0 5 10 15
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0 5 10 15
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0 5 10 15
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0 5 10 15
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1

81. 0 2 4 6 8 10 12 14
-6
-4
-2
0
2
4
6
8
10
12
Changing Absolute Value
Im
Re
1
0 5 10 15
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0 5 10 15
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0 5 10 15
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0 5 10 15
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0 2 4 6 8 10 12 14
-3
-2
-1
0
1
2
3
4

82. Conclusion for Complex Poles
 A complex pole appears in pair with its
complex conjugate
 The Z-1-transform generates a combination of
exponentially modulated sin and cos terms
 The exponential base is the absolute value of
the complex pole
 The frequency of the sinusoid is the angle of
the complex pole (divided by 2π)

83. Steady-State Analysis
 If a signal finally converges, what value does it converge to?
 When it does not converge
– Any |pj| is greater than 1
– Any |r| is greater than or equal to 1
 When it does converge
– If all |pj|’s and |r|’s are smaller than 1, it converges to 0
– If only one pj is 1, then the signal converges to cj
 If more than one real pole is 1, the signal does not converge … (e.g. the ramp signal)

 k
dr
k
br k
k
cos
sin 




 

m
1
j
1
-
k
j
j
impulse
0 p
c
(k)
u
c
y(k) 2
1
-1
)
z
(1
z



84. An Example
k
k
0.9)
(
3
0.5
2
u(k)
0.9
z
3z
0.5
z
z
1
z
2z
U(z)











0 10 20 30 40 50 60
-1
0
1
2
3
4
5
6
converge to 2

85. Final Value Theorem
 Enable us to decide whether a system has a
steady state error (yss-rss)

86. Final Value Theorem
1
Theorem: If all of the poles of (1 ) ( ) lie within the unit circle, then
lim ( ) lim ( 1) ( )
k z
z Y z
y k z Y z


 
2
1 1
0.11 0.11
( )
1.6 0.6 ( 1)( 0.6)
0.11
( 1) ( ) | | 0.275
0.6
z z
z z
Y z
z z z z
z
z Y z
z
 
 
 
   

   
 0 5 10 15
-0.35
-0.3
-0.25
-0.2
-0.15
-0.1
-0.05
0
k
y(k)
If any pole of (1-z)Y(z) lies out of or ON the
unit circle, y(k) does not converge!

87. What Can We Infer from TF?
 Almost everything we want to know
– Stability
– Steady-State
– Transients
 Settling time
 Overshoot
– …

88. Shift-Invariant System
h(n)
x(n) y(n)=x(n)*h(n)
X(z) Y(z)=X(z)H(z)
H(z)

89. Shift-Invariant System
H(z)
X(z) Y(z)
)
(
)
(
)
(
z
X
z
Y
z
H 

90. Nth-Order Difference Equation

 




M
r
r
N
k
k r
n
x
b
k
n
y
a
0
0
)
(
)
(

 




M
r
r
r
N
k
k
k z
b
z
X
z
a
z
Y
0
0
)
(
)
(







N
k
k
k
M
r
r
r z
a
z
b
z
H
0
0
)
(

91. Representation in Factored Form








 N
k
r
M
r
r
z
d
z
c
A
z
H
1
1
1
1
)
1
(
)
1
(
)
(
Contributes poles at 0 and zeros at cr
Contributes zeros at 0 and poles at dr

92. Stable and Causal Systems








 N
k
r
M
r
r
z
d
z
c
A
z
H
1
1
1
1
)
1
(
)
1
(
)
( Re
Im
Causal Systems : ROC extends outward from the outermost pole.

93. Stable and Causal Systems








 N
k
r
M
r
r
z
d
z
c
A
z
H
1
1
1
1
)
1
(
)
1
(
)
( Re
Im
Stable Systems : ROC includes the unit circle.
1

94. Example
Consider the causal system characterized by
)
(
)
1
(
)
( n
x
n
ay
n
y 


1
1
1
)
( 


az
z
H
Re
Im
1
a
)
(
)
( n
u
a
n
h n


95. Determination of Frequency Response
from pole-zero pattern
 A LTI system is completely characterized by its
pole-zero pattern.
)
)(
(
)
(
2
1
1
p
z
p
z
z
z
z
H




Example:
)
)(
(
)
(
2
1
1
0
0
0
0
p
e
p
e
z
e
e
H j
j
j
j



 



0

j
e
Re
Im
z1
p1
p2

96. Determination of Frequency Response
from pole-zero pattern
 A LTI system is completely characterized by its
pole-zero pattern.
)
)(
(
)
(
2
1
1
p
z
p
z
z
z
z
H




Example:
)
)(
(
)
(
2
1
1
0
0
0
0
p
e
p
e
z
e
e
H j
j
j
j



 



0

j
e
Re
Im
z1
p1
p2
|H(ej)|=? H(ej)=?

97. Determination of Frequency Response
from pole-zero pattern
 A LTI system is completely characterized by its
pole-zero pattern.
Example:
0

j
e
Re
Im
z1
p1
p2
|H(ej)|=? H(ej)=?
|H(ej)| =
| |
| | | | 1
2
3
H(ej) = 1(2+ 3 )

98. Example
1
1
1
)
( 


az
z
H
Re
Im
a
0 2 4 6 8
-10
0
10
20
0 2 4 6 8
-2
-1
0
1
2
dB