Output Feedback Stabilization of Nonlinear Systems: Min. Phase & Passive Maps, Slides of Nonlinear Control Systems

Download Output Feedback Stabilization of Nonlinear Systems: Min. Phase & Passive Maps and more Slides Nonlinear Control Systems in PDF only on Docsity! Nonlinear Systems and Control Lecture # 31 Stabilization Output Feedback – p. 1/12 Docsity.com In general, output feedback stabilization requires the use of observers. In this lecture we deal with three simple cases where an observer is not needed Minimum Phase Relative Degree One Systems Passive systems System with Passive maps from the input to the derivative of the output – p. 2/12 Docsity.com High-Gain Feedback: u = −ky, k > 0 η̇ = f0(η, y), ẏ = γ(x)[−ky − α(x)] V (η, y) = V1(η) + 1 2 y2 V̇ = ∂V1 ∂η f0(η, y) − kγ(x)y 2 − α(x)γ(x)y V̇ = ∂V1 ∂η f0(η, 0) + ∂V1 ∂η [f0(η, y) − f0(η, 0)] − kγ(x)y2 − α(x)γ(x)y V̇ ≤ −c3‖η‖ 2 + c4L1‖η‖ |y| − kγ0y 2 + L2‖η‖ |y| + L3y 2 – p. 5/12 Docsity.com V̇ ≤ −c3‖η‖ 2 + c4L1‖η‖ |y| − kγ0y 2 + L2‖η‖ |y| + L3y 2 V̇ ≤ − [ ‖η‖ |y| ]T [ c3 − 1 2 (L2 + c4L1) −1 2 (L2 + c4L1) (kγ0 − L3) ] ︸ ︷︷ ︸ Q [ ‖η‖ |y| ] det(Q) = c3(kγ0 − L3) − 1 4 (L2 + c4L1) 2 det(Q) > 0 for k > 1 c3γ0 [ c3L3 + 1 4 (L2 + c4L1) 2 ] The origin of the closed-loop system is exponentially stable If the assumptions hold globally, it is globally exp. stable – p. 6/12 Docsity.com Passive Systems Suppose the system ẋ = f(x, u), y = h(x, u) is passive (with a positive definite storage function) and zero-state observable Then, it can be stabilized by u = −ψ(y), ψ(0) = 0, yTψ(y) > 0, ∀ y 6= 0 V̇ ≤ uTy = −yTψ(y) ≤ 0 V̇ = 0 ⇒ y = 0 ⇒ u = 0 y(t) ≡ 0 ⇒ x(t) ≡ 0 – p. 7/12 Docsity.com For 1 ≤ i ≤ m zi is the output of bis s+ ai driven by yi ui = −ψi(zi) ai, bi > 0, ψi(0) = 0, ziψi(zi) > 0 ∀ zi 6= 0 żi = −aizi + biẏi Use Vc(x, z) = V (x) + m∑ i=1 1 bi ∫ zi 0 ψi(σ) dσ to prove asymptotic stability of the origin of the closed-loop system – p. 10/12 Docsity.com Example: Stabilize the pendulum mℓθ̈ +mg sin θ = u at θ = δ1 using feedback from θ x1 = θ − δ1, x2 = θ̇ ẋ1 = x2, ẋ2 = 1 mℓ [−mg sin θ + u] u = mg sin θ − kpx1 + v, kp > 0 ẋ1 = x2, ẋ2 = − kp mℓ x1 + 1 mℓ v y = x1, ẏ = x2 – p. 11/12 Docsity.com V = 1 2 kpx 2 1 + 1 2 mℓx22 V̇ = kpx1x2 +mℓx2 [ − kp mℓ x1 + 1 mℓ v ] V̇ = x2v = ẏv With v = 0, x2(t) ≡ 0 ⇒ x1(t) ≡ 0 - bs s+a - y z ξ̇ = −aξ + y, z = b(−aξ + y) v = −kdz, kd > 0 u = mg sin θ − kp(θ − δ1) − kdz – p. 12/12 Docsity.com