control systems analysis and design

8/13/2019 Control Systems Analysis and Design

1/61

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


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time

f(t)

Time kT

f(kT)

A/D

Time kT

f(kT)

D/A

Time kT

f(kT)


3/61


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 anddesign a controller. Can we have a similar

mathematical tool for discrete time system?


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2.1 Z-transform

For a continuous signal f(t), its sampled data can bewritten as,

Then we can define Z-transform of f(t) as

where z-1 represents one sampling period delay intime.

0 )()2()()0()()( ns nTfTfTffkTftf

21

0

)2()()0(

)()]([)]([)(

zTfzTff

znTfnTfZtfZzFn

n


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2.1 Z-transform

Solution:

Example 1: Find the Z-transform of unit step

function. f(t)

t

kT

f(kT)


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2.1 Z-transform

Apply the definition of Z-transform, we have

1

21

32

21

0

1

11)(

1

1

)()]([)]([)(

zzzzF

q

aaqaqaqa

zz

zkTfkTfZtfZzF k

k


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2.1 Z-transform

Another method

1

1

3212111

21

0

1

1)(1)()(

)1()(

1

)()]([)]([)(

zzFzFzzF

zzzzzzzFz

zz

zkTfkTfZtfZzF k

k


8/61

2.1 Z-transform

Example 2: Find the Z-transform of a

exponential decay.

Solution:

f(t)

t

1

221

0

1

1)(

1

)]([)(

)(

ze

zF

zeze

zekTfZzF

ekTf

aT

aTaT

k

kakT

akT


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2.1 Z-transform

Exercise 1: Find the Z-transform of a

exponential decay f(t)=e-at using other

method. f(t)

t


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2.1 Z-transform

Example 3: Find the Z-transform of a cosine

function.

Solution: As

2sin;2cos

sincos;sincos

j

eet

eet

tjtetjte

tjtjtjtj

tjtj


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2.1 Z-transform

21

1

21

1

211

11

11

11

11

1

cos21

cos1

)cos21(2cos22

)(1)(2

21

)1)(1(

11

2

1

1

1

1

1

2

1)(

1

1

][

])[][(2

1

2][cos)(

zTz

Tz

zTzTz

zzezezeze

zeze

zeze

zezezF

zeeZ

eZeZee

ZkTZzF

TjTj

TjTj

TjTj

TjTj

TjTj

aT

at

tjtjtjtj


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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


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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

zeeZzstepZZ

eZZeZkTfZzF

aT

aT

aT

aT

at

atat


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2.1 Z-transform

Exercise 3: Find the Z-transform for

attetf )(

21

1

)1()(

aT

aT

ez

eTzzF


15/61


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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)


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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


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If f(t)=0 for t


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Defining m=k-n, we have

Since f(mT)=0 for m


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Final value theorem:Suppose that f(t), where f(t)=0

for t


21/61

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)( 11

zFzfz

)(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


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s

111

1z

2

1

s 21

1

)1(

z

Tz

as

111

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

s21

1

cos21

sin

zTz

Tz

22s

s

21

1

cos21

cos1

zTz

Tz

22)(

as 221

1

cos21

sin

zeTez

TezaTaT

aT

22

)(

as

as

221

1

cos21

cos1

zeTez

TezaTaT

aT

f(t) F (s) F(z)

(t) 1 1

u(t)

t

e-at

1e-at

sint

cost

e-atsint

e-atcost


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2.3 Z-transform examples

Example 1: Assume that f(k)=0 for k


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)1()21(

2

13

211

)21(9]3[]2[)]2(9[)(

)21(

9

1)-(z/2

)2/(

2

9]2[

2

9]229[)]2(9[

211

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

2

11

1

221

1-

1-

11

zz

z

zzzzZZkZzF

z

zzkZkZkZ

zzzZZ

z

Tz

zz

z

a

zFtfaZ

ZZkZkZzF

kk

kkk

kk

t

kkkk


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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)


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Solution:From the figure, we have

K 0 1 2 3 4 5 6

f(k) 0 0 0 1/3 2/3 1 1Apply 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

zkfzF

k

k


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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

2

32

2

1

2

2

zez

zzaTeeeaT

zezz

aTz

asZ

sZ

s

aZsFZzF

asss

a

as

k

s

k

s

k

ass

asF

aT

aTaTaT

aT


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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 sequencef(kT) is called as the Inverse Z-transform. We

label inverse Z-transform as Z-1.

n

no

m

mo

n

no

m

mo

zazaza

zbzbzbbZzFZkTf

zazaza

zbzbzbbtfZkTfZzF

2

1

1

2

2

1

111

2

1

1

2

2

1

1

1)]([)(

1

)]([)]([)(


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Z-transform

=Inverse Z-transform


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The inverse Z-transform can yield the

corresponding time sequence f(kt) uniquely.

However, it says nothing about f(t). There mightbe numerous f(t) for a given f(kT).

f(t)

t0 T 2T 3T 4T 5T 6T


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x(kT) f(t)Zero-orderHold

Low-pass

Filter


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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,


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2.5 Inverse Z-transform examples

Example 2: Given ,

Find f(kT).

