Modern Optimal Control

Transcription

1 Modern Optimal Control Matthew M. Peet Arizona State University Lecture 19: Stabilization via LMIs

2 Optimization Optimization can be posed in functional form: min x F objective function : inequality constraints subject to which may have the form min f 0(x) : x F f i (x) 0 subject to i = 1, k Special Cases: Linear Programming If the fi(x) : R n R m are affine (f i(x) = Ax b) EASY: Simplex/Ellipsoid Algorithm Polynomial Programming The fi(x) : R n R m are polynomial. Semidefinite Programming The fi(x) : R n S m are affine. For semidefinite programming, what does f i (x) 0 mean? M. Peet Lecture 19: 2 / 30

3 Positive Matrices Definition 1. A symmetric matrix P S n is Positive Semidefinite, denoted P 0 if x T P x 0 for all x R n Definition 2. A symmetric matrix P S n is Positive Definite, denoted P > 0 if x T P x > 0 for all x 0 P is Negative Semidefinite if P 0 P is Negative Definite if P > 0 A matrix which is neither Positive nor Negative Semidefinite is Indefinite The set of positive or negative matrices is a convex cone. M. Peet Lecture 19: 3 / 30

4 Positive Matrices Lemma 3. P S n is positive definite if and only if all its eigenvalues are positive. Easy Proofs: A Positive Definite matrix is invertible. The inverse of a positive definite matrix is positive definite. If P > 0, then T P T T 0 for any T. If T is invertible, then T P T T > 0. Lemma 4. For any P > 0, there exists a positive square root, P 1 2 P = P 1 2 P 1 2. > 0 such that M. Peet Lecture 19: 4 / 30

5 Convex Optimization Why is Linear Programming easy and polynomial programming hard? min f 0(x) : x F f i (x) 0 subject to i = 1, k Consider the geometric representation of the optimization problem: min f 0(x) : subject to x F x S where S := {x : f i (x) 0, i = 1,, k}. or the more compact pure geometric representation min γ,x F γ : subject to (γ, x) S where S := {(γ, x) : γ f 0 (x) 0, f i (x) 0, i = 1,, k}. M. Peet Lecture 19: 5 / 30

6 Convexity Definition 5. A set is convex if for any x, y Q, {µx + (1 µ)y : µ [0, 1]} Q. The line connecting any two points lies in the set. M. Peet Lecture 19: 6 / 30

7 Convexity Example Convex Optimization: minimize x 1 + x 2 Definitionsubject 6. to 3x 1 + x 2 3 Consider the optimization problem x 2 1 x min γ : subject 1 4 to γ,x F x 1 + 5x 2 20 (γ, x) x S x 2 20 subject to Ax = b Cx d The problem is Convex Optimization if the set S is convex. Convex optimization problems have the property that scaled Newton iteration always converges to the global optimal. The question is, of course, when is the set S convex? For polynomial optimization, a sufficient condition is that all functions f i are convex. The set of convex functions is a convex cone. The level set of a convex function is a convex set. M. Peet Lecture 19: 7 / 30

8 Convexity Newton s Algorithm: Solve f(x) = 0 x k+1 = x k t f(x k) f (x k ) t is the step-size. For non-convex optimization, Newton descent may get stuck at local optima. For Newton descent, constraints are represented by barrier functions. M. Peet Lecture 19: 8 / 30

9 Semidefinite Programming minimize trace CX subject to trace A i X = b i for all i X 0 The variable X is a symmetric matrix X 0 means X is positive semidefinite The feasible set is the intersection of an affine set with the positive semidefinite cone { X S n X 0 } M. Peet Lecture 19: 9 / 30

10 SDPs with Explicit Variables We can also explicitly parameterize the affine set to give minimize c T x subject to F 0 + x 1 F 1 + x 2 F x n F n 0 where F 0, F 1,..., F n are symmetric matrices. The inequality constraint is called a linear matrix inequality (LMI); e.g., x 1 3 x 1 + x 2 1 x 1 + x 2 x x 1 which is equivalent to x x M. Peet Lecture 19: 10 / 30

