linear_system_notes_2015.pdf

I

Contents 1 Lecture 1 .......................................................................................................................... 1

Introduction to Design of Physical Systems ............................................................ 1 1.1

System Classification ............................................................................................... 2 1.2

Applications ............................................................................................................. 3 1.3

I/O Description of Dynamical Systems.................................................................... 4 1.4

Definition: Linearity ................................................................................................. 5 1.5

2 Lecture 2 .......................................................................................................................... 6

Definition: Causality ................................................................................................ 7 2.1

Definition: Time-Invariance ..................................................................................... 8 2.2

3 Lecture 3 ........................................................................................................................ 11

Impulse Response................................................................................................... 11 3.1

Computation of the Convolution Integral for Linear Systems ............................... 13 3.2

3.2.1 For a Causal, Linear, Relaxed, and Time-Varying System ............................ 13

3.2.2 For a Linear, Relaxed, Causal, Time-Invariant System .................................. 14

4 Lecture 4 ........................................................................................................................ 15

Computation of Impulse Response Using Differential Equation ........................... 15 4.1

Computation of Impulse Response Using Laplace Transform .............................. 18 4.2

5 Lecture 5 ........................................................................................................................ 21

State Diagram ......................................................................................................... 21 5.1

5.1.1 Concept of State .............................................................................................. 24

5.1.2 State Vector ..................................................................................................... 25

5.1.3 State Space ...................................................................................................... 25

5.1.4 State Trajectories ............................................................................................ 25

General State Space Representation ....................................................................... 26 5.2

II

6 Lecture 6 ....................................................................................................................... 27

State Models .......................................................................................................... 27 6.1

Canonical Representations .................................................................................... 31 6.2

6.2.1 Observable Canonical Form ........................................................................... 31

7 Lecture 7 ....................................................................................................................... 33

Controllable Canonical Form ................................................................................ 33 7.1

Jordan Canonical Form .......................................................................................... 34 7.2

7.2.1 Characteristic Polynomial D(s) = 0 Contains N Distinct Poles ..................... 35

7.2.2 Characteristic Polynomial D(s) = 0 Contains Multiply Repeated Poles ........ 36

8 Lecture 8 ....................................................................................................................... 39

Transformation From State Equations to Transfer Function ................................. 39 8.1

Examples for C.C.F., O.C.F., J.C.F. ...................................................................... 42 8.2

8.2.1 Observable C.F. ............................................................................................. 42

8.2.2 Controllable C.F. ........................................................................................... 42

8.2.3 Jordan Canonical Form .................................................................................. 43

9 Lecture 9 ....................................................................................................................... 46

Solution of State Equation ..................................................................................... 46 9.1

Fundamental Matrix .............................................................................................. 48 9.2

9.2.1 Definition: ...................................................................................................... 48

9.2.2 Examples: ....................................................................................................... 48

State Transition Matrix .......................................................................................... 50 9.3

9.3.1 Definition ....................................................................................................... 50

9.3.2 Proposition ..................................................................................................... 50

9.3.3 Properties of ),( 0tt .................................................................................. 50

10 Lecture 10 ..................................................................................................................... 53

State Transition Matrix for LTI Systems ............................................................... 53 10.1

III

10.1.1 Definition ........................................................................................................ 53

10.1.2 Approaches to Compute )(t ....................................................................... 54

Cayley-Hamilton Technique .................................................................................. 56 10.2

10.2.1 Cayley-Hamilton Theorem ............................................................................. 56

10.2.2 Cayley-Hamilton Technique ........................................................................... 57

10.2.3 Algorithm to Apply Cayley-Hamilton Technique .......................................... 57

11 Lecture 11 ...................................................................................................................... 61

Ф(t, t0) for LTI System ........................................................................................... 61 11.1

Ф(t, t0) for LTV Systems ........................................................................................ 65 11.2

Solution of the State Equations and Output Equation ............................................ 69 11.3

12 Lecture 12 ...................................................................................................................... 70

Definitions .............................................................................................................. 70 12.1

12.1.1 Controllability ................................................................................................. 70

12.1.2 Observability ................................................................................................... 70

12.1.3 Examples for Controllability and Observability ............................................. 71

Canonical Transformations .................................................................................... 72 12.2

12.2.1 A Canonical Decomposition of Dynamic System .......................................... 72

12.2.2 Controllable Canonical Form (C.C.F.) ........................................................... 73

12.2.3 Observable Canonical Form (O.C.F.) ............................................................. 74

13 Lecture 13 ...................................................................................................................... 78

Diagonalization of a Matrix ................................................................................. 78 13.1

13.1.1 Jordan Canonical Form (J.C.F.) – Distinct Eigenvalues Case ........................ 78

13.1.2 Jordan Canonical Form (J.C.F.) – Repeated Eigenvalues Case ...................... 81

13.1.3 Matrix Reduction to Jordan Form ................................................................... 83

Controllability of JCF and Observability of JCF ................................................... 87 13.2

13.2.1 Theorem .......................................................................................................... 87

IV

13.2.2 Examples ........................................................................................................ 88

14 Lecture 14 ..................................................................................................................... 92

Stable Feedback: General Case ............................................................................. 92 14.1

State Estimation: Full-Order Observer .................................................................. 95 14.2

14.2.1 Theorem ......................................................................................................... 95

14.2.2 Design of the State Estimator ......................................................................... 97

14.2.3 Closed-Loop System Characteristics ........................................................... 100

Reduced-Order Estimators .................................................................................. 101 14.3

Reduced-Order State Estimator: General Case ................................................... 104 14.4

15 Lecture 15 ................................................................................................................... 106

Tracking Problem ................................................................................................ 106 15.1

Closed-Loop Pole-Zero Assignment ................................................................... 108 15.2

General Case Using Controllable Canonical Form ............................................. 110 15.3

V

Table of Figures

Figure 1-1. Vehicle suspension system. ............................................................................... 2

Figure 1-2. System classification. ........................................................................................ 2

Figure 1-3. Applications #1. ................................................................................................ 3





Figure 1-8. Input/output system representation. .................................................................. 4

Figure 1-9. RC network. ...................................................................................................... 4

Figure 2-1. Time varying systems ....................................................................................... 8

Figure 3-1. Delta function. ................................................................................................. 11

Figure 3-2. Impulse at t = τ. ............................................................................................... 11

Figure 3-3. Property of step function. ................................................................................ 12

Figure 3-4. Behavior of rectangular pulse. ........................................................................ 12

Figure 3-5. Multiplication of a function with a delta function. ......................................... 12

Figure 3-6. Causal versus non-causal system. ................................................................... 13

Figure 3-7. Delta function and its shifted value. ................................................................ 14

Figure 4-1. Unit impulse at t = 0 , 0, 0+. .......................................................................... 15

Figure 4-2. Mass-spring-dashpot system. .......................................................................... 17

Figure 5-1. Basic state diagram operations. ....................................................................... 21

Figure 5-2. Non-minimal internal state representation. ..................................................... 21

Figure 5-3. External representation. ................................................................................... 22

Figure 5-4. Minimal internal state representation. ............................................................. 22

Figure 5-5. System state diagram for example 2. .............................................................. 22



Figure 5-8. RLC network. ................................................................................................. 24

Figure 5-9. System state trajectory for example 1. ............................................................ 25

Figure 6-1. State diagram for a linear time invariant system. ............................................ 27

VI

Figure 6-2. RC network. ................................................................................................... 28

Figure 6-3. Mass-Spring system. ....................................................................................... 29

Figure 6-4. State diagram. ................................................................................................. 31

Figure 7-1. Decomposition of the transfer function. ......................................................... 33



Figure 7-4. Parallel configuration. .................................................................................... 35

Figure 7-5. State diagram of the term in bracket ............................................................... 37

Figure 8-1. State diagram for example 2. .......................................................................... 42

Figure 8-2. Transfer function decomposition for example 3. ............................................ 42

Figure 8-3. System state diagram. ..................................................................................... 43


Figure 8-5. Armature controlled DC motor. ...................................................................... 45

Figure 9-1. Transition property. ........................................................................................ 51

Figure 10-1. System plot. .................................................................................................. 60

Figure 12-1. State trajectory as a result of the control u. .................................................. 70

Figure 12-2. System diagram. ........................................................................................... 70

Figure 12-3. System classification. ................................................................................... 72

Figure 13-1. Block diagram. .............................................................................................. 89

Figure 14-1. System state diagram. ................................................................................... 92

Figure 14-2. System plot for example 2. ........................................................................... 95

Figure 14-3. Block diagram of combined plant & observer. ............................................. 95

Figure 14-4. Plant and observer diagram. ......................................................................... 96

Figure 14-5. Block diagram for example 4. ...................................................................... 99

Figure 14-6. Signal flow graph. ......................................................................................... 99

Figure 14-7. Estimator response. ....................................................................................... 99

Figure 14-8. Block diagram of the combined plant and reduced order observer. ........... 102

Figure 14-9. System signal flow graph. .......................................................................... 103

Figure 14-10. Closed-loop combined plant and controller estimator. ............................. 104

Figure 14-11. Block diagram for system with reduced order estimator. ......................... 105

Figure 15-1. State feedback diagram. .............................................................................. 106

Figure 15-2. State diagram for example 1. ...................................................................... 107

VII

Figure 15-3. ADC motor system schematic. .................................................................... 108

Figure 15-4. System close-loop block diagram. .............................................................. 108

Figure 15-5. System block diagram. ................................................................................ 108

Figure 15-6. Root Locus. ................................................................................................. 109

Figure 15-7. State diagram for example 1. ...................................................................... 111

Figure 15-8. Block diagram of the closed-loop system for example 1. ........................... 111

Figure 15-9. Closed-loop diagram 3 for a unity feedback of example 1. ........................ 111

Figure 15-10. Root Locus for the standard feedback control system of example 2. ....... 112

Figure 15-11. Root Locus for the state feedback control system of example 2. .............. 112

Figure 15-12. System sate diagram for example 3. ......................................................... 113

Figure 15-13. Closed-loop block diagram for example 3. ............................................... 113

Figure 15-14. Root Locus for example 3. ........................................................................ 113


Figure 15-16. System state diagram for example 4. ........................................................ 115

Figure 15-17. Closed-loop block diagram for example 4. ............................................... 115


1

Objectives:

1) Introduction to design of physical systems

2) System classification

3) Applications

4) I/O description of systems

5) Linearity

1 Lecture 1

Introduction to Design of Physical Systems 1.1

a) Performance Specification

b) Modeling

A tool for study

A representation of the physical system

ODE or difference equations

c) Simulation

A tool for studying behavior of ODE’s or difference equations

d) Analysis

Qualitative: stability, controllability, observability

Quantitative: simulation

e) Optimization/control

Optimize parameters of the system

f) Realization

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2 Linear system: Lecture 1

Example 1: Vehicle suspension system

System Classification 1.2

Figure 1-2. System classification.

Seat Mass

Mass of Body

Wheel Mass

Road Profile

Seat springness

Seat damping

Spring Shock absorber

x1

x2

x3

Tires springness

f(t)

System classes

Distributed parameter Lumped parameter

Stochastic Deterministic

Continuous time Discrete time

Nonlinear Linear

Time-varying Time invariant

Figure 1-1. Vehicle suspension system.

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1.3 Applications 3

Linear spring: Nonlinear spring:

21

212,121

222

111

)(

ff

xxkfxx

kxfx

kxfx

21

3

212,121

3

222

3

111

)(

ff

xxkfxx

kxfx

kxfx

Applications 1.3

a) Deterministic Control Problem

Figure 1-3. Application #1.

b) Estimation Problems


Internal representation: State characterization

c) Stochastic Control Problem


Known plant x

Unknown

input

y

Output Sensor

Parameters known I/O description known

z

Measured

output

Known plant x

Known

input

y

States

(unknown)

Sensor z

Measured

output

Known plant x

Unknown

input

y

Output Sensor

z

Measured

output Noise input Noise input

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d) System Identification


e) Adaptive Control Problem


I/O Description of Dynamical Systems 1.4

Figure 1-8. Input/output system representation.

Figure 1-9. RC network.

oo

i edt

deRCe

Unknown

plant x

Unknown

input

y

Output Sensor

z

Measured

output

Noise Noise

Parameters unknown, however I/O description known

Unknown

plant

x

Unknown

input

y

Output Sensor

z

Measured

output

Noise Noise

Plant

(system)

L{.}

input u(t)

output y(t)

y(t) = L{u(t)} (*) ,

Under what conditions (*) holds.

R

C

+

-

+

-

i

dt

deci

c

i

dt

de oo

idtc

e

idtc

Rie

o

i

1

1

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1.5 Definition: Linearity 5

1) 0ie and 0)0( oe zero input response

2) 0ie and 0)0( oe zero state response

3) 0ie and 0)0( oe general response

Case 1： 00 oo

i edt

deRCe

RCt

oo eete /)0()(

(zero input response)

Case 2： 0)0( oe and Aei (constant) 0)0(; ooo eAe

dt

deRC

RCt

o eAte /1)(

(zero state response)

Case 3： RCtRCt

oo eAeete // 1)0()(

(general response)

Relaxed (at rest):

A system is relaxed (or at rest), if the initial conditions (i.c.) are set to zero.

