LMI Methods in Optimal and Robust Control
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1 LMI Methods in Optimal and Robust Control Matthew M. Peet Arizona State University Lecture 4: LMIs for State-Space Internal Stability
2 Solving the Equations Find the output given the input State-Space: ẋ = Ax(t) + Bu(t) y(t) = Cx(t) + Du(t) x(0) = 0 Input u State-Space System Output y Basic Question: Given an input function, u(t), what is the output? Definition 1. Define the Matrix Exponential: e A = I + A A A k! Ak + M. Peet Lecture 4: State-Space Theory 2 / 23
3 Properties of the Matrix Exponential The matrix exponential is similar to the scalar exponential with important differences. e 0 = I e 0 = I = I e M T = ( e M ) T e M T = I + M T (M T ) (M T ) k! (M T ) k + d dt eat = Ae At d dt eat = d ( I + (At) + 1 dt 2 (At) ) k! (At)k + ( = A I + (At) + 1 ) 1 2 (At)2 + + (k 1)! (At)k 1 + However, e M+N e M e N unless, MN = NM. M. Peet Lecture 4: State-Space Theory 3 / 23
4 State-Space Solutions and the Jordan Decomposition The equation ẋ(t) = Ax(t), x(0) = x 0 has solution Proof. Let x(t) = e At x 0, then ẋ(t) = Ae At x 0 = Ax(t). x(t) = e At x 0 x(0) = e 0 x 0 = x 0 Question: So what does e At look like? Let λ i be eigenvalues of A Definition 2. A Jordan Block, J i is a matrix which looks like 0 1 i 0 1 J i = λ i I + N = λ λ i 0 1 }{{} 0 λ ii }{{} N M. Peet Lecture 4: State-Space Theory 4 / 23
5 State-Space Solutions and the Jordan Decomposition Theorem 3. For and A R n n, there exists an invertible T such that A = T JT 1, J = J 1... J n where J i are Jordan Blocks A k = T J k T 1. Hence e At = T e Jt T 1 = T e J1t... e Jnt T 1 λ i I and N commute, hence e λii+n = e λii e N. Further N d = 0 for some d. e Jit = e λit [ N + 1 ] 2 N (k 1)! N k 1 t k 1 e λit lim t t i e λt = 0 if and only if Re λ < 0 M. Peet Lecture 4: State-Space Theory 5 / 23
6 Stability of Continuous and Discrete-Time Systems Definition 4. A is Hurwitz if Re λ i (A) < 0 for all i. Theorem 5. ẋ(t) = Ax(t) is stable if and only if A is Hurwitz. For Discrete-Time Systems: x k+1 = Ax k, x k = A k x 0 Definition 6. A is Schur if λ i (A) < 1 for all i. Theorem 7. x k+1 = Ax k is stable if and only if A is Schur. M. Peet Lecture 4: State-Space Theory 6 / 23
7 Lyapunov Functions Theorem 8 (Lyapunov). ẋ(t) = f(x(t)) V is a Lyapunov Function if V (0) = 0 and V (x) > 0 for x 0 and lim x V (x) =. If d V (x(t)) < 0 for ẋ(t) = f(x(t)). dt Then for any x(0) R the system ẋ(t) = f(x(t)) has a unique solution which is stable in the sense of Lyapunov. M. Peet Lecture 4: State-Space Theory 7 / 23
8 Lyapunov Functions: Local Stability ẋ(t) = f(x(t)) Theorem 9 (Lyapunov Stability). Suppose there exists a continuous V and α, β, γ > 0 where β x 2 V (x) α x 2 V (x) = V (x) T f(x) γ x 2 for all x X. Then any sub-level set of V in X is a Domain of Attraction. M. Peet Lecture 4: State-Space Theory 8 / 23
9 Lyapunov Functions Mass-Spring: Pendulum: ẍ = c mẋ k m x ẋ 2 = g l sin x 1 ẋ 1 = x 2 V (x) = 1 2 mẋ kx2 V (x) = (1 cos x 1 )gl l2 x 2 2 V (x) = ẋ( cẋ kx) + kxẋ = cẋ 2 kẋx + kxẋ V (x) = glx 2 sin x 1 glx 2 sin x 1 = 0 = cẋ 2 0 M. Peet Lecture 4: State-Space Theory 9 / 23
10 An Example of Global Stability Analysis A controlled model of a jet engine (Derived from Moore-Greitzer). ẋ = y 3 2 x2 1 2 x3 ẏ = 3x y This is feasible with V (x) = x xy y x x 2 y xy y x x 3 y x 2 y y 4 M. Peet Lecture 4: State-Space Theory 10 / 23
11 The Lyapunov Inequality (Our First LMI) Lemma 10 (An LMI for Hurwitz Stability). 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 4: State-Space Theory 11 / 23
12 The Lyapunov Inequality Proof. For the other direction, if A is Hurwitz, for any Q > 0, let P = Converges because A is Hurwitz. Furthermore P A = e AT s Qe As Ads = = e AT s QAe As ds = [ e AT s Qe As ] 0 = Q Thus P A + A T P = Q < 0. 0 e AT s Qe As ds e AT s Q d 0 ds d T 0 ds ea s Qe As A T e AT s Qe As = Q A T P ( e As ) ds M. Peet Lecture 4: State-Space Theory 12 / 23
13 Discrete-Time Lyapunov Functions Theorem 11 (Lyapunov). x k+1 = f(x k ) V is a Lyapunov Function if V (0) = 0 and V (x) > 0 for x 0 and lim x V (x) =. If V (x k+1 ) < V (x k ) where x k+1 = f(x k ). Then for any x 0 R n the system x k+1 = f(x k ) is stable in the sense of Lyapunov. M. Peet Lecture 4: State-Space Theory 13 / 23
