Nano System Control Lab.Chosun University 1

Improvement

• Understanding

• Problem solve, skill

• Job offer

Plant, Process Effective Teaching

• Preparation Time

• Knowledge

• Availability

• Exp.

Disturbances

(Health..)

Desired

Input

Control Box

Sensor

quiz, homeworks

Noise

(cheating)

Nano System Control Lab.Chosun University 2

1. Stability

2. Performance

• Command Following

• Disturbance Rejection

• Insensitivity to Sensor Noise

3. Robustness “Model”

fcbsasdscs 234

Unmodelled Dynamics

Uncertainties

Structured uncertaintyUnstructured uncertainty

Additive perturbationMultiplicative perturbation

Unmodelled DynamicsUncertainties

Nano System Control Lab.Chosun University 3

Ex1)

xq

u

y

e

f

Control vector

Output vector

State , , : q

v :velocity change

: angleof attack

: pitch angle: path angle

y

z

x

v

Flapperon Deflection

u xPitch

Dynamics

x G(x( ), u( ))t t

Elevator Deflection

Nano System Control Lab.Chosun University 4

Linearizing it

We can describe

If is knownx G(x (t),u (t))n n n

x x x G(x x, u u)n n n

Using Taylor Series Expansion

1 1

2 2

G(x x, u u) G(x , u ) G x G u

G Gx y

x , GG Gx y

x G x G u

x Ax+Bu

n n n n x un n

x n

n

x un n

xy

Linear Time Invariant Dynamic System

*Time varying x G(x ,u , )t

:

Nano System Control Lab.Chosun University 5

For F-8

1 2 e 3

1 2 3 e 4

f

f

qq a a a

q b q b b b

Altitude

Dynamic Pressure

Speed

0A

0P

0V

1 2 3

2 1 3 4

0 0 1 0 00 10

x=Ax+Bu

e

fa a a

q b b q b b

Ex2) 2-DOF Manipulator : Horizontal ArmY

X

2

1

Torque

1g

1

2

2222 ,,, Imlg2g

1 1 1 1, , ,g l m I2 2 2

1 1 1 1 2 2

22 1 2

1 1 1TE KE I2 2 2

1 I ( )2

m v m v

Nano System Control Lab.Chosun University 6

1 111 12 2

21 222 2

H H( , , )

H H i i i j i j

i : Gravity Force

ji : Coriolis Force

2i : Centrifugal Force

1 1

00 Ix x τ

0 H F Hx A x B u

2 2 21 1 1

1 1 1 1 1 1

1 1 1 1 1 1 1 1

2 2 21 1 1

Vcos sin

sin cos

V

x yx g y g

x g y g

g

1

2 1

21

2

x H F

Intertial Term

,

Nano System Control Lab.Chosun University 7

22 2 2 22 1 1 2 1 2 1 2 2 1 1 2

2 2 2 2 2 21 2 1 2 2 1 2 2 2 1 2 2 1 2

V 2 cos

2 cos 2 cos

l g l g

l g l g g g l g

2 2 22 2 2

2 1 1 2 1 2

2 1 1 2 1 2

2 1 1 1 2 1 2 1 2

2 1 1 1 2 1 2 1 2

V

cos cos

sin sin

sin sin

cos cos

x y

x l g

y l g

x l g

y l g

Nano System Control Lab.Chosun University 8

Inverted stick on the cart

2 2 2

2 2 2

2 2

1 1 1KE M( ) J mv2 2 2

PE (1 cos )v z y ( sin , cos )

( sin ) ( sin )

cg

cg m m m m

z

mglz z l y l

z l l

HW2.

Total Energy

2 2 2 2 21 1 1TE KE PE Mz J (z 2 z cos ) (1 cos )2 2 2

m l l mgl

Applying Lagrange Equation

Generalized coordinates

TE TE( ) ( )

TE TE( ) ( )

df tdt z zdtdt

),( z

Nano System Control Lab.Chosun University 9

Define

Physical Dynamic Systems

① Contains energy storage elements

② Energy storage elements are interconnected

③ Energy in e.s.e. changes as a function of time

• Key features :

Nano System Control Lab.Chosun University 10

Define

• System State : A minimum set of numbers needed to describe a system

• State variable :

• System dynamics :

Interconnection of all the states evaluation as a function of time.

