Linear Control Systems Lecture # 16 Observers and Output Feedback Control

Transcription

1 Linear Control Systems Lecture # 16 Observers and Output Feedback Control p. 1/2

2 p. 2/2 Observers: Consider the system ẋ = Ax + Bu y = Cx + Du where the initial state x(0) is unknown, but we can measure the output y. We want to estimate the state x from the available signals u and y We assume that we know the matrices {A, B, C, D} If x(0) was known we could have estimated x by solving the equation ˆx = Aˆx + Bu, ˆx(0) = x(0) on line and we would have obtained ˆx(t) x(t) because of the uniqueness of solutions

3 p. 3/2 This can also be seen by defining the estimation error e = x ˆx. Then, e satisfies the equation ė = Ax + Bu Aˆx Bu = Ae, e(0) = 0 whose solution is e(t) 0 If x(0) is unknown, we cannot choose ˆx(0) = x(0). Hence e(0) 0 and the estimation error will be different from zero. Notice, however, that if R e λ(a) < 0 lim e(t) = 0 t Hence, we can asymptotically estimate the state x

4 p. 4/2 What if A has some eigenvalues with nonnegative real parts? or if all the eigenvalues have negative real parts, but convergence is slow? Can we use feedback to stabilize the observer? What signal can we feedback? We measure y(t) = Cx(t) + Du(t). If an estimate of the state ˆx(t) is available on line, we can use it to estimate the output by ŷ(t) = C ˆx(t) + Du(t). Then y(t) ŷ(t) = Cx(t) + Du(t) C ˆx(t) Du(t) = Ce(t)

5 p. 5/2 Consider the observer (estimator): ˆx = Aˆx + Bu + Ky ŷ where K is the observer gain (an n p matrix, where p is the number of outputs) ˆx = Aˆx + Bu + Ky C ˆx Du e(t) = x(t) ˆx(t) ė = ẋ ˆx = Ax + Bu Aˆx Bu KCx + Du C ˆx Du = (A KC)e

6 p. 6/2 Design K such that When is that possible? R e λ(a KC) < 0 (A KC) T = A T C T K T Let à = A T, B = C T, F = K T (A KC) T = à + B F There is a matrix F such that R e λ(ã + B F ) < 0 if and only if (Ã, B) is stabilizable; i.e., if (Ã, B) is controllable or any uncontrollable eigenvalue has a negative real part

7 p. 7/2 There is a matrix K such that R e λ(a KC) < 0 if and only if (A, C) is detectable; i.e., (A, C) is observable or any unobservable eigenvalue has a negative real part How would you compute K? Find F to stabilize (Ã + B F ) using any of the methods studied under state feedback. Then, take K = F T

8 p. 8/2 Output Feedback Control: Design an output feedback controller to stabilize the system Separation principle: ẋ = Ax + Bu y = Cx + Du Design a state feedback controller u = F x + v such that R e λ(a + BF ) < 0 Design an observer ˆx = Aˆx + Bu + Ky C ˆx Du where K is chosen such that R e λ(a KC) < 0 Take the output feedback controller as u = F ˆx + v

9 p. 9/2 Closed-loop system under state feedback: ẋ = (A + BF )x + Bv y = (C + DF )x + Dv The system is asymptotically stable because R e λ(a + BF ) < 0. Hence, it is BIBO stable The closed-loop transfer function from v to y is given by H cl (s) = (C + DF )(si A BF ) 1 B + D

10 p. 10/2 Closed-loop system under output feedback: ẋ = Ax + BF ˆx + Bv ˆx = Aˆx + B(F ˆx + v) + KCx + Du C ˆx Du = KCx + (A + BF KC)ˆx + Bv y = Cx + DF ˆx + Dv Change of variables: x e = I 0 I I x ˆx ẋ = Ax + BF (x e) + Bv = (A + BF )x BF e + Bv ė = (A KC)e

11 p. 11/2 ẋ = (A + BF )x BF e + Bv ė = (A KC)e y = (C + DF )x DF e + Dv ẋ ė = y = (A + BF ) BF 0 (A KC) (C + DF ) DF x e x e + Dv + B 0 v The 2n closed-loop eigenvalues are the eigenvalues of (A + BF ) and (A KC)

12 p. 12/2 Since both (A + BF ) and (A KC) are designed to have eigenvalues with negative real parts, the closed-loop system is asymptotically stable and BIBO stable To find the closed-loop transfer function from v to y, take the Laplace transform of the state and output equations assuming zero initial conditions s x(s) = (A + BF ) x(s) BF ẽ(s) + Bṽ(s) sẽ(s) = (A KC)ẽ(s) ỹ(s) = (C + DF ) x(s) DF ẽ(s) + Dṽ(s) where x(s) = L{x(t)} ẽ(s) = 0 ỹ(s) = (C +DF )(si A BF ) 1 B+Dṽ(s) The same as in state feedback