Solution: Apply partial-fraction-expansion tosimplify F(z), then find the simpler terms from

the Z-transform table.

Then we need to determine k1and 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


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Multiply (1-z-1) to both side and let z-1=1, we have

1

2

1

1

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

11

2

1

11

1

1

2

1

11

11

11

1

z

aT

aT

aTaT

aT

aTaT

aT

ze

zek

ze

kzk

ze

ze

ze

k

z

kz

zez

zez


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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(

1

1

1

221

1

1

1

1

12

111

11

1

1

aTez

aTaTaT

aTaT

aT

aT

aT

z

zekk

z

kze

z

ze

ze

k

z

kzezez

zeze

0,1,2,k,1)(

1

1

1

1

)1)(1(

)1()(

1111

1

akT

aTaT

aT

ekTf

zezzez

zezF


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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


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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

1 1

1

11

z

z

zz-z

z

z

-

21

2

21

1

1

11

1

11zz

z

zz

-z

z

z

-


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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


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21

54

543

43

432

32

321

21

21

121

741

1013

102010

710

7147

47

484

4

21

2121

zz

zz

zzz

zz

zzz

zz

zzz

zz

zz

zzz


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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


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3) Computational method using Matlab

Example: Given find f(kT).

Solution:

num=[1 2 0]; den=[12 1]

Say we want the value of f(kT) for k=0 to 30u=[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


42/61



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


43/61

Reading

Study book

Module 2: The Z-transform and theorems

Textbook

Chapter 2 : The Z-transform (pp23-50)


44/61

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()


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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()


46/61

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.

4 8

Krad/s

F()

122 6 14 181610-4-8


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Tutorial

For 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


48/61

Tutorial

Exercise 1: Find the Z-transform of a

exponential decay f(t)=e-aT using other

method.f(t)

t


49/61

Tutorial

1

1

33221

22111

221

0

11)(1)()(

)1()(1

)()]([)]([)(

zezFzFzezF

zezeze

zezezezFzezeze

kTfkTfZtfZzF

aT

aT

aTaTaT

aTaTaTaT

aTaT

k


50/61

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

cos21cos1]cos[

)()]([);()]([

)(cos21

cos1cos

zeTze

Tze

zTzTzteZ

zeFtfeFzFtfZ

zFzTz

TztZ

aTaT

aT

zez

at

aTat

aT


51/61

Tutorial

Solution 2:

221

1

11

1

cos21

cos1

11

11

21)(

1

1][

])[][(2

1

2]cos[)(

zeTze

Tze

zezezF

zeeZ

eZeZ

eeZteZzF

aTaT

aT

TjaTTjaT

aT

at

tjattjat

tjattjatat


52/61

Tutorial

Exercise 3: Find the Z-transform for

Solution:

attetf )(

21

1

21

1

21

1

)1(

)1(][

)()]([);()]([

)()1(

ze

zTe

zTzteZ

zeFtfeFzFtfZ

zFz

TztZ

aT

aT

zez

at

aTat

aT


53/61

Tutorial


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 thefinal value theorem respectively, we have

)4.03.11)(1(

)(211

1

zzz

zzF


54/61


55/61

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


56/61

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


57/61

Tutorial


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

)150()50(11


58/61

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

2

1

3

2

212

2

2

5.0

21

32212

3

2

21

2

3

2

21

2

2211

11

z

z

z

zk

z

zkk

z

zk

z

z

zk

zk

zkz

zzzz

z

zk

z

zk

z

zkkz

z

z

k

z

k

z

kz

zz

zz

z

k

z

k

z

k

zz

z

z

zF

zz

zz

zz

zzzF

kkkz 150


59/61

kkk

zz

kf

zz

z

zzF

zzzzz

z

z

zF

zzz

zzk

zz

zkzk

z

zk

z

z

derivativez

zk

kz

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(5.05.015.0

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

2

8.0

2

8.0

'

3

33

'2

1

'

3

2

2

1

3

2

212

2

2


60/61

Tutorial

Partial fraction for inverse Z-transform

If F(z)/z involve s a multiple pole, eg. P1, thenipz

ii

n

n

n

mm

mm

nn

nn

mm

mm

z

zFpza

pz

a

pz

a

pz

apzpzpz

bzbzbzb

azazaz

bzbzbzb

z

zF

)()(;

)())((

)(

2

2

1

1

21

1

1

10

1

1

0

1

1

10

11

)()(;

)()(;

)()(

)()()()()(

2

12

2

11

3

3

1

22

1

1

3

2

1

1

1

10

pzpzpz

ii

n

n

n

mm

mm

z

zFpz

dz

dc

z

zFpzc

z

zFpza

pza

pza

pzc

pzc

pzpzpzbzbzbzb

zzF

i


61/61

Tutorial

Solution 2: Expand F(z) into a polynomial form

Num=[0 0.51 0];

Den=[12.1 1.440.32];

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

5.0

)8.01)(5.01(

)5.0()(

zzz

zz

zz

zzzF

control systems analysis and design

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