Nonlinear Control Lecture # 25 State Feedback Stabilization - PowerPoint PPT Presentation

Nonlinear Control Lecture # 25 State Feedback Stabilization Nonlinear Control Lecture # 25 State Feedback Stabilization

Backstepping η ˙ = f a ( η ) + g a ( η ) ξ ˙ η ∈ R n , ξ, u ∈ R ξ = f b ( η, ξ ) + g b ( η, ξ ) u, g b � = 0 , Stabilize the origin using state feedback View ξ as “virtual” control input to the system η = f a ( η ) + g a ( η ) ξ ˙ Suppose there is ξ = φ ( η ) that stabilizes the origin of η = f a ( η ) + g a ( η ) φ ( η ) ˙ ∂V a ∂η [ f a ( η ) + g a ( η ) φ ( η )] ≤ − W ( η ) Nonlinear Control Lecture # 25 State Feedback Stabilization

z = ξ − φ ( η ) η ˙ = [ f a ( η ) + g a ( η ) φ ( η )] + g a ( η ) z z ˙ = F ( η, ξ ) + g b ( η, ξ ) u 2 z 2 = V a ( η ) + 1 V ( η, ξ ) = V a ( η ) + 1 2 [ ξ − φ ( η )] 2 ∂V a ∂η [ f a ( η ) + g a ( η ) φ ( η )] + ∂V a ˙ V = ∂η g a ( η ) z + zF ( η, ξ ) + zg b ( η, ξ ) u � ∂V a � − W ( η ) + z ∂η g a ( η ) + F ( η, ξ ) + g b ( η, ξ ) u ≤ Nonlinear Control Lecture # 25 State Feedback Stabilization

� ∂V a � ˙ V ≤ − W ( η ) + z ∂η g a ( η ) + F ( η, ξ ) + g b ( η, ξ ) u 1 � ∂V a � u = − ∂η g a ( η ) + F ( η, ξ ) + kz , k > 0 g b ( η, ξ ) ˙ V ≤ − W ( η ) − kz 2 Nonlinear Control Lecture # 25 State Feedback Stabilization

Example 9.9 x 1 = x 2 1 − x 3 ˙ 1 + x 2 , x 2 = u ˙ x 1 = x 2 1 − x 3 ˙ 1 + x 2 x 2 = φ ( x 1 ) = − x 2 x 1 = − x 1 − x 3 1 − x 1 ⇒ ˙ 1 ˙ V a ( x 1 ) = 1 2 x 2 V a = − x 2 1 − x 4 1 , ∀ x 1 ∈ R ⇒ 1 z 2 = x 2 − φ ( x 1 ) = x 2 + x 1 + x 2 1 − x 1 − x 3 x 1 ˙ = 1 + z 2 u + (1 + 2 x 1 )( − x 1 − x 3 z 2 ˙ = 1 + z 2 ) Nonlinear Control Lecture # 25 State Feedback Stabilization

V ( x ) = 1 2 x 2 1 + 1 2 z 2 2 ˙ x 1 ( − x 1 − x 3 V = 1 + z 2 ) + z 2 [ u + (1 + 2 x 1 )( − x 1 − x 3 1 + z 2 )] ˙ − x 2 1 − x 4 V = 1 + z 2 [ x 1 + (1 + 2 x 1 )( − x 1 − x 3 1 + z 2 ) + u ] u = − x 1 − (1 + 2 x 1 )( − x 1 − x 3 1 + z 2 ) − z 2 ˙ V = − x 2 1 − x 4 1 − z 2 2 The origin is globally asymptotically stable Nonlinear Control Lecture # 25 State Feedback Stabilization

