1. State space model State space model: linear: where: u: input y: output
x: state vector
A,B,C,D, or F,G,H,J are const matrices
If dim(x) = n, we say system order = n
A is nxn matrix, called system matrix
System property largely determined by properies of A.

2. 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 = Inxn+At+ A2t A3t3 +
eAt is invertible: (eAt)-1= e(-A)t
eA0=I
eAt1 eAt2= eA(t1+t2)
def

3. Example

4. Example

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

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

7. >> n=[1 2 3];d=[ ];
>> [A,B,C,D]=tf2ss(n,d)
A =
B = 1
C =
D =
>> tf(n,d)
Transfer function:
s^2 + 2 s + 3
s^3 + 4 s^2 + 5 s + 6

8. 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

10. can ?
set t=0
∴No
can ? √
at t=0: ? √

11. 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)

12. S.S to T.F. X(s)=(sI-A)-1BU(s) Y(s)=C(sI-A)-1BU(s)+DU(s)
=(D+ C(sI-A)-1B)U(s)
∴ T.F. H(s)= D+ C(sI-A)-1B
In matlab: ss2tf
eig
roots
poly
use help to find out how to use these

13. In Matlab:
>> A=[0 1;-2 -3];
>> B=[0;1];
>> C=[1 3];
>> D=[0];
>> [n,d]=ss2tf(A,B,C,D)
n =
d =
>>tf(n,d)

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.
u= unit step

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

16. Eigenvalues, eigenvectors
Given a nxn square matrix A, p is an eigenvector of A if Ap∝p
i.e. λ s.t. Ap= λp
λis 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. In Matlab >> A=[2 0 1; 0 2 1; 1 1 4]; >> [P,D]=eig(A)
p p p3
D = λ1 λ2 λ3

18. If we use [P,D]=eig(A)
get approximate but wrong answer
Should use:
>>[P,J]=jordan(A)
P = J=
a 3x3 Jordan block assoc. w/. λ=-16

19. In general, if λ, P is an e-pair for A,
AP= λP
λP-AP=0
λIP-AP=0
(λI-A)P=0
∵ P≠0 ∴ det(λI-A)=0
∴ λ is a sol. of char. eq of A
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

20. If λ1 ≠λ2 ≠λ3⋯
then the corresponding P1, P2, ⋯ will be linearly independent, i.e., the matrix
P=[P1⋮P2 ⋮ ⋯Pn] will be invertible.
AP1= λ1P1
AP2= λ2P2 ⋮
A[P1⋮P2 ⋮ ⋯]=[AP1⋮AP2 ⋮ ⋯]
=[λ P1⋮ λ P2 ⋮ ⋯]
=[P1 P2 ⋯]

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

22. If A does not have n lin. ind. e-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