Full-State Feedback Control

Topic #11 16.31 Feedback Control Systems State-Space Systems • Full-state Feedback Control • How do we change the poles of the state-space system? • Or, even if we can change the pole locations. • Where do we change the pole locations to? • How well does this approach work? • Reading: FPE 7.3 Fall 2010 16.30/31 11–1 Full-state Feedback Controller • Assume that the single-input system dynamics are given by ẋ(t) = Ax(t) + Bu(t) y(t) = Cx(t) so that D = 0. • The multi-actuator case is quite a bit more complicated as we would have many extra degrees of freedom. • Recall that the system poles are given by the eigenvalues of A. • Want to use the input u(t) to modify the eigenvalues of A to change the system dynamics. u(t) r A, B, C y(t) − x(t) K • Assume a full-state feedback of the form: u(t) = r − Kx(t) where r is some reference input and the gain K is R1×n • If r = 0, we call this controller a regulator • Find the closed-loop dynamics: ẋ(t) = = = y(t) = October 17, 2010 Ax(t) + B(r − Kx(t)) (A − BK)x(t) + Br Acl x(t) + Br Cx(t) Fall 2010 16.30/31 11–2 • Objective: Pick K so that Acl has the desired properties, e.g., • A unstable, want Acl stable • Put 2 poles at −2 ± 2i • Note that there are n parameters in K and n eigenvalues in A, so it looks promising, but what can we achieve? • Example #1: Consider: � ẋ(t) = 1 1 1 2 � � x(t) + � 1 u 0 • Then det(sI − A) = (s − 1)(s − 2) − 1 = s2 − 3s + 1 = 0 so the system is unstable. � � • Define u = − k1 k2 x(t) = −Kx(t), then � � � � � 1 1 1 � Acl = A − BK = − k1 k2 0 1 2 � � 1 − k1 1 − k2 = 1 2 which gives det(sI − Acl ) = s2 + (k1 − 3)s + (1 − 2k1 + k2) = 0 • Thus, by choosing k1 and k2, we can put λi(Acl ) anywhere in the complex plane (assuming complex conjugate pairs of poles). October 17, 2010 Fall 2010 16.30/31 11–3 • To put the poles at s = −5, − 6, compare the desired characteristic equation (s + 5)(s + 6) = s2 + 11s + 30 = 0 with the closed-loop one s2 + (k1 − 3)s + (1 − 2k1 + k2) = 0 to conclude that � k1 − 3 = 11 k1 = 14 1 − 2k1 + k2 = 30 k2 = 57 � � so that K = 14 57 , which is called Pole Placement. • Of course, it is not always this easy, as lack of controllability might be an issue. • Example #2: Consider this system: � � � � 1 1 1 ẋ(t) = x(t) + u 0 0 2 with the same control approach � � � � � � � 1 1 1 − k1 1 − k2 1 � Acl = A − BK = − k1 k2 = 0 0 2 0 2 so that det(sI − Acl ) = (s − 1 + k1)(s − 2) = 0 So the feedback control can modify the pole at s = 1, but it cannot move the pole at s = 2. • System cannot be stabilized with full-state feedback. • Problem caused by a lack of controllability of the e2t mode. October 17, 2010 Fall 2010 16.30/31 11–4 • Consider the basic controllability test: �� � � �� � � � � 1 1 1 1 Mc = B AB = 0 0 0 2 So that rank Mc = 1 < 2. • Modal analysis of controllability to develop a little more insight � � 1 1 A= , decompose as AV = V Λ ⇒ Λ = V −1AV 0 2 where � Λ= 1 0 0 2 � � V = 1 1 � � V −1 = 0 1 1 −1 0 � 1 Convert ẋ(t) = Ax(t) + Bu � �T where z = z1 z2 . But: � V −1B = z=V −1 x(t) −→ 1 −1 0 1 �� 1 0 ż = Λz + V −1Bu � � = 1 0 � so that the dynamics in modal form are: � � � � 1 0 1 ż = z+ u 0 0 2 • With this zero in the modal B-matrix, can easily see that the mode associated with the z2 state is uncontrollable. • Must assume that the pair (A, B) are controllable. October 17, 2010 Fall 2010 16.30/31 11–5 Ackermann’s Formula • The previous outlined a design procedure and showed how to do it by hand for second-order systems. • Extends to higher order (controllable) systems, but tedious. • Ackermann’s Formula gives us a method of doing this entire design process is one easy step. � � −1 K = 0 . . . 0 1 Mc Φd(A) � � n−1 • Mc = B AB . . . A B as before • Φd(s) is the characteristic equation for the closed-loop poles, which we then evaluate for s = A. • Note: is explicit that the system must be controllable because we are inverting the controllability matrix. • Revisit Example # 1: Φd(s) = s2 + 11s + 30 �� � � � � �� � � � � 1 1 1 1 1 1 Mc = B AB = = 0 0 1 2 0 1 So ⎞ �−1 ⎛� �2 � � � � 1 1 1 1 1 1 ⎝ K = 0 1 + 11 + 30I ⎠ 0 1 1 2 1 2 �� �� � � � � 43 14 = 0 1 = 14 57 14 57 � • Automated in Matlab: place.m & acker.m (see polyvalm.m too) October 17, 2010 Fall 2010 16.30/31 11–6 Origins of Ackermann’s Formula • For simplicity, consider third-order system (case #2 on 6–??), but this extends to any order. ⎡ ⎤ ⎡ ⎤ −a1 −a2 −a3 1 � � A=⎣ 1 0 0 ⎦ B = ⎣ 0 ⎦ C = b1 b2 b3 0 1 0 0 • See key benefit of using control canonical state-space model • This form is useful because the characteristic equation for the system is obvious ⇒ det(sI − A) = s3 + a1s2 + a2s + a3 = 0 • Can show that ⎡ ⎤ ⎡ ⎤ −a1 −a2 −a3 1 � � ⎣ ⎦ ⎣ ⎦ Acl = A − BK = 1 0 0 − 0 k1 k2 k3 0 1 0 0 ⎡ ⎤ −a1 − k1 −a2 − k2 −a3 − k3 ⎦ = ⎣ 1 0 0 0 1 0 so that the characteristic equation for the system is still obvious: Φcl (s) = det(sI − Acl ) = s3 + (a1 + k1)s2 + (a2 + k2)s + (a3 + k3) = 0 October 17, 2010 Fall 2010 16.30/31 11–7 • Compare with the characteristic equation developed from the desired closed-loop pole locations: Φd(s) = s3 + (α1)s2 + (α2)s + (α3) = 0 to get that ⎫ a1 + k1 = α1 ⎬ k1 = α1 − a1 ... ... ⎭ an + kn = αn kn = αn − an • To get the specifics of the Ackermann formula, we then: • Take an arbitrary A, B and transform it to the control canonical form (x(t) � z(t) = T −1x(t)) � Not obvious, but Mc can be used to form this T • Solve for the gains K̂ using the formulas at top of page for the state z(t) u(t) = K̂z(t) • Then switch back to gains needed for the state x(t), so that ˆ −1 ⇒ u = Kz(t) ˆ = Kx(t) K = KT • Pole placement is a very powerful tool and we will be using it for most of this course. October 17, 2010 Fall 2010 16.30/31 11–8 Reference Inputs • So far we have looked at how to pick K to get the dynamics to have some nice properties (i.e. stabilize A) • The question remains as to how well this controller allows us to track a reference command? • Performance issue rather than just stability. • Started with ẋ(t) = Ax(t) + Bu u = r − Kx(t) y = Cx(t) • For good tracking performance we want y(t) ≈ r(t) as t → ∞ • Consider this performance issue in the frequency domain. Use the final value theorem: lim y(t) = lim sY (s) t→∞ s→0 Thus, for good performance, we want sY (s) ≈ sR(s) as s → 0 ⇒ � Y (s) �� =1 R(s) �s=0 • So, for good performance, the transfer function from R(s) to Y (s) should be approximately 1 at DC. October 17, 2010 Fall 2010 16.30/31 11–9 • Example #1 continued: For the system � � � � 1 1 1 x(t) + u ẋ(t) = 0 1 2 � � y = 1 0 x(t) � � • Already designed K = 14 57 so the closed-loop system is ẋ(t) = (A − BK)x(t) + Br y = Cx(t) which gives the transfer function Y (s) = C (sI − (A − BK))−1 B R(s) � �−1 � � � � s + 13 56 1 = 1 0 0 −1 s − 2 s−2 = 2 s + 11s + 30 • Assume that r(t) is a step, then by the FVT � Y (s) �� 2 �= 1 !! = − R(s) � 30 s=0 • So our step response is quite poor! October 17, 2010 Fall 2010 16.30/31 11–10 • One solution is to scale the reference input r(t) so that u = N r − Kx(t) • N extra gain used to scale the closed-loop transfer function • Now we have ẋ(t) = (A − BK)x(t) + BN r y = Cx(t) so that Y (s) = C (sI − (A − BK))−1 BN = Gcl (s)N R(s) If we had made N = −15, then Y (s) −15(s − 2) = 2 R(s) s + 11s + 30 so with a step input, y(t) → 1 as t → ∞. • Clearly can compute N = Gcl (0) −1 � −1 = − C(A − BK) B �−1 • Note that this development assumed that r was constant, but it could also be used if r is a slowly time-varying command. October 17, 2010 Fall 2010 16.30/31 11–11 • So the steady state step error is now zero, but is this OK? • See plots – big improvement in the response, but transient a bit weird. Fig. 1: Response to step input with and without the N correction. Code: Step Response (step1.m) 1 2 3 4 5 6 7 8 9 10 11 % full state feedback for topic 13 % reference input issues % a=[1 1;1 2];b=[1 0]';c=[1 0];d=0; k=[14 57]; Nbar=−15; sys1=ss(a−b*k,b,c,d); sys2=ss(a−b*k,b*Nbar,c,d); t=[0:.025:4]; [y,t,x]=step(sys1,t); [y2,t2,x2]=step(sys2,t); 12 13 14 15 16 plot(t,y,'−−',t2,y2,'LineWidth',2);axis([0 4 −1 1.2]);grid; legend('u=r−Kx','u=Nbar r−Kx','Location','SouthEast') xlabel('time (sec)');ylabel('Y output');title('Step Response') print −dpng −r300 step1.png October 17, 2010 Fall 2010 16.30/31 11–12 Pole Placement Examples • Simple example: G(s) = 8 · 14 · 20 (s + 8)(s + 14)(s + 20) • Target pole locations −12 ± 12i, −20 Fig. 2: Response to step input with and without the N correction. Gives the desired steady-state behavior, with little difficulty! Fig. 3: Closed-loop frequency response. Clearly shows unity DC gain October 17, 2010 Fall 2010 16.30/31 11–13 • Example system with 1 unstable pole 0.94 G(s) = 2 s − 0.0297 • Target pole locations −0.25 ± 0.25i Fig. 4: Response to step input with and without the N correction. Gives the desired steady-state behavior, with little difficulty! Fig. 5: Closed-loop frequency response. Clearly shows unity DC gain October 17, 2010 Fall 2010 16.30/31 11–14 • OK, so let’s try something challenging. . . 