1 / 22

State space model

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.

nita
Download Presentation

State space model

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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+ A2t2+ 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=4 a3 a2 a1 a0 b1 b0=b2=b3=0

  7. >> n=[1 2 3];d=[1 4 5 6]; >> [A,B,C,D]=tf2ss(n,d) A = -4 -5 -6 1 0 0 0 1 0 B = 1 0 0 C = 1 2 3 D = 0 >> 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

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

  10. 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)

  11. 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

  12. In Matlab: >> A=[0 1;-2 -3]; >> B=[0;1]; >> C=[1 3]; >> D=[0]; >> [n,d]=ss2tf(A,B,C,D) n = 0 3.0000 1.0000 d = 1 3 2 >>tf(n,d)

  13. 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

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

  15. 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 Let , ∴p2 is also an e-vector, assoc. with the λ =-2

  16. 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 p1 p2 p3 D =1.2679 0 0 0 2.0000 0 0 0 4.7321 λ1 λ2 λ3

  17. If we use [P,D]=eig(A) get approximate but wrong answer 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 assoc. w/. λ=-16

  18. 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

  19. 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 ⋯]

  20. ∴ 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.

  21. 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+1152λ2+10240λ+32768 λ1=-8 λ2=-16 λ3=-16 λ4=-16 by solving (λI-A)P=0

More Related