• State vector :

A set of state variables that defines a point in state space.

• State space :

A conceptual (mathematical) n-dimensional space with state

variables as coordinates.

1) An independent variable associated with each e.s.e.

2) Describes the evolution of energy in each e.s.e

Nano System Control Lab.Chosun University 11

x (x, )f t : unforced Dynamics

: forced Dynamics

: Autonomous System

- LTI System

Power Balance

in stored dissP E P

x

In general:

ddt

dE dF P F vdt dt

p e f

Power/dual (conjugate) variable

x (x,u, )f t

x (x,u)f

Ax+Bux

: effort:flow

ef

Mechanical: P=F vElectrical : P v iHydranlic: P p QRotational : P ω

Pressure ; volumetric flow

Nano System Control Lab.

Nano System Control Lab.Chosun University 13

Matrix Fundamentals

Addition : Commutative

Associative

(2) Multiplication : in general, non-commutative

Associative

(3) : Symmetric

(4) : Skew-symmetric

(5) : Hermitian *: complex conjugate

(6)

A+B B+A

(A+B)+C A+(B+C)

AB BA

(AB)C=A(BC)TA ATA A

*TA = A

1

tr(A) square matrix only

tr(A+B) tr(A) tr(B)tr(AB) tr(A) tr(B) , but tr(AB) tr(BA)

n

iii

a

State-space Representation

(1)

Nano System Control Lab.Chosun University 14

(7) Matrix InversionT

-1

-1

(A) CA , C ( 1) (A) : cofactorA A

A= A

ij i jij

Adj m

d ba b c ac d ad bc

(8) T T T

-1 -1 -1

(AB) =B A(AB) =B A A , B : Square matrix

Nano System Control Lab.Chosun University 15

Euclidean Norm

n2i2

i=1x = x = <x,x>

Inner product

T T

T

T T

<X,Y>=X Y=Y X<X,AY>=X AY<A X,Y>=X AY

Positive Definite Matrix A

TX AX 0 X 0Similally 0 negative definite

0 semi-negative definite

Nano System Control Lab.Chosun University 16

General Norm

x 3 properties 1. x 0 , iff x 0

2. x x : scalar

3. x+y x y

a a a

AB A B Triangle Inequality

Vector Norm

xx=

y

x

y

0

12 1

1

22

1

1 norm x x

2 norm x x

norm x max{ x }

n

ii

n

ii

ii

•

•

Nano System Control Lab.Chosun University 17

Matrix Norm :

11 1

12

22

1 1

1 norm A

2 norm A

n m

iji j

n m

iji j

a

a

1norm A max

m

iji ja

n mA

• Induced Norm

0 z 1

AxA Sup A Sup Azxx

• Spectral radius

r(A) max (A)ii

•

Nano System Control Lab.Chosun University 18

Eigen-value & Eigen-vector

Ax x (A I)x 0A i i i

0 1 1A , (λI A)

2 1 2 1

λI A ( 1) 2 0 1, 2

Ex1)

1

) 11 1 1 1

(A I)2 2 2 2

i

Adj

2

) 22 1 2 1

(A I)2 1 2 1

ii

Adj

1 2

2x

1x0

Nano System Control Lab.Chosun University 19

Similarity Transformation

n n

1 1

x=Ax+Bu y=Cx+Du

x R , u R , y R

Define z , x Pz , P is nonsingularPz AP z Buz P AP z P Bu

n m p

A B

-1 -1 -1 -1sI P AP sP P P AP P sI A P sI A

(A) (A )

sI A sI A

Because

Nano System Control Lab.Chosun University 20

Transfer Matrix

-1

x=Ax+BusIx=Ax+Bu(sI A)x=Bux=(sI A) Bu

-1

-1

y=Cx+Du

y=C(sI A) Bu+Du

y(s) =C(sI A) B+D=G(s)u(s)

-1 -1 -1 -1z x

-1 -1 -1 -1

G(s) = G(s) =CP[sI P AP ] P B+D

=CP[sI P AP] P B+D=C(sI A) B+D

(sI A)Two standard forms of x=Ax+Bu

Canonical form

find.Ps.t.

