State transition matrix: e At. e At is an nxn matrix e At = ℒ 1 ((sIA) 1 ), or ℒ (e At )=(sIA) 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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Recall: sX(s)x(0)=AX(s)+BU(s)
(sIA)X(s)=BU(s)+x(0)
X(s)=(sIA)1BU(s)+(sIA)1x(0)
x(t)=(ℒ1(sIA)1))*Bu(t)+ ℒ1(sIA)1) x(0)
x(t)= eA(tτ)Bu(τ)d τ+eAtx(0)
y(t)= CeA(tτ)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(λIA)=0 for λ
Then solve: (λIA)p=0 for p.
If λ1 ≠λ2 ≠λ3⋯
Let P=[p1⋮p2 ⋮ ⋯pn]
P1AP= D =diag(λ1, λ2, ⋯)
In Matlab: >> [P,D]=eig(A)
Or better: >>[P,J]=jordan(A)
>> 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)]
≠
√
√
same
system
as(#)
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
no pole/zero cancellation in D+C(sIA)1B
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
Thm: Controllability is unchanged after state feedback.
But observability may change!