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 ¸