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2. Z-transform and theorem
Gc(z) ZOH GP(s)R(z) E(z)M(z)
GHP(z)
Computer system
Y(z)
Plant
A/D D/A
Gc(s) GP(s)R(s) E(s) M(s) Y(s)
Controller Plant
2. Z-transform and theorem
time
f(t)
Time kT
f(kT)
A/D
Time kT
f(kT)
D/A
Time kT
f(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,
Then we can define Z-transform of f(t) as
where z-1 represents one sampling period delay in time.
0
)()2()()0()()(n
s nTfTfTffkTftf
21
0
)2()()0(
)()]([)]([)(
zTfzTff
znTfnTfZtfZzFn
n
2.1 Z-transform
Solution:
Example 1: Find the Z-transform of unit step function. f(t)
t
kT
f(kT)
2.1 Z-transform
Apply the definition of Z-transform, we have
121
32
21
0
1
11)(
1
1
)()]([)]([)(
zzzzF
q
aaqaqaqa
zz
zkTfkTfZtfZzF k
k
2.1 Z-transform
Another method
11
3212111
21
0
1
1)(1)()(
)1()(
1
)()]([)]([)(
zzFzFzzF
zzzzzzzFz
zz
zkTfkTfZtfZzF k
k
2.1 Z-transform
Example 2: Find the Z-transform of a exponential decay.
Solution: f(t)
t
1
221
0
1
1)(
1
)]([)(
)(
zezF
zeze
zekTfZzF
ekTf
aT
aTaT
k
kakT
akT
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
2sin ;
2cos
sincos ;sincos
j
eet
eet
tjtetjtetjtjtjtj
tjtj
2.1 Z-transform
21
1
21
1
211
11
11
11
11
1
cos21
cos1
)cos21(2
cos22
)(1
)(2
2
1
)1)(1(
11
2
1
1
1
1
1
2
1)(
1
1][
])[][(2
1
2][cos)(
zTz
Tz
zTz
Tz
zzeze
zeze
zeze
zeze
zezezF
zeeZ
eZeZee
ZkTZzF
TjTj
TjTj
TjTj
TjTj
TjTj
aTat
tjtjtjtj
2.1 Z-transform
Exercise 2: Find the Z-transform for decayed cosine function tetf at cos)(
221
1
cos21
cos1 )(
zeTez
TezzF
aTaT
aT
2.1 Z-transform
Example 4: Find the Z-transform for
Solution:
atetf 1)(
)1)(1(
)1(
1
1
1
1)(
1
1][ ;
1
1][]1[
][]1[1)]([)(
11
1
11
11
zez
ze
zezzF
zeeZ
zstepZZ
eZZeZkTfZzF
aT
aT
aT
aTat
atat
2.1 Z-transform
Exercise 3: Find the Z-transform forattetf )(
21
1
)1( )(
aT
aT
ez
eTzzF
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 Z-transform
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)
t
f(t-nT)
tnT
2.2 Z-transform theorems
If f(t)=0 for t<0 has the Z-transform F(z), then
Proving: By Z-transform definition, we have
1
0
)()()]([
and )()]([n
k
kn
n
zkTfzFznTtfZ
zFznTtfZ
0
)(
0
)(
00
)()(
)()()]([
k
nkn
k
nnk
k
nnk
k
k
znTkTfzzznTkTf
zznTkTfznTkTfnTtfZ
2.2 Z-transform theorems
Defining m=k-n, we have
Since f(mT)=0 for m<0, we can rewrite the above as
Thus, if a function f(t) is delayed by nT, its Z-transform 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.
nm
mn
k
nkn zmTfzznTkTfznTtfZ )()()]([0
)(
)()()()]([0
zFzzmTfzzmTfznTtfZ n
m
mn
nm
mn
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
There are other theorems for Z-transform. Please read the study book or textbook for more details.
