LINEAR CONTROL SYSTEMSkarimpor.profcms.um.ac.ir/imagesm/354/stories/lin_con/... · 2014-11-10 · Dr. Ali Karimpour Nov 2014 Lecture 4 4 Property. Poles and zeros and their physical

LINEAR CONTROLLINEAR CONTROLSYSTEMSSYSTEMS

Ali KarimpourAssociate Professor

Ferdowsi University of Mashhad

Dr. Ali Karimpour Nov 2014

Lecture 4

2

Lecture 4

Topics to be covered include:

External Description.(TF Model)

Internal Model Description. (SS Model)

External and Internal Description Model. (SS and TF Description)


Lecture 4

3






Lecture 4

4

Property.

Poles and zeros and their physical meaning.

SISO and MIMO.

Open loop and closed loop system.

Effect of feedback .

Characteristic equation for SISO system

External Description (Transfer function representation)


Lecture 4

5

TF model properties

خواص مدل تابع انتقال1- It is available just for linear time invariant systems. (LTI)

2- It is derived by zero initial condition.

3- It just shows the relation between input and output so it may lose some information.

4- It is just dependent on the structure of system and it is independent to the input value and type.

5- It can be used to show delay systems but SS can not.


Lecture 4

6

Table : Laplace transform table

جدول تبدیل الپالس

u(t) - u(t-τ)


Lecture 4

7

Table : Laplace transform properties.

خواص تبدیل الپالس

y(t-τ) u(t-τ)


Lecture 4

8

Laplace transform properties. خواص تبدیل الپالس

Example 1: Check F.V.T and I.V.T for following:

)3(3)()

ss

sYa Both F.V.T and I.V.T are valid.

2

1)()s

sYb Only I.V.T is valid.

12)()2

sssYc Only I.V.T is valid.

11)()

s

sYd Only I.V.T is valid.

1)()

sssYe I.V.T is valid with care.


Lecture 4

9

Poles of Transfer functionقطب تابع انتقال

Pole: s=p is a value of s that the absolute value of TF is infinite.

How to find it?

)()()(sdsnsG Transfer function

Let d(s)=0 and find the roots

Why is it important? It shows the behavior of the system.Why?????


Lecture 4

10

Example 2: Poles and their physical meaning.

656)( 2

ssssG

31

221

)65(6)(ofResponseStep 2

ssssssssc

S2+5s+6=0 p1= -2, p2=-3

Transfer functionroots([1 5 6])

)(1)(2)(1)( 32 tuetuetutc tt

p1= -2, p2=-3

قطبها و خواص فیریکی آنها


Lecture 4

11

656)( 2

ssssGTransfer function

step([1 6],[1 5 6])

0 0.5 1 1.5 2 2.5 30

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1Step Response

Time (sec)

Ampl

itude

Example 2: Poles and their physical meaning(Continue).


Lecture 4

12

Example 3: Derive the poles of Y(s).

)()2sin(1)( 22 tutteteeety tttt

-2 -1 1

2

-2

Remark1: Which terms go to infinity?

Remark 2: Poles on the imaginary axis?


Lecture 4

13

Zeros of Transfer function صفر تابع انتقال

Zero: s=z is a value of s that the absolute value of TF is zero.

Who to find it?

)()()(sdsnsG Transfer function

Let n(s)=0 and find the roots

Why is it important? It also shows the behavior of system.Why?????


Lecture 4

14

Example 4: Zeros and their physical meaning.

65610)(

656)( 2221

ssssG

ssssG

)(121)(3

12

21)65(

6)( 32121 tueetc

ssssssssc tt

Transfer functionsz1= -6 z2=-0.6

)(871)(3

82

71)65(

610)( 32222 tueetc

ssssssssc tt

صفرها و خواص فیریکی آنها

Remark: There is one more zero at infinity.

21

21

21 632

656)(

zzpp

ssssG Number of poles and zeros?


Lecture 4

15

Transfer function65

610)(65

6)( 2221

ssssG

ssssG

step([1 6],[1 5 6]) ;hold on;step([10 6],[1 5 6])


Example 4: Zeros and their physical meaning(Continue).


Lecture 4

16

uuyyy 65

Transfer function of the system is:

651

)()(

2

ss

ssusy

It has a zero at z=-1

What does it mean? u=e-1t and suitable initial condition leads to y(t)=0


Example 5: Zeros and their physical meaning.


