Topic # /31 Feedback Control Systems

Transcription

1 Topic # /31 Feedback Control Systems State-Space Systems What are the basic properties of a state-space model, and how do we analyze these? Time Domain Interpretations System Modes

2 Fall / Time Response Can develop a lot of insight into the system response and how it is modeled by computing the time response x(t) Homogeneous part Forced solution Homogeneous Part Take Laplace transform so that But can show ẋ = Ax, x(0) known X(s) = (si A) 1 x(0) x(t) = L 1 (si A) 1 x(0) I A A 2 (si A) 1 = s s2 s3 1 so L 1 (si A) 1 = I + At + (At) ! At = e x(t) = e At x(0) e At is a special matrix that we will use many times in this course Transition matrix or Matrix Exponential Calculate in MATLAB using expm.m and not exp.m 1 Note that e (A+B)t = e At e Bt iff AB = BA 1 MATLAB is a trademark of the Mathworks Inc.

3 Fall / Example: ẋ = Ax, with A = (si A) 1 = = = At e = [ s 2 1 s + 3 ] 1 s s (s + 2)(s + 1) 2 s s s s s s s s + 2 2e t e 2t e t e 2t 2e t + 2e 2t e t + 2e 2t We will say more about e At when we have said more about A (eigenvalues and eigenvectors) Computation of e At = L 1 {(si A) 1 } straightforward for a 2-state system More complex for a larger system, see this paper

4 Fall / Forced Solution Consider a scalar case: SS: Forced Solution ẋ = ax + bu, x(0) given t x(t) = e at x(0) + e a(t τ) bu(τ)dτ 0 where did this come from? 1. ẋ ax = bu 2. e at [ẋ ax] = d (e at x(t)) = e at dt bu(t) t 3. d e aτ x(τ)dτ = e at x(t) x(0) = t e aτ bu(τ)dτ 0 dτ 0 Forced Solution Matrix case: ẋ = Ax + Bu where x is an n-vector and u is a m-vector Just follow the same steps as above to get t x(t) = e At x(0) + e A(t τ) Bu(τ )dτ and if y = Cx + Du, then y(t) = Ce At x(0) + 0 t 0 Ce A(t τ) Bu(τ)dτ + Du(t) Ce At x(0) is the initial response Ce A(t) B is the impulse response of the system.

5 Fall / Have seen the key role of e At in the solution for x(t) Determines the system time response But would like to get more insight! Consider what happens if the matrix A is diagonalizable, i.e. there exists a T such that λ 1 T 1 AT = Λ which is diagonal Λ =... Then e At = T e Λt T 1 where e λ 1 t Λt e =... e λ n t λ n Follows since e At = I + At + 1 (At) and that A = T ΛT 1 2!, so we can show that 1 e At = I + At + (At) ! 1 = I + T ΛT 1 t + (T ΛT 1 t) ! = T e Λt T 1 This is a simpler way to get the matrix exponential, but how find T and λ? Eigenvalues and Eigenvectors

6 Fall / Eigenvalues and Eigenvectors Recall that the eigenvalues of A are the same as the roots of the characteristic equation (page 6 7) λ is an eigenvalue of A if det(λi A) = 0 which is true iff there exists a nonzero v (eigenvector) for which (λi A)v = 0 Av = λv Repeat the process to find all of the eigenvectors. Assuming that the n eigenvectors are linearly independent Av i = λ i v i i = 1,..., n λ 1 A v 1 v n = v 1 v n... AT = T Λ T 1 AT = Λ λ n

7 Fall / Jordan Form One word of caution: Not all matrices are diagonalizable A = det(si A) = s 0 0 only one eigenvalue s = 0 (repeated twice). The eigenvectors solve 0 1 r 1 = r 2 r eigenvectors are of the form 1, r 1 = 0 would only be one. 0 Need Jordan Form to handle the case with repeated roots 2 Jordan form of matrix A R n n is block diagonal: A λ j A 0 λ j 1 0 A = 2 0 with A j = A k λ j Observation: any matrix can be transformed into Jordan form with the eigenvalues of A determining the blocks A j. The matrix exponential of a Jordan form matrix is then given by 1 t t 2 t /2! n 1 (n 1)! e A 1t 0 0 t n 2 0 e A 2t t (n 2)! At e = with e A j t.. = e.. A k t... t 0 0 e λ j t 2 see book by Strang, Matlab command Jordan, or here

