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 = Inxn+At+ A2t2+ A3t3 +

• 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

def

!2

1

!3

1

Example

,00

00

A

s

ss

sAsI

10

01

0

0)(

1

1

)(0

0)(

)1

()0(

)0()1

())((

11

11

11

tu

tu

s

sAsIeAt

LL

LLL

,0

0

2

1

a

aA

2

1

1

2

11

10

01

0

0)(

as

asas

asAsI

)(0

0)(1

0

01

2

1

2

11

tue

tue

as

ase

ta

taAt L

Example

,10

11

A

1

10

)1(

1

1

1

10

11

)1(

1

10

11)(

2

2

1

1

s

sss

s

ss

sAsI

10

11

0

0

s

sAsI

)(0

)()(

1

10

)1(

1

1

12

1

tue

tutetue

s

sset

ttAt L

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

>> n=[1 2 3];d=[1 4 5 6];>> [A,B,C,D]=tf2ss(n,d)

A = -4 -5 -6 1 0 0 0 1 0B = 1 0 0C = 1 2 3D = 0

>> tf(n,d) Transfer function: s^2 + 2 s + 3---------------------s^3 + 4 s^2 + 5 s + 6

DuCxy

BuAxx

udt

du

dt

ud

ydt

dy

dt

yd

dt

yd

32

654

2

2

2

2

3

3

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

t

tt

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, p is 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

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 we use [P,D]=eig(A) get approximate but wrong answerShould 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 assoc. w/. λ=-16

• In general, if λ, P is an e-pair for A,AP= λP λP-AP=0 λIP-AP=0 (λI-A)P=0

∵ P≠0 det(∴ λI-A)=0 ∴ λ is a sol. of char. eq of A

• 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

• 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.AP1= λ1P1

AP2= λ2P2

⋮A[P1 P⋮ 2 ]=[AP⋮ ⋯ 1 AP⋮ 2 ]⋮ ⋯

=[λ P1 ⋮ λ P2 ]⋮ ⋯

=[P1 P2 ]⋯

0

0

0

0

2

1

• ∴ AP=PΛ P-1AP= Λ=diag(λ1, λ2, ⋯)

∴If A has n lin. ind. Eigenvectors then A can be diagonalized.

Note: Not all square matrices can be diagonalized.

If A does not have n lin. ind. e-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