)()1(lim)(lim 1
1zFztf
zt
0
)()]([)(k
kzkTftfZzF
)()()]()([ 22112211 zFkzFktfktfkZ
)()]([ aTat zeFtfeZ
)()]([a
zFtfaZ t
)()]([ zFzkTtfZ k
])()([)]([1
0
n
k
kk zkTfzFzkTtfZ
)()1()]1()([ 1 zFztftfZ
10 1
)()(
z
zFdfZ
t
)()1(lim)( 1
1zFzf
z
)(lim)0( zFfz
Theorem Name
Definition
Linearity
Multiply by e-at
Multiply by at
Time Shift 1
Time Shift 2
Differentiation
Integration
Final Value
Initial Value
s
111
1 z
2
1
s 21
1
)1(
z
Tz
as 1
11
1 ze aT
)( ass
a
)1)(1(
)1(11
1
zez
zeaT
aT
ab
ee btat
))((
1
bsas
)1)(1(
)(111
1
zeze
zee
ab bTaT
bTaT
22 s
21
1
cos21
sin
zTz
Tz
22 s
s21
1
cos21
cos1
zTz
Tz
22)(
as 221
1
cos21
sin
zeTez
TezaTaT
aT
22)(
as
as221
1
cos21
cos1
zeTez
TezaTaT
aT
f(t) F(s) F(z)
(t) 1 1
u(t)
t
e-at
1 – e-at
sint
cost
e-atsint
e-atcost
2.3 Z-transform examples
Example 1: Assume that f(k)=0 for k<0, find the Z-transform 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, sub-function 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
)1()21(
2
1
3
21
1
)21(
9]3[]2[)]2(9[)(
)21(
9
1)-(z/2
)2/(
2
9]2[
2
9]229[)]2(9[
21
1
12/
2/]12[]2[
)1()1(
Tz Z[t];
1z-1
1 Z[1];)]([
]3[]2[)]2(9[]32)2(9[)(
121
2
1121
11
21
1
211
1
221
1-
1-
11
zz
z
zzz
zZZkZzF
z
zzkZkZkZ
zz
zZZ
z
Tz
zz
z
a
zFtfaZ
ZZkZkZzF
kk
kkk
kk
t
kkkk
2.3 Z-transform examples
Example 2: Obtain the Z-transform of the curve x(t) shown below.
0 1 2 3 4 5 6 7 8
1
t
f(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
)1(313
2z
)1(3
2z
3
2
3000)()(
1
543
1
543-
21543-
6543
0
z
zzz
z
zz
zzzz
zzzz
zkfzFk
k
2.3 Z-transform examples
Example 3: Find the Z-transform of
Solution: Apply partial fraction to make F(s) as a sum of simpler terms.
)()(
2
2
ass
asF
)1()1(
])1()1[(
1
1
1
1
)1(
]1
[]1
[][)]([)(
11
)()(
121
11
1121
1
2
232
21
2
2
zez
zzaTeeeaT
zezz
aTz
asZ
sZ
s
aZsFZzF
asss
a
as
k
s
k
s
k
ass
asF
aT
aTaTaT
aT
2.4 Inverse Z-transform
The inverse Z-transform: When F(z), the Z-transform 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.
nno
mmo
nno
mmo
zazaza
zbzbzbbZzFZkTf
zazaza
zbzbzbbtfZkTfZzF
21
1
22
1111
21
1
22
11
1)]([)(
1)]([)]([)(
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)
t0 T 2T 3T 4T 5T 6T
2.4 Inverse Z-transform
x(kT) f(t)Zero-orderHold
Low-passFilter
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
Example 2: Given ,
Find f(kT).
Solution: Apply partial-fraction-expansion to simplify F(z), then find the simpler terms from the Z-transform table.
