Analysis of Systems with State-Dependent Delay

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Analysis of Systems with State-Dependent Delay Matthew M. Peet Arizona State University Tempe, AZ USA American Institute of Aeronautics and Astronautics Guidance, Navigation and Control Conference Boston, MA August 20, 2013

Systems with State-Dependent Delay State-Dependent Delay arises which communication distance changes with time. Any Moving System. Sonar (Speed of Sound) EM (Speed of Light) Example: Position Measurement using Sonar xt () ( ) Propagation Distance: 2 ˆx(t) = x 0 + x(t) Delay is τ(t) = 2 x0+x(t) c. Feedback Dynamics: neglecting inertia... ( ẋ(t) = ax t 2x 0 c x 0 2 c x(t) ) M. Peet State-Dependent Delay: Introduction 2 / 18

Stability of Systems with State-Dependent Delay Consider the general class of systems with state-dependent delay: ( ( ) ) ẋ(t) = f x(t), x t g(x(t)) Note: State x(t) enters into the independent argument, t. Contrast to the fixed-delay case: ẋ(t) = f (x ( t τ )) Linear/Affine Form: ẋ(t) = A 0 x(t) + A 1 x(t τ 0 b T x(t)) Question: Given A R n, τ 0 > 0 and τ R n, Determine whether the system is stable. Estimate the rate of decay. M. Peet State-Dependent Delay: Introduction 3 / 18

Why State-Dependent Delay? The most common source of delay is Propagation Time. A time-delay system is the interconnection of an ODE with a transport PDE in the feedback channel. ẋ(t) = A 0 x(t) + A i x(t τ i ) ODE: The system G 1 ẋ 1 (t) = Ax 1 (t) + Bu 1 (t) y 1 (t) = Cx 1 (t) + Du 1 (t) [ ] [ [ ] A B A0 A1 A n ] G 1 (A, B, C, D) C D I 0 PDE: The system G 2 t x 2(t, s) = s x 2(t, s) x 2 (t, 0) = u 2 (t), y 2 (t) = [ x 2 ( τ 1 ) x 2 ( τ K ) ] T Of course, the solution is just x 2 (t, s) = u 2 (t s). G 2 y 3 y 2 y 1 u M. Peet State-Dependent Delay: Introduction 4 / 18

Systems with State-Dependent Delay Restrictions on the Model: For physical systems: Delay is always positive Delay is usually affine in the state Lower and upper bounds for the delay Other types of systems are often ill-posed or otherwise pathological [Verriest, 2013] Global Stability is not a well-defined problem for this model. Assumptions: In this talk, we use scalar systems of the form ẋ(t) = ax(t b cx(t)) where a, c R, b > 0. To ensure b + cx(t) > 0, we bound the state as x(t) d. Also leads to the bound b + cx(t) τm. Note: The extension to R n is not hard. M. Peet State-Dependent Delay: Introduction 5 / 18

Lyapunov-Krasovskii Functionals For linear fixed delay systems: ẋ(t) = Ax(t) + Bx(t τ) Theorem 1. The discrete-delay system is stable if and only if there exist continuous functions M and N such that V (φ) α φ and V (φ) 0 where V (φ) = τ [ ] T [ ] φ(0) φ(0) M(s) ds + φ(s) T N(s, θ)φ(θ)dsdθ φ(s) φ(s) τ τ Note: The functional is parameterized by unknown functions M and M. Problem: How to numerically compute M and N such that Answer: Convex Optimization V (φ) > α φ V (φ) < 0 M. Peet State-Dependent Delay: Introduction 6 / 18

Tractable or Intractable? Convex Optimization Problem: max bx subject to Ax C The problem is convex optimization if C is a convex cone. b and A are affine. Computational Tractability: Convex Optimization over C is, in general, tractable if The set membership test for y C is in P. x is finite dimensional. M. Peet State-Dependent Delay: Introduction 7 / 18