Example 9.10 x 1 = x 2 1 − x 3 ˙ 1 + x 2 , x 2 = x 3 , ˙ x 3 = u ˙ x 1 = x 2 1 − x 3 ˙ 1 + x 2 , x 2 = x 3 ˙ def x 3 = − x 1 − (1 + 2 x 1 )( − x 1 − x 3 1 + z 2 ) − z 2 = φ ( x 1 , x 2 ) ˙ V a ( x ) = 1 2 x 2 1 + 1 2 z 2 V a = − x 2 1 − x 4 1 − z 2 2 , 2 z 3 = x 3 − φ ( x 1 , x 2 ) x 2 1 − x 3 x 1 ˙ = 1 + x 2 , x 2 = φ ( x 1 , x 2 ) + z 3 ˙ u − ∂φ 1 + x 2 ) − ∂φ ( x 2 1 − x 3 z 3 ˙ = ( φ + z 3 ) ∂x 1 ∂x 2 Nonlinear Control Lecture # 25 State Feedback Stabilization

V = V a + 1 2 z 2 3 ∂V a 1 + x 2 ) + ∂V a ˙ ( x 2 1 − x 3 V = ( z 3 + φ ) ∂x 1 ∂x 2 � u − ∂φ 1 + x 2 ) − ∂φ � ( x 2 1 − x 3 + z 3 ( z 3 + φ ) ∂x 1 ∂x 2 ˙ − x 2 1 − x 4 1 − ( x 2 + x 1 + x 2 1 ) 2 V = � ∂V a − ∂φ 1 + x 2 ) − ∂φ � ( x 2 1 − x 3 + z 3 ( z 3 + φ ) + u ∂x 2 ∂x 1 ∂x 2 u = − ∂V a + ∂φ 1 + x 2 ) + ∂φ ( x 2 1 − x 3 ( z 3 + φ ) − z 3 ∂x 2 ∂x 1 ∂x 2 The origin is globally asymptotically stable Nonlinear Control Lecture # 25 State Feedback Stabilization

Strict-Feedback Form x ˙ = f 0 ( x ) + g 0 ( x ) z 1 z 1 ˙ = f 1 ( x, z 1 ) + g 1 ( x, z 1 ) z 2 z 2 ˙ = f 2 ( x, z 1 , z 2 ) + g 2 ( x, z 1 , z 2 ) z 3 . . . z k − 1 ˙ = f k − 1 ( x, z 1 , . . . , z k − 1 ) + g k − 1 ( x, z 1 , . . . , z k − 1 ) z k z k ˙ = f k ( x, z 1 , . . . , z k ) + g k ( x, z 1 , . . . , z k ) u g i ( x, z 1 , . . . , z i ) � = 0 for 1 ≤ i ≤ k Nonlinear Control Lecture # 25 State Feedback Stabilization

Example 9.12 x = − x + x 2 z, ˙ z = u ˙ x = − x + x 2 z ˙ 2 x 2 ⇒ ˙ V a = 1 V a = − x 2 z = 0 ⇒ ˙ x = − x, 2 ( x 2 + z 2 ) V = 1 V = x ( − x + x 2 z ) + zu = − x 2 + z ( x 3 + u ) ˙ u = − x 3 − kz, k > 0 , ⇒ V = − x 2 − kz 2 ˙ Global stabilization Compare with semiglobal stabilization in Example 9.7 Nonlinear Control Lecture # 25 State Feedback Stabilization

Example 9.13 x = x 2 − xz, ˙ z = u ˙ x = x 2 − xz ˙ V 0 ( x ) = 1 z = x + x 2 ⇒ ˙ 2 x 2 ⇒ ˙ x = − x 3 , V = − x 4 V = V 0 + 1 2( z − x − x 2 ) 2 V = − x 4 + ( z − x − x 2 )[ − x 2 + u − (1 + 2 x )( x 2 − xz )] ˙ u = (1 + 2 x )( x 2 − xz ) + x 2 − k ( z − x − x 2 ) , k > 0 V = − x 4 − k ( z − x − x 2 ) 2 ˙ Global stabilization Nonlinear Control Lecture # 25 State Feedback Stabilization

Passivity-Based Control x = f ( x, u ) , ˙ y = h ( x ) , f (0 , 0) = 0 V = ∂V u T y ≥ ˙ ∂x f ( x, u ) Theorem 9.1 If the system is (1) passive with a radially unbounded positive definite storage function and (2) zero-state observable, then the origin can be globally stabilized by y T φ ( y ) > 0 ∀ y � = 0 u = − φ ( y ) , φ (0) = 0 , Nonlinear Control Lecture # 25 State Feedback Stabilization