11 Linear Matrix Inequalities Review Linear Matrix Inequalities are often a Simpler way to solve control problems. Semidefinite Programming min c T x : Y 0 + i x i Y i > 0 Here the Y i are symmetric matrices. Linear Matrix Inequalities Findx : Y 0 + i x i Y i > 0 Commonly Takes the Form FindX : A i XB i + Q > 0 i M. Peet Lecture 19: 11 / 30

12 Linear Matrix Inequalities Common Form: FindX : A i XB i + Q > 0 i The most important Linear Matrix Inequality is the Lyapunov Inequality. There are several very efficient LMI/SDP Solvers for Matlab: SeDuMi Fast, but somewhat unreliable. See LMI Lab (Part of Matlab s Robust Control Toolbox) Universally disliked See YALMIP (a parser for other solvers) See I recommend YALMIP with solver SeDuMi. M. Peet Lecture 19: 12 / 30

13 Semidefinite Programming(SDP): Common Examples in Control Some Simple examples of LMI conditions in control include: Stability Stabilization H 2 Synthesis We will go beyond these examples. A T X + XA 0 X 0 AX + BZ + XA T + Z T B T 0 X 0 min T r(w ) [ ] [ ] X A B2 + [ X Z ] [ ] Z T A T B2 T + B 1 B1 T 0 [ ] X (CX + DZ) T 0 (CX + DZ) W M. Peet Lecture 19: 13 / 30

14 Lyapunov Theory LMIs unite time-domain and frequency-domain analysis Theorem 7 (Lyapunov). Suppose there exists a continuously differentiable function V for which V (0) = 0 and V (x) > 0 for x 0. Furthermore, suppose lim x V (x) = and V (x(t + h)) V (x(t)) lim = d h 0 + h dt V (x(t)) < 0 for any x such that ẋ(t) = f(x(t)). Then for any x(0) R the system of equations ẋ(t) = f(x(t)) has a unique solution which is stable in the sense of Lyapunov. M. Peet Lecture 19: 14 / 30

15 The Lyapunov Inequality Lemma 8. A is Hurwitz if and only if there exists a P > 0 such that A T P + P A < 0 Proof. Suppose there exists a P > 0 such that A T P + P A < 0. Define the Lyapunov function V (x) = x T P x. Then V (x) > 0 for x 0 and V (0) = 0. Furthermore, V (x(t)) = ẋ(t) T P x(t) + x(t) T P ẋ(t) = x(t) T A T P x(t) + x(t) T P Ax(t) = x(t) T ( A T P + P A ) x(t) Hence V (x(t)) < 0 for all x 0. Thus the system is globally stable. Global stability implies A is Hurwitz. M. Peet Lecture 19: 15 / 30

16 The Lyapunov Inequality Proof. For the other direction, if A is Hurwitz, let P = Converges because A is Hurwitz. Furthermore P A = e AT s e As Ads = = e AT s Ae As ds = [ e AT s e As ] 0 = I Thus P A + A T P = I < 0. 0 e AT s e As ds e AT s d ( e As ) ds 0 ds d T 0 ds ea s e As A T e AT s e As = I A T P M. Peet Lecture 19: 16 / 30

17 The Lyapunov Inequality Other Versions: Lemma 9. (A, B) is controllable if and only if there exists a X > 0 such that A T X + XA + BB T 0 Lemma 10. (C, A) is stabilizable if and only if there exists a X > 0 such that AX + XA T + C T C 0 These also yield Lyapunov functions for the system. M. Peet Lecture 19: 17 / 30

18 The Static State-Feedback Problem Lets start with the problem of stabilization. Definition 11. The Static State-Feedback Problem is to find a feedback matrix K such that ẋ(t) = Ax(t) + Bu(t) u(t) = Kx(t) is stable We have already solved the problem earlier using the Controllability Grammian. Find K such that A + BK is Hurwitz. Can also be put in LMI format: Find X > 0, K : X(A + BK) + (A + BK) T X < 0 Problem: Bilinear in K and X. M. Peet Lecture 19: 18 / 30

