State transition matrix: e At

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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

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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,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

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)]

Example

diagonalized

decoupled

In Matlab:

>> S=ctrb(A,B)

>> r=rank(S)

If S is square (when B is nx1)

>> det(S)

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
• 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.
State Feedback

D

r

+

u

+

1

s

x

+

y

B

C

+

-

+

A

K

feedback from state x to control u

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