8 Fall / EV Mechanics 1 1 Consider A = 8 5 s (si A) = 8 s 5 det(si A) = (s + 1)(s 5) + 8 = s 2 4s + 3 = 0 so the eigenvalues are s 1 = 1 and s 2 = 3 Eigenvectors (si A)v = 0 s v (s 1 I A)v 1 = 11 = 0 8 s 5 v 21 s=1 2 1 v 11 = 0 2v 8 v 11 v 21 = 0, v 21 = 2v v 11 is then arbitrary (= 0), so set v 11 = 1 1 v 1 = 2 (s 2 I A)v 2 = 4 1 v 12 = 0 4v 12 v 22 = 0, v 22 = 4v v 22 Confirm that Av i = λ i v i 1 v 2 = 4

9 Fall / Dynamic Interpretation Since A = T ΛT 1, then e λ 1t T w 1 e At = T e Λt T 1 = v 1 v n.... e λ nt w T where we have written T w 1 T 1 =. w which is a column of rows. Multiply this expression out and we get that T n n e = e v i w i At λ i t T i=1 n Assume A diagonalizable, then ẋ = Ax, x(0) given, has solution x(t) = e At x(0) = T e Λt T 1 x(0) n = e λit T v i {w i x(0)} i=1 n λ i t = e v i β i i=1 State solution is linear combination of the system modes v i e λ it e λ it Determines nature of the time response v i Determines how each state contributes to that mode β i Determines extent to which initial condition excites the mode

10 Fall / Note that the v i give the relative sizing of the response of each part of the state vector to the response. 1 v 1 (t) = e t mode v 2 (t) = e 3t mode Clearly e λ it gives the time modulation λ i real growing/decaying exponential response λ i complex growing/decaying exponential damped sinusoidal Bottom line: The locations of the eigenvalues determine the pole locations for the system, thus: They determine the stability and/or performance & transient behavior of the system. It is their locations that we will want to modify when we start the control work

11 Fall / Diagonalization with Complex Roots If A has complex conjugate eigenvalues, the process is similar but a little more complicated. Consider a 2x2 case with A having eigenvalues a ± bi and associated eigenvectors e 1, e 2, with e 2 = ē 1. Then a + bi 0 1 A = e 1 e 2 e 1 e 2 0 a bi a + bi 0 1 = e 1 ē 1 T DT 1 ē 1 e 1 0 a bi Now use the transformation matrix 1 i M M = 0.5 = 1 i i i Then it follows that A = T DT 1 = (T M)(M 1 DM)(M 1 T 1 ) = (T M)(M 1 DM)(T M) 1 which has the nice structure: [ ] a b [ A = Re(e 1 ) Im(e 1 ) b a where all the matrices are real. Re(e 1 ) Im(e 1 ) ] 1 With complex roots, diagonalization is to a block diagonal form.

12 Fall / For this case we have that [ ] cos(bt) sin(bt) [ e At = Re(e 1 ) Im(e 1 ) e at sin(bt) cos(bt) ] Re(e 1 ) Im(e 1 ) 1 Note that [ ] Re(e 1 ) Im(e 1 ) Re(e 1 ) Im(e 1 ) = 0 1 So for an initial condition to excite just this mode, can pick x(0) = [Re(e 1 )], or x(0) = [Im(e 1 )] or a linear combination. Example x(0) = [Re(e 1 )] [ ] cos(bt) sin(bt) x(t) = e At x(0) = Re(e 1 ) Im(e 1 ) e at sin(bt) cos(bt) Re(e Im(e 1 ) 1 1 ) [Re(e 1 )] [ Im(e 1 ) ] e at cos(bt) sin(bt) 1 = Re(e 1 ) sin(bt) cos(bt) 0 [ ] cos(bt) = e at Re(e 1 ) Im(e 1 ) sin(bt) = e at (Re(e 1 ) cos(bt) Im(e 1 ) sin(bt)) which would ensure that only this mode is excited in the response

