Topic # Feedback Control. State-Space Systems Closed-loop control using estimators and regulators. Dynamics output feedback

Topic #17 16.31 Feedback Control State-Space Systems Closed-loop control using estimators and regulators. Dynamics output feedback Back to reality Copyright 21 by Jonathan How. All Rights reserved 1

Fall 21 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 = K ˆx Assume that we have designed a closed-loop estimator with gain L ˆx(t) = Aˆx(t)+Bu(t)+L(y ŷ) ŷ(t) = C ˆx(t) Then we have that the closed-loop system dynamics are given by: ẋ(t) = Ax(t)+Bu(t) ˆx(t) = Aˆx(t)+Bu(t)+L(y ŷ) y(t) = Cx(t) ŷ(t) = C ˆx(t) u = K ˆx Which can be compactly written as: ẋ A BK x = ˆx LC A BK LC ˆx ẋ cl = A cl x cl This does not look too good at this point not even obvious that the closed-system is stable. λ i (A cl )=??

Fall 21 16.31 17 2 Can fix this problem by introducing a new variable x = x ˆx and then converting the closed-loop system dynamics using the similarity transformation T x cl, x x Note that T = T 1 = I I I x ˆx = Tx cl Now rewrite the system dynamics in terms of the state x cl A cl TA cl T 1, Ācl Note that similarity transformations preserve the eigenvalues, so we are guaranteed that λ i (A cl ) λ i (Ā cl ) Work through the math: I A BK I Ā cl = I I LC A BK LC I I A BK I = A LC A + LC I I A BK BK = A LC Because Ā cl is block upper triangular, we know that the closed-loop poles of the system are given by det(si Ā cl ), det(si (A BK)) det(si (A LC)) =

Fall 21 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 21 16.31 17 4 The Compensator Dynamic Output Feedback Compensator is the combination of the regulator and estimator using u = K ˆx ˆx(t) = Aˆx(t)+Bu(t)+L(y ŷ) = Aˆx(t) BKˆx + L(y C ˆx) ˆx(t) = (A BK LC)ˆx(t)+Ly u = K ˆx Rewrite with new state x c ˆx ẋ c = A c x c + B c y u = C c x c where the compensator dynamics are given by: A c, A BK LC, B c, L, C c, 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. λ i (A BK LC) =??

Fall 21 16.31 17 5 For consistency in the implementation with the classical approaches, define the compensator transfer function so that u = G c (s)y From the state-space model of the compensator: U(s) Y (s), G c(s) = C c (si A c ) 1 B c = K(sI (A BK LC)) 1 L G c (s) =C c (si A c ) 1 B c Note that it is often very easy to provide classical interpretations (such as lead/lag) for the compensator G c (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) r e u y G c (s) G(s) 6 - - u = G c (s)e = G c (s)(r y) So we still have u = G c (s)y if r =. 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 21 16.31 17 6 Mechanics Basics: e = r y, u = G c e, y = Gu G c (s) : ẋ c = A c x c + B c e, u = C c x c G(s) : ẋ = Ax + Bu, y = Cx Loop dynamics L = G c (s)g(s) y = L(s)e ẋ = Ax +BC c x c ẋ c = +A c x c +B c e L(s) ẋ ẋ c A BCc x = A c y = C x x c x c + B c e To close the loop, note that e = r y, then µ ẋ A BCc x = + r C x ẋ c A c x c B c x c A BC = c x + r B c C A c x c B c y = C x x c A cl 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 21 16.31 17 7 Simple Example Let G(s) =1/s 2 with state-space model given by: 1 A =, B =, C = 1, D = 1 Design the regulator to place the poles at s = 4 ± 4j λ i (A BK) = 4 ± 4j K = 32 8 Time constant of regulator poles τ c =1/ζω n 1/4 =.25 sec Put estimator poles so that the time constant is faster τ e 1/1 Userealpoles,soΦ e (s) =(s + 1) 2 1 C L = Φ e (A) CA 1 Ã 2 1 1 1 = +2 + 1 1 2 2 = = 1 1 1! 1 1 1 1

Fall 21 16.31 17 8 Compensator: A c = A BK LC 1 2 = 32 8 1 1 2 1 = 132 8 B c = L = 2 1 C c = K = 32 8 1 Compensator transfer function: G c (s) = C c (si A c ) 1 B c, U E s +2.222 = 144 s 2 +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 21 16.31 17 9 1 2 Compensator Gc 1 1 1 5 Compensator Gc Phase (deg) 5 1 15 2 Figure 1: Plant is pretty simple and the compensator looks like a lead 2 1 rads/sec. 1 2 Loop L 1 1 1 Phase (deg) 12 14 16 18 2 22 24 26 28 Figure 2: Loop transfer function L(s) shows the slope change near ω c = 5 rad/sec. Note that we have a large PM and GM.

Fall 21 16.31 17 1 15 1 5 Imag Axis 5 1 15 15 1 5 5 Real Axis Figure 3: Freeze the compensator poles and zeros and look at the root locus of closed-loop poles versus an additional loop gain α (nominally α =1.) 1 2 closed loop Gcl 1 1 1 1 1 1 2 Figure 4: Closed-loop transfer function.

