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# Lect_32 - Nonlinear Systems and Control Lecture 32 Robust...

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Unformatted text preview: Nonlinear Systems and Control Lecture # 32 Robust Stabilization Sliding Mode Control – p. 1/1 Example x1 = x2 ˙ x2 = h(x) + g (x)u, ˙ g (x) ≥ g0 > 0 Sliding Manifold (Surface): s = a1 x1 + x2 = 0 s(t) ≡ 0 ⇒ x1 = −a1 x1 ˙ a1 > 0 ⇒ lim x1 (t) = 0 t→∞ How can we bring the trajectory to the manifold s = 0? How can we maintain it there? – p. 2/1 s = a1 x1 + x2 = a1 x2 + h(x) + g (x)u ˙ ˙ ˙ Suppose a1 x2 + h(x) g (x) ≤ ̺(x) V = 1 s2 2 ˙ V = ss = s[a1 x2 + h(x)]+ g (x)su ≤ g (x)|s|̺(x)+ g (x)su ˙ β (x) ≥ ̺(x) + β0 , s > 0, β0 > 0 u = −β (x) ˙ V ≤ g (x)|s|̺(x) − g (x)β (x)|s| ˙ V ≤ g (x)|s|̺(x) − g (x)(̺(x) + β0 )|s| = −g (x)β0 |s| – p. 3/1 s < 0, u = β (x) ˙ V ≤ g (x)|s|̺(x) + g (x)su = g (x)|s|̺(x) − g (x)β (x)|s| ˙ V ≤ g (x)|s|̺(x) − g (x)(̺(x) + β0 )|s| = −g (x)β0 |s| sgn(s) = 1, −1, s>0 s<0 u = −β (x) sgn(s) ˙ V ≤ −g (x)β0 |s| ≤ −g0 β0 |s| √ ˙ V ≤ −g0 β0 2V – p. 4/1 ˙ V ≤ −g0 β0 √ 2V √ dV √ ≤ −g0 β0 2 dt V 2 √ V (s(t)) V V (s(0)) V (s(t)) ≤ √ ≤ −g0 β0 2 t 1 V (s(0)) − g0 β0 √ t 2 |s(t)| ≤ |s(0)| − g0 β0 t s(t) reaches zero in ﬁnite time Once on the surface s = 0, the trajectory cannot leave it – p. 5/1 s=0 What is the region of validity? – p. 6/1 x1 = x2 ˙ x2 = h(x) − g (x)β (x)sgn(s) ˙ x1 = −a1 x1 + s ˙ s = a1 x2 + h(x) − g (x)β (x)sgn(s) ˙ ss ≤ −g0 β0 |s|, ˙ if β (x) ≥ ̺(x) + β0 V1 = 1 x2 21 ˙ V1 = x1 x1 = −a1 x2 + x1 s ≤ −a1 x2 + |x1 |c ≤ 0 ˙ 1 1 ∀ |s| ≤ c and |x1 | ≥ Ω= |x1 | ≤ Ω is positively invariant if c a1 c a1 , |s | ≤ c a1 x2 + h(x) g (x) ≤ ̺(x) over Ω – p. 7/1 Ω= |x1 | ≤ c a1 , |s | ≤ c x2 T s= rr 0 HH HH rr c H r HH rr H c/a1 E rr H HH rr x1 HH rr HH rr HH rr r H a1 x2 + h(x) g (x) ≤ k1 < k, ∀ x ∈ Ω u = −k sgn(s) – p. 8/1 Chattering f f Sliding manifold f f¡ ¡ f ! ¡f ¡ rf rr  s < 0 fr s>0 f¡ ¡f ! ¡f ra rr f f rr  rr f How can we reduce or eliminate chattering? – p. 9/1 Reduce the amplitude of the signum function s = a1 x2 + h(x) + g (x)u ˙ u=− ˆ [a1 x2 + h(x)] g (x) ˆ +v s = δ (x) + g (x)v ˙ δ (x) = a1 δ (x) g (x) g (x) ˆ x2 + h(x) − h(x) 1− g (x) ˆ g (x) ˆ g (x) ≤ ̺(x), β (x) ≥ ̺(x) + β0 v = −β (x) sgn(s) – p. 10/1 Replace the signum function by a high-slope saturation function s u = −β (x) sat ε sat(y ) = y, sgn(y ), sgn(y ) T 1 sat E y −1 if |y | ≤ 1 if |y | > 1 y ε 1 T ¢¢ ¢ ¢ε ¢ ¢ −1 E y – p. 11/1 How can we analyze the system? For |s| ≥ ε, u = −β (x) sgn(s) With c ≥ ε Ω = |x1 | ≤ c , a1 |s| ≤ c is positively invariant The trajectory reaches the boundary layer {|s| ≤ ε}in ﬁnite time The boundary layer is positively invariant – p. 12/1 Inside the boundary layer: x1 = −a1 x1 + s ˙ s = a1 x2 + h(x) − g (x)β (x) ˙ s ε x1 x1 ≤ −a1 x2 + |x1 |ε ˙ 1 0<θ<1 x1 x1 ≤ −(1 − ˙ θ )a1 x2 , 1 ∀ |x1 | ≥ ε θa1 The trajectories reach the positively invariant set Ωε = {|x1 | ≤ ε θa1 , |s| ≤ ε} in ﬁnite time – p. 13/1 What happens inside Ωε ? Find the equilibrium points 0 = −a1 x1 + s = x2 , φ(x1 ) = 0 = a1 x2 + h(x) − g (x)β (x) s ε h(x) a1 g (x)β (x) x2 =0 x1 = εφ(x1 ) Suppose x1 = εφ(x1 ) has an isolated root x1 = εk1 ¯ h(0) = 0 ⇒ x1 = 0 ¯ – p. 14/1 z1 = x1 − x1 , ¯ z2 = s − a1 x1 ¯ x2 = −a1 x1 + s = −a1 (x1 − x1 ) + s − a1 x1 = −a1 z1 + z2 ¯ ¯ z1 = −a1 x1 + s = −a1 z1 + z2 ˙ z2 = a1 x2 + h(x) − g (x)β (x) ˙ s ε = a1 (z2 − a1 z1 ) + h(x) − g (x)β (x) z2 = ℓ(z ) − g (x)β (x) ˙ ℓ(z ) = a1 (z2 − a1 z1 ) + a1 g (x)β (x) z2 + a1 x1 ¯ ε z2 ε h(x) a1 g (x)β (x) − x1 ¯ ε – p. 15/1 z 1 = − a1 z 1 + z 2 , ˙ z2 = ℓ(z ) − g (x)β (x) ˙ z2 ε ℓ(0) = 0, |ℓ(z )| ≤ ℓ1 |z1 | + ℓ2 |z2 | g (x)β (x) ≥ g0 β0 12 2 V = 2 z1 + 1 z2 2 ˙ V = z1 (−a1 z1 + z2 ) + z2 ℓ(z ) − g (x)β (x) 2 2 ˙ V ≤ −a1 z1 + (1 + ℓ1 )|z1 | |z2 | + ℓ2 z2 − z2 ε g0 β0 ε 2 z2 – p. 16/1 2 2 ˙ V ≤ −a1 z1 + (1 + ℓ1 )|z1 | |z2 | + ℓ2 z2 − ˙ V ≤− |z 1 | |z 2 | T g0 β0 ε 1 − 2 (1 + ℓ1 ) a1 g0 β0 ε − 1 (1 + ℓ1 ) 2 − ℓ2 2 z2 |z 1 | |z 2 | Q det(Q) = a1 g0 β0 ε h(0) = 0 ⇒ h(0) = 0 ⇒ − ℓ2 1 − 4 (1 + ℓ1 )2 lim x(t) = 0 t→∞ lim x(t) = t→∞ x1 ¯ 0 Read Section 14.1.1 – p. 17/1 ...
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