11

1

0A P AP , A

0

z z P Bun

Nano System Control Lab.Chosun University 21

1. Phase variable Form

Consider

1 2 1 2

1 1 1 11 2 1 2

n n n m m

n n m mn n n m m

d y d y d y d u d ua a a y c c c udt dt dt dt dt

1 1 11

2 1 2 1 11

2

3 2 3 2 11

-1( )

11

=

( )

( )

( )?

n nn

n nn

n nn

n nn n

n n n nn

ux y xs a s a

u sux y x x xs a s a

u s ux x x xs a s a

x x

u s ux xs a s a

, 1x y y u set is a component of y forced by

Nano System Control Lab.Chosun University 22

1 21 1

1 2 21 1 1 1

1 1 1 1

n n n nn n n nn n

n n n nn n n n

nn n n n

dx dx dx dxs a s a s a s udt dt dt dt

a s s x a s s x a s x s udx a x a x a x udt

1 1

2 2

1 2

0 1 0 00 0 1 0 00 0 0 0 00 0 0 00 0 0 0 0 1 1n n

n

x xx xd u

dtx x

a a a

1 1 2 2

1

21

x x x ,

0 0 0

m m

m

n

y c c c n mxx

y c c u

x

n m

Nano System Control Lab.Chosun University 23

Ex)

1

2 1

3 2

3

3 2 1

3 4 6 9 5

3 4 6 9 53 4 6 9 5

x x x x ux xx x xx x xx x x x u

x x x u

Poincare

1 1

2 2

3 3

0 1 0 0 00 0 1 0 06 4 3 5 9

x xd x x udt

x x

x

x

x

x

x

x

plotplanePhase

cyclelimit

Nano System Control Lab.Chosun University 24

2. Canonical Form

Find P 1

2-1

0 0 00 0 0

P AP0 0 00 0 0 n

If A is in phase variable form.

Then P is given by Vandermonde matrix

1 22 2 2

1 2 1 2 3

1 1 11 2

1 1 1

Pn

n n

n n nn

Nano System Control Lab.Chosun University 25

1

2

1 2

-1

1 1 2 2 n

z A z+P Bu

0 0 00 0 0

z( ) z(0) z ( ) th modal response0 0 00 0 0

x Pz z (0) z (0) z (0)

n

n

t

t

i

t

tt tn

ee

t t i

e

e e e

2 1x x

1 1 1

2 1 1 2

M u k b

k b 1M M M

x x x

x x x x u

Ex)

b

ku

1x

M

Nano System Control Lab.Chosun University 26

Assume M 1 , k 2 , b 3

1 1

2 2

1

2

0 1 0u

2 3 1

3 4

x xdx xdt

xy

x

2 2

1

0 1A 3 2 0 ; 1, 2

2 3

1

11 2 1

2

2

11 2 2

2

11 1 1

0 ;2 2 1

22 1 1

0 2 ;2 1 2

xx x

x

xx x

x

eigen vector

, ,

,

,

Nano System Control Lab.Chosun University 27

1 2

-1

1

2

1

1 1 2 2

1 1 2 1P , P

1 2 1 1

1 0 2 1 0 1 0 1z z u z u

0 2 1 1 1 0 2 1

CPz 1 5

1 2x(0) z(0) P x(0)

0 1

1 1x( ) (0) (0) 2

1 2t t t

zy

z

t e z e z e

2

21

22

( 1)

x ( ) 2x( )

x ( ) 2 2

t

t t

t t

e

t e et

t e e

Nano System Control Lab.Chosun University 28

1x

1x

slow mode

fast modeRead chap.7 of Brogan

1x

t

2

1

0

2x

t

2

0

2

Nano System Control Lab.Chosun University 29

Similar Matrices

-11 2 3A =P AP P n

For distinct eigenvalues si

1

T1

T-1 -1 -1 2

Tn

0A

0

A * : right eigenvector

A P =P A P Q

n

thi i i i i

T TA * : left eigenvectorthi i i i i

Nano System Control Lab.Chosun University 30

T

1T

T

A ( ) PA Q

0 Orthogonality

1 Co-linearity

n

i i ii

j i

i i

j i

Sylvester’s expansion (or Lagrangian interpolation)

