State transition matrix e at
This presentation is the property of its rightful owner.
Sponsored Links
1 / 30

State transition matrix: e At PowerPoint PPT Presentation


  • 70 Views
  • Uploaded on
  • Presentation posted in: General

State transition matrix: e At. e At is an nxn matrix e At = ℒ -1 ((sI-A) -1 ), or ℒ (e At )=(sI-A) -1 e At = Ae At = e At A e At is invertible: (e At ) -1 = e (-A)t e A0 =I e At1 e At2 = e A(t1+t2). I/O model to state space. Controller canonical form is not unique

Download Presentation

State transition matrix: e At

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


State transition matrix e at

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)


I o model to state space

I/O model to state space

  • Controller canonical form is not unique

  • This is also controller canonical form


Solution of state space model

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)


Eigenvalues eigenvectors

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

λis an eigenvalue of A

First solve: det(λI-A)=0 for λ

Then solve: (λI-A)p=0 for p.

If λ1 ≠λ2 ≠λ3⋯

Let P=[p1⋮p2 ⋮ ⋯pn]

P-1AP= D =diag(λ1, λ2, ⋯)

In Matlab: >> [P,D]=eig(A)

Or better: >>[P,J]=jordan(A)


More matlab examples

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)


State transition matrix e at

>> [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


State transition matrix e at

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)]


State transition matrix e at


Similarity transformation

Similarity transformation

same

system

as(#)


Example

Example

diagonalized

decoupled


Invariance

Invariance:


Controllability

Controllability:


Example1

Example:


State transition matrix e at

In Matlab:

>> S=ctrb(A,B)

>> r=rank(S)

If S is square (when B is nx1)

>> det(S)


Observability

Observability


Example2

Example:


State transition matrix e at

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

Lookfor controllability

Lookfor observability


Theorem

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


State transition matrix e at

  • (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.


State transition matrix e at

  • Controllability is invariant under similar transf.


State transition matrix e at

  • Observability is invariant under similar transf.


State feedback

State Feedback

D

r

+

u

+

1

s

x

+

y

B

C

+

-

+

A

K

feedback from state x to control u


State transition matrix e at

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


State transition matrix e at

Thm: Controllability is unchanged after state feedback.

But observability may change!


  • Login