Parametrization of Lyapunov-Krasovskii Functionals Problem 1: Is the set of decision variables finite-dimensional? Decision variables are the functions M and N ( 0 [φ(0) ] ) T ( [φ(0) ] ) V (φ) = ( ) Z M(s)Z ds τ φ(t) φ(s) φ(s) + ( ) τ φ(t) ( ) T ( ) ( ) Z φ(s) N(s, θ)z φ(θ) dsdθ τ φ(t) Solution: Suppose M and N are polynomials of bounded degree. M(s) = c 1 T Z(s), N(s, θ) = c 2 T Z(s, θ) Problem 2: How to enforce positivity of the L-K functional? Need constraints on c 1 and c 2. M. Peet State-Dependent Delay: Introduction 8 / 18

Positivity Constraints for Polynomials Sum-of-Squares: p(x) 0 if p(x) = i g i (x) 2, denoted p Σ s Lemma 2 (Parametrization of Sums-of-Squares). Given multivariate polynomial p of degree 2d, p Σ s (p(x) 0 for all x R n ) if and only if there exists a positive matrix M S q such that p(x) = Z d (x) T MZ d (x) where z d is the vector of monomials of degree d or less. Lemma 3 (Polynomial Positivity on a Subset of R n ). Given polynomial p, p(x) 0 for all x {x : g i (x) 0} if there exist s i Σ s such that p(s) = s 0 (x) + g i (x)s i (x) i M. Peet State-Dependent Delay: Introduction 9 / 18

Positivity Constraints for Lyapunov Functionals Lemma 4. Suppose there exist S Σ s and polynomial T such that V (φ(0), φ(s), s) φ(0) 2 = S(φ(0), φ(s), s) + T (φ(0), s) with T (φ(0), s)ds = 0. Then τ(x(t)) for any φ C[ τ m, 0] We can tighten this a bit Restrict s [ τ(φ(s)), 0] Restrict φ(s) d ( ) V (φ(0), φ(s), s)ds αφ(0) 2 τ φ(t) {s, φ : g 1 (s, φ(s)) = s(s + τ(φ(s))) 0} {s, φ : g 2 (s, φ(s)) = d φ(s) 2 0} M. Peet State-Dependent Delay: Introduction 10 / 18

Positivity Constraints for Lyapunov Functionals Lemma 5. Suppose there exist S, S 1, S 2, S 3 Σ s and polynomial T such that V (φ(0), φ(s), s) φ(0) 2 = S + g 1 (s, φ(s))s 1 + g 2 (s, φ(s))s 2 + g 2 (s, φ(0))s 3 + T (φ(0), s) with T (φ(0), s)ds = 0. Then τ(x(t)) for any φ d. ( ) V (φ(0), φ(s), s)ds αφ(0) 2 τ φ(t) Where Recall S, S 1, S 2, S 3, T and V are the decision variables Represented using positive matrices. Constraints are equalities and matrix positivity (SDP). M. Peet State-Dependent Delay: Introduction 11 / 18

Positivity Constraints for Lyapunov Functionals Lemma 6. Suppose there exists a positive matrix Q 0 such that V 2 (φ(s), φ(θ), s, θ) = Z(s, φ(s)) T QZ(θ, φ(θ)) Then ( ) τ φ(t) for any φ C[ τ m, 0]. Positivity Constraints ( ) V 2 (φ(s), φ(θ), s, θ) ds dθ 0 τ φ(t) Constraints are equalities and matrix positivity (SDP). Convex Optimization Positive Lyapunov Functionals are represented using vectors and matrix positivity constraints. M. Peet State-Dependent Delay: Introduction 12 / 18

The Derivative Now Recall the dynamics: ẋ(t) = ax(t τ 0 + bx(t)) τ(x(t)) = τ 0 + bx(t) With Lyapunov Functional V (t) = ) ( )V 1 (x(0), x(s), s ds+ τ x(t) The derivative is ( ) τ x(t) ) ( ) V 2 (x(s), s, x(θ), θ ds dθ τ x(t) ) V (t) = ( ) V 3 (x(t), x(t τ(x(t))), x(t + s), s ds τ x(t) ) + ( ) ( )V 4 (x(t + s), x(t + θ), s, θ ds dθ τ x(t) τ x(t) where V 4 (x θ, x ξ, θ, ξ) = θ V 2(x θ, x ξ, θ, ξ) + ξ V 2(x θ, x ξ, θ, ξ). (1) M. Peet State-Dependent Delay: Introduction 13 / 18

