2. Z-transform and theorem

Report
2. Z-transform and theorem
E(s)
R(s)
M(s)
Gc(s)
Controller
GP(s)
Y(s)
Plant
GHP(z)
R(z)
M(z)
E(z)
ZOH
Gc(z)
A/D
Y(z)
GP(s)
D/A
Computer system
Plant
2. Z-transform and theorem
f(kT)
f(t)
A/D
time
Time kT
f(kT)
f(kT)
D/A
Time kT
Time kT
2. Z-transform and theorem
How can we represent the sampled data
mathematically?
For continuous time system, we have a
mathematical tool Laplace transform. It
helps us to define the transfer function of a
control system, analyse system stability and
design a controller. Can we have a similar
mathematical tool for discrete time system?
2.1 Z-transform
For a continuous signal f(t), its sampled data can be
written as,

f s (t )  f (kT )  f (0)  f (T )  f (2T )     f (nT )
Then we can define Z-transform of f(t) as
n 0

F ( z )  Z [ f (t )]  Z [ f (nT )]   f (nT ) z n
n 0
 f (0)  f (T ) z 1  f (2T ) z 2  
where z-1 represents one sampling period delay in
time.
2.1 Z-transform
Solution:
Example 1: Find the Z-transform of unit step
f(t)
function.
f(kT)
t
kT
2.1 Z-transform
Apply the definition of Z-transform, we have

F ( z )  Z [ f (t )]  Z [ f (kT )]   f (kT )z  k
k 0
 1  z 1  z  2  
a
a  aq  aq  aq   
1 q
1
1
2
F ( z)  1  z  z   
1  z 1
2
3
2.1 Z-transform
Another method

F ( z )  Z [ f (t )]  Z [ f (kT )]   f ( kT )z  k
k 0
 1  z 1  z  2  
z 1 F ( z )  z 1 (1  z 1  z  2  )  z 1  z  2  z 3 
1
1
F ( z)  z F ( z)  1  F ( z) 
1  z 1
2.1 Z-transform
Example 2: Find the Z-transform of a
exponential decay.
f(t)
Solution:
f ( kT )  e  akT
t

F ( z )  Z [ f ( kT )]   e  akT z  k
k 0
 1  e  aT z 1  e  2 aT z  2  
1
F ( z) 
1  e  aT z 1
2.1 Z-transform
Exercise 1: Find the Z-transform of a
exponential decay f(t)=e-at using other
method.
f(t)
t
2.1 Z-transform
Example 3: Find the Z-transform of a cosine
function.
Solution: As
e jt  cost  j sin t; e  jt  cost  j sin t
cost 
e
jt
e
2
 jt
; sin t 
e
jt
e
j2
 jt
2.1 Z-transform
 e jt  e  jt  1
jt
 j t
F ( z )  Z [coskT ]  Z 

(
Z
[
e
]

Z
[
e
])

2

 2
1
 at
Z [e ] 
1  e  aT z 1
 jT 1
1
1
1
z  1  e jT z 1
 1 1 e
F ( z)  


jT 1
 jT 1 
2 1  e z
1 e
z  2 (1  e jT z 1 )(1  e  jT z 1 )
1 2  (e  jT z 1  e jT z 1 )
2  2 z 1 cosT


 jT 1
jT 1
2
2 1  (e
z e z ) z
2(1  2 z 1 cosT  z  2 )
1  z 1 cosT

1  2 z 1 cosT  z  2
2.1 Z-transform
Exercise 2: Find the Z-transform for decayed
cosine function f (t )  eat cost
1  z 1e  aT cosT
F ( z) 
1  2 z 1e aT cosT  e 2 aT z 2
2.1 Z-transform
Example 4: Find the Z-transform for
f (t ) 1  eat
Solution:


F ( z )  Z [ f (kT )]  Z 1  e  at  Z [1]  Z [e  at ]
1
1
 at
Z [1]  Z [ step] 
; Z [e ] 
1
1 z
1  e aT z 1
1
1
(1  e aT ) z 1
F ( z) 


