State space model: linear: or in some text: where: u: input y: output x: state vector A, B, C, D are const matrices.

State space model:

linear:

or in some text:

where: u: input y: output x: state vectorA, B, C, D are const matrices

),( :eqOutput

),( :eq State

uxhy

uxfx

DuCxy

BuAxx

JuHxy

GuFxx

Example

xy

uxx

31

1

0

32

10

1

0,

32

10

31,0

BA

CD

23

13

2)3(

13

131

2)3(

1

1

2)3(

131

1

0

2

13

2)3(

131

1

0

32

131

1

0

32

10

10

01310

)()(

2

1

1

1

ss

s

ss

s

sss

sss

s

s

ss

s

s

s

BAsICDsH

State transition, matrix exponential

)0(

:solution

:sHomogeniou

:caseScaler

xex(t)

axx

buaxx

at

matrixn transitiostate theiscalled

)0(

linearityby ),0(

:solution

:sHomogeniou

:caseMatrix

At

At

e

xex(t)

xx(t)

Axx

BuAxx

State transition matrix: eAt

• eAt is an nxn matrix • eAt =ℒ-1((sI-A)-1), or ℒ (eAt)=(sI-A)-1

• eAt= AeAt= eAtA

• eAt is invertible: (eAt)-1= e(-A)t

• eA0=I• eAt1 eAt2= eA(t1+t2)

dt

d

...!

1...

!3

1

!2

1

)0( :solution

3322

nnAt

At

tAn

tAtAAtIe

xex(t)

Axx

Example

)(222

2

2

2

1

1

2

2

1

22

1

1

1

2

1

1

2

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

2)2)(1(

1

)2)(1(

3

2

13

2)3(

1)(

32

1,

32

10

22

22

1

tueeee

eeeee

ssss

ssss

ss

s

ss

ssss

s

s

s

ssAsI

s

sAsIA

stttt

ttttAt

I/O model to state space• Infinite many solutions, all equivalent.

• Controller canonical form:

1 1

1 1 0 1 1 01 1

0 1 1

0 1 1

0 1 0 0 0

0 0 1 0 0

0

0 0 0 1 0

1

[0]

n n n

n nn n n

n

n

d d d d dy a y a y a y b u b u b u

dt dt dt dt dt

x x u

a a a

y b b b x u

I/O model to state space• Controller canonical form is not unique

• This is also controller canonical form

1

1 1 01

1

1 1 01

1 2 1 0

1 2 1 0

1

1 0 0 0 0

0

0 1 0 0 0

0 0 1 0 0

[0]

n n

nn n

n

n n

n n

n n

d d dy a y a y a y

dt dt dtd d

b u b u b udt dta a a a

x x u

y b b b b x u

Example

t

trydydt

dy

dt

yd

dt

yd02

2

3

3

)(235

dt

dry

dt

dy

dt

yd

dt

yd

dt

yd

dt

d 235:

2

2

3

3

4

4

n=4 a3 a2 a1 a0 b1 b0=b2=b3=0

xy

uxx

0010

1

0

0

0

5312

1000

0100

0010

Characteristic values• Char. eq of a system is

det(sI-A)=0the polynomial det(sI-A) is called char. pol.the roots of char. eq. are char. valuesthey are also the eigen-values of A

e.g.

∴ (s+1)(s+2)2 is the char. pol. (s+1)(s+2)2=0 is the char. eq.

s1=-1,s2=-2,s3=-2 are char. values or eigenvalues

uxx

0

1

0

200

120

001

2)2)(1(

200

120

001

det)det(

ss

s

s

s

AsI

2

100

)2(

1

2

10

001

1

)2)(1(00

1)2)(1(0

00)2(

)2)(1(

1)(

2

2

21

s

ss

s

ss

sss

s

ssAsI

)(00

)()(0

00)(

) (

2

22

1

tue

tutetue

tue

e

sts

ts

ts

t

At L

can At

t

t

ee

e

10

0

Set t=0 22I00

01

∴No

canAt

tt

t

eete

e

0

at t=0:

10

01

11

01 yes,

0

11

01

0

A

ete

e

etee

e

dt

d

tt

t

ttt

t

?

?

?

