Topic #17 16.31 Feedback Control tate-Space Systems Closed-loop control using estimators and regulators Dynamics output feedback “ Back to reality' Copyright[2001by JOnathan dHow.D
Topic #17 16.31 Feedback Control State-Space Systems • Closed-loop control using estimators and regulators. • Dynamics output feedback • “Back to reality” Copyright 2001 by Jonathan How. 1
Fall 2001 16.3117-1 Combined Estimators and Regulators Can now evaluate the stability and or performance of a controller when we design K assuming that u =-Ka, but we implement Assume that we have designed a closed-loop estimator with gain L a (t)=Ai(t)+ Bu(t)+l(y-y u(t)=Ci( Then we have that the closed-loop system dynamics are given by i(t)=Ax(t)+ Bu(t) (t)= Ac(t)+ Bu(t)+L(y-g t)=Ca(t) cal Which can be compactly written as BK LC A-BK-LCIi cl clcl This does not look too good at this point- not even obvious that the closed-system is stable 入(Aa)
Fall 2001 16.31 17—1 Combined Estimators and Regulators • Can now evaluate the stability and/or performance of a controller when we design K assuming that u = −Kx, but we implement u = −Kxˆ • Assume that we have designed a closed-loop estimator with gain L ˙ xˆ(t) = Axˆ(t) + Bu(t) + L(y − yˆ) yˆ(t) = Cxˆ(t) • Then we have that the closed-loop system dynamics are given by: x˙(t) = Ax(t) + Bu(t) ˙ xˆ(t) = Axˆ(t) + Bu(t) + L(y − yˆ) y(t) = Cx(t) yˆ(t) = Cxˆ(t) u = −Kxˆ • Which can be compactly written as: ∙ x˙ ˙ xˆ ¸ = ∙ A −BK LC A − BK − LC ¸ ∙ x xˆ ¸ ⇒ x˙ cl = Aclxcl • This does not look too good at this point — not even obvious that the closed-system is stable. λi(Acl) =??
Fall 2001 16.3117-2 Can fix this problem by introducing a new variable a a and then converting the closed-loop system dynamics using the similarity transformation T △ I 0 I-1| Note that T=T-1 Now rewrite the system dynamics in terms of the state cl TAct=A Note that similarity transformations preserve the eigenvalues, so we are guaranteed that 入(A4)≡A(Aa . Work through the math BK Acl I 0 A Ⅰ-1LCA-BK-LCI-I A BK I 0 A-LC-A+LCII-I A-BK BK 0 A-LC Because Ac is block upper triangular, we know that the closed DOD poles of the system are given by det(sI -Ac)= det(sI -(A- BK). det(sI-(A-LC))=0
Fall 2001 16.31 17—2 • Can fix this problem by introducing a new variable ˜x = x − xˆ and then converting the closed-loop system dynamics using the similarity transformation T x˜cl , ∙ x x˜ ¸ = ∙ I 0 I −I ¸ ∙ x xˆ ¸ = T xcl — Note that T = T −1 • Now rewrite the system dynamics in terms of the state ˜xcl Acl ⇒ T AclT −1 , A¯cl — Note that similarity transformations preserve the eigenvalues, so we are guaranteed that λi(Acl) ≡ λi(A¯cl) • Work through the math: A¯cl = ∙ I 0 I −I ¸ ∙ A −BK LC A − BK − LC ¸ ∙ I 0 I −I ¸ = ∙ A −BK A − LC −A + LC ¸ ∙ I 0 I −I ¸ = ∙ A − BK BK 0 A − LC ¸ • Because A¯cl is block upper triangular, we know that the closed-loop poles of the system are given by det(sI − A¯cl) , det(sI − (A − BK)) · det(sI − (A − LC)) = 0
Fall 2001 163117-3 Observation: The closed-loop poles for this system con- sist of the union of the regulator poles and estimator poles So we can just design the estimator/regulator separately and com- bine them at the end Called the Separation Principle Just keep in mind that the pole locations you are picking for these two sub-problems will also be the closed-loop pole locations Note: the separation principle means that there will be no ambi- guity or uncertainty about the stability and or performance of the close d-loop system The closed-loop poles will be exactly where you put them! And we have not even said what compensator does this amazin accomplishment!!!
