Pertemuan 13
Transformasi - Z
Z-Transform
Introduction
U(s)  u(t)
Y(s)  y(t)
G(s)
Linear system
Tools to analyse continuous systems : Laplace transform
It could be used for sampled or discrete systems
T
f (t)
f ' (t)
t
t
f (t)
X
f ' (t)
t
Z-Transform
f ' (t)
t


k 0
k 0
f ' ( t )   f ( t )( t  kT)   f (kT)
Apply Laplace transform of f’(t)
 


f
(
0
)
L[f ' ( t )]    f ( t )( t  kT)e st dt 
  f (kT)e skT
2
k 0
0 k 0
Factors like Exp(-sT) are involved
Unlike the majority of transfer functions of continuous systems
It will not lead to rational functions
Z-Transform
Definition


k 0
k 0
F(z)   f (kT)z  k   f (k )z  k
ze
sT
1
F (s)   f (0)  F' (k )
2
'
1
s
ln( z)    j
T
Re( z)  e T cos(T)
Im( z)  e T sin( T)
1
F(z)  F' (s)[s 
ln( z)]
T
Summary
The operation of taking the z-transform of a continuous-data
function, f(t), involves the following three steps:
1- f(t) is sampled by an ideal sampler to get f’(t)
2- Take the Laplace transform of f’(t)

F' (s)   f (kT)e  ksT
k 0

3- Replace
e
s T
by z in F’(s) to get
F(z)   f (kT)z  k
k 0
Mapping between the s-plane and the z-plane
ze
j
sT
S-plane
2
3

1
4
Imz
5
3
4
s
2
Primary strip
 s
2
5
2
2
s 
T
z-plane
1
Rez
1
The left half of the primary strip is mapped inside the unit circle
Mapping between the s-plane and the z-plane
ze
j
sT
S-plane
2
3
Primary strip
s
2
1
4
5
Imz
3
4
2
5

 s
2
Z-plane
1
Rez
1
The right half of the primary strip is mapped outside the unit circle
Mapping between the s-plane and the z-plane
e
( s  jks ) T
S-plane
e
sT
e
jks T
j
e
sT 2 jk
e
e
sT
1
s (  k )
2
Complementary strip
s (
Imz

1
 k)
2
Z-plane
Rez
1
The right half of the complementary strip is also mapped inside the unit circle
s-plane properties of F’(s)
j
Complementary s 0  j2s
strip
3s / 2
Complementary s 0  js
strip
Primary strip
s0
s / 2
 s / 2
Complementary s 0  js
strip
Complementary
strip
5s / 2
s 0  j2s
F' (s  jms )  F' (s)
 3s / 2
 5s / 2

s-plane properties of F’(s)
j
Complementary s 0  j2s
X
strip
Complementary s 0  js X
strip
Primary strip
s0
X
Complementary s 0  js
X
strip
Complementary
strip
5s / 2
3s / 2
s / 2
 s / 2
 3s / 2
s 0  j2s
X
X Poles of F’(s) in primary strip
 5s / 2

s-plane properties of F’(s)
j
Complementary s 0  j2s
X
strip
5s / 2
Folded back poles
Complementary s 0  js X
strip
Primary strip
s0
X
Complementary s 0  js
X
strip
Complementary
strip
3s / 2
s / 2
 s / 2
 3s / 2
s 0  j2s
X
X Poles of F’(s) in complementary strips
 5s / 2

The constant damping loci
z  e 1 T e jT
j
1
2

z  e   2 T e jT
s-plane
z-plane
The constant frequency loci
2 T
z  e j1 T
j
j2
1 T
j1
 j1

z  e  j1 T
s-plane
z-plane
The constant damping ratio loci
Imz
j
3
s
j
2
2
1
4
5
s-plane

2
5
3
4
s

2
z-plane
Rez
1
The constant damping ratio loci
s   tan   j

j
s
j
2
s
2
Imz
s
4
Rez

s

2
s-plane
s
3s
4
z-plane
Mapping between the s-plane and the z-plane
Conclusion:
All points in the left half of the s-plane are mapped into the
Region inside the unit circle in the z-plane.
The points in the right half of the s-plane are mapped into the
Region outside the unit circle in the z-plane
Example: discrete exponential function
f * (k )  e  k
1
k
f * (k )  0, k  0
Apply z-transform

F (z)   f * (k )z  k
*
k 0


k 0
k 0
F* (z)   e k z  k   (e  z 1 ) k
1
z
F (k ) 

  1
1 e z
z  e 
*
0
Series
Reminder

s n   y n  1  y  y 2  y 3  ........y n
n 0
s n  1  y  y 2  y3  ........y n
y.s n  y  y 2  y3  ........y n  y( n 1)
s n  y.s n  s n (1  y)  1  y( n 1)
1  y( n 1)
1
sn 

1 y
1 y
Example: discrete Cosine function
jk
 jk
e

e
f * (k )  cos(k ) 
2
1
z
z
F (z)  (

)
j
 j
2 ze
ze
*
 j
j
z
(
z

e
)

(
z

e
)
*
F ( z)  [
]
j
 j
2 (z  e )( z  e )
j
 j
z
2
z

(
e

e
)
*
F ( z)  ( 2
)
j
 j
2 z  z(e  e )  1
z(z  cos )
F (z)  2
z  2z cos   1
*
Z[e
 k
z
]
z  e 
Another approach
y(k)  cos(k)  j sin( k)  e
jk
z
z
Y( z) 

 j
ze
z  cos   j sin 
z(z  cos   j sin )
Y(z) 
(z  cos   j sin )( z  cos   j sin )
z(z  cos )  jz sin )
Y( z) 
z 2  2z cos   1
z(z  cos )
Z[cos]  2
z  2z cos   1
z sin( )
Z[sin]  2
z  2z cos   1
Dirac function
( t )
F[( t )]  1
Z[( t )]  1

F[( t )]   ( t )z sT  (0)  1
k 0
Sampled step function
u(t)
1
t
0
T 2T 3T 4T 5T

1
skT
U(s)   e

sT
1

e
k 0
1
z
U(z) 

1
1 z
z 1

U(s)   e skT
k 0

U(z)   z
NB: Equivalent to Exp(-k) as  0
k 0
k
1
z


1
1 z
z 1
Delayed pulse train
T   ( t  kT)
T
k
t
T
t
x 'e 

 x [(k  )T][t  (k  )T]
k  
e
X (s, )  e
'
e
sT

skT
x
[(
k


)
T
)
e
 e
k 0
Complete z-transform


k 0
k 0
F(z, )   f [( k  )T]z  k   f (k, )z  k
Example:exponential function
f (k, )  e ( k  ) ,   0

F(z, )   e
k 0
 ( k   )  k
z
e
 

e
 k
k 0
z
 
F(z, ) 
e
z  e 
z
k
z
 

e
z  e 
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