Then we need to determine k1 and k2
)1)(1(
)1()(
11
1
zez
zezF
aT
aT
12
11
11
1
11)1)(1(
)1()(
ze
k
z
k
zez
zezF
aTaT
aT
2.5 Inverse Z-transform examples
Multiply (1-z-1) to both side and let z-1=1, we have
12
11
11
1
11)1)(1(
)1()(
ze
k
z
k
zez
zezF
aTaT
aT
11
)1(
1
)1(
1
)1(
11)1(
)1)(1(
)1()1(
1
1
1
112
1
11
1
12
111
11
11
1
z
aT
aT
aTaT
aT
aTaT
aT
ze
zek
ze
kzk
ze
ze
ze
k
z
kz
zez
zez
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
Finally, we have
11
)1(
1
)1(
1
)1(
11)1(
)1)(1(
)1()1(
11
1
2211
1
1
1
12
111
11
11
aTez
aTaTaT
aTaT
aT
aTaT
z
zekk
z
kze
z
ze
ze
k
z
kze
zez
zeze
0,1,2,k ,1)(
1
1
1
1
)1)(1(
)1()(
1111
1
akT
aTaT
aT
ekTf
zezzez
zezF
2.5 Inverse Z-transform examples
Exercise 4: Given the Z-transform
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 Z-transform table, and finally applying initial & final value theorems of Z-transform.
)4.03.11)(1()(
211
1
zzz
zzF
2.5 Inverse Z-transform examples
2) Direct division method
Example 1: F(z)=1/(1+z-1), find f(kT).1
1
11
1
1
1
-z
z
z
1
2
21
1
1
11
1
11
z
z
zz
-z
z
z
-
21
2
21
1
1
11
1
11zz
z
zz
-z
z
z
-
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
Example 2: Given ,
Find f(kT).
Solution: Dividing the numerator by the denominator, we obtain
21
1
)1(
21)(
z
zzF
21
54
543
43
432
32
321
21
21
121
741
1013
102010
710
7147
47
484
4
21
2121zz
zz
zzz
zz
zzz
zz
zzz
zz
zz
zzz
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 5: ,
Find f(kT).Ans. :k 0 1 2 3 4 5…
f(kT) 0 0.3679 0.8463 1 1 1…
21
4321
3679.03679.11
05659.002221.0343.03679.0)(
zz
zzzzzF
2.5 Inverse Z-transform examples
3) Computational method using Matlab
Example: Given find f(kT).
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…
21
1
)1(
21)(
z
zzF
12
2
)1(
2
)1(
21)(
2
2
2
2
21
1
zz
zz
z
zz
z
zzF
2.5 Inverse Z-transform examples
Exercise 6: Given the Z-transform
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…
211
11
)8.01)(5.01(
)5.0()(
zz
zzzF
Reading
Study book
• Module 2: The Z-transform and theorems
Textbook
• Chapter 2 : The Z-transform (pp23-50)
TutorialExercise: The frequency spectrum of a continuous-
time 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.
-8 -4 4 8
Krad/s
F()
TutorialSolution: 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.
-8 -4 4 8
Krad/s
F()
Tutorial2) 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.
4 8
Krad/s
F()
122 6 14 181610-4-8
TutorialFor I=-1, 2,… we have
4 8
Krad/s
122 6 14 181610
-18
F()
4 8
Krad/s
122 6 14 181610-2-4-6-8-14
Tutorial
Exercise 1: Find the Z-transform of a exponential decay f(t)=e-aT using other method.