At (A)e f

1

1

1

A I

( ) ( )Z ( ) , ( )

n

jjni

i i i ni

i jji

f A f Z

Ex)2 2 3

A 1 1 11 3 1

only fordistinct eigenvalues

Nano System Control Lab.Chosun University 31

The eigenvalues are 1, 2, 3i

1

2

3

4 2 3 1 2 3 3 5 2(A 2I)(A 3I) 1 1z 1 3 1 1 2 1 3 5 2

(1 2)(1 3) 6 61 3 1 1 3 4 3 5 2

0 11 11(A I)(A 3I) 1z 0 1 1( 2 1)( 2 3) 15

0 14 14

5 1 4(A I)(A 2I) 1z 5 1 4

(3 1)(3 2) 105 1 4

A 2 3

3 5 2 0 11 11 5 1 41 1 13 5 2 0 1 1 5 1 4

6 15 103 5 2 0 14 14 5 1 4

t t t te e e e

Nano System Control Lab.Chosun University 32

2

kA

k 02

2 2 k k

2 2

A 1I A A2 k!

(PA Q)(PA Q)I PA Q2!

PA Q PA QI PA Q2! k!

AP I+A Q2!

t te t t

tt

t tt

tt

1

A A

T1

T1 2 2

T

T

1

P Q

0 00 00 0

i

t t

t

nt

n

nt

i ii

e e

e

e

e

Nano System Control Lab.Chosun University 33

• Special case : when these are multiplicities of eigenvalues, for each of which only one independent eigenvector exists, we can not have a diagonal . But, an almost diagonal can be obtained.A A

“Jordan Canonical Form”

A: CE, 3 2( 3) ( 4) 0 5 eigenvalue1 triple at 31 double at 4

We can find P s.t

3 1 0 0 00 3 1 0 0

A 0 0 3 0 00 0 0 4 10 0 0 0 4

Rank : The maximum number of linearly independent row vectors of a matrix is called the rank of A and is denoted by rank A. A jka

Nano System Control Lab.Chosun University 34

Ex1)3

1 2 3A 0 1 4 ( 1) 0 , 1 triple at 1

0 0 1

g (A I) #n rank of independent eigenvectors associated with i

3 2 1

1

0 0 8 1A I 0 0 0 0

0 0 0 0iAdj

We need two more independent eigenvectors 1 2 1 1 2g & g P g g

Generalized eigenvectors

1 1

2 1

3 2

A I

A I

A I

i

i

i

g

g g

g g

Nano System Control Lab.Chosun University 35

0 2 3 10 0 4 00 0 0 0

abc

2 14 0

0

b ccc

set 112

a

b

1

11g20

10 2 310 0 42

0 0 0 0

def

2 3 11421 5,8 16

e f

f

f e

2

15g

1618

1

1 1 1 1 2 3 1 1 01 5P= 0 P 0 2 5 A =P AP 0 1 12 16

0 0 8 0 0 110 08

,,

Nano System Control Lab.Chosun University 36

Ex2) 0 1 0A 0 0 1

8 12 6

4 4 1(A I) 8 8 2

16 16 4iAdj

2

3

1 12

1

( 6) 8 12 0( 2) 0

1 124

1 1

2 1 0 1 1g : 0 2 1 2 g 1

8 12 4 4 0

abc

2 2

342 1 0 11g : 0 2 1 1 g2

8 12 4 0 0

def

1,

Nano System Control Lab.Chosun University 37

1

31 1 10 04 41P 2 1 P 2 3 12

4 4 14 0 0

General Case

1 2 i j nA: CE ; ( P )( P ) ( P ) ( P ) ( P ) 0k l

,

-1

2 1 0A =P AP 0 2 1

0 0 2

Nano System Control Lab.Chosun University 38

1

2

i1

-1

1

n

PP

P 1 000 0 10 0 0 P

A =P AP

P 1 000 0 10 0 0 P

P

ik k k

j

jl l l

o o o o o o oo o o o o o oo o o o o o o

o o o o o o o

o o o o o o o

o o o o o o o

o o o o o o oo o o o o o o

Nano System Control Lab.Chosun University 39

1

x=Ax+Bu ; y=Cx+Du

(sI A)x u

x sI A Bu

B

1y(s) C(sI A) B D u(s)