Lecture 4

1717

Open loop and closed loop systems

Open-loopControl system Controller

سیستمهاي حلقه باز و حلقه بسته

Actuator


Lecture 4

1818

When is the open loop representation applicable?

The model on which the design of the controller has been based is a very good representation of the plant,

The model must be stable, and

Disturbances and initial conditions are negligible.

در چه شرایطی کنترل حلقه باز مناسب است؟


Lecture 4

1919

Open loop and closed loop systems

Controller

Open-loopControl system Controller

Closed-loopControl system

Sensor

ActuatorActuator


Lecture 4

2020

SISO and MIMO

سیستم تک ورودي تک خروجی و چند ورودي چند خروجی

Systemu1 y1

SISO

u1

u2

u3

y1

y2

y3

MIMO

System


Lecture 4

2121

Sensitivity حساسیت

Sensitivity is defined as:

MG

GM

GGMMS

//

M is the system + controller transfer function

G is the system transfer function


Lecture 4

2222

Sensitivity

)()( sKsGM

1)()(

)()(

sGsK

sGsKMG

GMS

)()(11

sGsKMG

GMS

حساسیت

G(s)K(s)

Open loop system

r c

G(s)K(s)+

-

r c

Closed loop system

)()(1)()()(sGsKsGsKsM


Lecture 4

2323

Feedback properties

ویژگیهاي فیدبکFeedback can effect:

a) System gain

b) System stability

c) System sensitivity

d) Noise and disturbance


Lecture 4

t (Time)

24

Example 6: A thermal system

یک سیستم حرارتی: 11مثال

Input signal is an step function so:

0001

)(tt

tus

su 1)(

000)(

ttkekty

t

/1)(

sk

sksy

Input signal1

Output signal

k

?)()()( susysg

1sk


Lecture 4

25

Example 7: A system with pure time delay

سیستمی با تاخیر ثابت: 12مثال

1)(

sksg

dsTesh )(

1)()(

skesgsh

dsT


Lecture 4

26






Lecture 4

27

State space solution. State transition matrix. State transition equation.

Controllability and pole placement by state feedback. Observability by state feedback. Similarity transformation and canonical forms. Eigenvalues of the matrix A and poles of transfer function. Controllability and observability from block diagram.

Internal Description Model. (SS Description)


Lecture 4

28

Topics to be covered include: State space solution.

State transition matrix. State transition equation.

Controllability and Pole placement by state feedback. Observability and estate estimation. Similarity transformation and canonical forms. Eigenvalues of the matrix A and poles of transfer function. Controllability and observability from block diagram.



Lecture 4

29

State Space ModelsGeneral form of state space model

LTI systems

مدل فضاي حالت

If initial condition and input are defined, then x(t), y(t) ?

و x(t)اگر شرائط اولیه و ورودي مشخص باشد مقدار چند است؟ y(t)

Start with homogeneous equation .با معادله همگن شروع کنیدAxx 00 )( xtx


Lecture 4

30

State transition matrix ماتریس انتقال حالت

Axx )0()0( 0 xtx

)0()()( xttx

matrixtransitionstateis:)(t ماتریس انتقال حالت

)()()( 00 txtttx If t0 is not zero


Lecture 4

31

Determination of state transition matrix

تعیین ماتریس انتقال حالتAxx 00 )( xtx

)()( 0 sAxxssx 01)()( xAsIsx

011 )()( xAsILtx

11 )()( AsILt

0)()( xttx

Method I


Lecture 4

32


تعیین ماتریس انتقال حالتAxx 00 )( xtx

0)( xetx At 0)()( xttx

Atet )(

...3.2.1

12.1

1 3322 tAtAAtIeAt

Method II


Lecture 4

33


تعیین ماتریس انتقال حالت

)0()0(

)()()()(

)()(

2

1

2221

1211

2

1

xx

tttt

txtx

)()(

01

)()()()(

)()(

21

11

2221

1211

2

1

tt

tttt

txtx

)()(

10

)()()()(

)()(

22

12

2221

1211

2

1

tt

tttt

txtx

So ith column of Φ is the response of system to

0..1..00 ith

position

Method III


Lecture 4

34

Example 8: Find state transition matrix and x(t) of following system.