13 p. 13/2 Remarks: 1. The controller is known as Observer-based controller 2. The observer-based controller can be represented as a dynamic compensator with cascade and feedback components u = F ˆx + v, ˆx = Aˆx + Bu + K(y C ˆx Du) ˆx = (A KC)ˆx + Ky + (B KD)u x(s) = (si A + KC) 1 Kỹ(s) + (B KD)ũ(s) ũ(s) = F (si A+KC) 1 Kỹ(s)+(B KD)ũ(s)+ṽ(s)

14 p. 14/2 ũ(s) = F (si A+KC) 1 Kỹ(s)+(B KD)ũ(s)+ṽ(s) Define G u (s) = F (si A + KC) 1 (B KD), G y (s) = F (si A + KC) 1 K ũ(s) = G u (s)ũ(s) + G y (s)ỹ(s) + ṽ(s) I G u (s)ũ(s) = G y (s)ỹ(s) + ṽ(s) ũ(s) = I G u (s) 1 G y (s)ỹ(s) + ṽ(s)

15 p. 15/2 ũ(s) = I G u (s) 1 G y (s)ỹ(s) + ṽ(s) 3. The eigenvalues of (A + BF ) are chosen to meet the design specifications on the transient response and any constraints on x or u (as in state feedback). The eigenvalues of (A KC) are chosen much faster than those of (A + BF ) (farther away to the left)

16 p. 16/2 Example: Stabilization of H(s) = 1/s 2 ẋ 1 = x 2, ẋ 2 = u, y = x 1 A = , B = 0 1, C = 1 0, D = 0 (A, B) is controllable and (A, C) is observable State feedback control: u = F x + v F = f 1 f 2 A + BF = 0 1 f 1 f 2 detsi (A + BF ) = s 2 f 2 s f 1

17 p. 17/2 Desired eigenvalues: A pair of complex eigenvalues with ω n = 2 and ζ = 0.5 s 2 + 2ζω n s + ω 2 n = s2 + 2s + 4 detsi (A + BF ) = s 2 f 2 s f 1 F = 4 2 Observer: K = k 1 k 2 A KC = k 1 1 k 2 0 detsi (A KC) = s 2 + k 1 s + k 2

18 p. 18/2 Desired eigenvalues: A pair of complex eigenvalues with ω n = 10 and ζ = 0.5 s 2 + 2ζω n s + ω 2 n = s2 + 10s detsi (A KC) = s 2 + k 1 s + k 2 10 K = 100 Output feedback controller: u = 4ˆx 1 2ˆx 2 + v ˆx 1 = ˆx (y ˆx 1 ) ˆx 2 = u + 100(y ˆx 1 )

19 p. 19/2 G u = 2(s + 12) s s (1 G u ) 1 = s2 + 10s s s G y = 80(3s + 5) s s + 100

20 p. 20/2 Given the system ẋ = Ax + Bu y = Cx + Du under what conditions can we stabilize it by output feedback? Theorem: The system {A, B, C, D} can be stabilized by linear output feedback if and only if (A, B) is stabilizable and (A, C) is detectable Proof of sufficiency: If (A, B) is stabilizable and (A, C) is detectable, we have seen how to design a stabilizing observer-based controller

21 p. 21/2 Proof of necessity: Show that uncontrollable and unobservable eigenvalues cannot be moved by output feedback It is clear that uncontrollable eigenvalues cannot be moved by feedback because they cannot be reached from the input; so it does not matter whether we deal with state or output feedback. They cannot be moved Â11 Â 12 ˆB 1 Â =, ˆB = 0 Â 22 0 ż 2 = Â 22 z 2

22 p. 22/2 Consider now unobservable eigenvalues Â11 0, Ĉ = Ĉ1 0 Â 21 Â 22 ż 1 = Â 11 z 1 + ˆB 1 u ż 2 = Â 21 z 1 + Â 22 z 2 + ˆB 2 u y = Ĉ 1 z 1 + Du Set v = 0 and consider any linear dynamic controller that can be represented by ũ(s) = G(s)ỹ(s). Let {A a, B a, C a, D a } be a state-space model of G(s)

23 p. 23/2 ż c = A a z c + B a y u = C a z c + D a y u = C a z c + D a (Ĉ 1 z 1 + Du) (I D a D)u = C a z c + Ĉ 1 z 1 The matrix (I D a D) must be nonsingular for the closed-loop system to be well defined u = (I D a D) 1 C a z c + Ĉ 1 z 1 Notice that u depends only on z 1 and z c. Also ż c = z 1 + z c

24 p. 24/2 The closed-loop equation takes the form ż 1 0 ż c = 0 ż 2 Â 22 z 1 z c z 2 The unobservable eigenvalues remain as closed-loop eigenvalues