1
 aT 1
1 z
1 e z
(1  z 1 )(1  e aT z 1 )
2.1 Z-transform
Exercise 3: Find the Z-transform for
f (t )  teat
Tz 1e  aT
F ( z) 
(1  z 1e aT ) 2
2.1 Z-transform
The functions can be given either in time
domain as f(t) or in S-domain as F(s). They
are equivalent. eg.
a) A unit step function: 1(t) or 1/s
b) A ramp function: t or 1/s2
c) f(t)=1-e-at or a/(s(s+a))
etc.
2.2 Z-transform theorems
Linearity: If f(t) and g(t) are Z-transformable
and  and  are scalar, then the linear
combination f(t)+g(t) has the Ztransform
Z[f(t)+g(t)]= F(z)+ G(z)
2.2 Z-transform theorems
Shifting Theorem:
Given that the Z-transform of f(t) is F(z), find
the Z-transform for f(t-nT).
f(t)
f(t-nT)
t
nT
t
2.2 Z-transform theorems
If f(t)=0 for t<0 has the Z-transform F(z), then
Z [ f (t  nT )]  z  n F ( z ) and
n 1

n
k 
Z [ f (t  nT )]  z  F ( z )   f (kT ) z 
k 0


Proving: By Z-transform definition, we have

Z [ f (t  nT )]   f (kT  nT ) z
k
k 0

  f (kT  nT ) z
k 0
( k  n )  n
z
z
n

  f (kT  nT ) z k z nn
k 0

( k  n )
f
(
kT

nT
)
z

k 0
2.2 Z-transform theorems
Defining m=k-n, we
have

Z [ f (t  nT )]  z
n
 f (kT  nT ) z
( k  n )
k 0
z
n

m
f
(
m
T
)
z

m n
Since f(mT)=0 for
the above as
 m<0, we can rewrite

Z [ f (t  nT )]  z n
m
n
m
n
f
(
m
T
)
z

z
f
(
m
T
)
z

z
F ( z)


m n
m 0
Thus, if a function f(t) is delayed by nT, its Ztransform would be multiplied by z-n. Or,
multiplication of a Z-transform by z-n has the
effect of moving the function to the right by nT
time. This is the so-called Shifting Theorem.
2.2 Z-transform theorems
Final value theorem:Suppose that f(t), where f(t)=0
for t<0, has the Z-transform of F(z), then the final
value of f(t) can be given by
1
lim f (t )  lim(1  z ) F ( z )
t 
z 1
There are other theorems for Z-transform. Please
read the study book or textbook for more details.
Theorem
Name

F ( z )  Z [ f (t )]   f (kT ) z k
Definition
k 0
Z[k1 f1 (t )  k 2 f 2 (t )]  k1 F1 ( z)  k 2 F2 ( z)
Z[e at f (t )]  F ( ze aT )
z
Z [ a t f (t )]  F ( )
a
Linearity
Multiply by e-at
Multiply by at
Z[ f (t  kT )]  z  k F ( z)
Time Shift 1
n 1
Time Shift 2
Z [ f (t  kT )]  z k [ F ( z )   f (kT ) z k ]
k 0
Z[ f (t )  f (t 1)]  (1  z 1 ) F ( z)
t
F ( z)
Z   f ( )d  
 0
 1  z 1
Differentiation
Integration
f ()  lim(1  z 1 ) F ( z )
Final Value
f (0)  lim F ( z )
Initial Value
z 1
z 
f(t)
F(s)
F(z)
(t)
1
1
u(t)
1
s
1
1  z 1
t
1
s2
Tz 1
(1  z 1 ) 2
e  at  e  bt
ba
e-at
1
( s  a)(s  b)
1
sa
1 – e-at
a
s( s  a)
sint