xx

3012

12

)0(x

011 )()( xAsILtx

1111

3012

)()(s

sLAsILt

310

)3)(2(1

21

1

s

sssL

t

ttt

eeee

t3

322

0)(

t

tt

t

ttt

eee

eeee

txtx

3

32

3

322

2

1 312

0)()(



Lecture 4

35

1- Φ(0) = I

LTIخواص ماتریس انتقال حالت در سیستمهاي

2- Φ-1(t)= Φ(-t)

3- Φ(t2-t1)Φ(t1-t0) = Φ(t2-t0)

5- Φk(t) = Φ(kt)

Property of state transition matrix for LTI systems

4- Φ(t2-t1) = Φ(t2) Φ-1(t1)


Lecture 4

36

State transition matrix for time varying systems

ماتریس انتقال حالت براي سیستمهاي متغیر با زمانx

tx

000

01

)0()0(

)0(2

1

xx

x

25.0

1)(

ttx

10

)0()0(

)0(2

1

xx

x

10

)(tx

15.001

)( 2tt

)()(0)(

12

1

ttxtxtx


Lecture 4

37

State transition equation معادله انتقال حالت

BuAxx 00 )( xtx

)()()( 0 sBusAxxssx

)()()()( 10

1 sBuAsIxAsIsx

dButxttxt

)()()()(00 Convolution

integral

State transition equation


Lecture 4

38

Example 9: Find state transition equation for unit step

uxx

10

3012

21

)0(x

011 )()( xAsILtx

1111

3012

)()(s

sLAsILt

310

)3)(2(1

21

1

s

sssL

t

ttt

eeee

t3

322

0)(



Lecture 4

39

uxx

10

3012

21

)0(x

t

ttt

eeee

t3

322

0)(

dbutxttxt

)()()()(00

21

0)(

3

322

t

ttt

eeee

tx

due

eeet

t

ttt

)(10

00 )(3

)(3)(2)(2

t

tt

eee

tx3

32

223

)(

de

eet

t

tt

0 )(3

)(3)(2

.....

.....


These are eigenvalues of A.


Lecture 4

40

Example 10: Find x1(s) and x2(s)

)()()0()()( 11 sBuAsIxAsIsx

uxx

10

3012

)0()0(

)0(2

1

xx

x

)(10

3012

)0(30

12)(

11

sus

sx

ss

sx

)(

31

)3)(2(1

)0()0(

310

)3)(2(1

21

)()(

2

1

2

1 su

s

ssxx

s

ssssxsx

)(3

1)0(3

1

)()3)(2(

1)0()3)(2(

1)0(2

1

2

21

sus

xs

suss

xss

xs



Lecture 4

41

Example 11: Derive x1(s) by using state diagram.

uxx

10

3012

1x 1xs -1 s -1

)0(1x

1

-21x2x

2x 2xs -1 s -1

)0(2x

1u(s)

-3

)()3)(2(

1)0()3)(2(

1)0(2

1)( 211 suss

xss

xs

sx



Lecture 4

42






Lecture 4

43

Controllability

Definition 1:The state equation (I) or the pair (A,b) is said to be controllable if

for any initial state x0 and any final state x1, there exists an input that transfers

x0 to x1 in a finite time. Otherwise (I) or (A,b) is said to be uncontrollable

، ورودي اي x1 براي هرو x0را کنترل پذیر گویند اگر براي هر (A,b)یا زوج (I)معادالت : 1تعریفیا زوج (I)معادالت در غیر اینصورت . برساند x1 را در زمان محدود به x0وجود داشته باشد که

(A,b) را کنترل ناپذیر گویند

، ورودي اي 1x براي هرو 0x براي هرگویند اگر کنترل پذیررا (A,b)یا زوج (I)معادالت : 1تعریفیا زوج (I)معادالت در غیر اینصورت . برساند x1 به زمان محدودرا در x0وجود داشته باشد که

(A,b) ناپذیر گویندرا کنترل

)(IducxybuAxx


Lecture 4

44

Controllability testتست کنترل پذیري

Theorem 1:The state equation (I) or the pair (A,b) is controllable if and only if

اگر فقط و اگر است پذیر کنترل (A,b) زوج یا (I) معادالت :1قضیه

)(IducxybuAxx

S= [b Ab A2b ….. An-1b], | S| ≠ 0


Lecture 4

45

Example 12: Check the controllability of following system

xy

uxx

]011[100

6116100010

2561610100

][ 2bAAbbS

12561610100

S

This kind of system is controllable so it is called controllable canonical form.