Ex)

2

1

0 1 1 0 1 0 0 0x x u , y x u

1 2 0 1 0 1 0 1

s 1(sI A) sI A s 2s 1

1 s 2

s 2 1 1sI A1 s sI A

1 0 s 2 1 1 0 0 0 sI A 1G(s)0 1 1 s 0 1 0 1sI A sI A

Basic LTI system

G(s) :Transfer Function

,

Nano System Control Lab.Chosun University 40

2 2

2

2 2

s+2 1s 2 11 s +2s+1 s +2s+1G(s)

1 det(sI A)det(sI A) 1 s +3s+1s +2s+1 s +2s+1

s

In General

11 1m

m1 mm

ij 1 2ij

1 2

C (sI A) B D det(sI A) QG(s)det(sI A) det(sI A)

Q Q1

sI A Q Q

Q (s) k (s z )(s z )G (s)

det(sI A) (s )(s ) (s )ij

m

Adj

Nano System Control Lab.Chosun University 41

CE: det(sI A) 0 Im

Re

Re( ) 0 unstableRe( ) 0 stable

i

i

For zeros

Re( ) 0 nonminimum phase system ( any of 1 ~ )Re( ) 0 minimum phase system ( 1 )

i

i

z i mz i m

2 2 2 2 2 2

( )2 2 2n n n n n n

K s z Ks Kzs s s s s s

① 1; input u=

s

Nano System Control Lab.Chosun University 42

2 2 2 2 2 2

( )2 2 2n n n n n n

K s z Ks Kzs s s s s

larger:

Impulse Step

②

Undershoot

Nano System Control Lab.Chosun University 43

First order Response

1 0y yy u y ud da a b K

dt dt

1

0 0

time constant , = input gaina bKa a

For unforced case

1y y, y y(0)t

e

y

t

increasing

0

Nano System Control Lab.Chosun University 44

Second Order Response

2

2 1 0 02

22

2

y y y u

y y2 y un n

d da a a bdt dt

d d Kdt dt

1

0 2

0

2

0

2

damping coefficient2

natural frequency

input gain

n

aa a

aa

bKa

21 2, 1n

Im

real

Re

ComplexConjugatepair

: ringing freq.d

Re

Im

21d n n

cos

0 0

Nano System Control Lab.Chosun University 45

General Response

Start with 0x=Ax+Bu : x(0)=x

(1) Homogeneous Response : x Ax

(unforced response, Zero-input response)

(2) Particular response(Zero-state)

1) Unforced Response

Assume that the solution can be obtained via power series.

H Hx Ax

Nano System Control Lab.Chosun University 46

2H 0 1 2

1H 1 2

H 0 1

x

x 2

Ax A A A

mm

mm

mm

c c t c t c t

c c t mc t

c c t c t

1 0

22 1 0

33 2 0

A1 1A A2 21 1A A3 3!

c c

c c c

c c c

0

2A At

H 0 0 H H

1 A!

Ax I A x (0) ( )x (0)2!

mm

t

c cm

tt c e c e t

H 0x (0) c

State Transition Matrix

Nano System Control Lab.Chosun University 47

Read chap.9 of Brogan

One correction

1 1

1

1

y(s) G(s) u(s)

C (sI A) B D u(s)

p p m m

n m p mp n n n m

(1) has all the poles of the MIMO system.1(sI A)

(2)A

H 0x xte1

00

Laplace Domainx=Ax

x(s) (sI A) xsIx( ) x Axs

AH 0

1H 0

Time Domain Solution : x x*

Freq. Domain Solution : x sI A x

te

p m

Nano System Control Lab.Chosun University 48

A 1

11 A

A A -1

nT

i=1

(sI A)

sI A

P Q P Q : Distinct eigenvalues

= i

t

t

t t

ti i

e

e

e e

e

* Modal Solution A TH 0 0

1x x xi

ntt

i ii

e e

Nano System Control Lab.Chosun University 49

Question

0

TH i

1

H1

x

1x

0

x

i

i

j

nt

i ji

nt

ii

K

if i jKe

i j

Ke

T

0

z z

z xi

i i

ti i

t t

t e

T0

1

T

1

z 0 x

x=Pz z

z=Qx= x

i i

n

i iin

i ii

i TH 0x e x

1s

i

ti i

t

i

e

Laplace model

TH 0

1

1x xs

n

i ii i

iz (0)

