Z-transform power point presentation.pdf

this is Ztransform ppt

1. The z-Transform
B Tech VIth Semester EIC-DSP

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

3. The z-Transform
INTRODUCTION

4. Why z- Transform?
A generalization of Fourier transform
Why generalize it?
◦FT does not convergeon all sequence
◦Notationgood for analysis
◦Bring the power of complex variable theorydeal with the discrete-time
signals and systems

5. The z-Transform
Z-TRANSFORM

6. Definition
The z-transform of sequence x (n) is defined by
∑
∞
−∞=
−
=
n
n
znxzX )()(
Let z = e
−jω
.
( ) ()
j jn
n
Xe xne
ωω
∞
−
=−∞
=∑
Fourier
Transform

7. z-Plane
Re
Im
z = e
−jω
ω
∑
∞
−∞=
−
=
n
n
znxzX )()(
( ) ()
j jn
n
Xe xne
ωω
∞
−
=−∞
=∑
Fourier Transform is to evaluate z-transform
on a unit circle.

8. z-Plane
Re
Im
X(z)
Re
Im
z = e
−jω
ω

9. Periodic Property of FT
Re
Im
X(z)
π−π ω
X(e
jω
)
Can you say why Fourier Transform is
a periodic function with period 2π ?

10. The z-Transform
ZEROS AND POLES

11. Definition
Give a sequence, the set of values of zfor which the z-transform converges,
i.e., |X(z)|<∞, is called the region of convergence.
∞<==∑∑
∞
−∞=
−
∞
−∞=
−
n
n
n
n
znxznxzX |||)(|)(|)(|
ROC is centered on origin and
consists of a set of rings.

12. Example: Region of
Convergence
Re
Im
∞<==∑∑
∞
−∞=
−
∞
−∞=
−
n
n
n
n
znxznxzX |||)(|)(|)(|
ROC is an annual ring centered
on the origin.
+−
<<
xx
RzR ||
r
}|{
+−
ω
<<==
xx
j
RrRrezROC

13. Stable Systems
A stable system requires that its Fourier transformis uniformly convergent.
Re
Im
1
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.

14. Example: A right sided
Sequence
)()( nuanx
n
=
12345678910-1-2-3-4-5-6-7-8
n
x(n)
. . .

15. Example: A right sided
Sequence
)()( nuanx
n
=
n
n
n
znuazX
−
∞
−∞=
∑= )()(
∑
∞
=
−
=
0n
nn
za
∑
∞
=
−
=
0
1
)(
n
n
az
For convergence of X(z), we
require that
∞<∑
∞
=
−
0
1
||
n
az
1||
1
<
−
az
||||az>
az
z
az
azzX
n
n
−
=
−
==
−
∞
=
−
∑ 1
0
1
1
1
)()(
||||az>

16. a−a
Example: A right sided
SequenceROC for x (n)=a
n
u(n)
|||| ,)( az
az
z
zX >
−
=
Re
Im
1
a−a
Re
Im
1
Which one is stable?

17. Example: A left sided
Sequence
)1()( −−−= nuanx
n
12345678910-1-2-3-4-5-6-7-8
n
x(n)
. . .

18. Example: A left sided
Sequence
)1()( −−−= nuanx
n
n
n
n
znuazX
−
∞
−∞=
∑ −−−= )1()(
For convergence of X(z), we
require that
∞<∑
∞
=
−
0
1
||
n
za
1||
1
<
−
za
||||az<
az
z
za
zazX
n
n
−
=
−
−=−=
−
∞
=
−
∑ 1
0
1
1
1
1)(1)(
||||az<
n
n
n
za
−
−
−∞=
∑−=
1
n
n
n
za∑
∞
=
−
−=
1
n
n
n
za∑
∞
=
−
−=
0
1

19. a−a
Example: A left sided Sequence
ROC for x(n)=−a
n
u(−n−1)
|||| ,)( az
az
z
zX <
−
=
Re
Im
1
a−a
Re
Im
1
Which one is stable?

20. The z-Transform
REGION OF
CONVERGENCE

21. Represent z-transform as a
Rational Function
)(
)(
)(
zQ
zP
zX=
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) = ∞

22. Example: A right sided
Sequence
)()( nuanx
n
= |||| ,)( az
az
z
zX >
−
=
Re
Im
a
ROC is bounded by the
poleand is the exterior
of a circle.

23. Example: A left sided
Sequence
)1()( −−−= nuanx
n
|||| ,)( az
az
z
zX <
−
=
Re
Im
a
ROC is bounded by the
poleand is the interior
of a circle.