فرم کانونی کنترل پذیراین نوع سیستم همواره کنترل پذیر است لذا به آن . گویند

Controllability test


Lecture 4

46

Use of state feedback to control a system

استفاده از فیدبک حالت براي کنترل سیستم

ducxybuAxx

x x

c

A

u

y

b

d

k

State feedback ? kxru

vector1nconstantn 1

r

drxdkcybrxbkAx

)()(

0A-sI of roots

0bkA-sI of roots


Lecture 4

47


ducxybuAxx

استفاده از فیدبک براي کنترل سیستم

Is it possible to assign the eigenvalues arbitrarily? drxdkcybrxbkAx

)()(

state feedbackkxru

)(I

Theorem: If the n-dimensional state equation (I) is controllable, then bystate feedback u=r-kx, where k is a 1×n real constant vector, theeigenvalues of A-bk can arbitrarily be assigned provided that complex

conjugate eigenvalues are assigned in pairs.

حالت فیدبک با آنگاه باشد، پذیر کنترل (I) بعدي- n حالت معادالت اگر :قضیهu=r-kx ، که k 1 ثابت بردار یک×n ویژه مقادیر توان می باشد، می A-bk بطور را

بصورت باید مختلط ویژه مقادیر که نمود توجه باید البته نمود تعیین دلخواه.شود انتخاب مزدوج


Lecture 4

48

Example 13: Is it possible to assign the eigenvalues of following system on arbitrarily position؟

uxx

021

7101401

800

132028121601

2bAAbbS

21132028121601

S System is controllable

So It is possible to assign the eigenvalues on arbitrarily position.

.لذا می توان مقادیر ویژه را در مکانهاي دلخواه قرار داد



Lecture 4

49

Example 14: Is it possible to assign eigenvalues of following system by state feedback.

So : s=2 is a fixed mode.

Methods of pole placement

Clearly no but why?


Lecture 4

50

Controllability test for each mode

تست کنترل پذیري براي هر مود

ducxybuAxx

Eigenvalues of A

Aمقادیر ویژه 1

2 3

n..

bAIi |

For the controllability of a mode it is sufficient that the following matrix has full row rank

i

......در حالت مقادیر ویژه تکراري : نکته


Lecture 4

51

Example 15: Check the controllability of each mode

xy

trxx

]11[

)(01

1012.

0

0021

SS It is not completely controllable

2,1)2)(1(10

12A-sI 21

sss

s

000111

:1 ofility Controllab 1

lecontrollabnot is1 1

010110

:2 ofility Controllab 2

rank row fullnot

rank row full lecontrollab is2 2

Controllability test for each mode


Lecture 4

52

Example 16: Consider following system.

uxx

021

300020001

System is not controllableI) Is it possible assign the eigenvalues on an arbitrarily places?

I (می توان مقادیر ویژه را در مکانهاي دلخواه قرار داد؟ آیاClearly No

واضح است که خیرII) Is it possible to assign the eigenvalues on -3, -4, -2 ?

II (قرار داد؟ 4- ,3- ,2-می توان مقادیر ویژه را در آیا Clearly Yes

واضح است که بلهIII) Is it possible to the eigenvalues on -13, -2±3j ?

III (3±2- ,13-می توان مقادیر ویژه را در آیاj قرار داد؟ Clearly No

واضح است که خیرIv) Is it possible to the eigenvalues on -3, -2+3j, -2-6j ?

Iv (3+2- ,3-می توان مقادیر ویژه را در آیاj, -2-6j قرار داد؟ Clearly No

واضح است که خیر



Lecture 4

53

Methods of pole placement

روشهاي تعیین محل قطبها

1- Direct method. 1- روش مستقیم

2- Use of similarity transformation. استفاده از تبدیالت همانندي -2


Lecture 4

54

Direct method of pole placement

bkA-sI:Find

روش مستقیم تعیین محل قطبها

= Desired characteristic equation

Then determine k1, k2, … , kn from above equation

ducxybuAxx

xkkkru n ]...[Let 21

Polynomial ofDegree n

Polynomial ofDegree n


Lecture 4

55

Example 17: Assign the poles on -2±j if it is possible.

uxx

11

1011

11

01S

01S

System is controllable

So it is possible to assign the poles on -2±j .

