Nonlinear Control Systems

Transcription

1 Nonlinear Control Systems António Pedro Aguiar 5. Input-Output Stability DEEC PhD Course

2 Input-Output Stability y = Hu H denotes a mapping or operator that specifies y in terms of u u is an input signal that map the time interval [0, ) into the Euclidean space R m, that is, u : [0, ) R m Typical spaces of signals: L m space of piecewise continuous, bounded functions, where the norm of u : [0, ) R m is defined as L m 2 u L = sup u(t) < t 0 space of piecewise continuous, square-integrable functions with s Z u L2 = u T (t)u(t)dt < 0 L m p for 1 p < is the set of all piecewise continuous functions u : [0, ) R m such that Z 1 u Lp = u(t) dt«p p < 0 2

3 Input-Output Stability For technical reasons, we have to introduce the extended space L m e = {u : uτ Lm, τ [0, )} where Note that u τ is a truncation of u. j u(t), 0 t τ u τ (t) = 0, t > τ Example u(t) = t / L but its truncation belongs to L for every finite τ. Hence u(t) = t L e 3

4 Input-Output Stability Definition A mapping H : L m e L p e is L stable if α K, β 0 : (Hu) τ L α( u τ L ) + β for all u L m e and τ [0, ). It is finite-gain L stable if γ, β 0 : (Hu) τ L γ u τ L + β for all u L m e and τ [0, ). In particular for L, the system is L stable if for every bounded input u(t), the output Hu(t) is bounded. Given the system when is it L stable? ẋ = f(t, x, u), x(0) = x 0 y = h(t, x, u) 4

5 Theorem 5.3 Theorem 5.3 Consider the system ẋ = f(t, x, u), x(0) = x 0 (1) y = h(t, x, u) (2) Suppose that (1) is ISS, that is h satisfies the inequality α 1, α 2 K, η > 0 x(t) β( x 0, t t 0 ) + γ( sup u(τ) ) τ [0,t] h(t, x, u) α 1 ( x ) + α 2 ( x ) + η Then, for each x 0 R n, system (1)-(2) is L stable. 5

6 Proof y(t) α 1 (β( x 0, t t 0 )) + γ( sup u(τ) ) + α 2 ( u ) + η τ [0,t] since α(a + b) α(2a) + α(2b), then y(t) α 1 (2β( x 0, t t 0 )) + α 1 (2γ( sup u(τ) )) + α 2 ( u ) + η τ [0,t] Thus where y τ L γ 0 ( u τ L ) + β 0 γ 0 = α 1 2γ + α 2, β 0 = α 1 (2β( x 0, 0)) + η 6

7 Example Is the following system L stable? Consider ẋ 1 = x g(t)x 2 ẋ 2 = g(t)x 1 x u y = x 1 + x 2 V = 1 2 x2 1 + x2 2 Thus, V = x g(t)x 1 x 2 g(t)x 1 x 2 x x 2 u = x 4 1 x4 2 + x 2u Since 1 2 x 4 = 1 2 (x2 1 + x2 2 )4/2 = 1 2 x x x2 1 x x x x x4 2 x x 4 2 7

8 Example Then V 1 2 x 4 + x u = 1 2 (1 θ) x θ x 4 + x u, 0 < θ < 1 1 «2 u 1/3 2 (1 θ) x 4, x θ Thus V is an ISS-Lyapunov function and the state equation is ISS. Moreover h(t) = ˆ 1 1» x 1 h(t) 2 x x 2 Thus the system is L stable. 8

9 L 2 Gain Theorem 5.4 Consider the linear time-invariant system ẋ = Ax + Bu y = Cx + Du where A Hurwitz. The L 2 gain of the linear system is given by q γ = sup G(jw) 2 = λ max(g T ( jw)g(jw)) = σ max(g(jw)) w R where G(s) = C(sI A) 1 B + D. Moreover, in this case, y L2 = γ u L2 9

10 L 2 Gain Theorem 5.5 Consider the nonlinear system ẋ = f(x) + G(x)u, x(0) = x 0 y = h(x) where f(x) is locally Lipschitz, G R n m and h : R n R q are continuous, and f(0) = 0, h(0) = 0. Let γ be a positive number and suppose that there is a C 1, positive semidefinite function V (x) that satisfies the Hamilton-Jacobi inequality V x f(x) + 1 «V T 2γ 2 G(x)GT (x) + 1 x 2 ht (x)h(x) 0 Then, x 0 R n, the system is finite-gain L 2 stable and its L 2 gain is less than or equal to γ. 10

11 Proof V (x) = V V f(x) + x x G(x)u 1 2γ 2 V x G(x)GT (x) «V T 1 x 2 ht (x)h(x) + V x G(x)u By completing the squares, the right hand-side term is equal to 1 2 γ2 u 1 «V T 2 γ 2 GT (x) + 1 x 2 γ2 u y 2 Hence V V f(x) + x x G(x)u 1 2 γ2 u y 2 Integrating V (x(τ)) V (x 0 ) 1 Z τ 2 γ2 u(τ) 2 dτ 1 Z τ y(τ) 2 dτ Since V (x) 0 Z 1 τ y(τ) 2 dτ 1 Z τ γ2 u(τ) 2 dτ + 2V (x 0 ) 0 Taking the square roots and using a 2 + b 2 a + b for a, b 0 yields y τ L2 γ u τ L2 + p 2V (x 0 ) 11