√

√

Solution of state space model

Recall: sX(s)-x(0)=AX(s)+BU(s)

(sI-A)X(s)=BU(s)+x(0)

X(s)=(sI-A)-1BU(s)+(sI-A)-1x(0)

x(t)=(ℒ-1(sI-A)-1))*Bu(t)+ ℒ-1(sI-A)-1) x(0)

x(t)= eA(t-τ)Bu(τ)d τ+eAtx(0)

y(t)= CeA(t-τ)Bu(τ)d τ+CeAtx(0)+Du(t)

DuCxy

BuAxx

t

0

t

0

But don’t use those for hand calculation

use:X(s)=(sI-A)-1BU(s)+(sI-A)-1x(0)

x(t)=ℒ-1{(sI-A)-1BU(s)}+{ℒ-1 (sI-A)-1} x(0)

& Y(s)=C(sI-A)-1BU(s)+DU(s)+C(sI-A)-1x(0)

y(t)= ℒ-1{C(sI-A)-1BU(s)+DU(s)}+C{ℒ-1 (sI-A)-1} x(0)

e.g.

xy

uxx

01

,0

1

20

01 If u= unit step

1

0)0(x

2

11

11

2

1

1

1

1

1

0

2

10

01

11

0

1

2

10

01

1

)(

s

ss

s

ss

s

ss

s

ssX

)(

)()()(

2 tue

tuetutx

t

t

1

11

10

2

11

11

01

)()()(

ss

ss

ss

sDUsCXsY

)()()( tuetuty t

Note: T.F.=D+ C(sI-A)-1B

1

1

0

1

2

10

01

1

010

s

s

s

Eigenvalues, eigenvectors

Given a nxn square matrix A, nonzero vector p is called an eigenvector of A if Ap p∝

i.e. λ s.t. Ap= λpλ is an eigenvalue of A

Example: ,

Let ,

∴p1 is an e-vector, & the e-value=1

Let ,

∴p2 is also an e-vector, assoc. with the λ =-2

20

01A

0

11p 11 0

1

0

1

20

01pAp

1

02p

1

02

2

0

1

0

20

012Ap

• For a given nxn matrix A, if λ, p is an eigen-pair, thenAp= λp λp-Ap=0 λIp-Ap=0 (λI-A)p=0

∵ p≠0 det(∴ λI-A)=0 ∴ λ is a solution to the char. eq of A: det(λI-A)=0

• char. pol. of nxn A has deg=n ∴ A has n eigen-values.e.g. A= , det(λI-A)=(λ-1)(λ+2)=0

⇒ λ1=1, λ2=-2

20

01

Eigenvalues, eigenvectors

• If λ1 ≠λ2 ≠λ3⋯then the corresponding p1, p2, will be ⋯linearly independent, i.e., the matrix

P=[p1 p⋮ 2 p⋮ ⋯ n] will be invertible.Then:

Ap1= λ1p1

Ap2= λ2p2

⋮A[p1 p⋮ 2 ]=[Ap⋮ ⋯ 1 Ap⋮ 2 ]⋮ ⋯

=[λ1p1 ⋮ λ2p2 ]⋮ ⋯

=[p1 p2 ]⋯

0

0

0

0

2

1

∴ AP=PΛ P-1AP= Λ=diag(λ1, λ2, ⋯)

∴If A has n linearly independent Eigenvectors, then A can be diagonalized.

Note: Not all square matrices can be diagonalized.

Example

ks wor1

1

0

0

312

11

0)(or

:1for

2,1

0)2)(1(

232)3(32

1det)det(

32

10

12

111

12

11

11

111111

21

2

P

PP

P

P

PAI

IPPAP

AI

A

11

12

11

12

12

1

21

11

21

11

ks wor2

1

0

0

322

12

0)( :2for

1

1

21

2

2

222

P

PPP

P

P

PAI

),(diag0

0

20

01

21

11

22

12

21

11

32

10

11

12

212

1

1

APP

121

211

),(diag

),(diag

PPA

APP

In Matlab

>> A=[2 0 1; 0 2 1; 1 1 4];

>> [P,D]=eig(A)