][11

1011

10 kkbkA

10

10

111kkkk

10

10

111

kskkks

bkAsI

110

2 1)( kskks

)2)(2(1)( 1102 jsjskskks 542 ss ]610[ k

Method I

Use of similarity transformation for pole placement


Lecture 4

56






Lecture 4

57

Observability

Definition 2:The state equation (I) or the pair (A,c) is said to be observable if

for any unknown initial state x0 , there exists a finite time t1 > 0 such that the

knowledge of the input u and the output y over [0,t1] suffices to determine

Uniquely the initial state x0. Other wise, the equation is unobservable.

x0 براي هر شرط اولیهگویند اگر رویت پذیررا (A,c)یا زوج (I)معادالت : 2تعریف ]t0,1[در بازه yو خروجی u، وجود داشته باشد که اطالعات ورودي 1t<0زمان محدود

.کافی باشد در غیر اینصورت سیستم را رویت ناپذیر گویند x0 منحصر بفردبراي محاسبه

)(IducxybuAxx


Lecture 4

58

Observability test

Theorem 2:The state equation (I) or the pair (A,c) is observable if and only if

اگر فقط و اگر است پذیر رویت (A,c) زوج یا (I) معادالت :2قضیه

0,

.

.

1

2

V

cA

cAcAc

V

n

)(IducxybuAxx


Lecture 4

59

Example 18: Check the observability of following system

xy

uxx

]011[100

6116100010

5116110011

2cAcAc

V

0)60(1)115(15116110011

VSo the system is not

observable.

پس سیستم رویت پذیر نیست

Observability test


Lecture 4

60

Observability test for each mode (system with distinct eigenvalues)

)سیستمهاي داراي مودهاي مجزا(تست رویت پذیري براي هر مود

ducxybuAxx

Eigenvalues of A

Aمقادیر ویژه 1

2 3

n..

c

AIi

For the observability of a mode it is sufficient that the following matrix has full column rank

i


Lecture 4

61

Use of state estimation to use in feedback loop

استفاده از تخمین زن حالت براي استفاده در مسیر فیدبک

x x

c

A

u

y

b

d

k

r

States are not available!

x Estate Estimator

x

Condition for xx ˆ


Lecture 4

62



x x

c

A

u

y

b

d

k

r

x Estate Estimator

x

)ˆ(ˆˆ xccxlbuxAx

)ˆ(ˆˆ xcylbuxAx


Lecture 4

63



)ˆ(ˆˆ xcylbuxAx

x x

c

A

u

y

b

d

k

r

x

x

A

b

c

l

Estimator


Lecture 4

64






Lecture 4

6565

Change of variables (Similarity transformation)

323

212

311

2xxwxxwxxw

Pxxw

110021101

wwPx

1

1

110021101

ducxybuAxx

duwcPyPbuwPAPw

1

1

udwcy

ubwAwˆˆ

ˆˆ

A b c d

ducxybuAxx

113113 33

isdiscisbisA

Let

wP 1

wAP 1 bu

y wcP 1 du


Lecture 4

6666

]06.04.0[ˆ 1 cPc

Example 19: A similarity transformation example

xy

uxx

]122[43

6

661221463

30

9.4ˆ Pbb

0ˆ dd

wy

uww

]06.04.0[30

9.4

400010002

21

42 13

Eigenvalues of AAمقادیر ویژه

21

42 13

Eigenvalues of Aمقادیر ویژه

A

New system is very simpler than older

400010002

ˆ 1PAPA

21-1-1-1121-2

1P

Eigenvector corresponding to -2



Lecture 4

67

Similarity transformation properties

ddcPc

PbbPAPA

ˆˆ

ˆˆ1

1ducxybuAxx

Pxw

ation transformSimilarity

udwcy

ubwAwˆˆ

ˆˆ

1- It can lead to a simpler system. امکان ساده سازي سیستم -12- It doesn’t change the eigenvalues. عدم تغییر مقادیر ویژه -23- Similar transfer functions. تابع انتقال یکسان - 3

5- It doesn’t change observability. عدم تغییر رویت پذیري - 54- It doesn’t change controllability. عدم تغییر کنترل پذیري -4


Lecture 4

68

Effect of similarity transformation on eigenvalues of system

تاثیر تبدیالت همانندي بر مقادیر ویژه

ddcPc

PbbPAPA

Pxw

ˆˆ

ˆˆ1

1

ducxybuAxx

udwcy

ubwAwˆˆ

ˆˆ

Similarity transformation doesn’t change the eigenvalues.تبدیل همانندي مقادیر ویژه را تغییر نمی دهد