s2   2
cost
s
s 2
e-atsint

e-atcost
2
( s  a) 2   2
sa
( s  a) 2   2

1 
(e  aT  e bT ) z 1


 aT 1
bT 1 
b  a  (1  e z )(1  e z ) 
1
1 e
 aT
z 1
(1  e  aT ) z 1
(1  z 1 )(1  e aT z 1 )
z 1 sin T
1  2 z 1 cosT  z  2
1  z 1 cosT
1  2 z 1 cosT  z  2
z 1e  aT sin T
1  2 z 1e  aT cosT  e 2 aT z 2
1  z 1e  aT cosT
1  2 z 1e  aT cosT  e 2 aT z 2
2.3 Z-transform examples
Example 1: Assume that f(k)=0 for k<0, find the Ztransform of f(k)=9k(2k-1)-2k+3, k=0,1,2….
Solution: Obvious f(k) is a combination of three
sub-function 9k(2k-1), 2k and 3. Therefore, first we
can apply linearity theorem to f(k). Second, subfunction 9k(2k-1) can be considered as a product of
k and 2-12k, then we can apply the theorem of
multiply by ak. Finally, we can find the answer by
combining these three together.
2.3 Z-transform examples
F ( z )  Z [9k (2 k 1 )  2 k  3]  Z [9k (2 k 1 )]  Z [2 k ]  Z [3]
-1
Tz
z
T
z
1
z




Z[t]
;

Z [a t f (t )]  F  ; Z[1]
(1  z 1 ) 2 ( z  1) 2
1 - z -1 z  1
a
1
z/2
k
k

Z [2 ]  Z [2 1] 
z / 2  1 1  2 z 1
9 z 1
9 ( z / 2)
9
k
k 1
k 1

Z [9k (2 )]  Z [9k 2 2 ]  Z [k 2 ] 
2
(1  2 z 1 ) 2
2 (z/2 - 1)
2
1
3
1
z
9


F ( z )  Z [9k (2 k 1 )]  Z [2 k ]  Z [3] 
(1  2 z 1 ) 2 1  2 z 1 1  z 1
2  z 2

(1  2 z 1 ) 2 (1  z 1 )
2.3 Z-transform examples
Example 2: Obtain the Z-transform of the
curve x(t) shown below.
f(t)
1
0
1
2 3 4
5
6
7
8
t
2.3 Z-transform examples
Solution: From the figure, we have
K
0
1
2
3
4
5
6…
f(k) 0
0
0
1/3 2/3 1
1…
Apply the definition of Z-transform, we have

F ( z )   f (k ) z  k
k 0
z 3 2 z 4
 000

 z 5  z  6  
3
3
z -3  2 z  4

 z 5 (1  z 1  z 2  )
3
z -3  2 z  4
z 5
z 3  z  4  z 5



1
3
1 z
3(1  z 1 )
2.3 Z-transform examples
a2
Example 3: Find the Z-transform of F ( s)  2
s ( s  a)
Solution: Apply partial fraction to make F(s) as a
sum of simpler terms.
k3
a2
k1 k 2
a 1
1
F (s)  2
 2 
 2

s (s  a) s
s sa s
s sa
a
1
1
F ( z )  Z [ F ( s )]  Z [ 2 ]  Z [ ]  Z [
]
s
s
sa
aTz1
1
1



1 2
1
(1  z ) 1  z
1  e  aT z 1
[(aT  1  e  aT )  (1  e  aT  aTe aT ) z 1 ] z 1

(1  z 1 ) 2 (1  e  aT z 1 )
2.4 Inverse Z-transform
The inverse Z-transform: When F(z), the Ztransform of f(kT) or f(t), is given, the operation
that determines the corresponding time sequence
f(kT) is called as the Inverse Z-transform. We
label inverse Z-transform as Z-1.
bo  b1 z 1  b2 z 2   bm z  m
F ( z )  Z [ f (kT )]  Z [ f (t )] 
1  ao z 1  a1 z  2   an z  n
1
2
m