Fall 21 16.31 17 11 1 1 1 Figure 5: Example #1: G(s) = 8 14 2 (s+8)(s+14)(s+2) Compensator Gc 1 2 closed loop Gcl 1 1 1 1 1 2 1 5 Compensator Gc Phase (deg) 5 1 15 1 1 2 1 1 1 1 1 Loop L 1 2 Bode Diagrams Gm=1.978 db (at 4.456 rad/sec), Pm=69.724 deg. (at 15.63 rad/sec) 5 5 1 2 5 Phase (deg); nitude (db) 1 15 5 1 Phase (deg) 1 15 2 25 15 2 25 3 35 4 1 1 1 1 2 1 3 Frequency (rad/sec)

Fall 21 16.31 17 12 5 Figure 6: Example #1: G(s) = 8 14 2 (s+8)(s+14)(s+2) 4 3 2 1 Imag Axis 1 2 3 4 5 8 7 6 5 4 3 2 1 1 2 Real Axis 25 2 15 1 5 Imag Axis 5 1 15 2 25 5 45 4 35 3 25 2 15 1 5 Real Axis 3 closed-loop poles, 5 open-loop poles, 2 Compensator poles, Compensator zeros

Fall 21 16.31 17 13 Two compensator zeros at -21.54±6.63j draw the two lower frequency plant poles further into the LHP. Compensator poles are at much higher frequency. Looks like a lead compensator.

Fall 21 16.31 17 14 1 1 1 Figure 7: Example #2: G(s) =.94 s 2.297 Compensator Gc 1 2 closed loop Gcl 1 1 1 1 1 2 1 5 Compensator Gc Phase (deg) 5 1 15 1 1 2 1 1 Loop L 1 2 Bode Diagrams Gm=11.784 db (at 7.493 rad/sec), Pm=36.595 deg. (at 2.7612 rad/sec) 5 1 1 1 1 2 5 Phase (deg); nitude (db) 5 1 1 15 Phase (deg) 1 15 2 25 2 25 3 1 2 1 1 1 1 1 1 2 Frequency (rad/sec)

Fall 21 16.31 17 15 6 Figure 8: Example #2: G(s) =.94 s 2.297 4 2 Imag Axis 2 4 6 8 6 4 2 2 Real Axis 2 1.5 1.5 Imag Axis.5 1 1.5 2 2 1.5 1.5.5 1 1.5 2 Real Axis 3 closed-loop poles, 5 open-loop poles, 2 Compensator poles, Compensator zeros

Fall 21 16.31 17 16 Compensator zero at -1.21 draws the two lower frequency plant poles further into the LHP. Compensator poles are at much higher frequency. Looks like a lead compensator.

Fall 21 16.31 17 17 1 1 1 Figure 9: Example #3: G(s) = 8 14 2 (s 8)(s 14)(s 2) Compensator Gc 1 2 closed loop Gcl 1 1 1 1 1 2 1 5 Compensator Gc Phase (deg) 5 1 15 1 1 2 1 1 Loop L 1 2 Bode Diagrams Gm=.942 db (at 24.221 rad/sec), Pm=6.674 deg. (at 35.813 rad/sec) 5 1 1 1 1 2 5 Phase (deg); nitude (db) 5 1 15 2 Phase (deg) 1 15 2 25 25 3 35 1 1 1 1 2 1 3 Frequency (rad/sec)

Fall 21 16.31 17 18 Figure 1: Example #3: G(s) = 8 14 2 (s 8)(s 14)(s 2) 1 8 6 4 2 Imag Axis 2 4 6 8 1 14 12 1 8 6 4 2 2 4 6 Real Axis 25 2 15 1 5 Imag Axis 5 1 15 2 25 25 2 15 1 5 5 1 15 2 25 Real Axis 3 closed-loop poles, 5 open-loop poles, 2 Compensator poles, Compensator zeros

Fall 21 16.31 17 19 Compensator zeros at 3.72±8.3j draw the two higher frequency plant poles further into the LHP. Lowest frequency one heads into the LHP on its own. Compensator poles are at much higher frequency. Note sure what this looks like.

Fall 21 16.31 17 2 1 1 1 Figure 11: Example #4: G(s) = Compensator Gc 1 2 (s 1) (s+1)(s 3) closed loop Gcl 1 1 1 1 1 2 1 5 Compensator Gc Phase (deg) 5 1 15 1 1 2 1 1 1 Loop L 1 2 Bode Diagrams Gm= 3.3976 db (at 4.5695 rad/sec), Pm= 22.448 deg. (at 1.464 rad/sec) 2 1 1 1 2 5 Phase (deg); nitude (db) 2 4 6 8 22 21 2 Phase (deg) 1 15 2 25 19 18 17 16 15 14 Frequency (rad/sec)

Fall 21 16.31 17 21 4 Figure 12: Example #4: G(s) = (s 1) (s+1)(s 3) 3 2 1 Imag Axis 1 2 3 4 1 4 3 2 1 1 2 Real Axis 8 6 4 2 Imag Axis 2 4 6 8 1 1 8 6 4 2 2 4 6 8 1 Real Axis 3 closed-loop poles, 5 open-loop poles, 2 Compensator poles, Compensator zeros

Fall 21 16.31 17 22 Compensator zero at -1 cancels the plant pole. Note the very unstable compensator pole at s =9!! Needed to get the RHP plant pole to branch off thereallineand head into the LHP. Other compensator pole is at much higher frequency. Note sure what this looks like. Separation principle gives a very powerful and simple way to develop a dynamic output feedback controller Note that the designer now focuses on selecting the appropriate regulator and estimator pole locations. Once those are set, the closed-loop response is specified. Canalmostconsiderthecompensatortobeaby-product. These examples show that the design process is extremely simple.