State space model:

linear:

or in some text:

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

⎩⎨⎧

=

=

),( :eqOutput ),( :eq Stateuxhyuxfx!

⎩⎨⎧

+=

+=

DuCxyBuAxx!

⎩⎨⎧

+=

+=

JuHxyGuFxx!

Back to state space

Example !x = 0 1

−2 −3

"

#$

%

&'x + 0

1

"

#$

%

&'u

y = 1 3"#

%&x

(

)**

+**

[ ]

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡

−−=

==

10

,32

10

31,0

BA

CD

H (s) = D+C(sI − A)−1B

= 0+ 1 3"#

$% s

1 00 1

"

#&

$

%'−

0 1−2 −3

"

#&

$

%'

"

#&&

$

%''

−1

01

"

#&

$

%'

= 1 3"#

$%

s −12 s+3

"

#&

$

%'

−101

"

#&

$

%'

= 1 3"#

$%

1s(s+3)+ 2

s+3 1−2 s

"

#&

$

%'01

"

#&

$

%'

= 1 3"#

$%

1s(s+3)+ 2

1s

"

#&

$

%'

=1

s(s+3)+ 21 3"

#$%1s

"

#&

$

%'

=3s+1

s(s+3)+ 2=

3s+1s2 +3s+ 2

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)

dtd

...!

1...!3

1!2

1)0( :solution

3322 ++++++=

=

=

nnAt

At

tAn

tAtAAtIe

xex(t)Axx!

Example

)(222

222

11

22

12

21

11

21

12

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

)2)(1(1

)2)(1(3

213

2)3(1)(

321

,3210

22

22

1

tueeeeeeee

e

ssss

ssss

sss

ss

sssss

ss

ssAsI

ss

AsIA

stttt

ttttAt

⎟⎟⎠

⎞⎜⎜⎝

⎛

+−+−

−−=

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

++

+

−

++

+

−+

−++

−+=

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

++++

−++++

+

=⎟⎟⎠

⎞⎜⎜⎝

⎛

−

+

++=−

⎟⎟⎠

⎞⎜⎜⎝

⎛

+

−=−⎟⎟

⎠

⎞⎜⎜⎝

⎛

−−=

−−−−

−−−−

−

I/O model to state space •  Infinite many solutions, all equivalent. •  Controller canonical form: d n

dtny + an−1

d n−1

dtn−1y +!+ a1

ddty + a0 y = bn−1

d n−1

dtn−1u+!+b1

ddtu+b0u

⇒ !x =

0 1 0 ! ! 00 0 1 ! ! 0! ! " " " !! ! " " 00 0 ! ! 0 1−a0 −a1 ! ! ! −an−1

#

$

%%%%%%%

&

'

(((((((

x +

00!!01

#

$

%%%%%%%

&

'

(((((((

u

y = b0 b1 ! ! ! bn−1#$%

&'(x + [0]u

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

d ndtny + an−1

d n−1dtn−1

y +!+ a1ddty + a0 y

= bn−1d n−1dtn−1

u+!+b1ddtu+b0u

⇒ !x =

−an−1 −an−2 ! ! −a1 −a01 0 0 ! ! 00 ! ! ! ! "" ! ! ! ! "0 ! " 1 0 00 ! ! 0 1 0

#

$

%%%%%%%

&

'

(((((((

x +

10!!00

#

$

%%%%%%%

&

'

(((((((

u

y = bn−1 bn−2 ! ! b1 b0#$%

&'(x + [0]u

Example

∫ =++++t

trydydtdy

dtyd

dtyd

02

2

3

3

)(235 τ

dtdry

dtdy

dtyd

dtyd

dtyd

dtd

=++++ 235: 2

2

3

3

4

4

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

!x =

0 1 0 00 0 1 00 0 0 1−2 −1 −3 −5

"

#

$$$$

%

&

''''

x +

0001

"

#

$$$$

%

&

''''

u

y = 0 1 0 0"#

%&x

(

)

***

+

***

Characteristic values •  Char. eq of a system is

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

⎦

⎤

⎢⎢⎢

⎣

⎡

+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−

−

=

010

200120001

!

2)2)(1(200120001

det)det( ++=⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

+

−+

+

=− sss

ss

AsI

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

+

++

+

=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

++

+++

+

++=− −

2100

)2(1

210

001

1

)2)(1(001)2)(1(0

00)2(

)2)(1(1)(

2

2

21

s

ss

s

sssss

s

ssAsI

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

=

=

−

−−

−

−

)(00)()(0

00)(

) (

2

22

1

tuetutetue

tue

e

sts

ts

ts

t

At L↓

can Att

t

ee

e

100

=⎟⎟⎠

⎞⎜⎜⎝

⎛

−

−−

−

Set t=0 22I0001

×≠⎟⎟⎠

⎞⎜⎜⎝

⎛−∴No

can At

tt

t

eete

e

0=⎟⎟

⎠

⎞⎜⎜⎝

⎛−−

−

at t=0: ⎟⎟⎠

⎞⎜⎜⎝

⎛

1001

( )

⎟⎟⎠

⎞⎜⎜⎝

⎛

−

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛

−

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛

−−

−=

−−

−

−−−

−

1101

yes,

011

01

0

A

etee

eteee

dtd

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)

⎩⎨⎧

+=

+=

DuCxyBuAxx!