Definition: Linearity 1.5

A relaxed system is said to be linear, if and only if

)}({)}({)}()({ 22112211 tuLatuLatuatuaL

(Superposition principle)

Condition for defining the I/O

map

6

Objectives:

1) Examples

2) Introduction to causality and time-invariance system

2 Lecture 2

Examples of linear systems :

Example 1 : RC net.

ioo ee

dt

deRC io

o eRC

eRCdt

de 11 (*)

Since Integrating factor o

RCtoRCt

o

RCt eeRCdt

deeee

dt

d /// 1 (**)

(*)(**) i

RCt

o

RCt eeRC

eedt

d // 1

0

/

0

/ 1dtee

RCdtee

dt

di

RCt

o

RCt

0

/0/ 1)0()( dtee

RCeeee i

RCt

oo

RC

Since the system is relaxed 0)0( oe

0

// 1)( dtee

RCee i

RCt

o

RC

0

//1)( dteee

RCe i

RCtRC

o

0

/)(1)( dtee

RCe i

RCt

o

(I/O representation)

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2.1 Definition: Causality 7

Q: Show if the above I/O representation is linear?

To show this, we use superposition principal

2

02

1

01 ; eeee LL

0

1

/)(1

0

1)( dtee

RCe RCt

;

0

2

/)(2

0

1)( dtee

RCe RCt

02211 eeaea L

0

2211

/)(

0 )(1

)( dteaeaeRC

e RCt

0

11

/)(1dteae

RC

RCt

0

22

/)(1dteae

RC

RCt

2

02

1

010 eaeae The system is linear.

Example 2: System t

duty0

)()( , is this system linear?

2211 yuyu ； t

duy0

11 ; t

duy0

22

yuauau 2211

2211

0

22

0

11

0

2211 )( yayaduaduaduauay

ttt

Therefore, the system is linear.

Example 3: t

duty0

2 )()( is nonlinear

Definition: Causality 2.1

The system is said to be causal if the output at time t does not depend on the input at times

after t.

A system which is not causal is said to be noncausal or anticipatory.

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Definition: Time-Invariance 2.2

Example 4: RC net.

t

i

RCt deeRC

te0

/)(

0 )(1

)(

Show if this system is time varying or not.

Change )()( Tee ii

t

i

RCt dTeeRC

te0

/)(

0 )(1

)(

Let TT , dd

Tt

T

i

RCTt deeRC

te )(1

)( /)(

0

Since 0)( ie for all 0 !

Tt

i

RCTt deeRC

te0

/)(

0 )(1

)(

)()(0 Ttete o

The system is time-invariant

L u y

u(t)

t

y(t)

t 0 0

t

t 0 0

Time-invariant

t

t 0 0

Time-varying

Figure 2-1. Time varying systems.

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2.2 Definition: Time-Invariance 9

Example 5: )()(1

)( tutyt

ty

Is this system time-invariant or time-varying?

Is this system linear or nonlinear?

In general, for ( )y a t y , by using integrating factor da

e)(

,

since

( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

a t dt a t dt a t dtdy t e y t a t e y t e

dt

y t ay t u t y t u t ay t

dttadtta

ueetydt

d )()(

)(

For this example t

ta1

)( teee tt

dtdtta

ln)(

)( )(1)( tutyttytytdt

d

so,

bb

utdttydt

ddt

b

utbbyy dt )()(

For a relaxed system, y(b) = 0

bb

ut

uty dt dt 1

)(

L u y

y(t)=L{u(t)}

b is the initial time

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Q: Linear or nonlinear?

t

b

dut

ty

)()(

11 yu t

b

dut

y

11 ; 22 yu t

b

dut

y

22

yuaua 2211

t

b

t

b

t

b

yayaduat

duat

duauat

y 221122112211 )(

Therefore, the system is linear.

Q: Time-invariant or time-varying?

t

b

dut

ty

)()( , y(t) = L{u(t)}

)()( Tuu Due to the shifted input, the resulting output y1 is

t

b

dTut

ty

)()(1 Question: Is ?)()(1 Ttyty

Tt

b

duTt

Tty

)()(

Chang of variable ddTT , (T is constant)

at Tbb ; at Ttt

Tt

Tb

dut

Tty

)()(1

)()(1 Ttyty

The system is time-varying

11

Objectives

1) Impulse Response of LTI and LTV systems

2) Computation of the Convolution Integral for Linear Systems

3 Lecture 3

ubdt

udb

dt

udbya

dt

yda

dt

yda

m

m

mm

m

mn

n

nn

n

n 01

1

101

1

1

This is a linear system since y… )(ny and u…

)(mu enter linearly.

If the coefficients 0aan and 0bbm are constant, then the system is time-invariant.

If they are time dependent, then the system is time-varying.

A nonlinear example is yy ,2y ,

2y , uy, yu , ....

Impulse Response 3.1

Let ),( th be the response of the system at time t due to an impulse applied at time .

),( th is called the impulse response of the system.

0

t

Unit impulse

0 t

Figure 3-1. Delta function.

Figure 3-2. Impulse at t = τ.

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Figure 3-3. Property of step function.

Figure 3-4. Behavior of rectangular pulse.

In the limit as )]()([1

lim)(00

tutut

For a given u , we can always write

dtutu )()()( , since

)()()()( 000 xxxfxxxf

)(

)()(

)()()(

1

tu

dttu

dttutu

Figure 3-5. Multiplication of a function with a delta function.

0 t

0 t

0 t

f(x)

x

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3.2 Computation of the Convolution Integral for Linear Systems 13

Computation of the Convolution Integral for Linear 3.2

Systems

The output y(t) due to u(t) is )}({)( tuLty

dtuLty )()()( L is an operator in time t

dutLdtuLty

th

)()()()()(

),(

dthuty ),()()(

(General Convolution Integral for a Linear, Relaxed LTI or LTV systems).

3.2.1 For a Causal, Linear, Relaxed, and Time-Varying System

For a causal, linear, relaxed, and time-varying system we now get

0

),()(),()()(

t

t

dthudthuty

t

dthuty ),()()(

For a causal system 0),( th for all t

Figure 3-6. Causal versus non-causal system.

0 t

0 t

noncausal

t

causal

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If the input is applied at time 0tt and is zero before that, then

t

t

dthuty

0

),()()(

Let h(t) be the impulse response due to )(t , ),()( tht

Figure 3-7. Delta function and its shifted value.

For a time-invariant system )()0,()( ththt

Example 1:

are time-varying

While 2)( t

, are time-invariant

3.2.2 For a Linear, Relaxed, Causal, Time-Invariant System

For a linear, relaxed, causal, time-invariant system, we now get

t

dthuty0

)()()(

We want to show that alternatively this is the same as

t

dtuhty0

)()()(

Let tddt ,

t

tt

dtuh

dtuhdhtuty

0

00

)()(

)()())(()()(

0

)()0,()( ththt

15

Objectives

1) Impulse response computation using differential equation

2) Impulse response computation using Laplace transform

4 Lecture 4

Computation of Impulse Response Using Differential 4.1

Equation

For a given general system given by,

ubdt

udb

dt

udbya

dt

yda

dt

yda

m

m

mm

m

mn

n

nn

n

n 01

1

101

1

1

to find the impulse response, let u(t) be an impulse.

Assuming a’s and b’s are constant this is a linear time-invariant system

System is assumed to be relaxed, that is initial conditions (I.C.’s) are zero at .

Figure 4-1. Unit impulse at t = 0 , 0, 0+.

0

u(t)

1

t

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Example 1: For RC network we have: ioo ee

dt

deRC . Let us find the impulse response:

For finding the impulse response )()( 0 thetei

(**) 0dt

dhRC 0 t

(*) )(dt

dhRC 0 t

hfor

thfor

From (**) the system is relaxed 0)0(

oe

)0(0)(0)0( tthh

from (*), integrate it from 0t to

0t . We do this to get the initial

condition at 0t , in other words, we need to find )0( h .

10)0()0()(

0

0

0

0

0

0

0

hhRCdtthdtdtdt

dhRC

RChRChRCh

1)0(1)0(0)0(

For t > 0, the system becomes 0 hdt

dhRC with

RCh

1)0(

Therefore, 01

)( / teRC

th RCt

Since h(t) does not contain an impulse, therefore 0

0

0-

dth

So, {

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4.1 Computation of Impulse Response Using Differential Equation 17

Example 2: For the system FKxxBxM ,

find impulse response.

Since this system is relaxed

0)0()0( xx

We need )0( x and )0( x for )(tF

00

)(0

KhhBhMtfor

tKhhBhMtfor

for t < 0 0)0()0( hh h(t) = 0, t < 0

for t ≥ 0 integrate from 0t to

0t

∫

∫

∫ ∫

10)0()0()0()0( hhBhhM

Since 1)0()0( BhhM

By double integrating the equation from 0t to

0t

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

dtdtKhdtdtdtdthBdtdthM

000)0()0( hhM

0)0( h , now since 1)0()0( BhhM M

h1

)0(

For t > 0 0 KhhBhM with 0)0( h M

h1

)0(

0 0

0

)(

t

t

th

The explicit solution depends on

particular value of M, B and K.

M

K

B

x

F

Figure 4-2. Mass-spring-dashpot system.

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Computation of Impulse Response Using Laplace 4.2

Transform

This method is only applicable to LTI systems.

Example 3: ioo ee

dt

deRC Find impulse response using Laplace transform (L.T.)

method.

Let )(tei )(theo thdt

dhRC

Take L.T. of both sides

hdt

dhRCL

0

)()()( dtetxsXtx stL 1)()0()( sHhssHRC

Since the system is relaxed 0)0( h 1HRCsH RCs

sH

1

1)(

01

)()( /1 teRC

sHLth RCt

For system

t

duthty0

)()()( , take L.T.

t

duthLtyL0

)()()(

0 0

)()()( dtduthesY

t

st

Since the system is causal

)(

00

0

)(

00 0

)()()()(

)()()()()(

sU

ss

esH

stst

duesHduesH

dudtethdtduthesY

s

Y(s) = H(s)∙U(s) )(

)()(

sU

sYsH

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4.2 Computation of Impulse Response Using Laplace Transform 19

Example 4: For system uuyyy 434

1) Find T.F. and impulse response.

Taking the L.T of the system,

)(5)(4)(3)(4)(2 sUssUsYssYsYs

54)(34)( 2 ssUsssY

34

54

)(

)()(..

2

ss

s

sU

sYsHFT

Impulse response )()( 1 sHLth

34

54)(

2

1

ss

sLth

Using partial fraction expansion

1

1

3

2/7)( 1

ssLth

02

1

2

7)( 3 teeth tt

2) Find the output for a step input using the convolution integral for a

relaxed system.

t

duthty0

)()()()(

00

01)()(

02

1

2

7)( 3

t

ttutu

teeth

s

tt

t

tt deety0

)()(3 12

1

2

7)(

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tt eety 2

1

6

7

3

5)( 3

Step Response

3) Find the output for a step input using T.F. for a relaxed system.

Y(s) = H(s)U(s)

34

54)(

2

ss

ssH ;

ssU

1)(

34

54)(

2

sss

ssY

tt eety 2

1

6

7

3

5)( 3

4) If y(0) = 1 and 0)0( y , find y(t) for a step input.

uuyyyL 5434

)0()0()(}{ 2 ysysYsyL

)0()(}{ yssYyL

)(5)(4)(31)(401)(2 sUssUsYssYssYs

54)(434)( 2 ssUssssY

2 2

. . 0

4 5 4( ) ( )

4 3 4 3due to i c

s sY s U s

s s s s

22

4 5 4( )

4 34 3

s sY s

s ss s s

tt eety 3

3

5

3

5)(

21

Objectives

1) State Diagram

2) State Space Representation

ʃ x Integrator

Differentiator

a ax Scalar

∑

Summary

×

Multiplier

+ +

x

x

5 Lecture 5

State Diagram 5.1

State diagram basic mathematical operations are

shown in figure 5-1.

Example1: Draw state diagram for system

cudt

dubay

dt

dy .

Solve for dt

dy in terms of the rest.

Minimal Representation:

aycudt

dub

dt

dy

Integrate both sides of the equation

to get

dtadtcudtdt

dubdt

dt

dyy y

dtaycubuy )( state diagram is shown as below.

Figure 5-2. Non-minimal internal state representation.

Figure 5-1. Basic state diagram operations.

b ʃ

c a

u y +

+ -

cu ay

of 115


Figure 5-3. External representation.

Figure 5-4. Minimal internal state representation.

Example 2: Single Input Single Output (SISO) uuyyy 4853

uuyyy 4853

uuyyy 4853

)38(54 yuyuy

State diagram is drawn as below:

Figure 5-5. System state diagram for example 2.

Example 3: For a multi - input - multi - output(MIMO) system

69

368532

211222

112111

uyyyyy

uuyyyyy y

draw state diagram for it.