b

b
z

b
z


b
z
1
1
o
1
2
m
f (kT )  Z [ F ( z )]  Z 
1
2
n 
1

a
z

a
z


a
z
o
1
n


2.4 Inverse Z-transform
Z-transform
=
Inverse Z-transform

2.4 Inverse Z-transform
The inverse Z-transform can yield the
corresponding time sequence f(kt) uniquely.
However, it says nothing about f(t). There might
be numerous f(t) for a given f(kT).
f(t)
0
T
2T
3T
4T
5T
6T
t
2.4 Inverse Z-transform
x(kT)
Zero-order
Hold
Low-pass
Filter
f(t)
2.5 Methods for Inverse Z-transform
How can we find the time sequence for a
given Z-transform?
1) Z-transform table
Example 1: F(z)=1/(1-z-1), find f(kT).
F(z)=1+z-1+z-2+z-3+…
f(kT)=Z-1[F(z)]=1, for k=0, 1, 2, …
2.5 Inverse Z-transform examples
(1  e  aT ) z 1
Example 2: Given F ( z ) 
,
1
 aT 1
(1  z )(1  e z )
Find f(kT).
Solution: Apply partial-fraction-expansion to
simplify F(z), then find the simpler terms from
the Z-transform table.
(1  e  aT ) z 1
k1
k2
F ( z) 


1
 aT 1
1
(1  z )(1  e z ) 1  z
1  eaT z 1
Then we need to determine k1 and k2
2.5 Inverse Z-transform examples
(1  e  aT ) z 1
k1
k2
F ( z) 


1
 aT 1
1
(1  z )(1  e z ) 1  z
1  eaT z 1
Multiply (1-z-1) to both side and let z-1=1, we have
 aT
1
(
1

e
)
z
k1
k2

1
1 
(1  z )

(
1

z
)


1
 aT 1
1
 aT 1 
(1  z )(1  e z )
1 e z 
1 z
(1  e  aT ) z 1
(1  z 1 )k 2
(1  e  aT ) z 1
 k1 
 k1 
 aT 1
 aT 1
1 e z
1 e z
1  e  aT z 1
1
z 1 1
2.5 Inverse Z-transform examples
Similar as the above, we let multiply (1-e-aTz-1) to
both side and let z-1 =eaT, we have
(1  e
 aT
(1  e  aT ) z 1
k1
k2

 aT 1 
z )

(
1

e
z
)


1
 aT 1
1
 aT 1 
(1  z )(1  e z )
1 e z 
1 z
1
(1  e  aT ) z 1 (1  e  aT z 1 )k1
(1  e  aT ) z 1

 k2  k2 
1
1
1 z
1 z
1  z 1
Finally, we have
 1
z 1  e aT
(1  e  aT ) z 1
1
1
F ( z) 


1
 aT 1
1
(1  z )(1  e z ) 1  z
1  e aT z 1
f (kT )  1  e akT , k  0,1,2,
2.5 Inverse Z-transform examples
Exercise 4: Given the Z-transform
1
z
F ( z) 
(1  z 1 )(1  1.3z 1  0.4 z 2 )
Determine the initial and final values of f(kT), the
inverse Z-transform of F(z), in a closed form.
Hint: Partial-fraction-expansion, then use Ztransform table, and finally applying initial &
final value theorems of Z-transform.
2.5 Inverse Z-transform examples
2) Direct division method
Example 1: F(z)=1/(1+z-1), find f(kT).
1
2
1  z 1
1
1

z

z


1
1 z 1
1  z 1 1
1  z 1 1
1  z 1
1  z 1
1  z 1
-z
1
-z 1
-z
 z 1  z  2
 z 1  z  2
z
-2
1
 z -2
2.5 Inverse Z-transform examples
Finally, we obtain: F(z)=1-z-1+z-2-z-3+…
K =
0
1
2
3
…
F(kT)=
1
-1
1
-1
1
1  2z
Example 2: Given
F ( z) 
1 2
(1  z )
Find f(kT).
Solution: Dividing the numerator by the
denominator, we obtain
,
1  4 z 1  7 z 2  
1  2 z 1  z  2 1  2 z 1
1  2 z 1  z  2
4 z 1  z  2
4 z 1  8 z  2  4 z 3
7 z  2  4 z 3
7 z  2  14z 3  7 z  4
10z 3  7 z  4
10z 3  20z  4  10z 5
13z  4  10z 5

