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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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)
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)
>> A=[0 1;-2 -3];
A = 0 1
[ -exp(-2*t)+2*exp(-t), exp(-t)-exp(-2*t)]
[ -2*exp(-t)+2*exp(-2*t), 2*exp(-2*t)-exp(-t)]
If S is square (when B is nx1)
rank must = n
Or if single output (ie V is square), can use
det must be nonzero
no pole/zero cancellation in D+C(sI-A)-1B
feedback from state x to control u
If it is n, then
③Select any n eigenvalues(must be in complex conjugate pairs)
ev=[λ1; λ2; λ3;…; λn]
A+Bk will have eigenvalues at
Thm: Controllability is unchanged after state feedback.
But observability may change!