H U Y

y 4

u

ʃ

8

+

5

-

∫

+

3

(8u-3y)

+

-

- ʃ

y + c

u

ʃ cu - ay

b bu

∫ +

a

-

ay

cu +

y(0)

of 115

5.1 State Diagram 23

69

368532

211222

1122111

uyyyyy

uuyyyyy

211222

1122111

69

368532

uyyyyy

uuyyyyy

96

652383

122212

1211211

yyyuyy

uyyuyyy

Therefore, state diagram of this system is drawn below:


Example 4: )sin()sin( yuyuyy (this is a nonlinear system)


3

ʃ

6

+

3

- +

2

+

-

- ʃ

ʃ +

+

-

ʃ

6

8

9

5

-

-

+

+

u +

sin

-

ʃ y

sin(y)

of 115


5.1.1 Concept of State

Internal Representation (Time) State Equations

External Representation I/O Representation

I/O is partitioned into (a) Convolution Integral (time) and (b) Transfer Function (frequency)

State: The minimal number of variables x(t) that their knowledge at time 0tt , and the

input u(t) for 0tt will completely characterize the response of the system.

Example 5: )(; 0teeedt

deRC oio

o oe is the state

response state zero

/)(

responseinput

/

0

0

)(1

)()(

t

t

i

RCt

zero

RCt

oo deeRC

etete

Example 6:

Figure 5-8. RLC network.

dt

de

dt

diRi

Cdt

idL i

12

2

)(tei ( 0tt ), )( 0ti , )( 0tdt

di should be specified !

States are i(t) and dt

di (*)

dt

dee

RCdt

ed

R

L

dt

de oo

i 1

2

0

2

States are oe and dt

deo (**)

Case (*) and (**) reveal that states are NOT UNIQUE , there is another

choice for the states !

Let 1

dtdv i

v iC dt C

dt

dvRCv

dt

vdLCei

2

2

R

+

-

+

-

i

C L

dt 1

Rie

RiiCdt

diLe

o

v

i

of 115

5.1 State Diagram 25

In order to solve for v(t) , we need )( 0tv , )( 0tdt

dv and ie , since

C

tit

dt

dv )()( 0

0

so, to solve the system we need )( 0tv and )( 0ti .

Therefore, for an RLC network the voltage across the capacitor and the current

through an inductor are the states.

5.1.2 State Vector

Given n states we define the state vector as an (n×1)-dimensional column vector

1

2

1

)(

)(

)(

)(

nn tx

tx

tx

tx

, nRtx )(

5.1.3 State Space

The space where the states are evolving in.

5.1.4 State Trajectories

0 0( ) ; , ( ), ( )x t f t t x t u t

Example 7: 023 yyy

Let

ytx

ytx

)(

)(

2

1,

2

1

x

xx is the state vector, 2Rx is the state space.

)()( 12

2

1 txececty tt with c1 and c2 depending on )( 0ty and )( 0ty

)(2)( 22

2

1 txececty tt

tt

tt

ecectx

ecectx

2

2

12

2

2

11

2)(

)(

let 00 t , y(0) =0 and 1)0( y

tt

tt

eetx

eetx2

2

2

1

2)(

)(

Figure 5-9. System state trajectory for example 1.

(

(t = 0)

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General State Space Representation 5.2

( ) ( ( ), ( ), ) Equations

( ) ( ( ), ( ), ) Equations

x t f x t u t t State

y t g x t u t t Output

n dim state

m dim output

diminput

n

m

q

x R

y R

u R q

This is a nonlinear, time varying, causal system.

Nonlinear, time invariant, causal:

))(),(()(

))(),(()(

tutxgty

tutxftx

Linear, time varying system:

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( )

x t A t x t B t u t

y t C t x t D t u t

MIMO system

nn x

x

x

xn

x

x

x

x

2

1

2

1

vector 1

( )

11 12 11 11 12 1 1

21 222 21 22 2 2

1 2

1

1

( ) ( ) ( )( ) ( ) ( )

( ) (( ) ( ) ( )

( ) ( ) ( )

A t

qn

n

n n n nn n

nn n

n

b t b t b tx a t a t a t x

b t b tx a t a t a t x

x a t a t a t x

( )

1

2 2

1 2

1

1

) ( )

( ) ( ) ( )

B t

q

n n nq q

qn q

n

u

b t u

b t b t b t u

1

1

2

1

)(

21

22221

11211

1

1

2

1

)(

21

22221

11211

2

1

)()()(

)()()(

)()()(

)()()(

)()()(

)()()(

m

q

q

qm

tD

mqmm

q

q

m

n

n

nm

tC

mnmm

n

n

mu

u

u

tdtdtd

tdtdtd

tdtdtd

x

x

x

tctctc

tctctc

tctctc

y

y

y

27

Objectives

1) State Models

2) Canonical Representations:

Observable Canonical Form (OCF)

6 Lecture 6

State Models 6.1

Linear Time Invariant System

Equations )()()(

Equations )()()(

OutputtDutCxty

StatetButAxtx

A, B, C and D are now constant matrices.

State Diagram

Figure 6-1. State diagram for a linear time invariant system.

Example 1: uytyty 2)1( , find the state space representation,

in another word, state equation and output equation

Let x1= y, 212 xxyx , now we need ?2 x

ytytuyx )1(2

2 21

2

2 )1( xtxtux

21

2

2

21

)1(

xtxtux

xx

State Equation

1xy Output Equation

y + B

u

ʃ x

D

+

A

+

+

C

of 115


R

C

+

- i

R

C

+

-

y u

i

In order to get A, B, C and D, define

2

1

x

xx

11

12122

1

22

2

122

1

1

0

)1(

10

u

x

x

ttx

x

BA

⏟

[

]

⏟

Example 2: For system shown as below, find the state and output equations.

1) dtiiC

Riu )(1

211

2) 0)(11

1222 dtiiC

dtiC

Ri

Let dtiC

x 11

1, dti

Cx 22

1 11 ixc , 22 ixc

1222

211

0 xxxxRC

xxxRCu

21

111

212

121

xRC

xRC

x

uRC

xRC

xRC

x

State Equation

22

1xdti

Cy 2xy Output Equation

Find A, B, C and D

22

21

11

RCRC

RCRCA ;

120

1

RC

B

21

10

C ; 110 D

This system is Time invariant since A, B, C and D are constant.

Now with the voltages across the capacitors as states, we get

Figure 6-2. RC network.

of 115

6.1 State Models 29

M K

M

K K

F

x x

1 1 1 2 1 2 1

1 1 2

2 2 2 2 2

1( )

1

x v i i dt i i CxC i Cx Cx

x v i dt i CxC

1 2 1

2 2 1

( ) 1) (*)

0 2) (**)

u RC x x x from

RCx x x from

212

11x

RCx

RCx

121 2(**)(*) xxxRCu 121

211x

RCx

RCu

RCx

11

112

212

211

xRC

xRC

x

uRC

xRC

xRC

x

State Equation

222 xyxvy Output Equation

RCRC

RCRCA

11

12

;

0

1

RCB ; 10C ; 0D

Example 3: System as below:

0)(

)(

1222

2111

yykkyym

Fyykkyym

Choose the states as 24231211 , , , yxyxyxyx

2

2

1324

43

3112

21

xm

kx

m

kyx

xx

xm

kx

m

k

m

Fyx

xx

State Equations, let

4

3

2

1

x

x

x

x

x

Figure 6-3. Mass-Spring system.

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(i)

3

1

2

1

x

x

y

y Output Equations or

(ii) 11 xyy Output Equation (valid) or

(iii) 32 xyy Output Equation (valid) or

(iv) 3121 xxyyy Output Equation (valid) etc.

44

02

0

1000

002

0010

m

k

m

k

m

k

m

k

A ;

140

0

1

0

mB

1114144414 FBxAx 1112144212 FDxCy

(i)

420100

0001

C ,

120

0

D , 12 42 12 DCy

(ii) 41

0001

C , 110 D , 11 41 11 DCy

(iii) 41

0100

C , 110 D

(iv) 41

0101

C , 110 D

Now choose the states as follows: 2142131211 , , , yyxyyxyxyx

3

3214

43

3112

21

xm

k

m

Fyyx

xx

xm

kx

m

k

m

Fyx

xx

State Equations

44

03

00

1000

00

0010

m

k

m

k

m

k

A

14

1

0

1

0

m

mB

Let as an example,

(Output Equation)

of 115

6.2 Canonical Representations 31

Canonical Representations 6.2

a) Observable Canonical Form (O.C.F)

b) Controllable Canonical Form (C.C.F) (also known as Phase Variable CF)

c) Jordan Canonical Form (J.C.F)

6.2.1 Observable Canonical Form

For system ubdt

udb

dt

udbya

dt

yda

dt

ydn

n

nn

n

nn

n

nn

n

01

1

101

1

1

we define dt

dD , then ubuDbuDbyayDayD n

n

n

n

n

n

n

0

1

10

1

1

ubuDbuDbyayDayD n

n

n

n

n

n

n

0

1

10

1

1

yaubD

yaubD

ubynnnn 0011

11

State Diagram:

Figure 6-4. State diagram.

Define the states as the output of each integrator.

00

111

2232

1121

yaubx

yaubxx

yaubxx

yaubxx

n

nn

nn

nn

and 1xuby n

)(

)(

)(

0010

223122

112111

ubabxax

ubabxxax

ubabxxax

nn

nnnn

nnnn

State Equations

ʃ

ʃ

ʃ

+

-

+

-

+

- + + +

+ y

u

of 115


100

11

22

11

nn

n

nnn

nnn

bab

bab

bab

bab

B

Outputubxy

bD

C

n

n

n

1

11

1001

The pair (A, C) is known as one of the Observable Canonical forms.

Alternatively, from the state diagram, reverse the assignment for the states, i.e. the output

of the left integral is denoted by 1x and consecutively increased to x then we get,

The pair (A, C) is now considered as the selected Observable Canonical Form (OCF) in

this course.

The coefficients of

the characteristic

polynomial

0 0 0 1 .C

33

Objectives

1) Controllable Canonical Forms (CCF)

2) Jordan Canonical Form (JCF)

7 Lecture 7

Controllable Canonical Form 7.1

0

1

1

1

1

0

1

1

1

1

)(

)(

asasas

bsbsbsb

sU

sYn

n

n

n

n

n

n

Figure 7-1. Decomposition of the transfer function.

0

1

1

1

)(

)(

asassU

sZn

n

n

uzazaz n

n

n

0

)1(

1

)(

Let

)1(

)2(

1

2

1

n

n

n

n

zx

zx

zx

zx

)(

1

32

21

n

n

nn

zx

xx

xx

xx

nn

n

n

n

xaxaxau

zazazauz

12110

)1(

110

)(

U Y

U Z Y

all pole system all zeros system

of 115


Therefore,

nnn

nn

xaxaxaux

xx

xx

xx

12110

1

32

21

State Equation

Output Equation

0

1

1 bsbsbZ

Y n

n

n

n zbzbzby n

n

n

n 0

)1(

1

)(

10112110 xbxbxaxaxauby nnnnn

0 0 1 1 1 2 1 1( ) ( ) ( )n n n n n n ny b b a x b b a x b b a x b u Output equation

)()()( 111100 nnnnn abbabbabbC nbD

The pair (A,B) is in the Controllable Canonical Form (CCF) also known as the Phase

Variable Canonical Form

Jordan Canonical Form 7.2

1 1

1 1 0

1 1

1 1 0

( )( ) ( ) ( ) ( )

( )

n n

n nnn n

n

b s b s b s b N sY s U s b U s U s

s a s a s a D s

Where )()()()( 00

2

22

1

11 n

n

nnn

n

nnn babsbabsbabsN

and

0

1

1)( asassD n

n

n

1

0

0

0

B

of 115

7.2 Jordan Canonical Form 35

γ ʃ

λ

+

+

u ẏ/λ y/λ y

y/λ

7.2.1 Characteristic Polynomial D(s) = 0 Contains N Distinct Poles

)())(()( 21 nssssD

then )()()(2

2

1

1 sUsss

sUbsYn

nn

Construct the state diagram.

Example 1: Consider Us

Y


yuyy

uy

Figure 7-4. Parallel configuration.

γ ʃ

λ

+

+

u y ẏ

ʃ

+

+

u

y

ʃ

+

+

ʃ

+

+

+

+ +

+

Figure 7-2. Alternative state diagram.

of 115


1 1 1

2 2 2

1 1 1n n n

n n n

x x u

x x u

x x u

x x u

Sate Equation

nnn xxxuby 2211 Output Equation

n

A

000

000

000

000

3

2

1

;

1

1

1

1

B ; nC 321 ; nnD

Jordan Canonical Form (JCF)

7.2.2 Characteristic Polynomial D(s) = 0 Contains Multiple Repeated

Poles

Suppose that D(s) can be expressed as

)())(()()()( 4321

nssssssD

UbY n Usss

)()()( 1

1

1

2

1

1

1 2

1

2 2 2

1 2

3 4

( ) ( ) ( )

n

n

Us s s

Us s s

This bracket is

represented in the next

state diagram

of 115

7.2 Jordan Canonical Form 37

First, construct sate diagram for )( 1s terms in

)( 1

1

sU

Yas shown below..

Then, similarly construct state diagram for

)( 2s and )()(),( 43 nsss

parallel with the above state diagram (detail omitted).