2.5 Inverse Z-transform examples
Finally, we obtain: F(z)=1+ 4z-1 + 7z-2 + 10z-3+…
K =
0
1
2
3
…
F(kT)=
1
4
7
10…
Exercise
0.3679z 1  0.343z 2  0.02221z 3  0.05659z 4
5:F ( z) 
,
1
1  1.3679z  0.3679z 2
Find f(kT).
Ans. :k
f(kT)
0
0
1
2
3
0.3679 0.8463 1
4
1
5…
1…
2.5 Inverse Z-transform examples
3) Computational method using Matlab
1
1

2
z
Example: Given F ( z ) 
find f(kT).
(1  z 1 ) 2
1  2 z 1
z 2  2z
z 2  2z
F ( z) 

 2
1 2
2
(1  z )
( z  1)
z  2z 1
Solution:
num=[1 2 0]; den=[1 –2 1]
Say we want the value of f(kT) for k=0 to 30
u=[1 zeros(1,30)]; F=filter(num, den, u)
1 4 7 10 13 16 19 22 25 28
31…
2.5 Inverse Z-transform examples
Exercise 6: Given the Z-transform
z 1 (0.5  z 1 )
F ( z) 
1
1 2
(1  0.5z )(1  0.8z )
Use 1) the partial-fraction-expansion method and 2)
the Matlab to find the inverse Z-transform of
F(z).
Answer: x(k)=-8.3333(0.5)k+8.333(0.8)k-2k(0.8)k-1
x(k)=0;0.5;0.05;–0.615;–1.2035;-1.6257;-1.8778…
Reading
Study book
• Module 2: The Z-transform and theorems
Textbook
• Chapter 2 : The Z-transform (pp23-50)
Tutorial
Exercise: The frequency spectrum of a continuoustime signal is shown below.
1) What is the minimum sampling frequency for
this signal to be sampled without aliasing.
2) If the above process were to be sampled at 10
Krad/s, sketch the resulting spectrum from –20
Krad/s to 20 Krad/s.
F()
 Krad/s
-8
-4
4
8
Tutorial
Solution: 1) From the spectrum, we can see that the
bandwidth of the continuous signal is 8 Krad/s.
The Sampling Theorem says that the sampling
frequency must be at least twice the highest
frequency component of the signal. Therefore,
the minimum sampling frequency for this signal
is 2*8=16 Krad/s.
F()
 Krad/s
-8
-4
4
8
Tutorial
2) Spectrum of the sampled signal is formed by
shifting up and down the spectrum of the
original signal along the frequency axis at i
times of sampling frequency. As s=10 Krad/s,
for i =0, we have the figure in bold line. For i=1,
we have the figure in bold-dot line.
F()
-8
-4
2 4
6 8 10 12 14 16 18
 Krad/s
Tutorial
For I=-1, 2,… we have
-18
-14
-8 -6 -4 -2
F()
2 4
6 8 10 12 14 16 18
 Krad/s
2 4
6 8 10 12 14 16 18
 Krad/s
Tutorial
Exercise 1: Find the Z-transform of a
exponential decay f(t)=e-aT using other
method.
f(t)
t
Tutorial