24. Example: Sum of Two Right Sided Sequences
)()()()()(
3
1
2
1
nununx
nn
−+=
3
1
2
1
)(
+
+
−
=
z
z
z
z
zX
Re
Im
1/2
))((
)(2
3
1
2
1
12
1
+−
−
=
zz
zz
−1/3
1/12
ROC is bounded by poles
and is the exterior of a circle.
ROC does not include any pole.

25. Example: A Two Sided
Sequence
)1()()()()(
2
1
3
1
−−−−= nununx
nn
2
1
3
1
)(
−
+
+
=
z
z
z
z
zX
Re
Im
1/2
))((
)(2
2
1
3
1
12
1
−+
−
=
zz
zz
−1/3
1/12
ROC is bounded by poles
and is a ring.
ROC does not include any pole.

26. Example: A Finite Sequence
10 ,)( −≤≤= Nnanx
n
n
N
n
n
N
n
n
zazazX )()(
1
1
0
1
0
−
−
=
−
−
=
∑∑==
Re
Im
ROC: 0 < z< ∞
ROC does not include any pole.
1
1
1
)(1
−
−
−
−
=
az
az
N
az
az
z
NN
N
−
−
=
−1
1
N-1 poles
N-1 zeros
Always Stable

27. Properties of ROC
A ringor diskin 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.

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

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

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

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

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

33. The z-Transform
IMPORTANT
Z-TRANSFORM
PAIRS

34. Z-Transform Pairs
Sequence z-Transform ROC
)(nδ 1
All z
)(mn−δ
m
z
− All z except 0 (if m>0)
or ∞(if m<0)
)(nu 1
1
1
−
−z
1||>z
)1(−−−nu 1
1
1
−
−z
1||<z
)(nua
n
1
1
1
−
−az
||||az>
)1(−−− nua
n
1
1
1
−
−az
||||az<

35. Z-Transform Pairs
Sequence z-Transform ROC
)(][cos
0nunω
21
0
1
0
]cos2[1
][cos1
−−
−
+ω−
ω−
zz
z
1||>z
)(][sin
0
nunω
21
0
1
0
]cos2[1
][sin
−−
−
+ω−
ω
zz
z
1||>z
)(]cos[
0
nunr
n
ω 221
0
1
0
]cos2[1
]cos[1
−−
−
+ω−
ω−
zrzr
zr
rz>||
)(]sin[
0
nunr
n
ω 221
0
1
0
]cos2[1
]sin[
−−
−
+ω−
ω
zrzr
zr
rz>||


 −≤≤
otherwise0
10 Nna
n
1
1
1
−
−
−
−
az
za
NN
0||>z

36. 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| < r
1
|z| > r
2
r
2< |z| < r
1

37. Some Common z- Transform Pairs
Sequence Transform ROC
1. δ[n] 1 all z
2. u[n] z/(z-1) |z|>1
3. -u[-n-1] z/(z-1) |z|<1
4. δ[n-m] z
-m
all z except 0 if m>0 or ฅif m<0
5. a
n
u[n] z/(z-a) |z|>|a|
6. -a
n
u[-n-1] z/(z-a) |z|<|a|
7. na
n
u[n] az/(z-a)
2
|z|>|a|
8. -na
n
u[-n-1] az/(z-a)
2
|z|<|a|
9. [cosω
0n]u[n] (z
2
-[cosω
0]z)/(z
2
-[2cosω
0]z+1) |z|>1
10. [sinω
0n]u[n] [sinω
0]z)/(z
2
-[2cosω
0]z+1) |z|>1
11. [r
n
cosω
0n]u[n] (z
2
-[rcosω
0]z)/(z
2
-[2rcosω
0]z+r
2
) |z|>r
12. [r
n
sinω
0n]u[n] [rsinω
0]z)/(z
2
-[2rcosω
0]z+r
2
) |z|>r
13. a
n
u[n] -a
n
u[n-N] (z
N
-a
N
)/z
N-1
(z-a) |z|>0

38. The z-Transform
INVERSE Z-
TRANSFORM

39. 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
◦B
ris 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
za
zb
zX
0
0
()
( )
∑∑∑
=
−
≠=
−
−
=
−
−
+
−
+=
s
1m
m
1
i
m
N
ik,1k
1
k
k
NM
0r
r
r
zd1
C
zd1
A
zBzX

40. Partial Fractional Expression
Coefficients are given as
Easier to understand with examples
()
( )
∑∑∑
=
−
≠=
−
−
=
−−
+
−
+=
s
1m
m
1
i
m
N
ik,1k
1
k
k
NM
0r
r
r
zd1
C
zd1
A
zBzX
( )()
kdz
1
kk
zXzd1A
=
−
−=
( )()
( ) ()[ ]
1
i
dw
1s
ims
ms
ms
i
m
wXwd1
dw
d
d!ms
1
C
−
=
−
−
−
−