Nano System Control Lab.Chosun University 50

2) Forced Response

AssumeA

p

A Ap

p

x p(t) p(0) 0

x ( ) p(t) A p(t)

=Ax +Bu

t

t t

e

t e e

A Ap(t) Bu p(t) But te e

A

0 0

A( )A At A

p 0

H P

At A( )0 0

p(t) p(0) Bu( ) : dummy variable

x p(t) e Bu( ) e Bu( )d

x( ) x x x(t) Ax(t)+Bu(t)

x Bu(τ)

t t

tt tt

o

t t

dp e d

e e d

t

e e d

Nano System Control Lab.Chosun University 51

“check”

A At

0

A A0 0

A

( ) ( )

x( ) x Bu( )

t

t

tt

d e edtd f d f tdt

t e e d

At A A A0 0

x( ) A x Bu( ) 0 Bu( )

x( ) Ax( ) Bu( )

t t td t e e d e e tdt

t t t

Nano System Control Lab.Chosun University 52

A A( )0 0

1 10

y=Cx(t)+Du(t)

Time : y( ) C x C Bu( ) Du( )

Laplace : y(s) C(sI A) x C(sI A) B+D u(s)

tt tt e e d t

Laplace Domain

01 1

0

x( ) Ax( ) Bu( )sIx(s) Ax(s) Bu(s)

x(s) (sI A) x (sI A) Bu(s)

t t tx

A A( )0 0

A( ) 1

0

x( ) Bu( ) : Time Domain

Bu( ) (sI A) Bu( )

tt t

t t

t e x e d

e d s

Nano System Control Lab.Chosun University 53

2 0

t

1 20

1 2

1 2

y ( ) ( )u( ) : Matrix Convolution

( ) ( ) ( ) : Matrix Convolution Integral

( ) ( )F (s) F (s)

tt t d

f t f t f d

f t f t

1F (s) 2F (s)

1 2F (s) F (s)

Note

1 11

0 for( ) ( )u( ) Unit step

( ) fort

f t f t tf t t

t A( )2 0

A(t- )

y ( ) C Bu( )

set Ce B ( )

tt e d

t

1 2 1 20 0( ) ( ) ( ) ( ) ( )

tf t f d f t u t f d

Nano System Control Lab.Chosun University 54

Taking Laplace Transform

1 20 0( ) ( ) ( ) ( )stf t e f t u t f d dt

Collecting and separatelyt

1 2 1 20 0 0 0

( ) ( )1 2 1 20 0 0 0

1 20 0

1 2

( )u( ) ( ) ( )u( ) ( )

set ( )

( ) ( ) ( ) ( )u( ) ( )

( ) ( )

F (s) F (s)

st st

s s

s s

f t t e dt f d e f t t f dt d

t

f u e d f d e f f d d

e f d e f d

Nano System Control Lab.Chosun University 55

0

1 00C V11 R

LL L

c c

LL

e eii

Assume C=1F , L 1H , R 3

1

A T

1

A A( )0 0

0 1 0A B

1 3 1CE : s(s+3)+1=0 s= 2.62 , 0.38

1 1 0.38 1P , Q=P / 2.24

2.62 0.38 2.62 1

( )x( ) x Bu( )

( )

i

ntt

i ii

tc t t

L

e e

e tt e e d

i t

L

0V C ceLi

Ex) R

,

Nano System Control Lab.Chosun University 56

0 0

0 0

2.62 0.38

2.62( )00

0.38( )