Finally, the combined state diagram leads to the following A and B matrices.

n

J

J

A

000

000

000

000

3

2

1

,

1

1

2

1

B

B

B

where

i

i

i

i

i

iJ

0000

1000

0000

0010

0001

,

1

0

0

0

0

iB for i = 1 and i = 2

ʃ

+

+

ʃ

+

+

u ʃ

+

+

+

terms

+

+

+ +

Jordan Block

Figure 7-5. State diagram of the term in bracket.

of 115


In other words,

1

1

1

1

1

1

0000

1000

0000

0010

0001

J ,

1

1

1

0

0

0

0

B

and

2

2

2

2

2

2

0000

1000

0000

0010

0001

J ,

1

2

1

0

0

0

0

B

39

Objectives

1) Transformation from state equations to Transfer Function

2) Examples for C.C.F., O.C.F., J.C.F

8 Lecture 8

Transformation From State Equations to Transfer Function 8.1

Given

DuCxy

BuAxx, find the transfer function (TF) ?

)(

)(

sU

sY

Take Laplace transform (L.T.) ( x(0) = 0 )

(**))()()(

(*))()()(

sDUsCXsY

sBUsAXssX

)()()((*) sBUsAXssX )()()( sBUsXAsI

nn

Assuming )( AsI is nonsingular, then 1)( AsI exists.

)()()( 1 sBUAsIsX

)()()()( 1 sDUsBUAsICsY

11

1

11

1

11

1

)()()(

qqpp

sUDBAsICsY

.).()()(

)( 1 FTDBAsICsU

sY

GUY

u

u

u

y

y

y

p

MatrixTransfer

p

***

***

***

2

1

2

1

T.F. for SISO system Transfer Matrix for MIMO system

of 115


Example 1: Find T.F. for system u

x

x

yu

x

x

x

x

2

1

2

1

2

1

11;

0

1

12

01

First approach: Brute Force

Write down equations for system:

uxxy

xxx

uxx

21

212

11

2

Since 1 2 1 2y x x u y x x u

uuxxuxxuxy 21211 2

uuxxuxuuxxy 21121 2

uuuxxy 213

Since uuxyy 22 2)(

2

1

2

12 yyuux

Since uxxy 21 uxyx 21

yyuyx 5.05.05.01

uuuyyuuyyuyy )5.05.05.0()5.05.05.0(3

uuuyyy 22

uuuyyy 22

Therefore, Transfer Function (T.F.):

usuusysyys 22 22

12

2

)(

)(2

2

ss

ss

sU

sY

of 115

8.1 Transformation From State Equations to Transfer Function 41

Second Approach: Using formula DBAsICsU

sY 1)(

)(

)(

With

12

01A ,

0

1B , 11 C , 1D to find ?)( AsI

s

ssI

0

0,

12

01

12

01

0

0

s

s

s

sAsI

12

01

0)1)(1(

1)( 1

s

s

ssAsI

10

01

)1)(1(

1

)1)(1(

2

0)1)(1(

1

12

01

ss

s

ss

ss

s

s

s

2

1

)1(

1

0

1

12

01

)1(

1)(

22

1s

ss

s

sBAsI

222

2

2

1

)1(

3

)1(

2

)1(

1

)1(

2

)1(

)1(

11)(

s

s

ss

s

s

s

s

BAsIC

2

2

2 )1(

)1(31

)1(

3

)(

)(

s

ss

s

s

sU

sY

12

2

12

123

)(

)(2

2

2

2

ss

ss

ss

sss

sU

sY

of 115


U X Y

Examples for C.C.F., O.C.F., J.C.F. 8.2

8.2.1 Observable C.F.

Example 2: uuyyy 265 , for observable C.F. draw the state diagram

uuyyyuuyyy 265265

Figure 8-1. State diagram for example 2.

212

21

52

6

xuxx

xux

State Equations 2xy Output Equation

2

1

x

xx ;

uDxCy

uBxAx

oo

oo

51

60oA ,

2

1oB , 10oC , 0oD Observable C.F.

8.2.2 Controllable C.F.

Example 3:

uxx

ccBA

1

0

56

10

uxy

ccDC

021

T

oc AA , T

oc CB , T

oc BC , oc DD

To get C.C.F, directly from the differential equation, we find the T.F.

65

12

)(

)(2

ss

s

sU

sY

Figure 8-2. Transfer function decomposition for example 3.

ʃ -

-

2

ʃ +

+

5

6

-

y u

of 115

8.2 Examples for C.C.F., O.C.F., J.C.F. 43

uxxxssU

X

65

65

12

Let

xx

xx

2

1, then

xx

xx

2

21 122 6565 xxuxxuxx

1 2

2 1 26 5

x x

x x x u

State Equations

Since (2 1)

2 (2 1) 21

Y sy x x Y s X sX X

X

122 xxy Output Equation

56

10

cA ,

1

0

cB , 21cC , 0cD

8.2.3 Jordan Canonical Form

For system 65

122

ss

s

U

Y , in order to find the J.C.F., follow the steps listed below:

1) Using partial fraction expansion, 3

7

2

5

ssU

Y

2) Draw the state diagram

Figure 8-3. System state diagram.

Therefore, for Jordan C.F.

30

02

A ,

1

1

B , 75C , 0D

ʃ +

-

2

-5

ʃ +

-

3

7

+

+

u y

21

22

11

75

3

2

xxy

uxx

uxx

of 115


Example 4: uuyyy 22 , show

21

10

cA ,

1

0

cB , 12cC , 0cD

21

10

oA ,

1

2

oB , 10oC , 0oD

For Jordan Canonical Form

2 2 2 2 2

( ) 2 2 3 3

( ) 2 1 ( 1) ( 1) ( 1) ( 1) ( 1) ( 1)

Y s s s a b a as a

U s s s s s s s s s

2

1 31

1 ( 1)

Ya

U s s


Jordan Block:

10

11

A ,

1

0

B , 13C , 0D

Example 5: Given poles: -1, -1, 3，what is A and B?

ʃ +

-

1

3 ʃ +

-

1

1

- +

u y

T.F.=

T.F.=

T.F.=

A=

, B=

21

22

121

3

xxy

xux

xxx

of 115

8.2 Examples for C.C.F., O.C.F., J.C.F. 45

Example 6: Armature Controlled DC Motor

Figure 8-5. Armature controlled DC motor.

From the above figure, the governing equations are

This is a 3rd

order system,

let 321 ,, xxix a

eKiRdt

diL

iKBJ

baaa

a

aT

eLL

Ki

L

R

dt

di

aa

ba

a

aa 1

1

313

32

311

xJ

Bx

J

Kx

xx

eL

xL

Kx

L

Rx

T

aa

b

a

a

Defining the output as the angular position of the motor

2xy , Jordan Canonical Form is:

J

B

J

K

L

K

L

R

A

T

a

b

a

a

0

100

0

,

0

0

1

aL

B , 010C , 0D

R

+

-

i L

+

R L

i

J

B T

bb

ba

aaa

aT

Ke

edt

diLiRe

iKT

TBJ

46

Objectives

1) Solution of State Equation

2) Fundamental Matrix

3) State Transition Matrix

9 Lecture 9

Solution of State Equation 9.1

Given

utDxtCy

utBxtAx

)()(

)()(

,

u , ,

q

mn

Ry

RRx

Example 1: RyRRx , u , ; utbxtax )()( , )( 0tx is given

[

]

∫

[

]

∫

∫

∫

⏟

∫

⏟

Integrating factor

of 115

9.1 Solution of State Equation 47

Example 2: Let ata )( ( constant ) utbaxx )(

∫

Example 3: Let

t

taat udbexetxt0

)(

0 )()0()(0

Show this by using Laplace Transform method !

We have , 0 t, )( utbaxx

b(t)u(t)LaX(s)x(0)-sX(s) )( utbaxxL

)()(1

)0(1

)( tutbLas

xsa

sX

taking 1L

)()(

1)0()( 1 tutbL

asLxetx at

)()()()()()( 21

0

2121

1 tgtgdgtgsGsGL

t

t

taat dubexetx0

)( )()()0()(

Example 4: tuxt

x

1

with u = t , x(1) =1

We know

0

1

1 1

1 1( ) ( ) ( ) ln ln

t t tt

t

da t b t t a d d t

t

，

0

1( ) ln

ln 1t

ta d

t te e et

, ln ln

1

1( ) 1 ( )

t

tx t e dt

for 1t

of 115


Fundamental Matrix 9.2

9.2.1 Definition:

GivennRxxtxxtAx ,(,)( 00） , an n×n matrix M(t) is said to be a fundamental

matrix of xtAx )( , if and only if the n columns of M(t) , denoted as

)(M , ),(M ),( n21 tttM are any n linearly independent solutions of xtAx )( .

9.2.2 Examples:

Example 5: Given x

t

x

20

01

, find M(t).

Let

12

2

12

1 )()()( tMtMtM ,

12

11

1

x

x

M , since 1M is a solution, it has to

satisfy the dynamic equation, namely 11 )( MtAM .

0,2

0x, x

12121212

11111111

2

xextxx

exx

t

t

let

01

te

M ;

2

0

2te

M

Why not

2

1t

t

e

e

M ,

0

0

2M ? Linear combination of M1 and M2 is

0

0

0

0

212

t

t

e

e

, since 02 these vectors are linearly dependent

NOT ACCEPTABLE FOR M(t) !

Now

0

00

02

21t

t

e

e

since 021 linearly independent ⇒M(t) √

2

0

0

)(t

t

e

e

tM is the fundamental matrix.

of 115

9.2 Fundamental Matrix 49

Fundamental matrix is not unique! Why?

Choose

2

40

02

)(t

t

e

e

tM , now it is still an acceptable fundamental

matrix as it satisfies the differential equations.

Example 6: Given x

tt

tx

22

01

2

, find M(t).

Let )()()( 21 tMtMtM ,

12

11

1

x

x

M

with 122

12121211 xt

xt

xtx

with 2

12121211

20 txx

txx

11

t

M ,

2

2

0

t

M

21

0

)(

t

t

tM

Check to see if M(t) satisfies the dynamic equation: MtAtM )()(

tt

t

tt

t

t 20

01

1

0

22

01

20

01

2

2

?

√

of 115


State Transition Matrix 9.3

9.3.1 Definition

Given xtAx )( , an n×n matrix ),( 0tt is a state transition matrix if and only if

)(),()( 00 txtttx is a solution of the state equation with )( 0tx given.

9.3.2 Proposition

Let M(t) be the fundamental matrix of xtAx )( , then

ttMtMtt

uniquenotunique

)()(),( 0

1

0

Example 7:

2

0

0

)(

20

01

t

t

e

e

tMx

t

x , find ),( 0tt .

We need

20

0

0

0

)( 0

t

t

e

e

tM , then

20

0

0

0

)( 0

1

t

t

e

e

tM

20

2

0

20

0

2

0

0

0

0

0

0

),( 0

tt

tt

t

t

t

t

e

e

e

e

e

e

tt

Moreover, )(),()( 00 txtttx

20

10

20

2

0

0

0

)(

x

x

e

e

txtt

tt

202

101

20

2

0

)(

)(

xetx

xetx

tt

tt

9.3.3 Properties of ),( 0tt

1) ),()(),(

00 tttA

t

tt

2) Itt ),( 00

of 115

9.3 State Transition Matrix 51

3) )(),(

00

0

tAt

tt

tt

4) Transition property: ),(),(),( 011202 tttttt

5) Inversion property: ),(),( 1001

1 tttt

1)

)(tA

t

Since )(),()( 00 txtttx and x(t) is a solution, therefore it should satisfy the differential

equation (d.e.), i.e.

)(),()()(),( 0000 txtttAtxttx

),()(),( 00 tttAtt √

2) Itt ),( 00

Since )()(),( 0

1

0 tMtMtt

ItMtMttttat )()(),(, 0

1

0000 √

3) Follows from (1) and (2), since ),()(),( 00 tttAtt

)(),(),()(),(, 000000000

tAtttttAttttattt

I

√

4) ),(),(),( 011202 tttttt

000 )(),()( tttxtttx (*) )(),()( 0011 txtttx for 01 tt

111 )(),()( tttxtttx (**) )(),()( 1122 txtttx for 12 tt

)(),(),()( 001122 txtttttx

From )(),()((*) 0022 txtttx

)(),()(),(),( 00200112 txtttxtttt √

)

Figure 9-1. Transition property.

of 115


5) ),(),( 1001

1 tttt

Since ),(),(),(),(),( 0110000101

1 ttttttItttt

),(),(),(),( 01100101

1 tttttttt

),(),( 1001

1 tttt √

Example 8:

x

t

x

20

01

20

2

0

0

0

),( 0

tt

tt

e

e

tt , check the above properties.

),()(),( 00 tttAtt

20

2

0

20

2

0

20

0

20

01

20

0

tt

tt

tt

tt

te

e

tte

e

√

Itt ),( 00

10

01

0

0

),(20

20

00

00

tt

tt

e

e

tt √

)(),( 000

tAtttt

)(

20

01

20

0

20

0

0

00

0

0

0

20

2

0

tA

tet

e

te

e

tt

tt

tt

√

53

Objectives

1) State Transition Matrix for LTI Systems

2) Cayley-Hamilton technique

10 Lecture 10

State Transition Matrix for LTI Systems 10.1

10.1.1 Definition

Assuming system Axx , )( 0tx is given, for a scalar system axx , Rx , )( 0tx

)()( 0

)( 0 txetxtta

, generalization to an n-th order system Axx , )( 0tx , nRx ,

suggests that the solution can be expressed as )()( 0

)( 0 txetxttA

. Check to see if x

satisfies the differential equation )()( 0

)( 0 txAtxeAxttA

.