0 AsI 0ˆ AsI

AsI ˆ 11 PAPsPP 1)( PAsIP 1 PAsIP

AsI


Lecture 4

69

Effect of similarity transformation on controllability

تاثیر تبدیالت همانندي بر کنترل پذیري

ddcPc

PbbPAPA

ˆˆ

ˆˆ1

1

ducxybuAxx

udwcy

ubwAwˆˆ

ˆˆ

Similarity transformation doesn’t change the controllability.تبدیل همانندي کنترل پذیري را تغییر نمی دهد

bAbAAbbS n 12 ...

bAbAbAbS n ˆˆ...ˆˆˆˆˆˆ

12

PbPPAPbPPAPbPAPPb n 11121 ... bAbAbAbS n ˆˆ...ˆˆˆˆˆˆ 12

bPAbPAPAbPb n 12 ... bAbAAbbP n 12 ...

Matrix ility Controllab

PS

SPPSS ˆ


Lecture 4

70

Effect of similarity transformation on observability

تاثیر تبدیالت همانندي بر رویت پذیري

ddcPc

PbbPAPA

ˆˆ

ˆˆ1

1

ducxybuAxx

udwcy

ubwAwˆˆ

ˆˆ

Similarity transformation doesn’t change the observability.تبدیل همانندي رویت پذیري را تغییر نمی دهد

1

.

.

ncA

cAc

VMatrix

ity Observabil

1ˆˆ

.

.

ˆˆ

ˆ

ˆ

nAc

Ac

c

V

With the same proof as pervious slide


Lecture 4

71

Important canonical forms

فرمهاي کانونی مهم

1- Controllable canonical form. فرم کانونی کنترل پذیر -1

2- Observable canonical form. فرم کانونی رویت پذیر -2

3- Jordan canonical form. فرم کانونی جردن -3


Lecture 4

72

Controllable canonical form

فرم کانونی کنترل پذیر

xy

uxx

]...[10.00

...1...000...........0...1000...010

...1

...10...........

...0001...000

S 0 S

xy

uxx

]021[100

7148100010

814712)( 23

sssssG


Lecture 4

73

Transforms to controllable canonical form

uxx

11

1011

تبدیل به فرم کانونی کنترل پذیر

0110ˆ 1PAPA

10ˆ Pbb

11

01S

01SSystem is

controllable

1

...nqA

qAq

P q is last rowof S-1

0111

P

uww

10

0110

Controllable canonical form


Lecture 4

74

Observable canonical form

فرم کانونی رویت پذیر

xy

uxx

]1...000[

.1000

..............010...001...000

...1...........

1...0010...00

V 0 V

xy

uxx

]100[021

7101401

800

814712)( 23

sssssG


Lecture 4

75

Transforms to observable canonical form

xy

uxx

]10[11

1221

تبدیل به فرم کانونی رویت پذیر

13ˆ Pbb

1210

VSystem is observable

qAAqqP n 11 ... q is last columnof V-1

1012

P

wy

uww

]10[13

2150

Observable

canonical form

2150ˆ 1PAPA

0ˆ

10ˆ 1

dd

cPc


Lecture 4

76

Jordan canonical form for systems with distinct eigenvalues

فرم کانونی جردن براي سیستم با مقادیر ویژه مجزا

xccccy

u

b

bb

xx

n

nn

]...[

.

.

...000......................0...000...00

321

2

1

2

1

xy

uxx

]33.05.017.0[137

100020004

Checking the controllability and observability is very simple

All modes are controllable and observable

تمام مودها کنترل پذیرو رویت پذیر است

4)7)(17.0()(

s

sg2

)3()5.0(

s 1)1()33.0(

s


Lecture 4

7777

]06.04.0[ˆ 1 cPc

Remember: A similarity transformation example

xy

uxx

]122[43

6

661221463

30

9.4ˆ Pbb

0ˆ dd

wy

uww

]06.04.0[30

9.4

400010002

21

42 13

Eigenvalues of AAمقادیر ویژه

400010002

ˆ 1PAPA

21-1-1-1121-2

1P

Eigenvector corresponding to -2


2)9.4)(4.0()(

s

sg1

)0()6.0(

s 4

)3()0(

s

296.1)(

s

sg

؟ Aمقادیر ویژه و قطبهاي تابع انتقالارتباط بین


Lecture 4

78

Eigenvalues of matrix A and poles of transfer function

xy

uxx

]011[100

6116100010

31

22 13

Eigenvalues of AAمقادیر ویژه 31 p

22 p

Poles of systemقطبهاي سیستم

What happened to -1 ?