1 10.38 1 2.62 12.62 0.382.24 2.24 2.24 2.24

1 00.38 1 V ( )2.62 12.24 2.24

1 2.620.38 2

c ct t

L L

t t

t

e ee e

i i

e d

e

00

01 V ( )1.24 2.24

td

Assume and unit step input0x 0 0V (t) 1

2.62 2.62 0.38 0.38

2.62 0.38

1 11 12.24 2.24x( ) 1 1

2.62 0.382.62 0.382.24 2.24

0.170 1.175x( ) 1 1

0.446 0.446

t t

t t

t e e e e

t e e

t

00

t

Nano System Control Lab.Chosun University 57

General Step Response

)(tec)(tiL

Let u(t) be a vector of step

t

1 1

1

u( )

m m

Kt

K

( )su t

t t

0

( )su t

Nano System Control Lab.Chosun University 58

A A( ) A A( )0 00 0

A( )

0

0 A A

0

2 2 2 2 3

0

x( ) x BKu u ( ) B K

1 1 1I A A I A A2! 2! 3!

t tt t t ts s

t t

tv v

t

t

t e e d e x e d

e d

e dv e dv

v v dv t t t

at 0,, 0

v t v tdv d t v

Multiply all terms by -1A A=I

1 2 2 1 A

A 1 A0

1

1A I A A I A I P2!

x ( ) A I B K

Steady-statex ( ) lim x ( ) A BK

t

t ts

ss st

t t e

t e x e

t t

Nano System Control Lab.Chosun University 59

0V ( )ce t0 1 0

A B1 3 1

Assume K=1 scalar0x 0 ,

2.62 0.38s

2.62 0.38

1 13 1 3 1 02.24 2.24x (t) 1

1 0 2.62 0.38 1 0 12.24 2.24

0.170 1.170 10.446 0.446 0

t t

t t

e e

e e

t

( )ce t

t

( )Li t

Nano System Control Lab.Chosun University 60

A( ) A A A 1 A 1 A 1 A

0 0* A A A I

t tt t t t t te d e e d e e e e

A 1 1 AA At te e

1 A A 1 Asince A I A It t te e e

A 1

A T

1

* (sI A)

i

t

ntt

i ii

e

e e

Nano System Control Lab.Chosun University 61

Cayley Hamilton Theory

1 21 2 1 0CE : A I ( ) C C C C ( ) 0n n n

n n

Then the corresponding matrix polynomial1 2

1 2 1 0Δ(A) ( 1) A C A C A C A C In n n nn n

Cayley Hamilton Theorem

Every matrix satisfies its own characteristic equation ;That is Δ(A)=0

Proof) A similarity transformation reduces A to the diagonal matrix A-1 2 2 -1 K K -1A=PA P , A =PA P , A =PA P

Therefore1 2 1

1 2 1 0Δ(A) P ( 1) A C A C A C A C I Pn n n nn n

Each term inside the brackets is a diagonal matrix.The sum of a typical , element is

11 1 0( ) C C Cn n

i n i i

i i

Nano System Control Lab.Chosun University 62

Which is zero because is a root of the characteristic equation.Therefore.

i

-1Δ(A) = P 0 P =0 (Q.E.D.)

Ex) 23 1Let A , A I (3 )(2 ) 1 5 5

1 2

Then2 10 5 3 1 1 0 0 0

Δ(A) A 5A 5I 5 55 10 1 2 0 1 0 0

Some use of the cayley-Hamilton Theorem

① Matrix inversion

11 1 0(A) ( 1) A C A C A C I 0n n n

n

Assuming exists, multiplication by gives1A1A

Nano System Control Lab.Chosun University 63

1 2 11 1 0

1 1 21 1

0

( 1) A C A C I C A 01A ( 1) A C A C I

C

n n nn

n n nnor

② Reduction of a polynomial in A to one of Degree n-1 or less

Ex)

2

2 1

3 1Let A , ( ) 5 5 Then

1 2

2 11 1(A) A 5A 5I 0, A A 5I1 35 5

The matrix polynomial can be writtenP(A)

1P(A)=Q(A)P (A)+R(A)

Since it is always defined in the same manner as its counterpart. If the arbitrary polynomial used above is selected as the characteristic polynomial of A, then the scalar version of is

1PP

P(x)=Q(x)Δ(x)+R(x)

Nano System Control Lab.Chosun University 64

Note that except for those specific values , the eigenvalues. The matrix version of is

(x) 0 P

x i

P(A)=Q(A)Δ(A)+R(A)

By the Caylay-Hamiton theorem, , so Δ(A) 0 P(A)=R(A) (residue)