Since )(),()( 00 txtttx

)(

00),(

ttAett

is a state transition matrix for a LTI system

)(),( 00 tttt

Now at 00 t , ( ,0) ( ) Att t e )0( 0 t , from property (5), ?)(1 t

1 1

0 0 0 0( , ) ( , ), ( ) ( )t t t t t t t t for 1

0 0 ( ) ( )t t t

)()(1 tt

of 115


10.1.2 Approaches to Compute )(t

1) 11 )()( AsILet At

2) Cayley-Hamilton technique

3) The Jordan Form technique

4)

0 !k

kkAt

k

tAe

For approach 1), given Axx with )( 0tx , take L-transform

)()()( 0 sAXtxssX )()()( 0txsXAsI

1

0

1

1

)()()(

nnnn

txAsIsX

)()()( 0

11 txAsILtx

Since )(),()( 00 txtttx 11

0 )(),( AsILtt

11 )( AsILeAt

Computation of 1)( AsI , using Leverrier’s algorithm:

01

1

1

01

2

2

1

11

)det(

)()(

asasas

PsPsPsP

AsI

AsIadjAsI

n

n

n

n

n

n

n

scalarsaresamatrixnnaresPwhere ii ','

Algorithm:

1) nnn IP 1

2) )(1 Atran

3) IaAPP nnn 112

of 115

10.1 State Transition Matrix for LTI Systems 55

4)

)(1

)(2

1

11

22

APtrkn

a

IaAPP

APtra

kk

kkk

nn

Final verification test is 00 IaAP o .

Example 1:

1

2

2 2

2

3 0 1

1 3 0

1 15

4

P P A a IP I

a

1 1 0 1 1

2 1 2

1 1 1

3

1( ) 7

22 4

a tr P A P P A a I

2

1 2 1 00 0 3 2

2 1 0

11

3

P s Ps Pa tr P A sI A

s a s a s a

2

2

3 2

2

3 2 1 21

1 3 1 15 7 1

3 2 4 4

s s s

s s ss s s

s s s s

Example 2: System xx

44

10 , xy 51 , i.c. y(0) = 0 , 1)0( y , find y(t).

1 2

1 2

( ) ( ) 5 ( )

( ) ( ) 5 ( )

y t x t x tt

y t x t x t

1 2

2 1 2

2 1 2 1 2

4 4

5( 4 4 ) 20 19

x x

x x x t

y x x x x x

11 2

1 22

5(0)

(0) (0) 5 (0) 0 81

(0) 20 (0) 19 (0) 1 1(0)

81

xy x x

y x xx

,

of 115


11 )( AsILeAt

44

1

44

10

0

0

)(

s

s

s

s

AsI

22

221

)2()2(

4

)2(

1

)2(

4

)(

s

s

s

ss

s

AsI

ttt

ttt

At

teete

teete

e222

222

24

2

)0()( xetx At

81/1

81/5

24

2

)(

)(

222

222

2

1

ttt

ttt

teete

teete

tx

tx

2

1 2( ) ( ) 5 ( ) ( ) ty t x t x t y t te

Cayley-Hamilton Technique 10.2

10.2.1 Cayley-Hamilton Theorem

For a given n×n matrix A with characteristic polynomial 0)det()( AIA , the

matrix A satisfies its characteristic polynomial (C.P.), i.e. 0)( AA .

In other words, if 0)( 1

1

1

nn

nn

A then

01

1

1

IAAA nn

nn .

Example 3:

21

10

A , 012

21

1

det)det()( 2

AIA

We want to show that (from C.H. Theorem): 0122 AA

00

00

10

01

42

20

21

10?

2

00

00

32

21

32

21

√

of 115

10.2 Cayley-Hamilton Technique 57

10.2.2 Cayley-Hamilton Technique

Given an n×n matrix A with characteristic polynomial (C.P.) 0)( A , the matrix

polynomial P(A) can be computed by considering the scalar polynomial )(P .

We compute ( ) ( )

( )( ) ( )A A

P RQ

at i ( i is the eigenvalue of A or the roots of the C.P.)

)()()()(

0

iiiAi RQP

1)()( ii RP i = 1,…, n (distinct si ' )

Now, from C.H. Theorem )()()()(

0

ARAQAAP A

)()( ARAP

10.2.3 Algorithm to Apply Cayley-Hamilton Technique

1) Find the eigenvalues of A.

2) If the eigenvalues are distinct, use equation (1) for the coefficients of )(R by

solving n simultaneous equations. If the eigenvalues are repeated, let’s say for an

eigenvalue with multiplicity m, then instead of (1) use:

3) Use the constants found in step (2) to get )()( ARAP .

[Examples of tAAt eAAeAP

2

;;; )( 2101 ]

Polynomial of degree n-1

with unknown coefficients

)()()()( RQP A

ii

ii

ii

m

m

m

m

ii

d

Rd

d

Pd

d

Rd

d

Pd

d

dR

d

dP

RP

1

1

1

1

2

2

2

2

)()(

of 115


Example 4: Find 101A for

43

10

A

1) 101)( AAP ; 101)( P

2) 2 ( )n R

3) ( ) det( ) 0 1, 3A I A

4) Apply ( ) ( ) 1,2, , . .i iP R for i i e

101

101

1 ( 1)

3 ( 3) 3

2

)3()1( 101101 ;

2

)3()1(3 101101

5) Apply )()( ARAP

IAARAAP )()( 101

10

01

2

)3()1(3

43

10

2

)3()1( 101101101101101A

2

)3(3)1(1

2

)3(3)1(3

2

)3()1(

2

)3()1(3

101101101101

101101101101

101A √

Example 5: Find Ate for

21

10

A

2, ( ) , ( ) , ( )At tn P A e P e R

1,10)1()( 2 A (Repeated)

of 115

10.2 Cayley-Hamilton Technique 59

( ) ( )

@ 1 1

t

t t

P R e

applydP dR

te at ted d

,t t tte e te

Apply P(A) = R(A) , hence

IeteteIAe tttAt

)(

21

10

ttt

ttt

At

teete

tetee

e √

Example 6:

22

10

A , find Ate .

21

( ) det( ) det 2 2 0 12 2

A I A j

jA

1

22

10

(distinct)

( ) ( ) ( )At tP A e P e R ，，

Apply ( ) ( ) 1P R at j

)1(

)1(

)1(

)1(

je

je

tj

tj

of 115


4/

4/

2)sin(cos

2)sin(cos

jt

jt

etjte

etjte

t

j

t

jt

t

j

t

jt

ee

ee

ee

ee

4/

4/

2

2

)2(2sin22 4/4/ je

tjeee

eet

jj

t

jtjt

tet

sin2

10

01

22

10sin

2t

ee

tAt

Figure 10-1. System plot.

.8

2

.6 .4 .2

5

-2

4

0

2

1

6

8

-4

-6

-8

1.2 1.4 1.6 1.8 2

xjxe jx sincos

4/21 jej

Use

√ to find β,

it will be a real number

61

11 Lecture 11

Ф(t, t0) for LTI System 11.1

From the scalar case: 22

!21 t

te t, we get

22

0

*2! !

k kAt

k

t t Ae I tA A

k

（）

Some of the properties of Ate are :

1) Ie 0

2) 2121 )( AtAtttA

eee

21,tt scalars

3) BtAttBA eee )(

if and only if (iff.) AB = BA (A & B commute)

4) AeAeedt

d AtAtAt

Proof of (1): from (*) with either t= 0 or A= 0!

Objectives

1) Using Cayley Hamilton technique to calculate 𝜱 𝒕 𝒕𝟎 for LTI Systems

2) Using Cayley Hamilton technique to calculate 𝜱 𝒕 𝒕𝟎 for LTV Systems

3) Solution of the state equations and output equation

of 115


Proof of (2): 2

21

2

21

)()(

!2)(21 tt

AttAIe

ttA

2 22 2 2

1 2 1 2 1 2 *2! 2!

A AI At At t t A t t （）

!2!2

2

2

2

2

2

1

2

121

tAAtI

tAAtIee

AtAt

2 2 2 222 1

2 1 1 2 **2! 2!

A t A tI At At A t t （）

Since (*) = (**) 1 2 1 2( )A t t At At

e e e

Proof of (3):

!2)()(

22)( t

BAtBAIe tBA

!2!2

2222 tBBtI

tAAtIee BtAt

2 2 2 3 2 2 2 32

2 22

2! 2! 2! 2!

( )2! 2!

B t AB t A t A BtI Bt At ABt

A BI A B t t AB

ABBABA

222

)( 22?2

)2(2

1)(

2

1 22?

22 ABBABAABBA

If and only if AB = BA AB+BA=2AB that is, A & B should commute!

of 115

11.1 Ф(t, t0) for LTI System 63

Proof of (4):

1 1

11

0 )!1(!! k k

kkkk

k

kkAt

k

At

k

Akt

k

At

dt

de

dt

d

Ak

At

k

AAt

k

AAt

k

At

k

kk

k

kk

k

kk

k

kk

0000

1

!!!!

AtAt

k

kk

AeAek

AtA

0 ! A &

Ate commute!

Jordan Canonical Representation Ate

1) A has distinct eigenvalues

2) A has repeated eigenvalues

Case 1: (Distinct λ) For LTI System

, find

{

[

]

[

]

{

[

]

of 115


Case 2: ( Repeated λ)

nn

J

1

1

1

1

1

0000

1000

0000

0010

0001

, find Jte

Using Cayley-Hamilton technique:

1) n

J JI )()det()( 1

2) JteJP )( ,

teP )(

1

11

2

12110 )()()()(

n

nR

3) )()( 11 RP 01

te √

11

d

dR

d

dP 1

1

tte √

11

2

2

2

2

d

Rd

d

Pd 2

2

1

2

t

et

√

11

1

1

1

1

n

n

n

n

d

Rd

d

Pd 1

1

1

)!1(

n

tn

en

t

√

of 115

11.2 Ф(t, t0) for LTV Systems 65

4) 1

11110 )()(

n

n

Jt IJIJIe

t

tt

tn

tn

t

tn

tn

tt

tn

tn

ttt

e

tee

en

te

n

te

en

te

n

ttee

en

te

n

te

ttee

1

11

111

1111

11111

0000

000

)!3()!4(00

)!2()!3(0

)!1()!2(!2

34

23

122

Ф(t, t0) for LTV Systems 11.2

1) )()(),( 1

0 tMtMtt

2) Direct solution of xtAx )( to get )(),()( 00 txtttx . Brute Force (Hyperlink)

reference to lecture 8 Example 1.

3) If A(t) and )( 0tA commute, then

t

t

dA

ett 0

)(

0 ),(

4) If ∑ are constant matrices where

are scalars and

, then

k

i

dM

t

t

ii

ett1

)(

00),(

of 115


Example 3: [

]

⏟

[

]

⏟

⏟

[

]

⏟

Since

[

]

[

]

[

] [

]

(This can be shown by applying Cayley-Hamilton technique -- details not shown)

Therefore, [

]

Example 4:

tt

ttA

22

01

)(

2

Check to see that A(t) and )( 0tA do not commute, therefore #3 does not apply.

Use method #2 11

1x

tx ; 2122

22x

tx

tx

)()()( 01

0

01

1

10 tx

t

ttxetx

t

t

d

)()(

01

0

2

)(

2 )(2

2)(

tutbta

txtt

xt

tx

of 115

11.2 Ф(t, t0) for LTV Systems 67

t

t

dd

dtxt

etxetx

tt

t

0

0 )(2

)()( 01

0

2

02

2

2

dtxt

etxe

t

t

ttt

)(

2)( 01

0

lnln

02

lnln

0

2220

2

t

t

dt

txttx

t

t

00

01

2

2

022

0

2 )(2)(

t

t

dtxt

ttx

t

t

0

301

0

2

022

0

2 2)()(

2

0

201

0

2

022

0

2

2

11)()()(

tttx

t

ttx

t

ttx

)(

)(

1

0

)(

)(

02

01

),(

2

0

2

3

0

2

0

0

2

1

0

tx

tx

t

t

t

t

t

t

t

tx

tx

tt

Example 5:

0

020

01)(

ttA )()()()( 00 tAtAtAtA , therefore, use method #3

[

]

∫[

]

[

]

[

]

[

] √

This follows from Cayley-Hamilton technique.

1 0( )

0 2A t

t

of 115


Example 6:

t

ttA

1

1)(

0

0

0

1

1)(

t

ttA

( )0 0 0

0 0

0 0 0

1 11( ) ( ) ( ) ( )

1 11

showst tt t ttA t A t A t A t

t t t ttt

√

Method #3 applies

dA

t

tett

0

)(

0 ),(

d

t

t01

1

B

tttt

tttt

2

220

2

0

0

20

2

, ?2

20

2

0

02

20

2

tttt

tttt

e

?Be

10)(,)(,)( RePeBP B

Find eigenvalues of B

)(2

0

2

2det)det( 0

2

0

2

2,12

0

2

0

0

2

0

2

tttt

tttt

tttt

BI

)sinh(

12

)sinh(

)()(

)()(

0

021

0

02

0

22

11

20

2

0

20

2

tt

tte

ttttee

RP

RP

tt

tt

tt

BIeB

10

2

00

00

0

20

2

)cosh()sinh(

)sinh()cosh(),(

tt

B etttt

ttttett

√

of 115

11.3 Solution of the State Equations and Output Equation 69

Solution of the State Equations and Output Equation 11.3

For system

DuCxy

BuAxx,

the complete solution is given by

respond state zero

t

trespondinput zero

00

0

dττ)B(τ)u(τ)Φ(t,+) tΦ(t,= )x(tx(t)

Proof:

∫

∫

Left=

{ ⏟

Since

{

∫

Leibnitz’s Rule

t

utuf

t

vtvfdx

t

txfdxtxf

t

tv

tu

tv

tu

),(),(

),(),(

)(

)(

)(

)(

∫

Given , constantx Ax Bu A and B are

t

t

tAttAdBuetxex(t)

0

0 )()( )(

0

)(

?