)3)(2(1)(

sssg

[num,den]=ss2tf(A,b,c,0), , g=minreal(g)g=tf(num,den)


Lecture 4

79

Extended Jordan canonical form

فرم کانونی جردن تعمیم یافته

xcccy

u

b

bb

xx

r

rr

]...[

....000

...........0...000...000...01

21

2

1

2

1

1

iff econrollabl are and 11 0 1 b

iff econrollabl is 2 0 2 b

0 rbiff econrollabl is r

.

...

iff observable are and 11 0 1 c

iff observable is 2 0 2 c

0 rciff observable is r

.

...


Lecture 4

80

Transfer function of controllable and observable systems.

تابع انتقال سیستمهاي کنترل پذیر و رویت پذیر

nnbeALetducxybuAxx

dbAsIcsG 1)()(

Suppose the system is controllable and observableفرض کنید سیستم کنترل پذیر و رویت پذیر باشد

Suppose the system is not controllable or is not observableنیستفرض کنید سیستم کنترل پذیر یا رویت پذیر


Lecture 4

81

Example 20: Transfer function of controllable and observable systems.

xy

uxx

]111[011

100021321

xy

uxx

]201[110

300010011

32)(

s

sG

375132)( 23

2

sss

sssGSystem is

controllable and

observable

System is not both controllable and

observable but 3 is controllable and

observableBut we can not know much more information about the system by this method

.اما نمی توان اطالعات بیشتري از این روش بدست آورد


Lecture 4

82






Lecture 4

83

Controllability and observability from block diagram and state diagram

کنترل پذیري و رویت پذیري از طریق بلوك دیاگرام و دیاگرام حالت1- Find the numbers of s-1in the block diagram or state diagram.

را در بلوك دیاگرام و یا دیاگرام حالت بدست آورید s-1ابتدا تعداد - 12 - Find the transfer function by Mason’s rule

تابع انتقال را با توجه به قانون گین میسون بدست آورید - 2

3 – If the numbers of s-1in the block diagram or state diagram is equal to degree of transfer function then the system is controllable and observable.

برابر در بلوك دیاگرام و یا معادالت حالت با درجه تابع انتقال s-1اگر تعداد - 3.است کنترل پذیر و رویت پذیرآنگاه سیستم با شد


Lecture 4

84

Example 21: Check the controllability and observability of following function.

TF

451

)()()( 2

sssrscsG

Mason’s rule

The numbers of s-1in the state diagram is 2 and the degree of transfer function is also 2 so the system is controllable and observable.

است لذا 2بوده و درجه تابع انتقال نیز 2در دیاگرام حالت برابر s-1تعداد .است کنترل پذیر و رویت پذیرسیستم

Controllability and observability from block diagram and state diagram


Lecture 4

85

Exercises

3334)( 23

2

ssS

ssG4-1 Find the poles and zeros of following system.

4-2 Is it possible to apply a nonzero input to the following systemfor t>0, but the output be zero for t>0? Show it. uuyyy 265

4-3 Find the step response of following system.333

4)( 23

2

sss

ssG

4-4 Find the step response of following system for a=1,3,6 and 9.

222)( 2

ssassG

)(sin1)( 2 tuttetty t 4-5 Find the poles of Y(s)

)1(1)(

ss

sY4-6 Check F.V.T and I.V.T for following system.


Lecture 4

86

Exercises

4-7) Find the state diagram of the following system

uxx

010

141230312

4-8) Find the transfer function of the following system.a) By use of the formula between the two representation.b) By use of state diagram

xy

uxx

]021[010

141230312

4-9) Find y(t) for initial condition [1 3 -1]T and the unit step as input.

xy

uxx

]021[010

141230312


Lecture 4

87

Exercises

4-10) The state transition matrix and state transition equation of following system.