∫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= λp λ 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

⎥⎦

⎤⎢⎣

⎡

−=

2001

A

⎥⎦

⎤⎢⎣

⎡=01

1p 11 01

01

2001

pAp =⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

−=

⎥⎦

⎤⎢⎣

⎡=10

2p ⎥⎦

⎤⎢⎣

⎡−=⎥

⎦

⎤⎢⎣

⎡

−=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡

−=

10

220

10

2001

2Ap

∃

•  For a given nxn matrix A, if λ, p is an eigen-pair, then Ap= 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

⎥⎦

⎤⎢⎣

⎡

− 2001

Eigenvalues, eigenvectors

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

P=[p1⋮p2 ⋮ ⋯pn] will be invertible. Then: Ap1= λ1p1 Ap2= λ2p2 ⋮ A[p1⋮p2 ⋮ ⋯]=[Ap1⋮Ap2 ⋮ ⋯] =[λ1p1⋮ λ2p2 ⋮ ⋯] =[p1 p2 ⋯]

λ1 00 λ2! 0 "0 ! "

!

"

#####

$

%

&&&&&

∴ 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

00

31211

0)(or :1for

2,10)2)(1(

232)3(32

1det)det(

3210

12

111

12

11

11

111111

21

2

⎟⎟⎠

⎞⎜⎜⎝

⎛

−=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛

+−

−−

=−

==−=

−=−=⇒

=++=

++=++=⎟⎟⎠

⎞⎜⎜⎝

⎛

+

−=−

⎥⎦

⎤⎢⎣

⎡

−−=

PP

P

PP

PAIIPPAP

AI

A

λ

λλλ

λλ

λλ

λλλλλ

λλ

[ ]

⎥⎦

⎤⎢⎣

⎡

−−=⎥

⎦

⎤⎢⎣

⎡ −−

+−=⎥

⎦

⎤⎢⎣

⎡

−−=

⎥⎦

⎤⎢⎣

⎡

−−==

⎟⎟⎠

⎞⎜⎜⎝

⎛

−=

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛

+−

−−

=−−=

−

−

1112

1112

121

2111

2111

ks wor2

1

00

32212

0)( :2for

11

21

2

2

222

P

PPP

P

P

PAIλλ

),(diag0

0

2001

2111

2212

2111

3210

1112

212

1

1

λλλ

λ=Λ=⎥

⎦

⎤⎢⎣

⎡=

⎥⎦

⎤⎢⎣

⎡

−

−=

⎥⎦

⎤⎢⎣

⎡

−−⎥⎦

⎤⎢⎣

⎡ −−=

⎥⎦

⎤⎢⎣

⎡

−−⎥⎦

⎤⎢⎣

⎡

−−⎥⎦

⎤⎢⎣

⎡

−−=− APP

121

211

),(diag

),(diag −

−

⋅⋅=

=∴

PPAAPP

λλ

λλ

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=-8 λ2=-16 λ3=-16 λ4=-16 by solving (λI-A)P=0

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−−

−−−

−

−−−

99149311634424444812

,

0101

1

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=p⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

2010

2p

There are only two linearly independent eigen-vectors

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 ↑

Similarity transformation

DDCPCBPBAPPAuDxCyDuxCPyuBxAx

BuPxAPPxBuxAPxP

xPxxPx

DuCxyBuAxx

====

+=

+=

+=

+=

+=⇒

==

⎩⎨⎧

+=

+=

−−

−−

,,,

,let weIf

)(#

11

11

!

!

!

!!

!

same

system

as(#)

Example

[ ]

[ ]

⎪⎩

⎪⎨

⎧

−=⎩⎨⎧

+−=

−−=

⇒

⎪⎩

⎪⎨

⎧

−=

⎥⎦

⎤⎢⎣

⎡−+⎥

⎦

⎤⎢⎣

⎡

−

−=

⎥⎦

⎤⎢⎣

⎡

−

−==

⎪⎩

⎪⎨

⎧

=

⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡

−−=

21

22

11

222

1221

2001

2212

,let

0110

3210

xxyuxx

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

⇒

∴

−=

−=

−=

−=

−=−

−

−

−−

−

AsIPAsIP

PAsIPAPPPsP

APPsIAsI

changed rseigenvectoBut

))(()(

)()()(

])([)(

)()()(

:functionTransfer

111

1

11111

111

1111

111

1

−−−

−

−−−−−

−−−

−−−−

−−−

−

=

=

−+=

−+=

−+=

−+=

−+=

−+=

ABABsH

BAsICDBPPAsICPPD

BPPAsIPCPDBPAPPPsPCPD

BPAPPsICPDBAsICDsH

Controllability:

[ ] λλ ∀=

×≠

=

=

∃

∀⎩⎨⎧

+=

+=

−

−

nnBBAABB

nABAABB

xtxtux

DuCxyBuAxx

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

31

||

10

10

3210

||

10

][

210

,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)

12211

rank e.g. =⎟⎟⎠

⎞⎜⎜⎝

⎛

]||[ 2 !BAABBS =

Observability

λλ

∀=⎥⎦

⎤⎢⎣

⎡

×≠

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎩⎨⎧

+=

+=

−−

,)C

A-Irank(or

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

0set can ,generality of lossWithout (0). determine tous enablecan timefinite aover

)(),( of knowledge theif obserrable completely is

11

n

nC

CA

CAC

n

CA

CAC

ux

tytuDuCxyBuAxx

nn

!!

"

Example: [ ]

[ ]

c.o.

01det(

1001

3210

01

012

01 ,32

10

∴

≠=⎟⎟⎠

⎞⎜⎜⎝

⎛

⎥⎦

⎤⎢⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎦

⎤⎢⎣

⎡

−−=⎥

⎦

⎤⎢⎣

⎡

=

=⎥⎦

⎤⎢⎣

⎡

−−=

CAC

CAC

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

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

←

!

2CACAC

V

Lookfor controllability

Lookfor observability

[ ]

)1)(()0()()(;001

)0()

))(()

0001output

100

331100010

)

3333

;;,,

33

11

11

1

3

2

1

3

2

1

3213

32

21

321

2

2

3

3

sBAsIxttxxc

AsIbDCuxxyyBA

rxxx

xxx

dtda

rxxxyyyrydtdx

xydtdxx

dtdxyxyxyx

rydtdy

dtyd

dtyd

-

-

−

−

−+=⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

=

−=

+===

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

+⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

−−−

=⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

+−−−=−−−==

======

=+++

L

L

φ

φ

!!!!!!

!!!!!

•  Recall linear transformation:

•  Controllability=being able to use u(t) to drive any state to origin in finite time

•  Observability=being able to computer any x(0) from observed y(t)

•  After transformation, eigenvalues, char. poly, char. eq, char. values, T.F., poles, zeros un-changed, but eigenvector changed

xPxxPx 1 , −==

⎪⎩

⎪⎨⎧

==+=

==+=⇔

⎩⎨⎧

+=

+=−

−

CPCBPBuDxCyDDAPPAuBxAx

DuCxyBuAxx

, ,

1

1!!

nBABAABBQS nC ===⇔ − rank has ]|||[ 12 !

n

CA

CAC

QV

n

O =

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

==⇔

−

rank full has

1

!

•  Controllability is invariant under 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

QQn

QPQQPQ

BABAABBPBAPBAPABPBPBAPPAPPPBAPPPBP

BABABABQBABAABBQ

≤

≤

≤∴

=

=

=

=

=

=

−

−

−−

−−−−−

−−−−−−

−

−

!

!

!

!

!

{ }{ }

changednot ility Controllab)(rank)(rank)(rank)(rank

)(rank,min )(rank),(rankmin

)(rank)(rank

But

)(rank)(rank1

∴

=∴

≤∴

≤

≤

=∴

=⇒

=

≤∴−

CC

CC

C

C

CC

CC

CC

CC

QQQQ

QnQP

QPQQPQ

QPQQQ

•  Observability invariant under transf.

1

22

11

1

2

2

after

, before :Proof

−

−−

−

==∴

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

PQQPQQ

PCACAC

PCACAPCP

APAPPCPPAPCPP

CP

ACACC

Q

CACAC

Q

OOOO

O

O

!!

!!

!

{ }

{ }

changednot ity Observabil

)(rank)(rank)(rank),(rankmin

)(rank)(rank

)(rank )(rank),(rankmin)(rank

1

1

∴

=∴

≤

=

≤

≤

−

−

OO

O

OO

O

OO

QQPQ

PQQQ

PQQ

State Feedback

law controlfeedback state a called is :law the

Given

rkxuDuCxyBuAxx

+−=⎩⎨⎧

+=

+=!

B 1

s C

D

A

K

r u x! x y

+ +

+ +

+

-

feedback from state x to control u

BkA

BkAA

DuCxyBrxBkAx

BrBkxAxrkxBAx

BuAxx

−

⇒

−

⎩⎨⎧

+=

+−=

+−=

+−+=

+=

of thoseofeedback t state by changed valuess/char.eigenvalue

tochangedMatrix only the

)(

)(

equation space state loop-closed

!

!

kBkAnQC

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!