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

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Presentation on theme: "State space model: linear: or in some text: where: u: input y: output x: state vector A, B, C, D are const matrices."— Presentation transcript:

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

3

4 State transition, matrix exponential

5 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)

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=4 a 3 a 2 a 1 a 0 b 1 b 0 =b 2 =b 3 =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. s 1 =-1,s 2 =-2,s 3 =-2 are char. values or eigenvalues

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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) -1 BU(s)+(sI-A) -1 x(0) x(t)=( ℒ -1 (sI-A) -1 ))*Bu(t)+ ℒ -1 (sI-A) -1 ) x(0) x(t)= e A(t- τ ) Bu( τ )d τ +e At x(0) y(t)= Ce A(t- τ ) Bu( τ )d τ +Ce At x(0)+Du(t)

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

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, ∴ p 1 is an e-vector, & the e-value=1 Let, ∴ p 2 is also an e-vector, assoc. with the λ =-2

17 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 ⇒ λ 1 =1, λ 2 =-2 Eigenvalues, eigenvectors

18 If λ 1 ≠λ 2 ≠λ 3 ⋯ then the corresponding p 1, p 2, ⋯ will be linearly independent, i.e., the matrix P=[p 1 ⋮ p 2 ⋮ ⋯ p n ] will be invertible. Then: Ap 1 = λ 1 p 1 Ap 2 = λ 2 p 2 ⋮ A[p 1 ⋮ p 2 ⋮ ⋯ ]=[Ap 1 ⋮ Ap 2 ⋮ ⋯ ] =[λ 1 p 1 ⋮ λ 2 p 2 ⋮ ⋯ ] =[p 1 p 2 ⋯ ]

19 ∴ AP=PΛ P -1 AP= Λ=diag(λ 1, λ 2, ⋯ ) ∴ If A has n linearly independent Eigenvectors, then A can be diagonalized. Note: Not all square matrices can be diagonalized.

20 Example

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23 In Matlab >> A=[2 0 1; 0 2 1; 1 1 4]; >> [P,D]=eig(A) P = 0.6280 0.7071 0.3251 0.6280 -0.7071 0.3251 -0.4597 -0.0000 0.8881 p 1 p 2 p 3 D = 1.2679 0 0 0 2.0000 0 0 0 4.7321 λ1λ1 λ2λ2 λ3λ3

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)= λ 4 +56λ 3 +1152λ 2 +10240λ+32768 λ 1 =-8 λ 2 =-16 λ 3 =-16 λ 4 =-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] A = -12 8 -4 -4 4 -24 4 4 -3 -6 -11 3 9 -14 9 -9 >> [P,D]=eig(A) P = -0.7071 0.0000 + 0.0000i 0.0000 - 0.0000i 0.0000 -0.0000 0.4472 - 0.0000i 0.4472 + 0.0000i -0.4472 0.7071 -0.0000 + 0.0000i -0.0000 - 0.0000i -0.0000 -0.0000 0.8944 0.8944 -0.8944 D = -8.0000 0 0 0 0 -16.0000 + 0.0000i 0 0 0 0 -16.0000 - 0.0000i 0 0 0 0 -16.0000

26 Should use: >>[P,J]=jordan(A) P = 0.3750 0 1 0.625 0 8 4 0 -0.375 0 0 0.375 0 16 9 0 J= -8 0 0 0 0 -16 1 0 0 0 -16 1 0 0 0 -16 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) ans = (s+2)*(s+1)

28 >> [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

29 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)] ≠

30 √ √

31 Similarity transformation same system as(#)

32 Example diagonalized decoupled

33 Invariance:

34


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