)(21

)(4031

)( trtXtX

t

ttt

eeee

s

sssLs

sLAsILt

4

421

1111

04

10

453

11

4031

)()(

dtReeee

Xeeee

tXt

t

ttt

t

ttt

0 )(4

)(4)()(

4

4

)(21

0)0(

0)(

Answer is:

4-11) Suppose x1(0)=1, x2(0)=3 and x3(0)=2 and r(t)=u(t) . Find x(t) and c(t) for t≥0

)(]001[)()(011

)(300122041

)( tXtctrtXtX


Lecture 4

88

Exercises

4-13) a)Find the eigenvalues of following system.b) Find the pole of transfer function.c) Are the eigenvalues controllable and observable?

xy

uxx

]021[010

141230312

4-14) Check the controllability and observability of every mode in the following system.

xy

uxx

]122[43

6

661221463

4-12) Find the modes of following system.uxx

010

141230312


Lecture 4

89

Exercises4-15) Check the controllability and observability of every mode in the following system.

wy

uww

]011[102

400010002

4-16) Examine the controllability and observability of each mode of the following system.

)()(]101[)()(101

)(400320001

)( trtXtctrtXtX

Answer: All modes are controllable. -1 and -4 are observable but -2 is not observable.4-17) The eigenvalues of a system are -3,4,-5 and the poles of its transfer function are -3 and -5.(Midterm spring 2008)a) What is your idea about the observability of each eigenvalue.b) What is your idea about the controllability of each eigenvalue.


Lecture 4

90

Exercises

4-19) Check the controllability and observability of every mode in the following system by direct inspection (without calculation). wy

uww

]011[102

400010002

S-1

-2

11 -3

S-1 -21

S-1

-1 31

r(s) c(s)4-20) Check the controllability and observability of the following system.

4-18) Consider the following system.a) Find the controllable canonical form. b) Find the observable canonical form.

xy

uxx

]122[43

6

661221463


Lecture 4

91

4-21) Check the controllability and observability of the following system by transfer function method.

21s

s13

+

+

++

Exercises

4-22) Find a state space realization for following system such that it is neither observable nor controllable.(Final 1391)

12)(

sssg

4-23) Is it possible to assign the eigenvaluesof following system on arbitrarily places?

xy

uxx

]021[010

141230312


Lecture 4

92

xy

uxx

]122[43

6

661221463

4-24) Is it possible to assign the eigenvalues of following system on arbitrarily places?

wy

uww

]011[102

400010002

4-25) Use a state feedback to assign the eigenvalues of following system on -4, -1 , -6?b) Use a state feedback to assign the eigenvalues of following system on -4, -2 , -6?c) Use a state feedback to assign the eigenvalues of following system on -1, -2 ±6j?

Exercises


Lecture 4

93

Exercises4-26) Is it possible to assign the eigenvalues of following system on arbitrarily places. If yes, put them on -2, -3 and -4.

uxx

021

310020001

4-27) Is it possible to assign the eigenvalues of following system on arbitrarily places (Final 1391)? If yes, put them on -2 and -4.

xy

uxx

]11[5

1123510


Lecture 4

94


استفاده از تبدیالت همانندي در تعیین محل قطبها

buAxx Pxw uw

aaaa

w

n

10..00

...1...000

...............................................0...1000...010

1210

?ˆˆˆ kbA

11221100ˆ...ˆˆˆ

1...000...............................................................................0...1000...010

nn kakakaka

012

21

1 ... :is system ofequation sticCharacteri asasasas nn

nn

n

012

21

1 ... :isequation sticcharacteri Desired bsbsbsbs nn

nn

n

2b 1 nb1b0b


Lecture 4

95

Use of similarity transformation for pole placement استفاده از تبدیالت همانندي در تعیین محل قطبها

11221100ˆ...ˆˆˆ

1...000...............................................................................

0...1000...010

ˆˆˆ

nn kakakaka

kbA

2b 1 nb1b0b

11221100 ....ˆ nn ababababk

wkru ˆ Pkk ˆPxkr ˆ kxr

x x

c

A

u

y

b

d

Pw

k

k

r


Lecture 4

96

01S

System is controllable

So it is possible to assign the poles on -2±j .

Method II

qAq

P 1]10[ Sq

0111

P

]46[]04)1(5[ˆ1100 ababk

System characteristic equation: s2+0s-1Desired characteristic equation: (s+2+j)(s+2-j)=s2+4s+5

]610[ˆ Pkk

uxx

11

1011

11

01S

place(a,b,[-2+i -2-i])


Example 22: Assign the poles on -2±j if it is possible.

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LINEAR CONTROL SYSTEMSkarimpor.profcms.um.ac.ir/imagesm/354/stories/lin_con/... · 2014-11-10 · Dr. Ali Karimpour Nov 2014 Lecture 4 4 Property. Poles and zeros and their physical