Fall 2001 16.31 17—3 • Observation: The closed-loop poles for this system consist of the union of the regulator poles and estimator poles. • So we can just design the estimator/regulator separately and combine them at the end. — Called the Separation Principle. — Just keep in mind that the pole locations you are picking for these two sub-problems will also be the closed-loop pole locations. • Note: the separation principle means that there will be no ambiguity or uncertainty about the stability and/or performance of the closed-loop system. — The closed-loop poles will be exactly where you put them!! — And we have not even said what compensator does this amazing accomplishment!!!
Fall 2001 6.3117-4 The Compensator e Dynamic Output Feedback Compensator is the combina- tion of the regulator and estimator using u=-K3 i(t)=Ai(t)+ Bu(t)+L(y-y) A i(t)-BK2+L(y-Ca =a(t)=(A-BK-LCi(t)+Ly F Rewrite with new state c= S c= Acrc+ Bcy C where the compensator dynamics are given by A≌A-BK-LC,B全L,C≌K Note that the compensator maps sensor measurements to ac tutor commands, as expected Closed-loop system stable if regulator/estimator poles placed in the LHP, but compensator dynamics do not need to be stable Ai(A- BK-LC)=??
Fall 2001 16.31 17—4 The Compensator • Dynamic Output Feedback Compensator is the combination of the regulator and estimator using u = −Kxˆ ˙ xˆ(t) = Axˆ(t) + Bu(t) + L(y − yˆ) = Axˆ(t) − BKxˆ + L(y − Cxˆ) ⇒ ˙ xˆ(t)=(A − BK − LC)ˆx(t) + Ly u = −Kxˆ • Rewrite with new state xc ≡ xˆ x˙ c = Acxc + Bcy u = −Ccxc where the compensator dynamics are given by: Ac , A − BK − LC , Bc , L, Cc , K — Note that the compensator maps sensor measurements to actuator commands, as expected. • Closed-loop system stable if regulator/estimator poles placed in the LHP, but compensator dynamics do not need to be stable. λi(A − BK − LC) =??
Fall 2001 16.3117-5 For consistency in the implementation with the classical approaches define the compensator transfer function so that Gc(sy From the state-space model of the compensator U(s) Cc(sI-AcB K(sI-(A-BK-LCDL Gc(s)=Cc(SI-Ac-B Note that it is often very easy to provide classical interpretations (such as lead lag)for the compensator Gc(s) e One way to implement this compensator with a reference command (t)is to change the feedback to be on e(t)=r(t-y(t)rather than just -y(t e ul Ge(s Gc(se=gas(r-y So we still have u=-Gc(sy if r=0 Intuitively appealing because it is the same approach used for the classical control, but it turns out not to be the best approach More on this later
Fall 2001 16.31 17—5 • For consistency in the implementation with the classical approaches, define the compensator transfer function so that u = −Gc(s)y — From the state-space model of the compensator: U(s) Y (s) , −Gc(s) = −Cc(sI − Ac) −1 Bc = −K(sI − (A − BK − LC))−1 L ⇒ Gc(s) = Cc(sI − Ac) −1Bc • Note that it is often very easy to provide classical interpretations (such as lead/lag) for the compensator Gc(s). • One way to implement this compensator with a reference command r(t) is to change the feedback to be on e(t) = r(t) − y(t) rather than just −y(t) Gc(s) G(s) - - 6 — re y u ⇒ u = Gc(s)e = Gc(s)(r − y) — So we still have u = −Gc(s)y if r = 0. — Intuitively appealing because it is the same approach used for the classical control, but it turns out not to be the best approach. More on this later
