Loading in 5 sec....

State transition matrix: e AtPowerPoint Presentation

State transition matrix: e At

- By
**fred** - Follow User

- 123 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about ' State transition matrix: e At' - fred

**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: 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

- Controller canonical form is not unique
- This is also controller canonical form

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

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

>> 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)

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 -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)]

≠

√

>> 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

- 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

- (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.

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!

Download Presentation

Connecting to Server..