Proof V = ∂V ˙ ∂x f ( x, − φ ( y )) ≤ − y T φ ( y ) ≤ 0 ˙ V ( x ( t )) ≡ 0 ⇒ y ( t ) ≡ 0 ⇒ u ( t ) ≡ 0 ⇒ x ( t ) ≡ 0 Apply the invariance principle A given system may be made passive by (1) Choice of output, (2) Feedback, or both Nonlinear Control Lecture # 25 State Feedback Stabilization

Choice of Output ∂V x = f ( x ) + G ( x ) u, ˙ ∂x f ( x ) ≤ 0 , ∀ x No output is defined. Choose the output as � T � ∂V def y = h ( x ) = ∂x G ( x ) V = ∂V ∂x f ( x ) + ∂V ˙ ∂x G ( x ) u ≤ y T u Check zero-state observability Nonlinear Control Lecture # 25 State Feedback Stabilization

Example 9.14 x 2 = − x 3 x 1 = x 2 , ˙ ˙ 1 + u V ( x ) = 1 4 x 4 1 + 1 2 x 2 2 ˙ V = x 3 1 x 2 − x 2 x 3 With u = 0 1 = 0 Take y = ∂V ∂x G = ∂V = x 2 ∂x 2 Is it zero-state observable? with u = 0 , y ( t ) ≡ 0 ⇒ x ( t ) ≡ 0 u = − (2 k/π ) tan − 1 ( x 2 ) u = − kx 2 or ( k > 0) Nonlinear Control Lecture # 25 State Feedback Stabilization

Feedback Passivation Definition The system x = f ( x ) + G ( x ) u, ˙ y = h ( x ) ( ∗ ) is equivalent to a passive system if ∃ u = α ( x ) + β ( x ) v such that x = f ( x ) + G ( x ) α ( x ) + G ( x ) β ( x ) v, ˙ y = h ( x ) is passive Theorem [20] The system (*) is locally equivalent to a passive system (with a positive definite storage function) if it has relative degree one at x = 0 and the zero dynamics have a stable equilibrium point at the origin with a positive definite Lyapunov function Nonlinear Control Lecture # 25 State Feedback Stabilization

Example 9.15 ( m -link Robot Manipulator) M ( q )¨ q + C ( q, ˙ q ) ˙ q + D ˙ q + g ( q ) = u M = M T > 0 , ( ˙ M − 2 C ) T = − ( ˙ M − 2 C ) , D = D T ≥ 0 Stabilize the system at q = q r e = q − q r , e = ˙ ˙ q M ( q )¨ e + C ( q, ˙ q )˙ e + D ˙ e + g ( q ) = u ( e = 0 , ˙ e = 0) is not an open-loop equilibrium point ( K p = K T u = g ( q ) − K p e + v, p > 0) M ( q )¨ e + C ( q, ˙ q )˙ e + D ˙ e + K p e = v Nonlinear Control Lecture # 25 State Feedback Stabilization

M ( q )¨ e + C ( q, ˙ q )˙ e + D ˙ e + K p e = v 2 e T K p e V = 1 e T M ( q )˙ e + 1 2 ˙ e T ( ˙ ˙ V = 1 e T D ˙ e T K p e + ˙ e T v + e T K p ˙ e T v 2 ˙ M − 2 C )˙ e − ˙ e − ˙ e ≤ ˙ y = ˙ e Is it zero-state observable? Set v = 0 e ( t ) ≡ 0 ⇒ ¨ ˙ e ( t ) ≡ 0 ⇒ K p e ( t ) ≡ 0 ⇒ e ( t ) ≡ 0 e T φ (˙ v = − φ (˙ e ) , [ φ (0) = 0 , ˙ e ) > 0 , ∀ ˙ e � = 0] u = g ( q ) − K p e − φ (˙ e ) K d = K T Special case: u = g ( q ) − K p e − K d ˙ e, d > 0 Nonlinear Control Lecture # 25 State Feedback Stabilization