F ( z )  Z [ f (t )]  Z [ f (kT )]   f (kT )
k 0
 1  e  aT z 1  e  2 aT z  2  
e  aT z 1 F ( z )  e  aT z 1 (1  e  aT z 1  e  2 aT z  2  )
 e  aT z 1  e  2 aT z  2  e 3aT z 3 
F ( z)  e
 aT
1
z F ( z)  1  F ( z) 
1
1  e  aT z 1
Tutorial
Exercise 2: Find the Z-transform for a
decayed cosine function f (t )  eat cost
Solution 1: Z cost   1  z 1 cosT  F ( z )
1  2 z 1 cosT  z  2
Z [ f (t )]  F ( z ); F [e  at f (t )]  F ( ze aT )
1
1

z
cosT
 at
Z [e cost ] 
1  2 z 1 cosT  z  2
1  e  aT z 1 cosT

1  2e  aT z 1 cosT  e  2 aT z  2
z  e aT z
Tutorial
Solution 2:
 at  jt
 at  jt


e

e
 at
F ( z )  Z [e cost ]  Z 

2


1
 ( Z [e  at  jt ]  Z [e  at  jt ])
2
1
 at
Z [e ] 
1  e  aT z 1
1
1
1

F ( z)  

 aT  jT 1
2 1  e
z
1  e  aT  jT z 1 
1  e  aT z 1 cosT

1  2e  aT z 1 cosT  e  2 aT z  2
Tutorial
Exercise 3: Find the Z-transform for
f (t )  teat
Solution:
Tz 1
Z t  
 F ( z)
1 2
(1  z )
Z [ f (t )]  F ( z ); F [e  at f (t )]  F ( ze aT )
1
Tz
Z [te  at ] 
(1  z 1 ) 2
Te  aT z 1

(1  e  aT z 1 ) 2
z  e aT z
Tutorial
Exercise 4: Given the Z-transform
1
z
F ( z) 
(1  z 1 )(1  1.3z 1  0.4 z 2 )
Determine the initial and final values of f(kT), the
inverse Z-transform of F(z), in a closed form.
Solution: Apply the initial value theorem and the
final value theorem respectively, we have
Tutorial
f (0)  lim F ( z )  lim
z 
z 
z 1
0
1
1
2
(1  z )(1  1.3z  0.4 z )
f ()  lim [(1  z 1 ) F ( z )]  lim
z 1
z 
(1  z 1 ) z 1
1

(1  z 1 )(1  1.3z 1  0.4 z  2 ) 2.7
z 1
z 1
F ( z) 

1
1
2
(1  z )(1  1.3z  0.4 z ) (1  z 1 )(1  0.8 z 1 )(1  0.5 z 1 )
k3
k1
k2
0.37
1.11
1.48






1  z 1 1  0.5 z 1 1  0.8 z 1 1  z 1 1  0.5 z 1 1  0.8 z 1
1
f (k ) 
(1  3(0.5) k  4(0.8) k )
27
Tutorial
Exercise 5: Given
0.3679z 1  0.343z 2  0.02221z 3  0.05659z 4
F ( z) 
1  1.3679z 1  0.3679z 2
Find f(kT) using direct-division method.
Solution:
1  1.3679z 1  0.3679z  2
0.3679z 1
0.3679z 1  0.343z  2  0.02221z 3  0.0565z  4
0.3679z 1  0.5033z  2  0.1354z 3
0.8463z  2  0.1576z 3  0.0565z  4
Tutorial
Continuous
1  1.3679z 1  0.3679z  2
0.3679z 1  0.8463z 2  z 3  z 4 
0.3679z 1  0.343z  2  0.02221z 3  0.0565z 4
0.3679z 1  0.5033z  2  0.1354z 3
0.8463z 2  0.1576z 3  0.0565z 4
0.8463z 2  1.1576z 3  0.3114z 4
z 3  0.3679z  4
z 3  1.3679z 4  0.3679z 5
z 4  0.3679z 5

f (k )  0.3679z 1  0.8463z  2  z 3  z 4  
Tutorial
Exercise 6: Given the Z-transform
z 1 (0.5  z 1 )
F ( z) 
1
1 2
(1  0.5z )(1  0.8z )
Use 1) the partial-fraction-expansion method and 2)
the Matlab to find the inverse Z-transform of
F(z).
Solution1: To make the expanded terms more
recognizable in the Z-transform table, we
usually expand F(z)/z into partial fractions.
z 1 (0.5  z 1 )
z (0.5 z  1)
F ( z) 