8 · 14 · 20 G(s) = (s − 8)(s − 14)(s − 20) • Target pole locations −12 ± 12i, −20 Fig. 6: Response to step input with and without the N correction. Gives the desired steady-state behavior, with little difficulty! Fig. 7: Closed-loop frequency response. Clearly shows unity DC gain October 17, 2010 Fall 2010 16.30/31 11–15 • The worst possible. . . Unstable, NMP!! G(s) = (s − 1) (s + 1)(s − 3) • Target pole locations −1 ± i Fig. 8: Response to step input with and without the N correction. Gives the desired steady-state behavior, with little difficulty! Fig. 9: Closed-loop frequency response. Clearly shows unity DC gain October 17, 2010 Fall 2010 16.30/31 11–16 FSFB Summary • Full state feedback process is quite simple as it can be automated in Matlab using acker and/or place • With more than 1 actuator, we have more than n degrees of freedom in the control → we can change the eigenvectors as desired, as well as the poles. • The real issue now is where to put the poles. . . • And to correct the fact that we cannot usually measure the state → develop an estimator. October 17, 2010 Fall 2010 Code: Step Response (step3.m) 1 2 3 4 5 6 7 8 9 10 % Examples of pole placement with FSFB % demonstrating the Nbar modifcation to the reference command % % Jonathan How % Sept, 2010 % close all;clear all set(0,'DefaultLineLineWidth',2) set(0,'DefaultlineMarkerSize',10);set(0,'DefaultlineMarkerFace','b') set(0, 'DefaultAxesFontSize', 14);set(0, 'DefaultTextFontSize', 14); 11 12 13 14 15 % system [a,b,c,d]=tf2ss(8*14*20,conv([1 8],conv([1 14],[1 20]))); % controller gains to place poles at specified locations k=place(a,b,[−12+12*j;−12−12*j;−20]); 16 17 18 % find the feedforward gains Nbar=−inv(c*inv(a−b*k)*b); 19 20 21 sys1=ss(a−b*k,b,c,d); sys2=ss(a−b*k,b*Nbar,c,d); 22 23 24 25 t=[0:.01:1]; [y,t,x]=step(sys1,t); [y2,t2,x2]=step(sys2,t); 26 27 28 29 30 31 32 33 34 figure(1);clf plot(t,y,'−−',t2,y2,'LineWidth',2);axis([0 1 0 1.2]);grid; legend('u=r−Kx','u=Nbar r−Kx');xlabel('time (sec)');ylabel('Y output') title('Step Response') hold on plot(t2([1 end]),[.1 .1]*y2(end),'r−−'); plot(t2([1 end]),[.1 .1]*9*y2(end),'r−−'); hold off 35 36 37 38 text(.4,.6,['k= [ ',num2str(round(k*1000)/1000),' ]'],'FontSize',14) text(.4,.8,['Nbar= ',num2str(round(Nbar*1000)/1000)],'FontSize',14) export fig triple1 −pdf 39 40 41 42 43 44 45 46 47 48 49 figure(1);clf f=logspace(−1,2,400); gcl1=freqresp(sys1,f); gcl2=freqresp(sys2,f); loglog(f,abs(squeeze(gcl1)),f,abs(squeeze(gcl2)),'LineWidth',2);grid xlabel('Freq (rad/sec)') ylabel('G {cl}') title('Closed−loop Freq Response') legend('u=r−Kx','u=Nbar r−Kx') export fig triple11 −pdf 50 51 52 53 54 %%%%%%%%% % example 2 % clear all 55 56 57 58 59 [a,b,c,d]=tf2ss(8*14*20,conv([1 −8],conv([1 −14],[1 −20]))) k=place(a,b,[−12+12*j;−12−12*j;−20]) % find the feedforward gains Nbar=−inv(c*inv(a−b*k)*b); 60 61 62 sys1=ss(a−b*k,b,c,d); sys2=ss(a−b*k,b*Nbar,c,d); 63 64 65 66 t=[0:.01:1]; [y,t,x]=step(sys1,t); [y2,t2,x2]=step(sys2,t); 67 68 69 70 71 72 73 74 figure(2);clf plot(t,y,'−−',t2,y2,'LineWidth',2);axis([0 1 0 1.2]) grid; legend('u=r−Kx','u=Nbar r−Kx') xlabel('time (sec)');ylabel('Y