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...

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nnAt

At

tAn

tAtAAtIe

xex(t)

Axx

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

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

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 AFirst solve: det(λI-A)=0 for λThen solve: (λI-A)p=0 for p.

If λ1 ≠λ2 ≠λ3⋯Let P=[p1 p⋮ 2 p⋮⋯ n] P-1AP= D =diag(λ1, λ2, ⋯)

In Matlab: >> [P,D]=eig(A)Or better: >>[P,J]=jordan(A)

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

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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

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changed rseigenvectoBut

))((

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111

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111

1

ABAB

sH

BAsICD

BPPAsICPPD

BPPAsIPCPD

BPAPPPsPCPD

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BAsICDsH

Controllability:

n

nBBAABB

nABAABB

x

txtu

x

DuCxy

BuAxx

n

n

)BA-Irank(or

)1 is (if 0]|det[or

B])|||[rank( iff c.c. :Thm

time.finitein 0

to)( bringcan which )( control

,)0(any if lecontrollab completely is

1

12

Example:

01)det(or

2rank ind.linearly

231

10rank

3

1

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0

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0

32

10

|

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0][

2

1

0 ,

32

10

ABB

ABB

n

BA

In Matlab:

>> S=ctrb(A,B)

>> r=rank(S)

If S is square (when B is nx1)

>> det(S)

122

11rank e.g.

]||[ 2 BAABBS

Observability

,)C

A-Irank(or

)1 is (if 0detor ,rank iff c.o. :Thm

0set can ,generality of lossWithout

(0). determine tous enablecan timefinite aover

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11

n

nC

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C

n

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C

u

x

tytu

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BuAxx

nn

Example:

c.o.

01det(

10

01

32

1001

01

2

01 ,32

10

CA

C

CA

C

n

CA

In Matlab:

>> V=obsv(C,A)>> r=rank(V) rank must = n

Or if single output (ie V is square), can use>> det(V) det must be nonzero

2CA

CA

C

V

Lookfor controllability

Lookfor observability

Theorem• A state space model with A, B, C, D

matrices is both controllable and observable if and only if:

no pole/zero cancellation in D+C(sI-A)-1B

• If there is pole/zero canvellation– Either controllability is lost– Or observability is lost– Or both lost

)1

)(()0()()(;

0

0

1

)0()

))(()

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sBAsIxttxxc

AsIb

DC

uxxyy

BA

r

x

x

x

x

x

x

dt

da

rxxxyyyrydt

dx

xydt

dxx

dt

dxyxyxyx

rydt

dy

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yd

dt

yd

-

-

L

L

• (A, B) in controller canonical form, cntr

• (C, A) in observer canonical form, obsv

• But if we change C to [1 1 0]– H(s) = (s+1)/(s^3+3s^2+3s+1) = 1/(s+1)^2– Pole/zero cancellation.– But (A, B) still in contr canonical form, cntr– (C, A) no longer in obsv canonical form– Must have lost observability

– Can check obs. Matrix to verify.

• Controllability is invariant under similar transf.

)(rank

)(rank,min

)(rank),(rankmin)(rank

]|||[

]|||[

]|||[

]|||[after

]|||[ before :Proof

1

1

121

112111

111111

12

12

C

C

CC

CC

n

n

nC

nC

Q

Qn

QPQ

QPQ

BABAABBP

BAPBAPABPBP

BAPPAPPPBAPPPBP

BABABABQ

BABAABBQ

changednot ility Controllab

)(rank)(rank

)(rank)(rank

)(rank,min

)(rank),(rankmin

)(rank)(rank

But

)(rank)(rank1

CC

CC

C

C

CC

CC

CC

CC

QQ

QQ

Qn

QP

QPQ

QPQ

QPQ

QQ

• Observability is invariant under similar transf.

1

22

11

1

2

2

after

, before :Proof

PQQPQQ

PCA

CA

C

PCA

CAP

CP

APAPPCPP

APCPP

CP

AC

AC

C

Q

CA

CA

C

Q

OOOO

O

O

changednot ity Observabil

)(rank)(rank

)(rank),(rankmin

)(rank)(rank

)(rank

)(rank),(rankmin)(rank

1

1

OO

O

OO

O

OO

QQ

PQ

PQQ

Q

PQQ

State Feedback

law controlfeedback state a called is

:law the

Given

rkxu

DuCxy

BuAxx

B 1

s C

D

A

K

r u x xy

+ +

+ +

+

-

feedback from state x to control u

BkA

BkAA

DuCxy

BrxBkAx

BrBkxAx

rkxBAx

BuAxx

of thoseofeedback t state

by changed valuess/char.eigenvalue

tochangedMatrix only the

)(

)(

equation space state loop-closed

k

BkAnQC of choiceby any tochanged

becan of seigenvalue)(rank

i.e.

true.also is converse The

location.arbitrary

any toeigenvalueor valueschar.

thechangecan feedback statethen

lecontrollab completely is system theIf :Thm

In Matlab:

Given A,B,C,D

①Compute QC=ctrb(A,B)

②Check rank(QC)

If it is n, then

③Select any n eigenvalues(must be in complex conjugate pairs)

ev=[λ1; λ2; λ3;…; λn]

④Compute:

K=place(A,B,ev)

A+Bk will have eigenvalues at

Thm: Controllability is unchanged after state feedback.

But observability may change!