1
1 2
(1  0.5 z )(1  0.8 z )
( z  0.5)(z  0.8) 2
k3
F ( z)
0.5 z  1
k1
k2




z
( z  0.5)(z  0.8) 2 ( z  0.5) ( z  0.8) 2 z  0.8
 k1
k3 
0.5 z  1
k2

( z  0.5)
 ( z  0.5)


2
2
( z  0.5)(z  0.8)
z  0.8 
 ( z  0.5) ( z  0.8)
0.5 z  1
k 2 ( z  0.5) k3 ( z  0.5)
 k1 

2
2
( z  0.8)
( z  0.8)
z  0.8
let z  0.5  k1 
( z  0.8) 2
0.5 z  1
( z  0.8) 2
 8.333
z  0.5
k3 
0.5 z  1
k1
k2
2



(
z

0
.
8
)


2
2

( z  0.5)(z  0.8)
z  0.8 
 ( z  0.5) ( z  0.8)
k ( z  0.8)
0.5 z  1 k1 ( z  0.8) 2
0.5 z  1

 k2  3
 k2 
 2
z  0.5
z  0.5
z  0.5
( z  0.5) z 0.8
k3 
0.5 z  1
k1
k2
2

( z  0.8)
 ( z  0.8) 


2
2
( z  0.5)(z  0.8)
z  0.8 
 ( z  0.5) ( z  0.8)
2
k3 ( z  0.8)
0.5 z  1 k1 ( z  0.8) 2

 k2 
 derivative
z  0.5
z  0.5
z  0.5
 0.5 z  1   k1 ( z  0.8)

  
 z  0.5   z  0.5
'
 0.5 z  1 
k3  

 z  0.5 
'

z  0.8
2
'

k3 ( z  0.5)  k3 ( z  0.8)
  0 
; let z  0.8
( z  0.5)

0.5( z  0.5)  (0.5 z  1)
0.5 * 0.3  0.6

 8.333
2
2
( z  0.5)
0.3
z  0.8
F ( z)
0.5 z  1
8.333
2
8.333




2
2
z
( z  0.5)(z  0.8)
( z  0.5) ( z  0.8)
z  0.8
8.333
2 z 1
8.333
F ( z)  


(1  0.5 z 1 ) (1  0.8 z 1 ) 2 1  0.8 z 1
f (k )  8.333(0.5) k  2(0.8) k 1  8.333(0.8) k
Tutorial
Partial fraction for inverse Z-transform
F ( z ) b0 z m  b1 z m 1   bm 1 z  bm b0 z m  b1 z m 1  bm 1 z  bm
 n

n 1
z
z  a0 z   an 1 z  an
( z  p1 )(z  p2 )  ( z  pn )
an
a1
a2
F ( z)



; ai  ( z  pi )
z  p1 z  p2
z  pn
z
z  pi
If F(z)/z involve s a multiple pole, eg. P1, then
a3
an
F ( z ) b0 z m  b1 z m1  bm1 z  bm
c1
c2






z
( z  p1 ) 2 ( z  p3 )  ( z  pn ) ( z  p1 ) 2 z  p1 z  p3
z  pn
ai  ( z  pi )
F ( z)
F ( z)
d 
F ( z) 
; c1  ( z  p1 ) 2
; c2  ( z  p1 ) 2
z z  pi
z z  p1
dz 
z  z  p1
Tutorial
Solution 2: Expand F(z) into a polynomial form
z 1 (0.5  z 1 )
0.5z 1  z 2
F ( z) 

1
1 2
(1  0.5z )(1  0.8z ) 1  2.1z 1  1.44z 2  0.32z 3
Num=[0 0.5 –1 0];
Den=[1 –2.1 1.44 –0.32];
U==[1 zeros(1,40)];
F=filter(Num, den,U)
0
0.5 0.05 -0.615
-1.2035…

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