output');title('Step Response') hold on plot(t2([1 end]),[.1 .1]*y2(end),'r−−'); October 17, 2010 16.30/31 11–17 Fall 2010 75 76 plot(t2([1 end]),[.1 .1]*9*y2(end),'r−−'); hold off 77 78 79 80 text(.4,.6,['k= [ ',num2str(round(k*1000)/1000),' ]'],'FontSize',14) text(.4,.8,['Nbar= ',num2str(round(Nbar*1000)/1000)],'FontSize',14) export fig triple2 −pdf 81 82 83 84 85 86 87 88 89 90 91 figure(2);clf f=logspace(−1,2,400); gcl1=freqresp(sys1,f); gcl2=freqresp(sys2,f); loglog(f,abs(squeeze(gcl1)),f,abs(squeeze(gcl2)),'LineWidth',2);grid xlabel('Freq (rad/sec)') ylabel('G {cl}') title('Closed−loop Freq Response') legend('u=r−Kx','u=Nbar r−Kx') export fig triple21 −pdf 92 93 94 95 %%%%%%%%%%%%%% % example 3 clear all 96 97 98 99 100 [a,b,c,d]=tf2ss(.94,[1 0 −0.0297]) k=place(a,b,[−1+j;−1−j]/4) % find the feedforward gains Nbar=−inv(c*inv(a−b*k)*b); 101 102 103 sys1=ss(a−b*k,b,c,d); sys2=ss(a−b*k,b*Nbar,c,d); 104 105 106 107 t=[0:.1:30]; [y,t,x]=step(sys1,t); [y2,t2,x2]=step(sys2,t); 108 109 110 111 112 113 114 115 116 117 figure(3);clf plot(t,y,'−−',t2,y2,'LineWidth',2);axis([0 30 0 2]) grid; legend('u=r−Kx','u=Nbar r−Kx') xlabel('time (sec)');ylabel('Y output');title('Step Response') hold on plot(t2([1 end]),[.1 .1]*y2(end),'r−−'); plot(t2([1 end]),[.1 .1]*9*y2(end),'r−−'); hold off 118 119 120 121 text(15,.6,['k= [ ',num2str(round(k*1000)/1000),' ]'],'FontSize',14) text(15,.8,['Nbar= ',num2str(round(Nbar*1000)/1000)],'FontSize',14) export fig triple3 −pdf 122 123 124 125 126 127 128 129 130 131 132 figure(3);clf f=logspace(−3,1,400); gcl1=freqresp(sys1,f); gcl2=freqresp(sys2,f); loglog(f,abs(squeeze(gcl1)),f,abs(squeeze(gcl2)),'LineWidth',2);grid xlabel('Freq (rad/sec)') ylabel('G {cl}') title('Closed−loop Freq Response') legend('u=r−Kx','u=Nbar r−Kx') export fig triple31 −pdf 133 134 135 136 %%%%%%%%%%%% % example 4 clear all 137 138 139 140 141 [a,b,c,d]=tf2ss([1 −1],conv([1 1],[1 −3])) k=place(a,b,[[−1+j;−1−j]]) % find the feedforward gains Nbar=−inv(c*inv(a−b*k)*b); 142 143 144 sys1=ss(a−b*k,b,c,d); sys2=ss(a−b*k,b*Nbar,c,d); 145 146 147 148 t=[0:.1:10]; [y,t,x]=step(sys1,t); [y2,t2,x2]=step(sys2,t); 149 150 151 figure(3);clf plot(t,y,'−−',t2,y2,'LineWidth',2);axis([0 10 −1 1.2]) October 17, 2010 16.30/31 11–18 Fall 2010 152 153 154 155 156 157 158 159 grid; legend('u=r−Kx','u=Nbar r−Kx') xlabel('time (sec)');ylabel('Y output') title('Unstable, NMP system Step Response') hold on plot(t2([1 end]),[.1 .1]*y2(end),'r−−'); plot(t2([1 end]),[.1 .1]*9*y2(end),'r−−'); hold off 160 161 162 163 text(5,.6,['k= [ ',num2str(round(k*1000)/1000),' ]'],'FontSize',14) text(5,.8,['Nbar= ',num2str(round(Nbar*1000)/1000)],'FontSize',14) export fig triple4 −pdf 164 165 166 167 168 169 170 171 172 173 174 figure(4);clf f=logspace(−2,2,400); gcl1=freqresp(sys1,f); gcl2=freqresp(sys2,f); loglog(f,abs(squeeze(gcl1)),f,abs(squeeze(gcl2)),'LineWidth',2);grid xlabel('Freq (rad/sec)') ylabel('G {cl}') title('Closed−loop Freq Response') legend('u=r−Kx','u=Nbar r−Kx') export fig triple41 −pdf October 17, 2010 16.30/31 11–19