P = 0.6280 0.7071 0.3251 0.6280 -0.7071 0.3251 -0.4597 -0.0000 0.8881

p1 p2 p3

D = 1.2679 0 0 0 2.0000 0 0 0 4.7321

λ1

λ2

λ3

If A does not have n linearly independent eigen-vectors(some of the eigenvalues are identical),then A can not be diagonalized

E.g. A=

det(λI-A)= λ4+56λ3+1152λ2+10240λ+32768

λ1=-8λ2=-16λ3=-16λ4=-16

by solving (λI-A)P=0

99149

31163

44244

44812

,

0

1

0

1

1

p

2

0

1

0

2p

There are only two linearly independent eigen-vectors

>> A=[-12 8 -4 -4; 4 -24 4 4; -3 -6 -11 3; 9 -14 9 -9]A = -12 8 -4 -4 4 -24 4 4 -3 -6 -11 3 9 -14 9 -9>> [P,D]=eig(A)P = -0.7071 0.0000 + 0.0000i 0.0000 - 0.0000i 0.0000 -0.0000 0.4472 - 0.0000i 0.4472 + 0.0000i -0.4472 0.7071 -0.0000 + 0.0000i -0.0000 - 0.0000i -0.0000 -0.0000 0.8944 0.8944 -0.8944 D = -8.0000 0 0 0 0 -16.0000 + 0.0000i 0 0 0 0 -16.0000 - 0.0000i 0 0 0 0 -16.0000

Should use: >>[P,J]=jordan(A)P =

0.3750 0 1 0.625 0 8 4 0 -0.375 0 0 0.375 0 16 9 0

J=

-8 0 0 0 0 -16 1 0 0 0 -16 1 0 0 0 -16

a 3x3 Jordan block associated with λ=-16

More Matlab Examples

>> s=sym('s');>> A=[0 1;-2 -3];>> det(s*eye(2)-A) ans = s^2+3*s+2 >> factor(ans)ans =(s+2)*(s+1)

22

I

>> [P,D]=eig(A)P = 0.7071 -0.4472 -0.7071 0.8944

D = -1 0 0 -2

>> [P,D]=jordan(A)P = 2 -1 -2 2

D = -1 0 0 -2

2,1 21

2

1

toscale

8944.0

4472.0

1

1

toscale

7071.0

7071.0

2

2

1

1

P

P

P

P

A = 0 1 -2 -3

>> exp(A)ans = 1.0000 2.7183 0.1353 0.0498

>> expm(A)ans = 0.6004 0.2325 -0.4651 -0.0972

>> t=sym('t') >> expm(A*t) ans = [ -exp(-2*t)+2*exp(-t), exp(-t)-exp(-2*t)][ -2*exp(-t)+2*exp(-2*t), 2*exp(-2*t)-exp(-t)]

32

10

ee

ee

32

10

e

At

tttt

tttt

eeeee

eeee

22

22

222

2

≠

tttt

tttt

tt

tt

tt

tt

tttt

At

tttt

ttttAt

eeee

eeee

ee

ee

ee

eeeeee

e

eeee

eeeee

dt

d

22

22

2

2

2

2

22

22

22

442

222

)}2(3

)(2{

)}22(3

)2(2{222

)(32

10

442

222)(

10

01

1222

1121 :0 t:check √

√

Similarity transformation

DDCPCBPBAPPA

uDxCy

DuxCPy

uBxAx

BuPxAPPx

BuxAPxP

xPxxPx

DuCxy

BuAxx

,,,

,let weIf

)(#

11

11

same

system

as(#)

Example

21

22

11

2

22

12

2

1

20

01

22

12,let

01

1

0

32

10

xxy

uxx

uxx

xy

uxx

PxPx

xy

uxx

diagonalized

decoupled

Invariance:

changednot seigenvalue & valueschar.

sformationafter tran changednot eq. char.or poly char.

)det(

)det()det()det(

))(det(

)det(

)det()det(

1

1

11

1

AsI

PAsIP

PAsIP

APPPsP

APPsIAsI

changed rseigenvectoBut

))((

)(

)(

)()(

])([

)(

)(

)()(

:functionTransfer

111

1

11111

111

1111

111

1

ABAB

sH

BAsICD

BPPAsICPPD

BPPAsIPCPD

BPAPPPsPCPD

BPAPPsICPD

BAsICDsH