f(t)
t
Tutorial
11
33221
22111
221
0
1
1)(1)()(
)1()(
1
)()]([)]([)(
zezFzFzezF
zezeze
zezezezFze
zeze
kTfkTfZtfZzF
aTaT
aTaTaT
aTaTaTaT
aTaT
k
Tutorial
Exercise 2: Find the Z-transform for a decayed cosine function
Solution 1:
tetf at cos)(
221
1
21
1
21
1
cos21
cos1
cos21
cos1]cos[
)()]([ );()]([
)(cos21
cos1cos
zeTze
Tze
zTz
TzteZ
zeFtfeFzFtfZ
zFzTz
TztZ
aTaT
aT
zez
at
aTat
aT
Tutorial
Solution 2:
221
1
11
1
cos21
cos1
1
1
1
1
2
1)(
1
1][
])[][(2
1
2]cos[)(
zeTze
Tze
zezezF
zeeZ
eZeZ
eeZteZzF
aTaT
aT
TjaTTjaT
aTat
tjattjat
tjattjatat
Tutorial
Exercise 3: Find the Z-transform for
Solution:
attetf )(
21
1
21
1
21
1
)1(
)1(][
)()]([ );()]([
)()1(
ze
zTe
z
TzteZ
zeFtfeFzFtfZ
zFz
TztZ
aT
aT
zez
at
aTat
aT
Tutorial
Exercise 4: Given the Z-transform
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
)4.03.11)(1()(
211
1
zzz
zzF
Tutorial
))8.0(4)5.0(31(27
1)(
8.01
48.1
5.01
11.1
1
37.0
8.015.011
)5.01)(8.01)(1()4.03.11)(1()(
7.2
1
)4.03.11)(1(
)1()]()1[()(
0)4.03.11)(1(
)()0(
11113
12
11
111
1
211
1
211
111
1
211
1
limlim
limlim
kk
zz
zz
kf
zzzz
k
z
k
z
k
zzz
z
zzz
zzF
zzz
zzzFzf
zzz
zzFf
Tutorial
Exercise 5: Given
Find f(kT) using direct-division method.
Solution:
21
4321
3679.03679.11
05659.002221.0343.03679.0)(
zz
zzzzzF
1
432
321
432121
3679.0
0565.01576.00.8463
1354.05033.03679.0
0565.002221.0343.03679.03679.03679.11
z
zzz
zzz
zzzzzz
Tutorial
Continuous
4321
4321
54
543
43
432
432
321
432121
8463.03679.0)(
8463.03679.0
3679.0
3679.03679.1
3679.0
3114.01576.18463.0
0565.01576.00.8463
1354.05033.03679.0
0565.002221.0343.03679.03679.03679.11
zzzzkf
zzzz
zz
zzz
zz
zzz
zzz
zzz
zzzzzz
Tutorial
Exercise 6: Given the Z-transform
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.
211
11
)8.01)(5.01(
)5.0()(
zz
zzzF
2)5.0(
15.0
5.0
)8.0(
5.0
)8.0(
5.0
15.0
8.0)8.0()5.0()8.0(
)8.0)(5.0(
15.0)8.0(
333.8)8.0(
15.00.5zlet
8.0
)5.0(
)8.0(
)5.0(
)8.0(
15.0
8.0)8.0()5.0()5.0(
)8.0)(5.0(
15.0)5.0(
8.0)8.0()5.0()8.0)(5.0(
15.0)(
)8.0)(5.0(
)15.0(
)8.01)(5.01(
)5.0()(
8.0
23
2
21
32
2122
2
5.021
32
212
32
212
32
212
2211
11
z
z
z
zk
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z
z
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zz
zz
zzzF
kkk
zz
kf
zz
z
zzF
zzzzz
z
z
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z
zz
z
zk
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zkzk
z
zk
z
z
derivativez
zkk
z
zk
z
z
z
k
z
k
z
kz
zz
zz
)8.0(333.8)8.0(2)5.0(333.8)(
8.01
333.8
)8.01(
2
)5.01(
333.8)(
8.0
333.8
)8.0(
2
)5.0(
333.8
)8.0)(5.0(
15.0)(
333.83.0
6.03.0*5.0
)5.0(
)15.0()5.0(5.0
5.0
15.0
8.0let ;)5.0(
)8.0()5.0(0
5.0
)8.0(
5.0
15.0
5.0
)8.0(
5.0
)8.0(
5.0
15.0
8.0)8.0()5.0()8.0(
)8.0)(5.0(
15.0)8.0(
1
121
1
1
22
28.0
2
8.0
'
3
33
'21
'
32
21
32
2122
2
Tutorial
Partial fraction for inverse Z-transform
If F(z)/z involve s a multiple pole, eg. P1, thenipz
iin
n
n
mmmm
nnnn
mmmm
z
zFpza
pz
a
pz
a
pz
a
pzpzpz
bzbzbzb
azazaz
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212
211
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Tutorial
Solution 2: Expand F(z) into a polynomial form
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…
321
21
211
11
32.044.11.21
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