Fall 2001 6.3117-6 Mechanics ● basics: y, y Gcs): ic= Acic+ C G(s) A Bu · Loop dynamics l=G(S)G(s)→y=L(s)e A +bc +A +B e ABC1「x 0 A B C O To "close the loop, note that e=r-y, then A BC 0 C 0 0 A B A BC Bc A B C 0 Ac is not exactly the same as on page 17-1 because we have re- arranged where the negative sign enters into the problem. Same result though
Fall 2001 16.31 17—6 Mechanics • Basics: e = r − y, u = Gce, y = Gu Gc(s) : x˙ c = Acxc + Bce, u = Ccxc G(s) : x˙ = Ax + Bu , y = Cx • Loop dynamics L = Gc(s)G(s) ⇒ y = L(s)e x˙ = Ax +BCc xc x˙ c = +Ac xc +Bce L(s) ∙ x˙ x˙ c ¸ = ∙ A BCc 0 Ac ¸ ∙ x xc ¸ + ∙ 0 Bc ¸ e y = £ C 0 ¤ ∙ x xc ¸ • To “close the loop”, note that e = r − y, then ∙ x˙ x˙ c ¸ = ∙ A BCc 0 Ac ¸ ∙ x xc ¸ + ∙ 0 Bc ¸ µr − £ C 0 ¤ ∙ x xc ¸¶ = ∙ A BCc −BcC Ac ¸ ∙ x xc ¸ + ∙ 0 Bc ¸ r y = £ C 0 ¤ ∙ x xc ¸ — Acl is not exactly the same as on page 17-1 because we have rearranged where the negative sign enters into the problem. Same result though
Fall 2001 16.3117-7 Simple example Let G(s)=1/s2 with state-space model given by A B 00 D=0 Design the regulator to place the poles at s=-4+ 43 A(A-BK)=-4±4→K=[328] Time constant of regulator poles Tc =1/Swn A 1/4=0.25 sec Put estimator poles so that the time constant is faster Te N 1/10 Use real poles, so e(s)=(s+10) L=重2(A CA 01 01 10 00/+20 1000 00 01 10020 100
Fall 2001 16.31 17—7 Simple Example • Let G(s)=1/s2 with state-space model given by: A = ∙ 0 1 0 0 ¸ , B = ∙ 0 1 ¸ , C = £ 1 0 ¤ , D = 0 • Design the regulator to place the poles at s = −4 ± 4j λi(A − BK) = −4 ± 4j ⇒ K = £ 32 8 ¤ — Time constant of regulator poles τc = 1/ζωn ≈ 1/4=0.25 sec • Put estimator poles so that the time constant is faster τe ≈ 1/10 — Use real poles, so Φe(s)=(s + 10)2 L = Φe(A) ∙ C CA ¸−1 ∙ 0 1 ¸ = Ã∙ 0 1 0 0 ¸2 + 20 ∙ 0 1 0 0 ¸ + ∙ 100 0 0 100 ¸! ∙ 1 0 0 1 ¸−1 ∙ 0 1 ¸ = ∙ 100 20 0 100 ¸ ∙ 0 1 ¸ = ∙ 20 100 ¸
Fall 2001 6.3117-8 Compensator A-BK-LC 328 10 201 132-8 L 100 C=K=[328] Compensator transfer function G(s)=C(SI-A)-1B全 E 1440-+222 +285+292 Note that the compensator has a low frequency real zero and two higher frequency poles Thus it looks like a"lead"compensator
Fall 2001 16.31 17—8 • Compensator: Ac = A − BK − LC = ∙ 0 1 0 0 ¸ − ∙ 0 1 ¸ £ 32 8 ¤ − ∙ 20 100 ¸ £ 1 0 ¤ = ∙ −20 1 −132 −8 ¸ Bc = L = ∙ 20 100 ¸ Cc = K = £ 32 8 ¤ • Compensator transfer function: Gc(s) = Cc(sI − Ac) −1 Bc , U E = 1440 s + 2.222 s2 + 28s + 292 • Note that the compensator has a low frequency real zero and two higher frequency poles. — Thus it looks like a “lead” compensator
Fall 2001 6.3117-9 Plant G Freq(rad/sec) PAT= -200 Figure 1: Plant is pretty simple and the compensator looks like a lead 2-10 rads/ Freq(rad/sec) Figure 2: Loop transfer function L(s) shows the slope change near 'c=5 rad sec. Note that we have a large PM and Gm
Fall 2001 16.31 17—9 10−1 100 101 102 103 100 101 102 Freq (rad/sec) Mag Plant G Compensator Gc 10−1 100 101 102 103 −200 −150 −100 −50 0 50 Freq (rad/sec) Phase (deg) Plant G Compensator Gc Figure 1: Plant is pretty simple and the compensator looks like a lead 2—10 rads/sec. 10−1 100 101 102 103 100 101 102 Freq (rad/sec) Mag Loop L 10−1 100 101 102 103 −280 −260 −240 −220 −200 −180 −160 −140 −120 Freq (rad/sec) Phase (deg) Figure 2: Loop transfer function L(s) shows the slope change near ωc = 5 rad/sec. Note that we have a large PM and GM