Ex) 2 4 3 23 1Let A , ( ) 5 5 compute P(A)=A +3A +2A +A+I

1 2

Method 1 :By long division,

2 22

P(x) 146x-184x +8x+37+ or P(x) (x +8x+37) (x) (146x 184)Δ(x) x -5x+5

Therefore, and the Cayley-Hamilton theorem guarantees that

R(x) 146x 184 P(A)=R(A) 146A 184I

Method 2 : From the Cayley-Hamilton theorem,

2 2A -5A+5I 0 or A 5(A I)

Nano System Control Lab.Chosun University 65

Hence,

4 2 2 2

3 2 2

A A A 25 A I A I 25 A 2A I 25 5 A I 2A I

25 3A 4I

A A(A ) 5(A A) 5 4A 5I

Thus, P(A)=25 3A 4I 15 4A 5I 10 A I A I 146A 184I

Nano System Control Lab.Chosun University 66

③ Closed Form solution for Analytic Functions of Matrices

Let be a function which is analytic in a region of the complex plane and let A be an matrix whose eigenvalues . Then has a power series representation.

( )f x n n i

( )f x

0

0

(x) x

(x) (x) x R(x)

kk

k

kk

k

f

f

The remainder R will have degree less than or equal to n-1 . The analytic function of the square matrix A is defined by the same series as its scalar counterpart, but with A replacing x, therefore , since is always the null matrix.

(A)=R(A)f Δ(A)

Although the form of is known to beR(x)

2 10 1 2 1R(x) x x xn

n

Nano System Control Lab.Chosun University 67

It is clearly impossible to find the coefficients by long division as in previous example. However, if the n eigenvalues are distinct, equations for determining the terms are available. Since setting gives

ii

inn

( ) 0,i x i

( ) R( ), 1, 2,i if i n

Find the closed form expression for if Ate

0 1A

1 3

( ) ( 3) 1 0, so 1 22.62, 0.38

Since A is a matrix, it is known that R is of degree one (or less):2 2

0 1 1 2R(x) x Also using x and x gives

1

2

2.620 1 1 0 1

0.380 1 2 0 1

2.62

0.38

t t

t t

e e

e e

From two eqns, 2.62 0.38 0.38 2.62 0.381 0

0.38 2.62

1 0.38,2.24 2.24

2.62 0.382.24 2.24

t t t t t

t t

e e e e e

e e

A0 1; I Ate

Ex)

Nano System Control Lab.Chosun University 68

When is a repeated root, this procedure must be modified. Some of the equations will be repeated, so they do not form a set of n linearly independent equations.

However, for a repeated root, also,

i( ) R( )i if

i

0

( ) ( )i i i i

k kk i i k

k

df d d dR dRd d d d d

For an eigenvalue with algebraic multiplicity the first derivatives of all vanish and thus

,im

1im

From a set of linealy independent equations. Thus a full set of n equations is always available for finding the coefficients of the remainder term R

imi

( ) 0i

ddt

and so

Find the closed form expression for if Ate 0 1 0

A 0 0 127 27 9

Ex)

2 2

2 2

R R( ) R( ) ,i i i i

i idf d d f dfd d d d

, ,1 1

1 1Ri i

i i

i i

m m

m md f dd d

•

Nano System Control Lab.Chosun University 69

We have

33 2

1 2 3At 2

0 1 2

A I ( ) 9 27 27 33,

e R(A) I A A

Where3

0 1 2

2 30 1 2 1 2

33

2 22 2 3

0 1 2 22 23 3

3 9

or 6

or 2

t

tt

tt

e

de d ted d

d e d t ed d

Therefore and

Solving for and gives 0 1, 22 3 3 2 3

2 1 2

3 2 30 1 2

1 , 6 ( 3 )2

93 9 1 32

t t t

t t

t e te t t e

e t t e

Nano System Control Lab.Chosun University 70

Using these coefficients in givesR(A)

2 2 2

A 2 2 2 3

2 2 2

9 11 3 32 2

27 31 3 92 2

81 927 27 27 1 62 2

t t

t t t t t

e t t t t t e

t t t t t

When some eigenvalues are repeated and others are simple roots, a full set of n independent equations are still available for computing the coefficients.

i