Download Control Lyapunov Functions-Control of Non Linear Systems-Lecture Slides and more Slides Nonlinear Control Systems in PDF only on Docsity! Nonlinear Systems and Control Lecture # 30 Stabilization Control Lyapunov Functions – p. 1/12 Docsity.com ẋ = f(x) + g(x)u, f(0) = 0, x ∈ Rn, u ∈ R Suppose there is a continuous stabilizing state feedback control u = ψ(x) such that the origin of ẋ = f(x) + g(x)ψ(x) is asymptotically stable By the converse Lyapunov theorem, there is V (x) such that ∂V ∂x [f(x) + g(x)ψ(x)] < 0, ∀ x ∈ D, x 6= 0 If u = ψ(x) is globally stabilizing, then D = Rn and V (x) is radially unbounded – p. 2/12 Docsity.com Theorem: Let V (x) be a CLF for ẋ = f(x) + g(x)u, then origin is stabilizable by u = ψ(x), where ψ(x) = − ∂V ∂x f+ q (∂V ∂x f) 2 +(∂V ∂x g) 4 (∂V ∂x g) , if ∂V ∂x g 6= 0 0, if∂V ∂x g = 0 The function ψ(x) is continuous for all x ∈ D0 (a neighborhood of the origin) including x = 0. If f and g are smooth, then ψ is smooth for x 6= 0. If V is a global CLF, then the control u = ψ(x) is globally stabilizing Sontag’s Formula – p. 5/12 Docsity.com Proof: For properties of ψ, see Section 9.4 of [88] ∂V ∂x [f(x) + g(x)ψ(x)] If ∂V ∂x g(x) = 0, V̇ = ∂V ∂x f(x) < 0 for x 6= 0 If ∂V ∂x g(x) 6= 0 V̇ = ∂V ∂x f − [ ∂V ∂x f + √ ( ∂V ∂x f )2 + ( ∂V ∂x g )4 ] = − √ ( ∂V ∂x f )2 + ( ∂V ∂x g )4 < 0 for x 6= 0 – p. 6/12 Docsity.com How can we find a CLF? If we know of any stabilizing control with a corresponding Lyapunov function V , then V is a CLF Feedback Linearization ẋ = f(x) +G(x)u, z = T (x), ż = (A−BK)z P (A−BK) + (A−BK)TP = −Q, Q = QT > 0 V = zTPz = T T (x)PT (x) is a CLF Backstepping – p. 7/12 Docsity.com Method Expression FL-u −ax+ bx3 − kx FL-CLS ẋ = −kx CLF-u −ax+ bx3 − x √ (a− bx2)2 + 1 CLF-CLS −x √ (a− bx2)2 + 1 Method Small |x| Large |x| FL-u (−a+ k)x bx3 FL-CLS ẋ = −kx ẋ = −kx CLF-u −(a+ √ a2 + 1)x −ax CLF-CLS − √ a2 + 1x −bx3 – p. 10/12 Docsity.com Lemma: Let V (x) be a CLF for ẋ = f(x) + g(x)u and suppose ∂V ∂x (0) = 0. Then, Sontag’s formula has a gain margin [1 2 ,∞); that is, u = kψ(x) is stabilizing for all k ≥ 1 2 Proof: Let q(x) = 1 2 − ∂V ∂x f + √ ( ∂V ∂x f )2 + ( ∂V ∂x g )4 q(0) = 0, ∂V ∂x g 6= 0 ⇒ q > 0 ∂V ∂x g = 0 ⇒ q = −∂V ∂x f > 0 for x 6= 0 q(x) is positive definite – p. 11/12 Docsity.com u = kψ(x) ⇒ ẋ = f(x) + g(x)kψ(x) V̇ = ∂V ∂x f + ∂V ∂x gkψ ∂V ∂x g = 0 ⇒ V̇ = ∂V ∂x f < 0 for x 6= 0 ∂V ∂x g 6= 0, V̇ = −q + q + ∂V ∂x f + ∂V ∂x gkψ q + ∂V ∂x f + ∂V ∂x gkψ = − ( k − 1 2 ) ∂V ∂x f + √ ( ∂V ∂x f )2 + ( ∂V ∂x g )4 – p. 12/12 Docsity.com