14 Discrete-Time Lyapunov Functions Lemma 12 (An LMI for Schur Stability). A is Schur if and only if there exists a P > 0 such that A T P A P < 0 Proof. Suppose there exists a P > 0 such that A T P A P < 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 k+1 ) = x T k+1p x k+1 = x T k A T P Ax k < x T k P x k = V (x k ) Hence V (x k+1 ) < V (x k ) for all k 0. Thus the system is Stable. Stability implies A is Schur. M. Peet Lecture 4: State-Space Theory 14 / 23
15 Lyapunov Functions Proof. For the other direction, if A is Hurwitz, for any Q > 0, let P = (A T ) k QA k k=0 A T P A P = (A T ) k QA k (A T ) k QA k k=1 k=0 = (A T ) 0 QA 0 = Q < 0 Thus A T P A P < 0. YALMIP Code: > P = sdpvar(n); eta=.1; > F=[P>=eta*eye(n)]; > F=[F; A P A - P<=0]; > optimize(f); M. Peet Lecture 4: State-Space Theory 15 / 23
16 Pole Locations AKA D-stability Some people still care about pole locations. For these people, we have D-stability. To begin, you have to define the acceptable region of the complex plane using inequality constraints. Rise Time: ω n 1.8 t r Settling Time: σ 4.6 t s ln Mp Percent Overshoot: σ π ω d Recall that if z is the complex pole location: ω 2 n = z 2 = z z ω d = Im z = (z z )/2 σ = Re z = (z + z )/2 Which yields Rise Time: z z 1.82 t 0 2 r Settling Time: (z + z ) t s 0 Percent Overshoot: z z + π ln M p z + z 0 M. Peet Lecture 4: State-Space Theory 16 / 23
17 An LMI for Pole Locations Gutman proposed a nice LMI for D-stability with a single constraint Theorem 13 (Gutman). The pole locations, z C of A satisfy z {z C : k,l c kl z k (z ) l < 0} if and only if there exists some P > 0 such that c kl A k P (A T ) l < 0 k,l But this has some disadvantages There can only be one constraint. The LMI is not linear in A. So controller synthesis is not an LMI. M. Peet Lecture 4: State-Space Theory 17 / 23
18 An LMI for Convex Regions of the Complex Plane To get around the limitations of Gutman s result, we introduce the concept of LMI regions. These are regions which can be represented using LMIs in the z and z variables Definition 14. An LMI Region of the complex plane has the form {z C : F 0 + zf 1 + z F 2 < 0} Such regions are hard to visualize, but Are convex e.g. Minimum rise time is not allowed! Can combine multiple convex regions M. Peet Lecture 4: State-Space Theory 18 / 23
19 An LMI for Convex Regions of the Complex Plane Examples Rise Time: z z 1.82 t 2 0 r }{{} [ r 2 ] [ ] [ ] [ ] r z r z = + z + z < 0 r 0 r }{{}}{{}}{{} F 0 F 1 F 2 Which by the Schur complement is equivalent to r z r 1 z > 0. Settling Time: 4.6 t s + z + z 0 Percent Overshoot: z z + π ln M p (z + z ) 0 [ ] ln Mp (z + z ) π(z z ) π(z z ) ln M p (z + z = ) [ ] [ ] 0 0 ln Mp π π ln M p }{{}}{{} F 0 F 1 Which by the Schur complement is equivalent to z z < 0 and (z z ) 2 > ( π ln M p ) 2 z + z 2 0. [ ln Mp π z+ π ln M p } {{ } F 2 ] z < 0 M. Peet Lecture 4: State-Space Theory 19 / 23
20 An LMI for Convex Regions of the Complex Plane Theorem 15 (Chilali + Gahinet). The pole locations, z C of A satisfy z {z C : F 0 + zf 1 + z F 2 < 0} if and only if there exists some P > 0 such that F 0 P + F 1 (AP ) + F 2 (AP ) T < 0 The notation F P is Kronecker notation and means for each element of F z, replace the scalar z with the matrix P. So, e.g. [ ] [ ] f11 f 12 f11 P f P := 12 P f 12 f 22 f 12 P f 22 P M. Peet Lecture 4: State-Space Theory 20 / 23
21 An LMI for Sector Regions of the Complex Plane Rise Time: [ r becomes Lemma 16. z ] [ ] [ ] [ ] z r = + z + z < 0 r 0 r }{{}}{{}}{{} F 0 F 1 F 2 The pole locations, z C of A satisfy z z r 2 if and only if there exists some P > 0 such that [ ] rp AP (AP ) T < 0 rp Settling Time: 4.6 t s + z + z 0 becomes Lemma 17. The pole locations, z C of A satisfy 2 Re x α if and only if there exists some P > 0 such that AP + (AP ) T + αp < 0 M. Peet Lecture 4: State-Space Theory 21 / 23
22 An LMI for Sector Regions of the Complex Plane Percent Overshoot: z + z ln Mp π z z Lemma 18. The pole locations, z C of A satisfyz + z ln Mp π z z if and only if there exists some P > 0 such that [ ] ln Mp (AP + (AP ) T ) π(ap (AP ) T ) π(ap (AP ) T ) T ln M p (AP + (AP ) T < 0 ) M. Peet Lecture 4: State-Space Theory 22 / 23
23 Discrete-Time Lyapunov Functions Next Time: LMIs for Controllability and Observability M. Peet Lecture 4: State-Space Theory 23 / 23
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