−
−−
=

41. Example: 2
nd
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
11
>






−





−
=
−−
zz
zX
()






−
+






−
=
−− 1
2
1
1
z
2
1
1
A
z
4
1
1
A
zX
() 1
4
1
2
1
1
1
zXz
4
1
1A
1
4
1
z
1
1
−=














−
=





−=
−
=
−
() 2
2
1
4
1
1
1
zXz
2
1
1A
1
2
1
z
1
2
=














−
=





−=
−
=
−

42. Example Continued
ROC extends to infinity
◦Indicates right sided sequence
()
2
1
z
z
2
1
1
2
z
4
1
1
1
zX11
>






−
+






−
−
=
−−
[] [] []nu
4
1
-nu
2
1
2nx
nn












=

43. Example #2
Long division to obtain B
o
()
( )
( )
1z
z1z
2
1
1
z1
z
2
1
z
2
3
1
zz21
zX
11
2
1
21
21
>
−





−
+
=
+−
++
=
−−
−
−−
−−
1z5
2z3z
2
1z2z1z
2
3
z
2
1
1
12
12
12
−
+−
+++−
−
−−
−−
−−
()
( )
11
1
z1z
2
1
1
z51
2zX
−−
−
−





−
+−
+=
()
1
2
1
1
z1
A
z
2
1
1
A
2zX
−
−
−
+
−
+=
() 9zXz
2
1
1A
2
1
z
1
1
−=





−=
=
−
( ) ()8zXz1A
1z
1
2
=−=
=
−

44. Example #2 Continued
ROC extends to infinity
◦Indicates right-sides sequence
() 1z
z1
8
z
2
1
1
9
2zX
1
1
>
−
+
−
−=
−
−
[] [] [][]n8u-nu
2
1
9n2nx
n






−δ=

45. An Example – Complete Solution
3
86zz
1414z3z
limU(z)limc
2
2
zz
0
=
+−
+−
==
∞→∞→
4-z
1414z3z
86zz
1414z3z
2)(z(z)U
2
2
2
2
+−
=
+−
+−
−=
2-z
1414z3z
86zz
1414z3z
4)(z(z)U
2
2
2
4
+−
=
+−
+−
−= 1
4-2
1421423
(2)Uc
2
21
=
+⋅−⋅
==
86zz
1414z3z
U(z)
2
2
+−
+−
=
4z
c
2z
c
cU(z)
21
0
−
+
−
+=
3
2-4
1441443
(4)Uc
2
42 =
+⋅−⋅
==
4z
3
2z
1
3U(z)
−
+
−
+=



>⋅+
=
=
−−
0k,432
0k3,
u(k)
1k1k

46. 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 nxzX
() [] [] [][] []  ++++−+−+=
−− 2112
2 1 0 1 2 zxzxxzxzxzX
() ( )( )
12
1112
z
2
1
1z
2
1
z
z1z1z
2
1
1z zX
−
−−−
+−−=
−+





−=
[][ ] [][] []1n
2
1
n1n
2
1
2nnx −δ+δ−+δ−+δ=
[]









=
=
=−
−=−
−=
=
2n0
1n
2
1
0n1
1n
2
1
2n1
nx

47. 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
RROC zXnx =→←
[] [] () ()
21 xx21
Z
21
RRROC zbXzaXnbxnax ∩=+→←+
[] [] []N-nua-nuanx
nn
=

48. Z-Transform Properties: Time Shifting
Here n
ois 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
nZ
o
RROC zXznnx
o
=→←−
−
()
4
1
z
z
4
1
1
1
z zX1
1
>












−
=
−
−
[] []1-nu
4
1
nx
1-n






=

49. Z-Transform Properties: Multiplication by Exponential
C is scaled by |z
o|
All pole/zero locations are scaled
If z
ois a positive real number: z-plane shrinks or expands
If z
ois a complex number with unit magnitude it rotates
Example: We know the z-transform pair
Let’s find the z- transform of
[] ()
xoo
Zn
oRzROCzzXnxz =→← /
[] 1z:ROC
z-1
1
nu
1-
Z
>→←
[] () []() []( ) []nure
2
1
nure
2
1
nuncosrnx
n
j
n
j
o
n
ooω−ω
+=ω=
() rz
zre1
2/1
zre1
2/1
zX
1j1j
oo
>
−
+
−
=
−ω−−ω