19 The Static State-Feedback Problem The bilinear problem in K and X is a common paradigm. Bilinear optimization is not convex. To convexify the problem, we use a change of variables. Problem 1: Problem 2: Definition 12. Find X > 0, K : X(A + BK) + (A + BK) T X < 0 Find P > 0, Z : AP + BZ + P A T + ZB T < 0 Two optimization problems are equivalent if a solution to one will provide a solution to the other. Theorem 13. Problem 1 is equivalent to Problem 2. M. Peet Lecture 19: 19 / 30

20 The Dual Lyapunov Equation Problem 1: Find X > 0, : Lemma 14. XA + A T X < 0 Problem 1 is equivalent to problem 2. Problem 2: Find Y > 0, : Y A T + AY < 0 Proof. First we show 1) solves 2). Suppose X > 0 is a solution to Problem 1. Let Y = X 1 > 0. If XA + A T X < 0, then Hence X 1 (XA + A T X)X 1 < 0 X 1 (XA + A T X)X 1 = AX 1 + X 1 A T = AY + Y A T < 0 Therefore, Problem 2 is feasible with solution Y = X 1. M. Peet Lecture 19: 20 / 30

21 The Dual Lyapunov Equation Problem 1: Find X > 0, : XA + A T X < 0 Problem 2: Find Y > 0, : Y A T + AY < 0 Proof. Now we show 2) solves 1) in a similar manner. Suppose Y > 0 is a solution to Problem 1. Let X = Y 1 > 0. Then XA + A T X = X(AX 1 + X 1 A T )X = X(AY + Y A T )X < 0 Conclusion: If V (x) = x T P x proves stability of ẋ = Ax, Then V (x) = x T P 1 x proves stability of ẋ = A T x. M. Peet Lecture 19: 21 / 30

22 The Stabilization Problem Thus we rephrase Problem 1 Problem 1: Find P > 0, K : (A + BK)P + P (A + BK) T < 0 Theorem 15. Problem 1 is equivalent to Problem 2. Problem 2: Find X > 0, Z : AX + BZ + XA T + ZB T < 0 Proof. We will show that 2) Solves 1). Suppose X > 0, Z solves 2). Let P = X > 0 and K = ZP 1. Then Z = KP and (A + BK)P + P (A + BK) T = AP + P A T + BKP + P K T B T = AP + P A T + BZ + Z T B T < 0 Now suppose that P > 0 and K solve 1). Let X = P > 0 and Z = KP. Then AP + P A T + BZ + Z T B T = (A + BK)P + P (A + BK) T < 0 M. Peet Lecture 19: 22 / 30

23 The Stabilization Problem The result can be summarized more succinctly Theorem 16. (A, B) is static-state-feedback stabilizable if and only if there exists some P > 0 and Z such that AP + P A T + BZ + Z T B T < 0 with u(t) = ZP 1 x(t). Standard Format: [ ] [ ] P A B + [ P Z ] [ ] Z T A T B T < 0 There are a number of general-purpose LMI solvers available. e.g. SeDuMi - Free LMI Lab - in Matlab Robust Control Toolbox (in Computer lab) YALMIP - A nice front end for several solvers (Free) M. Peet Lecture 19: 23 / 30

24 The Schur complement Before we get to the main result, recall the Schur complement. Theorem 17 (Schur Complement). For any S S n, Q S m and R R n m, the following are equivalent. [ ] M R 1. R T > 0 Q 2. Q > 0 and M RQ 1 R T > 0 A commonly used property of positive matrices. Also Recall: If X > 0, then X ɛi > 0 for ɛ sufficiently small. M. Peet Lecture 19: 24 / 30

