Download Robust Stabilization of Nonlinear Systems: Sliding Mode Control and more Slides Nonlinear Control Systems in PDF only on Docsity! Nonlinear Systems and Control Lecture # 32 Robust Stabilization Sliding Mode Control – p. 1/17 Docsity.com Example ẋ1 = x2 ẋ2 = h(x) + g(x)u, g(x) ≥ g0 > 0 Sliding Manifold (Surface): s = a1x1 + x2 = 0 s(t) ≡ 0 ⇒ ẋ1 = −a1x1 a1 > 0 ⇒ lim t→∞ x1(t) = 0 How can we bring the trajectory to the manifold s = 0? How can we maintain it there? – p. 2/17 Docsity.com V̇ ≤ −g0β0 √ 2V dV √ V ≤ −g0β0 √ 2 dt 2 √ V ∣ ∣ ∣ ∣ V (s(t)) V (s(0)) ≤ −g0β0 √ 2 t √ V (s(t)) ≤ √ V (s(0)) − g0β0 1 √ 2 t |s(t)| ≤ |s(0)| − g0β0 t s(t) reaches zero in finite time Once on the surface s = 0, the trajectory cannot leave it – p. 5/17 Docsity.com s=0 What is the region of validity? – p. 6/17 Docsity.com ẋ1 = x2 ẋ2 = h(x) − g(x)β(x)sgn(s) ẋ1 = −a1x1 + s ṡ = a1x2 + h(x) − g(x)β(x)sgn(s) sṡ ≤ −g0β0|s|, if β(x) ≥ ̺(x) + β0 V1 = 1 2x 2 1 V̇1 = x1ẋ1 = −a1x21 + x1s ≤ −a1x21 + |x1|c ≤ 0 ∀ |s| ≤ c and |x1| ≥ c a1 Ω = { |x1| ≤ c a1 , |s| ≤ c } Ω is positively invariant if ∣ ∣ ∣ ∣ a1x2 + h(x) g(x) ∣ ∣ ∣ ∣ ≤ ̺(x) over Ω – p. 7/17 Docsity.com Reduce the amplitude of the signum function ṡ = a1x2 + h(x) + g(x)u u = − [a1x2 + ĥ(x)] ĝ(x) + v ṡ = δ(x) + g(x)v δ(x) = a1 [ 1 − g(x) ĝ(x) ] x2 + h(x) − g(x) ĝ(x) ĥ(x) ∣ ∣ ∣ ∣ δ(x) g(x) ∣ ∣ ∣ ∣ ≤ ̺(x), β(x) ≥ ̺(x) + β0 v = −β(x) sgn(s) – p. 10/17 Docsity.com Replace the signum function by a high-slope saturation function u = −β(x) sat ( s ε ) sat(y) = { y, if |y| ≤ 1 sgn(y), if |y| > 1 - 6 −1 1 y sgn(y) - 6 −1 1 yε sat (y ε ) – p. 11/17 Docsity.com How can we analyze the system? For |s| ≥ ε, u = −β(x) sgn(s) With c ≥ ε Ω = { |x1| ≤ ca1 , |s| ≤ c } is positively invariant The trajectory reaches the boundary layer {|s| ≤ ε}in finite time The boundary layer is positively invariant – p. 12/17 Docsity.com z1 = x1 − x̄1, z2 = s − a1x̄1 x2 = −a1x1 +s = −a1(x1 − x̄1)+s−a1x̄1 = −a1z1 +z2 ż1 = −a1x1 + s = −a1z1 + z2 ż2 = a1x2 + h(x) − g(x)β(x) s ε = a1(z2 − a1z1) + h(x) − g(x)β(x) z2 + a1x̄1 ε ż2 = ℓ(z) − g(x)β(x) z2 ε ℓ(z) = a1(z2 − a1z1) + a1g(x)β(x) [ h(x) a1g(x)β(x) − x̄1 ε ] – p. 15/17 Docsity.com ż1 = −a1z1 + z2, ż2 = ℓ(z) − g(x)β(x) z2 ε ℓ(0) = 0, |ℓ(z)| ≤ ℓ1|z1| + ℓ2|z2| g(x)β(x) ≥ g0β0 V = 1 2 z21 + 1 2 z22 V̇ = z1(−a1z1 + z2) + z2 [ ℓ(z) − g(x)β(x)z2 ε ] V̇ ≤ −a1z21 + (1 + ℓ1)|z1| |z2| + ℓ2z22 − g0β0 ε z22 – p. 16/17 Docsity.com V̇ ≤ −a1z21 + (1 + ℓ1)|z1| |z2| + ℓ2z22 − g0β0 ε z22 V̇ ≤ − [ |z1| |z2| ]T [ a1 −12(1 + ℓ1) −12(1 + ℓ1) ( g0β0 ε − ℓ2 ) ] ︸ ︷︷ ︸ Q [ |z1| |z2| ] det(Q) = a1 ( g0β0 ε − ℓ2 ) − 14(1 + ℓ1) 2 h(0) = 0 ⇒ lim t→∞ x(t) = 0 h(0) 6= 0 ⇒ lim t→∞ x(t) = [ x̄1 0 ] Read Section 14.1.1 – p. 17/17 Docsity.com