50. 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
RROC
dz
zdX
znnx =−→←
() ( ) az az1logzX
1
>+=
−
() ()
1
1
1
2
az1
1
az
dz
zdX
z
az1
az
dz
zdX
−
−
−
−
+
=−⇒
+
−
=
[] () []1nuaannx
1n
−−=
−
[]() []1nu
n
a
1nx
n
1n
−−=
−

51. Z-Transform Properties: Conjugation
Example
[] ()
x
**Z*
RROC zXnx =→←
() []
() [] []
() []() [] [] {}nxZz nxz nxzX
z nxz nxzX
z nxzX
n
n
n
n
n
n
n
n
n
n
∗
∞
−∞=
−∗
∞
−∞=
∗
∗∗∗
∞
−∞=
∗
∗
∞
−∞=
−∗
∞
−∞=
−
===
=








=
=∑∑
∑∑
∑

52. Z-Transform Properties: Time Reversal
ROC is inverted
Example:
Time reversed version of
[] ()
x
Z
R
1
ROC z/1Xnx =→←−
[] [] nuanx
n
−=
−
[]nua
n
()
1
11-
1-1
az
za-1
za-
az1
1
zX
−
−
−
<=
−
=

53. Z-Transform Properties: Convolution
Convolution in time domain is multiplication in z-domain
Example:Let’scalculate 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
x1x21
Z
21
RR:ROC zXzXnxnx ∩→←∗
[] [] [][]nunx and nuanx
2
n
1 ==
() az:ROC
az1
1
zX
11
>
−
=
− () 1z:ROC
z1
1
zX
12
>
−
=
−
() ()()
( )( )
1121
z1az1
1
zXzXzY
−−
−−
==
() 1z :ROC asume
az1
1
z1
1
a1
1
zY
11
>





−
−
−−
=
−−
[] [] []( ) nuanu
a1
1
ny
1n+
−
−
=

54. The z-Transform
Z-TRANSFORM THEOREMS
AND PROPERTIES

55. Linearity
xRzzXnx ∈= ),()]([Z
y
RzzYny ∈= ),()]([Z
yxRRzzbYzaXnbynax ∩∈+=+ ),()()]()([Z
Overlay of
the above two
ROC’s

56. Shift
xRzzXnx ∈= ),()]([Z
x
n
RzzXznnx ∈=+ )()]([
0
0
Z

57. Multiplication by an Exponential Sequence
+
<<=
xx-
RzRzXnx || ),()]([Z
x
nRazzaXnxa ⋅∈=
−
|| )()]([
1
Z

58. Differentiation of X(z)
xRzzXnx ∈= ),()]([Z
xRz
dz
zdX
znnx ∈−=
)(
)]([Z

59. Conjugation
xRzzXnx ∈= ),()]([Z
x
RzzXnx ∈= *)(*)](*[Z

60. Reversal
xRzzXnx ∈= ),()]([Z
xRzzXnx /1 )()]([
1
∈=−
−
Z

61. Real and Imaginary Parts
xRzzXnx ∈= ),()]([Z
x
RzzXzXnxe ∈+= *)](*)([)]([
2
1
R
xj
RzzXzXnx ∈−= *)](*)([)]([
2
1
Im

62. Initial Value Theorem
0for ,0)( <= nnx
)(lim)0( zXx
z∞→
=

63. Convolution of Sequences
xRzzXnx ∈= ),()]([Z
y
RzzYny ∈= ),()]([Z
yx
RRzzYzXnynx ∩∈= )()()](*)([Z

64. Convolution of Sequences
∑
∞
−∞=
−=
k
knykxnynx )()()(*)(
∑∑
∞
−∞=
−
∞
−∞=






−=
n
n
k
zknykxnynx )()()](*)([Z
∑∑
∞
−∞=
−
∞
−∞=
−=
k
n
n
zknykx )()(
∑ ∑
∞
−∞=
−
∞
−∞=
−
=
k
n
n
k
znyzkx )()(
)()(zYzX=

65. The z-Transform
SYSTEM
FUNCTION

66. 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(kb...1)u(kbn)y(ka...1)y(kay(k)
m1n1−++−+−++−=
∏
∏
=
=
−
−
==
m
1j
j
n
1i
i
)p(z
)z(z
D(z)
N(z)
Y(z)
zeros
poles

67. Why are poles important?
∑
∏
∏
=
=
=−
+=
−
−
==
m
1j j
j
0m
1j
j
n
1i
i
pz
c
c
)p(z
)z(z
D(z)
N(z)
Y(z)
∑
=
×+×=
m
1j
1-k
jjimpulse0
pc(k)ucY(k)
Z
-1
Z domain
Time domain
poles
component
s

