1. 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)

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

3. 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)

4. 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
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)

5. More Matlab Examples >> s=sym('s'); >> A=[0 1;-2 -3];
>> det(s*eye(2)-A)
ans =
s^2+3*s+2
>> factor(ans)
(s+2)*(s+1)

6. >> [P,D]=eig(A)
P =
D =
>> [P,D]=jordan(A)

7. ≠ A = 0 1 -2 -3 >> exp(A) ans = 1.0000 2.7183 0.1353 0.0498
>> exp(A)
ans =
>> expm(A)
>> t=sym('t')
>> expm(A*t)
[ -exp(-2*t)+2*exp(-t), exp(-t)-exp(-2*t)]
[ -2*exp(-t)+2*exp(-2*t), 2*exp(-2*t)-exp(-t)] ≠

8. √ √

9. Similarity transformation
same
system
as(#)

10. Example
diagonalized
decoupled

11. Invariance:

13. Controllability:

14. Example:

15. In Matlab:
>> S=ctrb(A,B)
>> r=rank(S)
If S is square (when B is nx1)
>> det(S)

16. Observability

17. Example:

18. Or if single output (ie V is square), can use >> det(V)
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

19. 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

21. (A, B) in controller canonical form,  cntr
(C, A) in observer canonical form,  obsv
But if we change C to [ ]
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.

22. Controllability is invariant under similar transf.

24. Observability is invariant under similar transf.

26. State Feedback D r + u + 1 s x + y B C + - + A K
feedback from state x to control u

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

30. Thm: Controllability is unchanged after state feedback.
But observability may change!