12 The Small-Gain Theorem Let H 1 : L m e Lq e and H 2 : L q e L m e and Suppose both systems are finite-gain L stable, that is, y 1τ L γ 1 e 1τ L + β 1, e 1 L m e, τ [0, ) y 2τ L γ 2 e 2τ L + β 2, e 2 L q e, τ [0, ) Suppose that the feedback is well defined and let»»» u1 y1 e1 u =, y =, e = u 2 y 2 e 2 Then, the feedback connection is finite-gain L stable if γ 1 γ 2 < 1. 12

13 Proof Let e 1τ = u 1τ (H 2 e 2 ) τ and e 2τ = u 2τ (H 1 e 1 ) τ. Then Thus Since γ 1 γ 2 < 1, then e 1τ L u 1τ L + (H 2 e 2 ) τ L u 1τ L + γ 2 e 2τ L + β 2 u 1τ L + γ 2 ( u 2τ L + γ 1 e 1τ L + β 1 ) + β 2 = γ 1 γ 2 e 1τ L + ( u 1τ L + γ 2 u 2τ L + β 2 + γ 2 β 1 ) (1 γ 1 γ 2 ) e 1τ L = u 1τ L + γ 2 u 2τ L + β 2 + γ 2 β 1 e 1τ L for all τ [0, ). Similarly, Thus, using e 2τ L 1 1 γ 1 γ 2 ( u 1τ L + γ 2 u 2τ L + β 2 + γ 2 β 1 ) 1 1 γ 1 γ 2 ( u 2τ L + γ 1 u 1τ L + β 1 + γ 1 β 2 ) e L e 1 L + e 2 L we conclude the result. 13

14 Example 1 Suppose we have two systems: H1 : linear time-invariant system, where G(s) is a Hurwitz square transfer function matrix H2: y 2 = φ(t, e 2 ) such that φ(t, y) γ 2 y Some remarks H1 is finite gain L 2 stable with L 2 gain given by γ 1 = sup w R G(jw) 2 H2 is finite gain L 2 stable with L 2 gain less that or equal to γ 2 Thus, the interconnection (assuming that is well defined) is finite-gain L 2 stable if γ 1 γ 2 < 1 14

15 Example 2: Robustness of the controller with respect to unmodeled actuator dynamics Consider the system ẋ = f(t, x, v + d 1 (t)) plant dynamics ɛż = Az + B[u + d 2 (t)] fast actuator dynamics v = Cz where f is a smooth function, ɛ > 0 is a small parameter (fast dynamics), A is Hurwitz, CA 1 B = I, and d 1, d 2 L are disturbance signals with d 1, d 2 L. Goal: Attenuate the effect of the disturbance on the state x This can be achieved if the feedback control law can be designed such that the closed-loop input-output map from (d 1, d 2 ) to x is finite-gain L stable with L gain less than some given tolerance δ > 0. 15

16 Example 2 Neglecting the actuator dynamics by setting ɛ = 0 we have z = A 1 B(u + d 2 ) and v = CA 1 B(u + d 2 ) = u + d 2 Thus, we obtain ẋ = f(t, x, u + d), d = d 1 + d 2 Suppose we design a control law u = K(t, x) such that x L γ d L + β (3) for some γ < δ. Is the controller robust with respect to the unmodeled actuator dynamics? 16

17 Closed-loop dynamics: Example 2 ẋ = f(t, x, cz + d 1 (t)) ɛż = Az + B[K(t, x) + d 2 (t)] Let η be the error, that is, Then, η = z z ɛ=0 = z + A 1 B[K(t, x) + d 2 (t)] ɛ η = Az + B[K( ) + d 2 ] + ɛa 1 B[ K( ) + d 2 ] = Aη AA 1 B[K( ) + d 2 ] + B[K( ) + d 2 ] + ɛa 1 B[ K( ) + d 2 ] Let e 2 := K( ) + d 2 then Notice also that η = 1 ɛ Aη + A 1 Be 2 ẋ = f(t, x, Cη CA 1 B[K( ) + d 2 ] + d 1 ) = f(t, x, K( ) + Cη + d) = f(t, x, K( ) + e 1 ) where CA 1 B = I and e 1 := Cη + d 17

18 Example 2 Assume that Then using (3) K t + K x f(t, x, K(t, x) + e 1) c 1 x + c 2 e 1 y 1 L c 1 γ e 1 L + c 1 β + c 2 e 1 L = γ e 1 L + β 1 where γ 1 = c 1 γ + c 2 and β 1 = c 1 β. Since H2 is a linear system then y 2 L ɛγ f e 2 L + β 2 Thus, we can conclude that if ɛγ 1 γ f < 1, then the closed-loop system is L stable. 18