25 The KYP Lemma (AKA: The Bounded Real Lemma) The most important theorem in this class. Lemma 18. Suppose Then the following are equivalent. G H γ. [ A B Ĝ(s) = C D There exists a X > 0 such that [ ] A T X + XA XB B T + 1 [ ] C T [C ] D < 0 X γi γ D T ]. Can be used to calculate the H -norm of a system Originally used to solve LMI s using graphs. (Before Computers) Now used directly instead of graphical methods like Bode. The feasibility constraints are linear Can be combined with other methods. M. Peet Lecture 19: 25 / 30

26 The KYP Lemma Proof. We will only show that ii) implies i). The other direction requires the Hamiltonian, which we have not discussed. We will show that if y = Gu, then y L2 γ u L2. From the 1 x 1 block of the LMI, we know that A T X + XA < 0, which means A is Hurwitz. Because the inequality is strict, there exists some ɛ > 0 such that [ ] A T X + XA XB B T + 1 [ ] C T [C ] D X (γ ɛ)i γ D T [ ] A = T X + XA XB B T + 1 [ ] [ ] C T [C ] 0 0 D + X γi γ D T < 0 0 ɛi Let y = Gu. Then the state-space representation is y(t) = Cx(t) + Du(t) ẋ(t) = Ax(t) + Bu(t) x(0) = 0 M. Peet Lecture 19: 26 / 30

27 The KYP Lemma Proof. Let V (x) = x T Xx. Then the LMI implies [ ] [ T [A ] x(t) T X + XA XB u(t) B T + 1 [ ] C T [C ] ] [ ] x(t) D X (γ ɛ)i γ D T u(t) [ ] T [ ] [ ] x A = T X + XA XB x u B T + 1 [ ] T [ ] x C T [C ] [ ] x D X (γ ɛ)i u γ u D T u [ ] T [ [ ] x A = T X + XA XB x u B T + X (γ ɛ)i] 1 u γ yt y = x T (A T X + XA)x + x T XBu + u T B T Xx (γ ɛ)u T u + 1 γ yt y = (Ax + Bu) T Xx + x T X(Ax + Bu) (γ ɛ)u T u + 1 γ yt y = ẋ(t) T Xx(t) + x(t) T Xẋ(t) (γ ɛ) u(t) γ y(t) 2 = V (x(t)) (γ ɛ) u(t) γ y(t) 2 < 0 M. Peet Lecture 19: 27 / 30

28 The KYP Lemma Proof. Now we have V (x(t)) (γ ɛ) u(t) γ y(t) 2 < 0 Integrating in time, we get T ( V (x(t)) (γ ɛ) u(t) γ y(t) 2) dt 0 = V (x(t )) V (x(0)) (γ ɛ) T 0 u(t) 2 dt + 1 γ Because A is Hurwitz, lim t x(t) = 0. Hence lim t V (x(t)) = 0. Likewise, because x(0) = 0, we have V (x(0)) = 0. T 0 y(t) 2) dt < 0 M. Peet Lecture 19: 28 / 30

29 The KYP Lemma Proof. Since V (x(0)) = V (x( )) = 0, [ lim V (x(t )) V (x(0)) (γ ɛ) T = 0 0 (γ ɛ) 0 u(t) 2 dt + 1 γ = (γ ɛ) u 2 L γ y 2 L 2 dt < 0 T 0 0 u(t) 2 dt + 1 γ y(t) 2 dt T 0 y(t) 2) ] dt Thus y 2 L 2 dt < (γ 2 ɛγ) u 2 L 2 By definition, this means G 2 H (γ 2 ɛγ) < γ 2 or G H < γ M. Peet Lecture 19: 29 / 30

30 The Positive Real Lemma A Passivity Condition A Variation on the KYP lemma is the positive-real lemma Lemma 19. Suppose Then the following are equivalent. [ A B Ĝ(s) = C D G is passive. i.e. ( u, Gu L2 0). There exists a P > 0 such that [ ] A T P + P A P B C T B T P C D T 0 D ]. M. Peet Lecture 19: 30 / 30