1. State space model:
linear:
or in some text:
where: u: input
y: output
x: state vector
A, B, C, D are const matrices

2. Example

4. State transition, matrix exponential

5. 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)

6. Example

7. I/O model to state space
Infinite many solutions, all equivalent.
Controller canonical form:

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

9. Example
n= a a a a0 b b0=b2=b3=0

10. Characteristic values
Char. eq of a system is
det(sI-A)=0
the polynomial det(sI-A) is called char. pol.
the roots of char. eq. are char. values
they are also the eigen-values of A
e.g.
∴ (s+1)(s+2)2 is the char. pol.
(s+1)(s+2)2=0 is the char. eq.
s1=-1,s2=-2,s3=-2 are char. values or eigenvalues

12. can ?
Set t=0
∴No
can ? √
at t=0: ? √

13. 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)

14. But don’t use those for hand calculation
use:X(s)=(sI-A)-1BU(s)+(sI-A)-1x(0)
x(t)=ℒ-1{(sI-A)-1BU(s)}+{ℒ-1 (sI-A)-1} x(0)
& Y(s)=C(sI-A)-1BU(s)+DU(s)+C(sI-A)-1x(0)
y(t)= ℒ-1{C(sI-A)-1BU(s)+DU(s)}+C{ℒ-1 (sI-A)-1} x(0)
e.g.
If u= unit step

15. Note: T.F.=D+ C(sI-A)-1B

16. 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
Example: ,
Let ,
∴p1 is an e-vector, & the e-value=1
∴p2 is also an e-vector, assoc. with the λ =-2

17. Eigenvalues, eigenvectors
For a given nxn matrix A, if λ, p is an eigen-pair, then
Ap= λp
λp-Ap=0
λIp-Ap=0
(λI-A)p=0
∵ p≠0 ∴ det(λI-A)=0
∴ λ is a solution to the char. eq of A: det(λI-A)=0
char. pol. of nxn A has deg=n
∴ A has n eigen-values.
e.g. A= , det(λI-A)=(λ-1)(λ+2)=0
⇒ λ1=1, λ2=-2

18. If λ1 ≠λ2 ≠λ3⋯
then the corresponding p1, p2, ⋯ will be linearly independent, i.e., the matrix
P=[p1⋮p2 ⋮ ⋯pn] will be invertible.
Then:
Ap1= λ1p1
Ap2= λ2p2 ⋮
A[p1⋮p2 ⋮ ⋯]=[Ap1⋮Ap2 ⋮ ⋯]
=[λ1p1⋮ λ2p2 ⋮ ⋯]
=[p1 p2 ⋯]

19. ∴ AP=PΛ P-1AP= Λ=diag(λ1, λ2, ⋯)
∴If A has n linearly independent Eigenvectors, then A can be diagonalized.
Note: Not all square matrices can be diagonalized.

20. Example

23. In Matlab >> A=[2 0 1; 0 2 1; 1 1 4]; >> [P,D]=eig(A) P =; ];
>> [P,D]=eig(A)
P =
p p p3
D = λ1 λ2 λ3

24. If A does not have n linearly independent eigen-vectors
(some of the eigenvalues are identical),
then A can not be diagonalized
E.g. A=
det(λI-A)= λ4+56λ3+1152λ λ+32768
λ1=-8
λ2=-16
λ3=-16
λ4=-16
by solving (λI-A)P=0
There are only two linearly independent eigen-vectors

25. >> A=[-12 8 -4 -4; 4 -24 4 4; -3 -6 -11 3; 9 -14 9 -9]
>> [P,D]=eig(A)
P =
i i
i i
i i
D = i i

26. >>[P,J]=jordan(A)
Should use:
>>[P,J]=jordan(A)
P = J=
a 3x3 Jordan block associated with λ=-16

27. 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)

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

29. ≠ 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)] ≠

30. √ √

31. Similarity transformation
same
system
as(#)

32. Example
diagonalized
decoupled

33. Invariance:

35. Controllability:

36. Example:

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

38. Observability

39. Example:

40. 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

42. Recall linear transformation:
Controllability=being able to use u(t) to drive any state to origin in finite time
Observability=being able to computer any x(0) from observed y(t)
After transformation, eigenvalues, char. poly, char. eq, char. values, T.F., poles, zeros un-changed, but eigenvector changed

43. Controllability is invariant under transf.

45. Observability invariant under transf.

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

50. 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

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