70

Objectives

1) Controllability; Observability

2) Canonical Transformations: C.C.F., O.C.F

12 Lecture 12

Definitions 12.1

12.1.1 Controllability

Given

DuCxy

BuAxx, RuyRx n ,, (SISO), the system is controllable if and only if

the n×n matrix (known as the controllability matrix)

is nonsingular; i.e. nCRank x .

Figure 12-1. State trajectory as a result of the control u.

12.1.2 Observability

Given the above system, we say it is observable if and only if the n×n matrix Ox (known

as the observability matrix) is nonsingular, in another word, nORank x , where

nn

n

x

CA

CA

C

O

1

Figure 12-2. System diagram.

System

Observer

y u

x

Controller

Control loop

of 115

12.1 Definitions 71

12.1.3 Examples for Controllability and Observability

Example 1:

01

10A , 1 2

1 1

1 1B B B

, is this system controllable ?

[

]

2)( xCRank full rank controllable !

Q: check to see if the system is controllable from individual inputs.

[

] 21)( 1 xCRank not controllable from u1

[

] 21)( 2 xCRank not controllable from u2

Example 2:

40

21A ,

2

1

10

01

C

CC , is this system observable?

CA

COx

40

21

10

01

xO 2)( xORank Observable

Q: is the system observable from each output alone?

21

01

1

11

AC

COx 2)( 1 xORank Observable from y1

40

10

2

22

AC

COx 21)( 2 xORank Unobservable from y2

of 115


Canonical Transformations 12.2

12.2.1 A Canonical Decomposition of Dynamic System

Figure 12-3. System classification.

If Ox singular system is not observable

If Cx singular system is not controllable

Example 3: uxx

0

1

20

01 , xy 11 (n = 2)

For Controllability

21)(

00

11

xx CRankABBC Uncontrollable

{

⇒

For Observability

2)(

21

11

xx ORank

CA

C

O Observable

Let us set C as 01C , then

21)(

01

01

xx ORankO Unobservable

2

1

22

11

xy

xx

uxx

C/O

C/NO NC/O

NC/NO

u y

Uncontrollable mode is also unobservable

(in addition to being unstable)

of 115

12.2 Canonical Transformations 73

12.2.2 Controllable Canonical Form (C.C.F.)

12.2.2.1 Theorem

DuCxy

BuAxx

RuyRx n ,, , assume that the system is controllable i.e.

is nonsingular. We want to find a nonsingular transformation

xTx cc

1 , so that in the new coordinate system we have

uDxCy

uBxAx

ccc

cccc

where

[

]

[ ]

cC does not have any particular form and DDc .

For xTx cc

1 , BuTAxTBuAxTxTx ccccc

1111 )( .

Since ccxTx

uBTxATTx

ccB

cc

A

ccc

11

, uDxCTy

cc

D

c

C

c

DD

CTC

TB

ATTA

c

cc

cc

ccc

B1

1

The controllability matrix for the new system

)()())(( 1111111 BTATTBTTATBTBABABC c

n

cc

I

ccccc

n

ccccxc

xc

C

n

c

n

ccc CTBAABBTBATABTBT

x

1111111 )

)1)(1()1)(1()1(1)1(

1

B

I

cT

cT

A

A

I

cT

I

cTA

I

cT

cTA

cTB

cTn

cAT

cT

n

of 115


By assumption, original system is controllable, nCRank x )(

existCandCnCTRankCRankcc xxccx

111 )(

111 xxcxxc CCTorCCTcc

12.2.2.2 Procedure for Controllable Canonical Form

0) Check to see if the system is controllable.

If yes, proceed to the following steps; if not, stop as there is no CCF possible.

1) Determine the coefficients of the C.P. of matrix A.

2) Form cA and cB .

3) Calculate xC andcxC , using A, B and cA , cB .

4) Find the transformation1cxxc CCT .

5) Use cc CTC to get cC .

6) As a check, verify that ccc ATTA 1 and BTB cc

1 .

7) DDc .

12.2.3 Observable Canonical Form (O.C.F.)

12.2.3.1 Theorem

DuCxy

BuAxx

Given this system, find a transformation that takes the system into an

observable canonical form. The transformed system takes the form

uDxCy

uBxAx

ooo

oooo

Where

[

]

oB does not have any particular structure, DDo

of 115


Define xTx oo

1 , ooxTx BuTxATTBuAxTxTx ooooooo

1111 )(

DuxCTDuCxy oo

DuxCTy

BuTxATTx

oo

ooooo

11

DD

CTC

BTB

ATTA

o

oo

oo

ooo

1

1

Form xO and oxO as

1n

x

CA

CA

C

O

and

1n

oo

oo

o

x

AC

AC

C

Oo

, it follows that oxx TOO

o .

)()( oxx TORankORanko . Since oT is full rank, xO is full rank nORank

ox )(

oxO is nonsingular! oo xxoxxo OOTorOOT 111

12.2.3.2 Procedure for Observable Canonical Form

0) Check to see if the system is observable.

If yes, proceed to the following steps; if not, stop as there is no possible OCF.

1) Find the coefficients of the C.P. of A.

2) Form oA and oC .

3) Calculate xO and 0xO using A, C and oA , oC .

4) Find the transformation xxo OOTo

11 .

5) Use BTo

1 to get BTB oo

1 .

6) As a check, verify that ooo ATTA 1 and oo CTC .

7) DDo .

of 115


Example 4: System given as uxx

1

1

2

020

113

021

, xy 100

transform it into C.C.F and O.C.F.

Controllable form

0) Check that xC is nonsingular? √

1) C.P. = 0290)det( 3 AI

2)

092

100

010

cA ;

1

0

0

cB then ?cC

1221

861

1642

2BAABBCx ;

901

010

100

2

cccccx BABABCc

123

161

242

1

cxxc CCT

123

123

161

242

100

cc CTC

123

1

0

0

092

100

010

c

cc

xy

uxx

of 115


Observable Canonical Form

0)

126

020

100

2CA

CA

C

Ox nonsingular Observable

T

co AA

010

901

200

; T

co BC 100

901

010

100

2

oo

oo

o

x

AC

AC

C

Oo

T

cooxxo CBTBOOTo

1

2

3

11

78

Objectives

1) Diagonalization of a matrix – Jordan canonical form

2) Controllability of J.C.F; Observability of J.C.F.

13 Lecture 13

Diagonalization of a Matrix 13.1

13.1.1 Jordan Canonical Form (J.C.F.) – Distinct Eigenvalues Case

Given an nxn matrix A with distinct eigenvalues, the diagonal form of A denoted by A

takes the form

n

A

00

00

00

ˆ2

1

where AMMA 1ˆ , with 1 2 nM x x x

( known as modal transformation).

Given

DuCxy

BuAxx, transformation into Jordan Canonical Form is achieved through

JJ MxxxMx ,1, BuMAMxMBuMAxMxMx JJ

11111

11

DuCMxy

BuMAMxMx

J

JJ

, if A has n distinct eigenvalues, then choose M as the modal

matrix, then

00

00

00

21

2

1

2

1

Duxcccy

u

b

b

b

xx

Jn

n

J

n

J

Eigenvectors corresponding to eigenvalues

of 115

13.1 Diagonalization of a Matrix 79

The above system in Jordan Canonical Form is

Controllable if and only if niibi ,...,2,1,0 and

Observable if and only if niici ,...,2,1,0

Example 1: uxx

0

1

43

10

, √ √

Transform it into J.C.F. and also check its controllability and observability

3

10)det(

43

10

2

1

AIA

Given a matrix A and eigenvalue λ, the eigenvalue and eigenvector v

relations are given by: reigenvectoisvwherevAv ,

1

1

1

1

033

0

1

43

10

1

2

1

21

21

2

1

2

1

1

1

1

vx

x

xx

xx

x

x

x

x

vvA

Similarly,

Without loss of generality, normalize to a unit length, named

√

√

3

1

3

1

03

03

3

43

10

2

2

1

21

21

2

1

2

1

2

2

2

vx

x

xx

xx

x

x

x

x

vvA

Normalizing to unit length

√

√

1 1 2 2

1 1 1 31 3

2 2 10 10

TT

nv nv

，

of 115


Modal matrix:

2

10

2

10

2

2

2

23

2

1

2

1

10

1

10

3

2

20

10

3

2

1

10

1

2

1

1MM

30

011 AMM

uxx JJ

2

10

2

23

30

01

Jxy

10

220

43

10A ,

0

1B , √ √

30

01

ABBCx rank = 2 Controllable

2323

22

CA

COx rank = 1 Unobservable

≠

𝑖 𝑜𝑛 𝑟𝑜𝑙𝑙 𝑙

≠

𝑖 𝑜𝑛 𝑟𝑜𝑙𝑙 𝑙

𝑖 𝑛𝑜 𝑟𝑣 𝑙

≠ 𝑖 𝑜 𝑟𝑣 𝑙

of 115


13.1.2 Jordan Canonical Form (J.C.F.) – Repeated Eigenvalues Case

13.1.2.1 Diagonalization of Matrices

An n×n matrix A may be diagonalized by a similarity transformation where AMMA 1ˆ

if and only if nullity of )( IA i (which is )( IArankn i ) is equal to the multiplicity

of the eigenvalues, and moreover the modal matrix M is the matrix of eigenvectors.

Therefore, if multiple eigenvalues, then same number of linearly independent

eigenvectors. If this is not the case, then a diagonal form is not possible, and Jordan form is

used as the standard form.

Example 2:

21

10A 1,10)1()det( 2 IA

111

11)(

rankIArank ,

therefore, nullity of one linearly independent eigenvector.

For 1 A cannot be diagnolized.

We will show later in example3 that the proper form is

10

11A

13.1.2.2 Generalized Eigenvectors

A nonzero vector x for which 0)( 1 xIA k but 0)( xIA k is called a generalized

eigenvector of rank k of A associated with the eigenvalue .

Generalized eigenvectors are defined as

xxk

kk xIAxIAx )()(1

1

2

2 )()( kk xIAxIAx

2

1

1 )()( xIAxIAx k

generalized eigenvectors kxxx ,,, 21

of 115


Theorem:

The set of generalized eigenvectors kxxx ,...,, 21 for the same eigenvalue are

linearly independent.

The generalized eigenvectors of A associated with different eigenvalues are

linearly independent.

Example 3:

21

10A , 1,1

Here k = 2, 2xx is computed from 0)( 2 xIA and 0)( 2

2 xIA

0

00

000

11

11

22

2

xx Any vector that satisfies 0)( 2 xIA

can be used. Choose Tx 012

011)( 2 T

xIA

Calculate

1

1

0

1

11

11)( 2

12

1 xIAx

21

01

11xxM

10

11ˆˆ 1 AAMMA

Example 4:

201

010

001

A

1,1,20)2()1()det( 2 IA

201

010

001

IA

of 115


Q : Nullity ? IA i when 1i

101

000

000

IA , 1)( IArank ,

1 of 213)( tymultipliciIArankn

2 linearly independent eigenvectors for 1

0

1

0

,

1

0

1

0

101

000

000

)( 2111 xxxxIA

For

1

0

0

0

001

010

001

2 33 xx

200

010

001

ˆ

101

010

001

1

321 AMMAxxxM

13.1.3 Matrix Reduction to Jordan Form

13.1.3.1 Definition

The Jordan form J of an n×n matrix A is an upper triangular matrix whose diagonal entries

are of the matrix form

j

j

j

j

j

jijL

1

1

1

1

)(

0

0

of 115


The general case of the Jordan Matrix J is

[

]

)( jijL matrices are called Jordon blocks.

13.1.3.2 General Rules for Constructing the Jordan Matrices

1) There is one and only one linearly independent eigenvector associated with each

Jordan block. In general, there are )( IArankn i linearly independent

eigenvectors for a general n×n matrix A with its associated eigenvalue i .

2) The number of 1’s directly above the diagonal terms equal the order of A minus the

number of linearly independent eigenvectors r, in another words, rnns

'1

3) The number of Jordan blocks associated with each eigenvalue is equal to the

number of linearly independent eigenvectors associated with each eigenvalue r.

Example 5:

[

]

[

]

For JCF.in one 1 134)# r(with 1 onlyrnblocksofJ

For JCF.in sone' 2 224)#(2 onlyrnblocksofrwithJ

1 generalized eigenvector; 1 linearly independent eigenvector for

1 l.i.e.v. for

1 l.i.e.v. for

2 g.e.v.;

1 l. i.e.v. for

1 l.i.e.v. for

of 115


Example 6: Transform A into Jordan form

5972

1000

0100

0010

A

2,1,1,102795)det( 234 IA

)(....# IAranknveilof i

3

0000

0100

0010

0001

rank

4-9-7-2-

1100

0110

0011

rank)( 1

IArank

-1 with associated l.i.e.v. 13-4only .

