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# Lect_25 - Nonlinear Systems and Control Lecture 25...

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Nonlinear Systems and Control Lecture # 25 Stabilization Basic Concepts & Linearization – p.1/17

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We want to stabilize the system ˙ x = f ( x, u ) at the equilibrium point x = x ss Steady-State Problem: Find steady-state control u ss s.t. 0 = f ( x ss , u ss ) x δ = x x ss , u δ = u u ss ˙ x δ = f ( x ss + x δ , u ss + u δ ) def = f δ ( x δ , u δ ) f δ (0 , 0) = 0 u δ = γ ( x δ ) u = u ss + γ ( x x ss ) – p.2/17
State Feedback Stabilization: Given ˙ x = f ( x, u ) [ f (0 , 0) = 0] find u = γ ( x ) [ γ (0) = 0] s.t. the origin is an asymptotically stable equilibrium point of ˙ x = f ( x, γ ( x )) f and γ are locally Lipschitz functions – p.3/17

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Linear Systems ˙ x = Ax + Bu ( A, B ) is stabilizable (controllable or every uncontrollable eigenvalue has a negative real part) Find K such that ( A BK ) is Hurwitz u = Kx Typical methods: Eigenvalue Placement Eigenvalue-Eigenvector Placement LQR – p.4/17
Linearization ˙ x = f ( x, u ) f (0 , 0) = 0 and f is continuously differentiable in a domain D x × D u that contains the origin ( x = 0 , u = 0) ( D x R n , D u R p ) ˙ x = Ax + Bu A = ∂f ∂x ( x, u ) vextendsingle vextendsingle vextendsingle vextendsingle x =0 ,u =0 ; B = ∂f ∂u ( x, u ) vextendsingle vextendsingle vextendsingle vextendsingle x =0 ,u =0 Assume ( A, B ) is stabilizable. Design a matrix K such that ( A BK ) is Hurwitz u = Kx – p.5/17

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Closed-loop system: ˙ x = f ( x, Kx ) ˙ x = bracketleftbigg ∂f ∂x ( x, Kx ) + ∂f ∂u ( x, Kx ) ( K )
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Lect_25 - Nonlinear Systems and Control Lecture 25...

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