The Derivative ẋ(t) = ax(t τ 0 + bx(t)) τ(x(t)) = τ 0 + bx(t) ) V (t) = ( ) V 3 (x(t), x(t τ(x(t))), x(t + s), s ds τ x(t) ) + ( ) ( ) V 4 (x(t + s), x(t + θ), s, θ ds dθ τ x(t) τ x(t) V 3 (x t, x τ, x s, s) = 1 τ(x t ) (abx τ 1)V 1 (x t, x τ, τ(x τ )) + 1 τ(x t ) V 1(x t, x t, 0) + ax τ x V 1(x t, x s, s) θ V 1(x t, x s, θ) + (2abx τ 2)V 2 (x τ, x s, τ(x t ), θ) + 2V 2 (x t, x s, 0, θ) M. Peet State-Dependent Delay: Introduction 14 / 18

Stability Test Suppose there exist S i Σ s for i = 1, 2, 3, L i Σ s for i = 1, 2, 3, 4, some ɛ j > 0 for j = 1, 2, polynomial R 1 (x 0, θ), R 2 (x 0, x 1, θ) and matrices M, N > 0, such that 1) V 1 (x 0, x 2, θ) + R 1 (x 0, θ) ɛ 1 x 2 0 3 i=1 S i(x 0, x 2, θ)g i (x 0, x 2, θ) Σ s, 2) V 3 (x 0, x 1, x 2, θ) + R 2 (x 0, x 1, θ) ɛ 2 x 2 0 4 i=1 L i(x 0, x 1, x 2, θ)g i (x 0, x 1, x 2, θ) Σ s, 3) V 2 (x 2, x 3, θ, ξ) = Z T d (x 2, θ)mz d (x 3, ξ), 4) V 4 (x 2, x 3, θ, ξ) = Z T d (x 2, θ)nz d (x 3, ξ), 5) τ(x 0) R 1(x 0, θ)dθ = 0, 6) τ(x 0) R 2(x 0, x 1, θ)dθ = 0, Then the system is asymptotically stable for all x t Ω, where Ω is defined as Ω := {x t C : x t τ 0 /(2b)}. M. Peet State-Dependent Delay: Introduction 15 / 18

Numerical Validation xt () x 0 ẋ(t) = ax(t τ 0 + bx(t)) a= 0.1 a= 0.5 a= 1 τ 0 =0.1 b [4e-4, 1] b [1e-4, 1] b [2e-4, 1] τ 0 =0.5 b [6e-4, 2] b [3e-4, 2] b [3e-3, 0.02] τ 0 =1 b [7e-4, 3] b [8e-4, 2] b [3e-3, 0.02] Table: The minimum and maximum stable values of b for a fixed a and τ 0. M. Peet State-Dependent Delay: Introduction 16 / 18

Numerical Validation xt () x 0 ẋ(t) = ax(t τ 0 + bx(t)) 0.9 0.8 0.7 0.6 0.5 x(t) 0.4 0.3 0.2 0.1 0 0.1 0 10 20 30 40 50 60 70 t Figure: Simulation Results using a = 0.1, b = 1, τ 0 = 6, and initial condition φ(θ) = 0.5 sin(θ) M. Peet State-Dependent Delay: Introduction 17 / 18

Conclusions: A Difficult Problem: Lyapunov Stability Test Convexifies the problem Relies on SDP. Complexity depends on Accuracy. Numerical Code Produced: Not currently posted Must generalize to multidimensional systems Practical Implications The effect is small, but finite. Future Work Joint Positivity Controller Synthesis Will be Available for download at http://control.asu.edu M. Peet State-Dependent Delay: Conclusion 18 / 18