Thus

[ ]

( # of 1’s in J are n - 2 = 4-2 = 2,

where A has 2 l.i.e.v.) Check: Let 4321 xxxxM

Define 3x by

0)(

0)(

3

2

1

3

3

1

xIA

xIA

and subsequently

312 )( xIAx , 211 )( xIAx and 0)( 42 xIA .

For 3x , we have: 0)( 3

3 xIA 0

824248

412124

2662

1331

3

x

only 1 linearly independent vector

Txdcba 1001033 3

TxIAx 2101)( 32 ,

TxIAx 1111)( 21

4x is calculated from TxxIA 84210)2( 44

AMMAxxxxM 1

4321ˆ √

of 115


13.1.3.3 A Procedure for Generating a JCF

1) Compute the eigenvalues of A by solving 0)det( IA . Let n ,, 21 be these

distinct eigenvalues with multiplicity nnnn ,, 21 , respectively.

2) Compute the number of linearly independent eigenvectors for1 . This is also the

number of Jordan blocks for1 .

3) For eigenvalue 1 determine its Jordan blocks from the information in (1) and (2)

and the general rules.

4) Determine the linearly independent eigenvector for 1 associated with each Jordan

block and the remaining generalized eigenvectors for each block. Denote the

linearly independent one with the subscript of the order of the Jordan block. Thus if

the Jordan block were of order 3, then denote the linearly independent vector as 3x

and the generalized eigenvectors as 2x and

1x ..

5) Repeat steps (2), (3), and (4) for eigenvectors n ,, 32 to generate a Modal

matrix by placing the eigenvectors as columns. The order of the column placement

is important and is directly related to how they are numbered in step (4).

As an example, suppose that A is 55 with eigenvalues 32111 ,,,, . Also, suppose

there are two linearly independent eigenvectors associated with1 , that is

2)-(A ofnullity 3)( 11 IIArank . Therefore, there are two Jordan blocks

associated with 1 as 1111 )( J &

1

1

1210

1)(

J

The Jordan form is

[

]

[

]

of 115

13.2 Controllability of JCF and Observability of JCF 87

Assume we have found for block )( 111 J an eigenvector1x from

11 xAx .For block

)( 121 J , we need two generalized eigenvectors 3x from 0)( 3

2

1 xIA and

231 )( xxIA . Eigenvalues 2 and 3 will contribute the two eigenvectors

4x and 5x ,

respectively from 424 xAx and 535 xAx .

The Modal matrix is formed by 54321 xxxxxM

Controllability of JCF and Observability of JCF 13.2

13.2.1 Theorem

DuCxy

BuAxx qpn RyRuRx ,,

m

nn

A

A

A

A

2

1

,

m

pn

B

B

B

B2

1

, mnq CCCC 21

A has m distinct eigenvalues m ,, 21 ; iA denotes the Jordan blocks associated with i .

r(i) is the number of Jordan blocks in iA ; ijA is the j-th Jordan block in iA .

1 2 1 2 ( ), , , ; , , ,m i i i ir iA diagn A A A A diagn A A A

)(

2

1

)(

iir

i

i

nni

A

A

A

Aii

;

)(

2

1

)(

iir

i

i

pni

B

B

B

Bi

( ) 1 2 ( )ii q n i i ir iC C C C

m

i

ir

j

ij

m

i

i nnn1

)(

11

i

i

i

i

nnij ijijA

1

1

1

)( ,

lij

ij

ij

pnij

b

b

b

Bij

2

1

)(,

T

lij

ij

ij

pnij

c

c

c

Cij

2

1

)(

of 115


The n-dimensional LTI Jordan form is controllable if and only if for each i =1, 2,…,m,

the rows of the pir )( matrix

)(

2

1

ilir

li

li

l

i

b

b

b

B

are linearly independent.

and the system is observable if and only if for each i = 1, 2,…,m the columns of the

)(irq matrix are linearly independent.

13.2.2 Examples

Example 7:

[

]

[ ]

lecontrollabBrankfullB ll mode

100

010

001

111

lecontrollabBl modetindependenlinealry 100 22

observable modeCrank full

321

211

021

1

1

1

1

1

C

𝑙

𝑙

𝑙

𝑙

of 115


leunobservab mode0 rank

0

0

0

2

1

2

C

Block diagram of this system:

Figure 13-1. Block diagram.

If 021lb all modes in the chain can be controlled;

If 0121c all modes in the chain can be observed.

Example 8:

[ ]

[ ]

What is the JCF for a system with complex conjugate eigenvalues?

𝑙

+

u

u

y

u y

u y

u

u u

y y

y

𝑙

𝑙

𝑙

+

+

𝑙

uncontrollable.

≠ controllable.

≠ observable. ≠ observable.

of 115


Let u

b

b

x

x

A

A

x

x

2

1

2

1

1

1

2

1

0

0

,

2

1

11

x

x

ccy

where 1A is the Jordan block associated with and 1A is the complex

conjugate of 1A , in another word, 1A is the Jordan block associated with 1 .

Introduce the equivalence transformation Pxx

where

jIjI

II

P and

jII

jII

P2

11

It can be verified that

1 11 1 1

2 21 1 1

1

1 12

Re Im 2Re

Im Re 2Im

Re Im

A A bx xu

A A bx x

xy c c

x

This system can now be used in computer simulations.

Example 9: jjA 21 , 21 , 2

[

]

[

]

Let Pxx , where

[ ]

of 115


[

]

[ ]

All the coefficients are real!

Example 10: jjA 32 , 1

[

]

[

]

A

��

92

Objectives

State Space Control Method

1) State Feedback: General Case

2) State Estimation: Full Order Observer

3) State Estimation: Reduced Order Observer

14 Lecture 14

Stable Feedback: General Case 14.1

State: RuRxBuAxx n ,,

Control: Kxu with nnKKKK 121

Figure 14-1. Closed-loop system state diagram.

matrix loop-closed

A )( f

BKAA

xxBKAx

f

,

the characteristic polynomial for the closed-loop system is

0)det()det( BKAsIAsI f

Let the Design Specification require closed-loop eigenvalues at n ,,, 21 .

0)())(()( 01

1

121

sssssss n

n

n

nc

Pole-placement design procedure is achieved by setting

01

1

1)det(

sssBKAsI n

n

n

y + B

r ʃ

D

+

A

+

Ax

+

+

K

-

C u

of 115

14.1 Stable Feedback: General Case 93

Solve for n unknown nKK ,...,1 . The equations are linear!

K can be computed directly from this formula if and only if (A,B) is controllable.

Consider the transfer function 01

1

1

01

1

1)(asasas

bsbsbsG

n

n

n

n

np

The controllable canonical form is

u

aaaa

x

n

1

0

0

0

1000

0100

0010

1210

xbbby n 110

The control law is Kxu

Closed - loop fA matrix is

nn

f

KaKaKaKa

BKAA

1322110

1000

0100

0010

Characteristic polynomial 0)det( BKAsI

0)()()( 1021

1

1

KasKasKas n

nn

n

Since desired characteristic polynomial is

0)( 01

1

1

ssss n

n

n

c

niaK iii ,...2,111

Alternatively, K can be computed according to the Ackermann’s Formula as follows:

of 115


1

2 1

1

1 1 0

0 0 0 1 ( )

( )

x

n n

c

C

n n

c n

K B AB A B A B A

where A A A A I

Example 1: uxx

1

0

00

10

Let the desired characteristic polynomial be

0)()( 2121

2 sssc

Based on Ackermann’s formula, need to compute xC

01

10

01

101

xx CABBC

21

2121

2121

2

0)()(

IAAAc

212121

1)(10

KKKAABBK c

Example 2: Let the design specifications require a critically damped system with a

settling time of 1 sec., i.e, 14 .

4damped),y (criticall141

nnn

168)4)(4()( 2 sssssc

8,16 212211 KK

If we compute the closed-loop transfer function, we get the figure 14-2.

Also, 168

101)(

2

1

ssBBKAsIsT as desired!

As a different example, let2.707, .25 4 4 ( ) 8 32cs j s s s

1 1 2

2 1 2

3232 8

8

KK

K

of 115

14.2 State Estimation: Full-Order Observer 95

Plant

x(t)

State

Estimator u(t)

y(t)

Figure 14-2. System plot for example 2.

State Estimation: Full-Order Observer 14.2

14.2.1 Theorem

RyCxy

RuRxBuAxx n ,

The estimator equation is expressed as

yGuHxFx nnnn 11ˆˆ

We need to choose matrices F, G, and H such that )(ˆ tx is accurate estimate of x(t). Then

in the control system, the estimated states are used to generate the feedback control, i.e.

xKu ˆ . To do this, the transfer function from u to ix must be equal to the transfer

function from u to ix , i =1,2,…,n, that is

nisU

sX

sU

sX

i

i

i

i ,...2,1)(

)(

)(

)(ˆ

The Laplace Transform of state equations are

( ) (0) ( ) ( ), 0

( ) ( ),

sX s x AX s BU s x( ) is unknown

Y s CX s A, B, C, and D known

By neglecting i.c. )()()( 1 sBUAsIsX

)(

)(

sU

sX i can be obtained.

The Laplace Transform of estimator is (neglecting i.c.)

)()()(ˆ)()()(ˆ)(ˆ sGCXsHUsXFsGYsHUsXFsXs

Since )()( sCXsY

)()()()()()()(ˆ 111 sUBAsIGCHFsIsGCXsHUFsIsX

t .5 1

1.5

Figure 14-3. Block diagram of combined plant & observer.

of 115


We want )(

)(ˆ

)(

)(

sU

sX

sU

sX ii

BAsIGCHFsIBAsI 111 )()()(

Collecting BAsI 1)( terms

HFsIBAsIGCFsII 111 )()()(

Since IFsIFsI )()( 1 HFsIBAsIGCFsIFsI 111 )()()(

HGCFsIBAsIHBAsIGCFsI 111 )()()(

This equation is satisfied if we choose AGCFandBH

BHandGCAF

G is chosen such that an acceptable transient response or frequency response for the state

estimator is achieved.

xGCGyBuxAx ˆˆˆ

xCyGyBuxGCAx ˆˆ,ˆ)(ˆ

ˆˆ

)ˆ(ˆˆ

xCy

yyGBuxAx

Figure 14-4. Plant and observer diagram.

Let )(ˆ)()( txtxte GyBuxGCABuAxxxe ˆ)(

Since Cxy , hence )ˆ()(ˆ)( xxGCAxGCxGCAAxe

eGCAe )(

y + B

u

ʃ x

A

+

C

observer gain

+ B ʃ

A

+ +

C

+

G -K

Plant

of 115


The error in the estimation of the states is governed by the above dynamic equation with

the characteristic polynomial: 0)det( GCAsI

The gain vector G is chosen to make the dynamics of the estimator faster than the open-

loop system dynamics (~ 2 to 4 times faster).

14.2.2 Design of the State Estimator

1) GyBuxGCAx ˆ)( , the characteristic polynomial is 0)det( GCAsI

2) Let the desired char. poly. be )(se , that reflects the desired transient behavior:

0)( 01

1

1

ssss n

n

n

e

3) The gain matrix G is calculated to satisfy )()det( sGCAsI e

4) Using the transformation xTx oo

1 transforms the system into observable

Canonical form

{

[

]

[

]

Now GCA becomes

[

]

01

1

1)()det(

ssssGCAsI n

n

n

e

niag iii ,...2,111

Now using Ackermann’s formula we have [if and only if (A, C) is observable]

)(

1

0

0

)(

01

1

1

1

1

1

IAAAA

CA

CA

C

AG

n

n

n

e

n

en

of 115


Example 3:

xy

uxx

01

1

0

00

10

A controller was designed for the characteristic polynomial 328)( 2 sssc ,

which yielded a time constant 707.sec,25.

We choose the estimator to be critically damped with a .sec1.

10020)10()( 22 sssse

1000

2010010020)( 2 IAAAe

Now

10

01,

10

011

xx OCA

CO

100

20

1

0)(

1

CA

CAG e

2

1

21

g ; 100g , 20

gGg

Example 4: Combining previous two examples:

100

20,832 GK

The estimator equations GyxBKGCAGyBuxGCAx ˆ)(ˆ)(

Since xKu ˆ , now

8132

120BKGCA

Observer/Controller

The T.F. from output u to input y is GGCBKAsIKsGec

1)()(

29228

32001440)(

2

ss

ssGec

The transfer function of the controller-estimator is

GGCBKAsIKsGec

1)()(

ˆ832

100

20ˆ

8132

120ˆ

xu

yxx

acting as input

acting as output

of 115


Figure 14-5. Block diagram for example 4.

The characteristic polynomial of the closed-loop system is

0)()(1 sGsG pec

Example 5: xyuxx 01,1

0

00

10

The controller-estimator is given by

Signal flow graph of the controller-estimator

Figure 14-6. Signal flow graph.

Defining

[

]

[

]

[

]

[

]

Figure 14-7. Estimator response.

