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.
Fall 2001 16.3113-5 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 K=[0…01]Ma(A) nere M=[BAB…A-B] pa(s) is the characteristic equation for the closed-loop poles which we then evaluate for s= A It is explicit that the system must be controllable because we are inverting the controllability matrix · Revisit example#1:Φa(s)=s2+11+30 A=1:211]-1 11 11 +11 +301 12 4314 1457 Automated in Matlab: place. m &z acker. m(see polyvalm m too
Fall 2001 16.31 13–5 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. K = 0 ... 0 1 M−1 c Φd(A) where – Mc = B AB . . . An−1B – Φd(s) is the characteristic equation for the closed-loop poles, which we then evaluate for s = A. – It is explicit that the system must be controllable because we are inverting the controllability matrix. • Revisit example # 1: Φd(s) = s2 + 11s + 30 Mc = B AB = 1 0 1 1 1 2 1 0 = 1 1 0 1 So K = 0 1 1 1 0 1 −1 1 1 1 2 2 + 11 1 1 1 2 + 30I = 0 1 43 14 14 57 = 14 57 • Automated in Matlab: place.m & acker.m (see polyvalm.m too)
Fall 2001 16.3113-6 . Where did this formula come from? For simplicity, consider a third-order system(case #2), but this extends to any order 100B b1 b2b3 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)=s+a1s+a2S+a3=0 ● Can show that Ac=A-BK 100 0k k2 ks 0 a1-k1-a2-k2-a3-k3 0 o that the characteristic equation for the system is still obvious 重a(s)=det(s-Aa) s3+(a1+k1)s2+(a2+k2)s+(a3+ka)=0
Fall 2001 16.31 13–6 • Where did this formula come from? • For simplicity, consider a third-order system (case #2), but this extends to any order. A = −a1 −a2 −a3 100 010 B = 1 0 0 C = b1 b2 b3 – 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 Acl = A − BK = −a1 −a2 −a3 100 010 − 1 0 0 k1 k2 k3 = −a1 − k1 −a2 − k2 −a3 − k3 100 010 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
Fall 2001 163113-7 We then compare this with the desired characteristic equation de- veloped from the desired closed-loop pole locations 重(s)=s3+(a1)s2+(a2)s+(a3)=0 to get that +k1 k1 1 +k To get the specifics of the Ackermann formula, we then Take an arbitrary A, B and transform it to the control canonical form(az=T- r) Solve for the gains K using the formulas above for the state z (=Kz) Then switch back to gains needed for the state so that K- KT-I (u=Kz=Kr) Pole placement is a very powerful tool and we will be using it for most of this course
Fall 2001 16.31 13–7 • We then compare this with the desired 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 . . . an + kn = αn k1 = α1 − a1 . . . 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 ❀ z = T −1x) – Solve for the gains Kˆ using the formulas above for the state z (u = Kz ˆ ) – Then switch back to gains needed for the state x, so that K = KTˆ −1 (u = Kz ˆ = Kx) • Pole placement is a very powerful tool and we will be using it for most of this course