68. 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

69. 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

70. 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

71. How fast does it converge?
U(k)=a
k
, 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
3.912ln0.02|a|kln
0.02|a|
k
−
≈
−==
=
11
0.36
4
|0.7|ln
4
k
0.7a
≈
−
−
≈
−
≈
=
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)=0.7
k
y(11)=0.0198

72. When There Are Complex Poles …
U(z)
za...za1
zb...zb
Y(z)
n
n
1
1
m
m
1
1
−−
−−
−−−
++
= c)...bz(az
2
++
2a
4acbb
z
2
−±−
=
0,4acb
2
≥−
)
2a
4acbb
)(z
2a
4acbb
a(zcbzaz
22
2
−−−
−
−+−
−=++
0,4acb
2
<− )
2a
b4acib
)(z
2a
b4acib
a(zcbzaz
22
2 −−−
−
−+−
−=++If
If
Or in polar coordinates,
)irr)(zirra(zcbzaz
2
θθθθ sincossincos +−−−=++

73. 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
1j
22
j
j
0
l
1j j
j
0
r)z(2rz
)rdz(zbzr
pz
c
c
irrz
c'
irrz
c
pz
c
cY(z)
θ
θθ
θθθθ
cos
cossin
sincossincos
coskθdrsinkθbrpc(k)ucy(k)
kk
m
1j
1-k
jjimpulse0 ++×+×= ∑
=
Z
-1
Time domain

74. 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(0.82y(k)
0.640.8zz
zz
Y(z)
kk
2
2
⋅+⋅⋅=
+−
+
=
Z-Domain: Complex Poles
Time-Domain:
Exponentially Modulated Sin/C

75. Poles Everywhere

76. 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

77. 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

78. 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

79. 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)

80. Steady-State Analysis
If a signal finally converges, what value does it converge to?
When it does not converge
◦Any |p
j|is greater than 1
◦Any |r| is greater than or equal to 1
When it does converge
◦If all |p
j|’s and |r|’s are smaller than 1, it converges to 0
◦If only one p
jis 1, then the signal converges to c
j
◦If more than one real pole is 1 , the signal does not converge … (e.g. the ramp signal)
θθkdrkbr
kk
cossin++×+×=∑
=
m
1j
1-k
jjimpulse0
pc(k)ucy(k)
21
-1
)z(1
z
−
−

81. An Example
kk
0.9)(30.52u(k)
0.9z
3z
0.5z
z
1z
2z
U(z)
−⋅++=
+
+
−
+
−
=
0 10 20 30 40 50 60
-1
0
1
2
3
4
5
6
converge to 2

82. Final Value Theorem
Enable us to decide whether a system has a steady state error (y
ss-r
ss)

83. Final Value Theorem
1
Theorem: If all of the poles of (1 ) ( ) lie within the unit circle, then
lim ( ) lim ( 1) ( )
kz
zYz
yk z Yz
∞
−
= − 
2
11
0.11 0.11
()
1.6 0.6 ( 1)( 0.6)
0.11
( 1) ( )| | 0.275
0.6
zz
zz
Yz
z z zz
z
z Yz
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!

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

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

86. Shift-Invariant System
H(z)
X(z) Y(z)
)(
)(
)(
zX
zY
zH=

87. N
th
-Order Difference Equation
∑∑
==
−=−
M
r
r
N
k
k
rnxbknya
00
)()(
∑∑
=
−
=
−
=
M
r
r
r
N
k
k
k
zbzXzazY
00
)()(
∑∑=
=
−
=
−
N
k
k
k
M
r
r
r
zazbzH
00
)(

88. Representation in Factored
Form
∏
∏
=
−
=
−
−
−
=
N
k
r
M
r
r
zd
zcA
zH
1
1
1
1
)1(
)1(
)(
Contributes polesat 0 and zerosat c
r
Contributes zerosat 0 and polesat d
r

89. Stable and Causal Systems
∏
∏
=
−
=
−
−
−
=
N
k
r
M
r
r
zd
zcA
zH
1
1
1
1
)1(
)1(
)(
Re
Im
Causal Systems : ROC extends outward from the outermost pole.

90. Stable and Causal Systems
∏
∏
=
−
=
−
−
−
=
N
k
r
M
r
r
zd
zcA
zH
1
1
1
1
)1(
)1(
)(
Re
Im
Stable Systems : ROC includes the unit circle.
1

91. Thanks!
Reference :
https://akademik.adu.edu.tr/fakulte/muhendislik/personel/uploads/yilmaz.kalkan/dsp3-1577359228.ppt