Topic #13 16.31 Feedback Control State-Space Systems e Ful-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? Copyright 2001 by Jonathan How
Topic #13 16.31 Feedback Control 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? Copyright 2001 by Jonathan How. 1
Fall 2001 16.3113-1 Full-state Feedback Controller Assume that the single-input system dynamics are given by Ax+ Bu 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 . Assume a full-state feedback of the form K where r is some reference input and the gain K is R xn If r=0, we call this controller a regulator Find the closed-loop dynamics Ac+B(r-kr) (A-BK)C+ B Act+ Br
Fall 2001 16.31 13–1 Full-state Feedback Controller • Assume that the single-input system dynamics are given by x˙ = Ax + Bu y = Cx 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. • Assume a full-state feedback of the form: u = r − Kx 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: x˙ = Ax + B(r − Kx) = (A − BK)x + Br = Aclx + Br y = Cx
Fall 2001 16.3113-2 Objective: Pick K so that Ad has the desired properties, e. g A unstable, want Ac stable Put2 poles at-2±2j it looks promising, but what can we achieve> eig Note that there are n parameters in K and n eigenvalues in A,so Example #1: Consider .+ 12 Ther det(I-A)=(s-1)(s-2)-1=s2-3s+1=0 so the system is unstable -Define u=-k1 k2 I Ka. then Ac=A-BK= 12-0[]=1一1一k2 11 2 So then we have that det(I-Aa)=s2+(k1-3)s+(1-2k1+k2)=0 Thus, by choosing k1 and k2, we can put Ai(Ad) anywhere in he complex plane(assuming complex conjugate pairs of poles
Fall 2001 16.31 13–2 • Objective: Pick K so that Acl has the desired properties, e.g., – A unstable, want Acl stable – Put 2 poles at −2 ± 2j • 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: x˙ = 1 1 1 2 x + 1 0 u – Then det(sI − A)=(s − 1)(s − 2) − 1 = s2 − 3s +1=0 so the system is unstable. – Define u = − k1 k2 x = −Kx, then Acl = A−BK = 1 1 1 2 − 1 0 k1 k2 = 1 − k1 1 − k2 1 2 – So then we have that 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).
Fall 2001 6.3113-3 To put the poles at s=-5,-6, compare the desired characteristic equation (s+5)(s+6)=s2+11s+30=0 ith the closed-loop one +(k1-3)x+(1-2k1+k2)=0 to conclude that k1-3=11 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 the issue of controllability must be addressed Example #2: Consider this system C+ 02 0 with the same control approach Ad=A- BK 11 02 [k1k2] 1-k11-k2 that det(s-Aa)=(s-1+k)(s-2)=0 So the feedback control can modify the pole at s= 1, but it cannot move the pole at s=2 The system cannot be stabilized with full-state feed back control
Fall 2001 16.31 13–3 • 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)x + (1 − 2k1 + k2)=0 to conclude that k1 − 3 = 11 1 − 2k1 + k2 = 30 k1 = 14 k2 = 57 so that K = 14 57 , which is called Pole Placement. • Of course, it is not always this easy, as the issue of controllability must be addressed. • Example #2: Consider this system: x˙ = 1 1 0 2 x + 1 0 u with the same control approach Acl = A − BK = 1 1 0 2 − 1 0 k1 k2 = 1 − k1 1 − k2 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. • The system cannot be stabilized with full-state feedback control.
Fall 2001 16.3113-4 What is the reason for this problem? It is associated with loss of controllability of the e2t mode Consider the basic controllability test BAB 02|0 So that rank M=1<2 Consider the modal test to develop a little more insight 02/, decompose as AV=VA=A=V-AV he where 11 01 Convert A Bu A+v- Bu where z=[21 2272 But V-B 0 so that the dynamics in modal form are 10 0 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
Fall 2001 16.31 13–4 • What is the reason for this problem? – It is associated with loss of controllability of the e2t mode. • Consider the basic controllability test: Mc = B AB = 1 0 1 1 0 2 1 0 So that rank Mc = 1 < 2. • Consider the modal test to develop a little more insight: A = 1 1 0 2 , decompose as AV = V Λ ⇒ Λ = V −1 AV where Λ = 1 0 0 2 V = 1 1 0 1 V −1 = 1 −1 0 1 Convert x˙ = Ax + Bu z=V −1x −→ z˙ = Λz + V −1 Bu where z = z1 z2 T . But: V −1 B = 1 −1 0 1 1 0 = 1 0 so that the dynamics in modal form are: z˙ = 1 0 0 2 z + 1 0 u • 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.