Controller-Estimator Plant

+

-

R = 0 y

ˆ832

100

20ˆ

8132

120ˆ

xu

yxx

R 1 1 1 u 20

100

32

8

-20

+1

-132

-8 -1

t .5 1 1.5

y

≠

of 115


14.2.3 Closed-Loop System Characteristics

The characteristic polynomial with full state feedback is

and the characteristic polynomial of a state estimator is

We now derive the characteristic polynomial of the closed-loop system.

First consider , the plant equations are {

Hence,

The error dynamic equation is , therefore the closed-loop system is

[

] [

] [

]

The characteristic polynomial of the closed-loop system is

[

]

The 2n roots of the closed-loop characteristic polynomial are then the n roots of the pole-

placement design plus the n roots of the estimator. This is fortunate, since otherwise the

roots from the pole-placement would have been shifted by the addition of the estimator.

This is called the separation principle.

Example 6: From the previous examples we have

We have

or equivalently from ( )

[

]

The above characteristic polynomial is obtained.

of 115

14.3 Reduced-Order Estimators 101

Reduced-Order Estimators 14.3

Estimators developed so far are called full-order estimators since all states of the plant are

estimated. However, usually from output measurement some states are directly available.

Hence it is not logical to estimate states that are directly measured.

Assume without loss of generality that

Partition x into [

], where are the states to be estimated.

Partition also[

] [

] [

] [

]

The equation of the states to be estimated

And the equation of the state that is measured where is

unknown and to be estimated and and u are known.

To derive the equation of the estimator observer that for the full state estimation

{

For the reduced order observer we have

unknown

ee

knownknownknown

eeeee

unknown

e

xAubxax

uBxAxAx

1111111

11

)(

)(

Comparing we have

e

ee

ee

ee

AC

ubxaxy

uBxABu

AA

xwithxreplacetxtx

1

11111

1

)()(

Making these substitutions for the full-order observer equations yield

]ˆ)(ˆ[ GyBuxGCAx

)(ˆ)(ˆ11111 ubyayGuByAxAGAx eeeeeeeee

of 115


Thus the error dynamic is ee xxe ˆ eAGAe eeee )( 1

Thus, the characteristic polynomial of the reduced-order estimator is

0)det()( 1 eeeee AGAsIs

We then choose to satisfy this equation where is chosen to give the estimator desired

dynamics. This can be achieved if and only if ),( ieee AA is observable.

Ackermann’s formula yields

1

0

0

)(

1

2

1

1

1

n

eee

eee

e

eeee

AA

AA

A

AG

The above estimator requires measurement of y . To relax this requirement, define a new

variable yGxxoryGxx eeeeee 11ˆˆˆˆ

Substituting in observer dynamic yields

)()ˆ)((ˆ1111111 ubyayGuByAyGxAGAyGx eeeeeeeeeee

ubGByGAGGAaGAxAGAx eeeeeeeeeeeeeeee )()(ˆ)(ˆ11111111

The control u is

)ˆ(ˆ111 yGxKyKxKyKu eeeee

Where ， .

Figure 14-8. Block diagram of the combined plant and reduced order observer.

Plant

x

Reduced Order

Observer

u y

- -

+ +

+

of 115

14.3 Reduced-Order Estimators 103

Example 7: { [

] [

]

the controller designed earlier yielded

As for the estimator of reduced order (1st order), let

From Ackermann’s formula:

The estimated value of is then

Example 8: If we combine plant equations with the estimator and the control we have

The signal flow graph becomes

Figure 14-9. System signal flow graph.

The characteristic polynomial is

⏟

⏟

Now opening the graph from u we calculate

1 -8 u 1

-32 -10 1

-100 10

of 115


Figure 14-10. Closed-loop combined plant and controller estimator.

√

Reduced-Order State Estimator: General Case 14.4

Consider

q

pn

RyCxy

RuRxBuAxx

,

,,

Assumption: C has full rank,

Define: [

] , where arbitrary as long as P is nonsingular.

Define

[

] [

] [

]

Define the equivalence transformation

{

Partition above system as

{

[

] [

] [

] [

]

Therefore, only the last (n - q) elements of need to be estimated.

We need only an (n - q)-dimensional state estimator.

R = 0

Controller - estimator plant

+

-

of 115

14.4 Reduced-Order State Estimator: General Case 105

Using {

Define

Lemma:

The pair (A, C) or equivalently is observable, iff. the pair is observable.

Therefore, a (n - q)-dimensional estimator of in the form

)()(ˆ)(ˆ221111212222 uByAuByAyGxAGAx

such that the eigenvalues of can be arbitrarily assigned by proper choice

of

To eliminate define

yGxz 2ˆ

Define

by proper choice of .

Now 1

2

ˆˆ

ˆ

yxx

Gy zx

, since

1

1 2 1 2

0

ˆˆ [ ] [ ]

q

n q

Iy yx Px x P x Qx x Qx Q Q Q Q

zGy z G I

which is an estimate of the original vector x.

Figure 14-11. Block diagram for system with reduced order estimator.

��

ʃ

�� 𝐴

+ + + u

y

z

106

Objectives

1) Tracking Problem

2) Closed-Loop Pole-Zero Assignment

3) General Case Using Phase Variable Canonical Form

15 Lecture 15

Tracking Problem 15.1

1xy

rKKxu

BuAxx

r

Figure 15-1. State feedback diagram.

It is assumed that K is designed to meet a desired closed-loop characteristics. Problem is to

design to satisfy the tracking requirement.

⏟

, where a choice for is to defined as

Plant

r e u y

+ -

+

- -

of 115

15.1 Tracking Problem 107

Example 1: System { [

] [

]

from lecture 14

Example 2. (Hyperlink)


For the general case, if ≠ , instead ,

then we express

since we want the system to be driven by the difference between the input

and output.

Comparing with

{

Hence one gain has to be determined from other design criteria.

32

8

r e y +

-

+

-

of 115


Closed-Loop Pole-Zero Assignment 15.2

Figure 15-3. ADC motor system schematic.

{

Figure 15-5. System block diagram.

( ) ( )

( ) 1 ( ) ( )eq

Y s G s

R s G s H s

,

2

200( )

( 6 5)

AG s

s s s

3 2

3 2 3 1

( ) 200

( ) (6 100 ) (5 200 100 ) 200

Y s A

R s s k A s k A k A s k A

𝑖 + e

-

Amplifier

A

Current

Sensor

+

-

+

-

+ -

𝑖

Amp.

Fixed Field

B

Velocity Sensor

(Tachometer)

Position Sensor

(Potentiometer)

+

-

+

-

A

R U E

0.1

y

+

+ -

+

-

+ +

+

Figure 15-4. System close-loop block diagram.

of 115

15.2 Closed-Loop Pole-Zero Assignment 109

2

3 2 3 1(2 ) 2( )

2eq

k s k k s kH s

must be designed such that | and other design specifications are

satisfied.

Observations:

1) To have zero steady state error due to

.

2) Poles of are poles of G.

3) Zeros of are zeros of

4) Use of state feedback produces additional zeros in loop transfer function without

adding poles! Additional zeros move the root locus to the left system is more

stable and time response is improved.

Figure 15-6. Root Locus.

Let the desired closed-loop poles be located at

Therefore, solve for

-3

j4

-j4

s - plane

-8

of 115


General Case Using Controllable Canonical Form 15.3

Let

1

1 1 0

1

1 1 0

( )( )( ) , m n

( )

m m

m

n n

n

K s c s c s cY sG s

U s s a s a s a

0 1 2 1

0 1

0 1 0 0 0

0 0 1 0 0

0 0 0 1 0

1

0 0

n

m

x x u

a a a a

y c c c x Cx

1 1

0 1 1

( ) ( )( ) ( )( )

( ) ( ) ( ) ( )

T T

n neq

m

k X s k X sk X s k X sH s

Y s CX s c X s X s

Since

1 21 1 1 2 1

1 1

1 1 0

( ) ( ) ( ) ( )n n

j j n nj eq n m

m

k s k s k s kX s s X s s Y s H s

s c s c s c

1 2 1 2

1 2 1 2 1 0

1 1

1 1 0 1 1 0

( )

n n n n

n n n n

eq n n n n

n n

K k s k s k s k k K s s sGH s

s a s a s a s a s a s a

Now

1

1 0

1

1 0 1

( )( )

( ) ( ) ( )

m m

m

n n

n n

K s c s cY s

R s s a Kk s a Kk

Example 2: Given system as below,

design a state feedback such that due to a unit step is zero and

%OS = 4.3% and

A

U(s) Y(s)

of 115

15.3 General Case Using Controllable Canonical Form 111


Figure 15-8. Block diagram of the closed-loop system for example 1.

To get 0 10, ( ) 1ss sunit step

e with R R u t k

Given 1,2% 4.3 0.708 0.7064

5.65, 0.708s n

OS s j

T

Now we place the 3rd

pole arbitrarily at 3 100s

Therefore, the desired closed–loop transfer function is

3 2

( ) 10 10

( ) ( 100)( 0.708 0.7064) 101.4 142.6 100

Y s A A

R s s s j s s s

3

1

2 3

2

3

106 2 101.4

15 ( )10 142.6

3.39310 100

4.77

AAk

kk k A

kA

k

2

100( ) ( 1 )

( 101.4 142.7)eqG s type systm

s s s

Figure 15-9. Closed-loop diagram 3 for a unity feedback of example 1.

A

R U

+

+ + -

+ + +

Y(s)

𝐴

R U Y

+

-

G

R E Y

of 115


Figure 15-10. Root Locus for the standard feedback control system of example 2.

1 0

1 1

( )( 1)( 5)

2 , 0

(2 1) , 0

KGH

GH GHKK K G s

s s sGH k k

GH k k

Figure 15-11. Root Locus for the state feedback control system of example 2.

9.54 ( 0.721 0.726)( )

( 1)( 5)eq

A s jGH s

s s s

As a result of vicinity of desired pole locations with the zero of the controller,

a system is insensitive to variations in A.

s - plane

-2 -3 -4 -5 -6

-1

-2

2

0

3

×

×

×

1

Root Locus

-1

s - plane

-2 -5 -100

-1

-2

2

0

3

×

× 𝐴 𝐴

×

1

Root Locus

𝐴

𝐴

-1

𝐴

𝐴

𝐴

𝐴

of 115


Example 3: Given system as below,

Figure 15-12. System sate diagram for example 3.

Figure 15-13. Closed-loop block diagram for example 3.

3 2

3 2 3 2 1 1

( ) ( 2)

( ) [6 ( ) ] [5 ( 2 ) ] 2

Y s A s

R s s k k A s k k k A s k A

A desired closed-loop T.F. is selected as

1.22 2

3 2

( ) 2( 2) 2; 1 , 2, 0.707

( ) ( 2 2)( 2) 2 2

2( 2)

4 6 4

n

Y s ss j

R s s s s s s

s

s s s

1 2 31, 0.5, 1.5k k k

Figure 15-14. Root Locus for example 3.

A

R U

+

+ + -

+ + +

Y(s)

𝐴

R U Y +

-

×

s - plane

-2 -5

-1

-2

2

0

3

× 𝐴 𝐴

×

1

Root Locus

𝐴

-1

𝐴 𝐴

-1.18

1.68

𝐴

𝐴

𝐴

No non-dominant pole.

of 115


2( 0.5 2)( )

( 1)( 5)eq

A s sGH s

s s s

Unfortunately the system is sensitive to variations in A. In order to achieve

this we must have one non-dominant pole. Let the desired closed-loop T.F.

be 1 2 32

( ) 2( 2), 50, 0.5, 47.5

( ) ( 2 2)( 100)

Y s sk k k

R s s s s

2 248 98.5 100 48 98.5 100, ,

( 2) ( 1)( 5)eq eq

s s s sH Hence GH

s s s s


Example 4: Alternatively, since in no dominant pole cases there are 3 dominant poles

at must have 3 zeros to be insensitive to A.

However, to have 3 zeros in requires having 4 poles in G. Therefore, one

pole is added to G by using a cascade compensator

Then

can have one non-dominant pole p and

2

( ) ( 2)

( ) ( 2 2)( 2)( )desired

Y s A s

R s s s s s p

Selection of A and p are independent since

The closer poles and zeros are the less sensitivity of changes in A.

Selecting is non-dominant.

4 3 2

( ) 100( 2)

( ) 54 206 304 200

Y s s

R s s s s s

s - plane

-1 -5

-1

1

0

×

× 𝐴 𝐴

×

Root Locus

𝐴

𝐴

𝐴

Sensitivity to variations in

A is reduced significantly

of 115



Figure 15-17. Closed-loop block diagram for example 4.

To guarantee a controllable and observable plant ≠ , let a = 1.

Therefore, with A = 100, a =1

4 3 2

4 4 3 2 4 3 2 1 1

( )

( )

100( 2)

(7 100 ) [11 100(6 )] [5 100(5 )] 200

Y s

R s

s

s k s k k k s k k k k s k

Due to a unit step reference input, with

3 2

4 3 2

0.47 ( 4.149 6.362 4.255)

7 11 5eq

A s s sGH

s s s s


A

R U

+

+

-

+ + +

Y(s)

+

+

𝐴

R U Y +

-

×

s - plane

-1 -2

-1

1

0

× × 𝐴 𝐴

Root Locus

𝐴 𝐴

𝐴

-5 ×

-50

𝐴

𝐴

-2.07

𝐴

𝐴

𝐴

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