13 Fall / Example: Spring Mass System Classic example: spring mass system consider simple case first: m i = 1, and k i = 1 k 5 Wall k 1 k m 2 k 1 m 3 k 3 m 4 2 Wall x A K = = = [ z1 z 2 z 3 ż 1 ż 2 ż 3 ] [ ] 0 I M 1 M = diag(m K 0 i ) k 1 + k 2 + k 5 k 5 k 2 k 5 k 3 + k 4 + k 5 k 3 k 2 k 3 k 2 + k 3 Eigenvalues and eigenvectors of the undamped system λ 1 = ±0.77i λ 2 = ±1.85i λ 3 = ±2.00i v 1 v 2 v ±0.77i ±0.77i ±1.08i ±1.85i ±1.85i 2.61i ±2.00i 2.00i 0.00

14 Fall / Initial conditions to excite just the three modes: x i (0) = α 1 Re(v i ) + α 2 Im(v i ) α j R Simulation using α 1 = 1, α 2 = 0 Visualization important for correct physical interpretation Mode 1 λ 1 = ±0.77i m 1 m 3 m 2 Lowest frequency mode, all masses move in same direction Middle mass has higher amplitude motions z 3, motions all in phase

15 Fall / Mode 2 λ 2 = ±1.85i m 1 m 3 m 2 Middle frequency mode has middle mass moving in opposition to two end masses Again middle mass has higher amplitude motions z 3

16 Fall / Mode 3 λ 3 = ±2.00i m 1 m 3 m 2 Highest frequency mode, has middle mass stationary, and other two masses in opposition Eigenvectors that correspond to more constrained motion of the system are associated with higher frequency eigenvalues

17 Fall / Result if we use a random input is a combination of all three modes

18 Fall / Code: Simulation of Spring Mass System Modes % Simulate modal response for a spring mass system % Jonathan How, MIT % Fall 2009 alp1=1; % weighting choice on the IC m=eye(3); % mass k=[3 1 1; 1 3 1; 1 1 2]; % stiffness a=[m*0 eye(3); inv(m)*k m*0]; [v,d]=eig(a); t=[0:.01:20]; l1=1:25:length(t); G=ss(a,zeros(6,1),zeros(1,6),0); % use the following to cll the function above close all set(0, 'DefaultAxesFontSize', 14, 'DefaultAxesFontWeight','demi') set(0, 'DefaultTextFontSize', 14, 'DefaultTextFontWeight','demi') set(0,'defaultaxesfontname','arial'); set(0,'defaulttextfontname','arial');set(0,'defaultlinemarkersize',10) figure(1);clf x0=alp1*real(v(:,1))+(1 alp1)*imag(v(:,1)) [y,t,x]=lsim(g,0*t,t,x0); plot(t,x(:,1),' ','LineWidth',2);hold on plot(t(l1),x(l1,2),'rx','linewidth',2) plot(t,x(:,3),'m ','LineWidth',2);hold off xlabel('time');ylabel('displacement') title(['mode with \lambda=', num2str(imag(d(1,1))),' i']) legend('z1','z2','z3') print dpng r300 v1.png figure(2);clf x0=alp1*real(v(:,5))+(1 alp1)*imag(v(:,5)) [y,t,x]=lsim(g,0*t,t,x0); plot(t,x(:,1),' ','LineWidth',2);hold on plot(t(l1),x(l1,2),'rx','linewidth',2) plot(t,x(:,3),'m ','LineWidth',2);hold off xlabel('time');ylabel('displacement') title(['mode with \lambda=', num2str(imag(d(5,5))),' i']) legend('z1','z2','z3') print dpng r300 v3.png figure(3);clf x0=alp1*real(v(:,3))+(1 alp1)*imag(v(:,3)) [y,t,x]=lsim(g,0*t,t,x0); plot(t,x(:,1),' ','LineWidth',2);hold on plot(t,x(:,2),'r.','linewidth',2) plot(t,x(:,3),'m ','LineWidth',2);hold off xlabel('time');ylabel('displacement') title(['mode with \lambda=', num2str(imag(d(3,3))),' i']) legend('z1','z2','z3') print dpng r300 v2.png figure(4);clf x0=[ ]'; [y,t,x]=lsim(g,0*t,t,x0); plot(t,x(:,1),' ','LineWidth',2) hold on plot(t,x(:,2),'r.','linewidth',2) plot(t,x(:,3),'m ','LineWidth',2) hold off xlabel('time');ylabel('displacement') title(['random Initial Condition']) legend('z1','z2','z3') print dpng r300 v4.png

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