Nonlinear Control Lecture # 22 Special nonlinear Forms Nonlinear Control Lecture # 22 Special nonlinear Forms
Observer Form Definition A nonlinear system is in the observer form if x = Ax + ψ ( y, u ) , ˙ y = Cx where ( A, C ) is observable Observer: ˙ x = A ˆ ˆ x + ψ ( y, u ) + H ( y − C ˆ x ) x = x − ˆ ˜ x ˙ x = ( A − HC )˜ ˜ x Design H such that ( A − HC ) is Hurwitz Nonlinear Control Lecture # 22 Special nonlinear Forms
Example 8.15 (A single link manipulator with flexible joints) x 2 0 − a sin x 1 − b ( x 1 − x 3 ) 0 x = ˙ + y = x 1 u, x 4 0 c ( x 1 − x 3 ) d x = Ax + ψ ( u, y ) , ˙ y = Cx 0 1 0 0 0 − b 0 0 − a sin y b A = ψ = , 0 0 0 1 0 c 0 − c 0 du � 1 0 � 0 0 C = . . . , ( A, C ) is observable Nonlinear Control Lecture # 22 Special nonlinear Forms
Example 8.16 (Inverted pendulum) x 1 = x 2 , ˙ x 2 = a (sin x 1 + u cos x 1 ) , ˙ y = x 1 x = Ax + ψ ( u, y ) , ˙ y = Cx � 0 � � � 1 0 A = , ψ = 0 0 a (sin y + u cos y ) � 1 0 � C = Nonlinear Control Lecture # 22 Special nonlinear Forms
m � x = f ( x ) + ˙ g i ( x ) u i , y = h ( x ) i =1 Is there z = T ( x ) such that m � z = A c z + φ ( y ) + ˙ γ i ( y ) u i , y = C c z i =1 0 1 0 . . . 0 0 0 1 0 . . . . . ... . . � � A c = . . , C c = 1 0 . . . 0 0 . . . 0 1 0 0 0 . . . . . . Nonlinear Control Lecture # 22 Special nonlinear Forms
x = f ( x ) , ˙ y = h ( x ) h ( x ) y L f h ( x ) ˙ y Φ( x ) = = . . . . . . L n − 1 y ( n − 1) h ( x ) f z 1 z 2 + F 1 ( z 1 ) ˜ Φ( z ) = Φ( x ) | x = T − 1 ( z ) = . . . z n + F n − 1 ( z 1 , . . . , z n − 1 ) Nonlinear Control Lecture # 22 Special nonlinear Forms
∂ ˜ ∂T − 1 ∂z = ∂ Φ Φ ∂x ∂z 1 0 · · · 0 ∗ 1 0 0 ∂ ˜ Φ . . . . ∂z = . . ∗ · · · ∗ 1 0 ∗ · · · ∗ 1 ∂ Φ ∂x must be nonsingular Nonlinear Control Lecture # 22 Special nonlinear Forms
∂ Φ � � ∂x τ = b, b = col 0 , · · · 0 , 1 L τ L n − 1 L τ L k f h ( x ) = 0 , 0 ≤ k ≤ n − 2 , h ( x ) = 1 f Equivalently τ h ( x ) = ( − 1) n − 1 L ad k f τ h ( x ) = 0 , 0 ≤ k ≤ n − 2 , L ad n − 1 f τ k = ( − 1) n − k ad n − k Define τ, 1 ≤ k ≤ n f � τ 1 ∂T � τ 2 · · · τ n = I ∂x Nonlinear Control Lecture # 22 Special nonlinear Forms
0 . . . 0 ∂T def ∂x τ k = e k = 1 ← k th row 0 . . . 0 ( − 1) n − k ∂T τ = ( − 1) n − k ∂T ∂x ad n − k ∂x [ f, ad n − k − 1 = τ ] e k f f f ( z ) , ( − 1) n − k − 1 e k +1 ] = ∂ ˜ f ( − 1) n − k [ ˜ = ∂z e k +1 Nonlinear Control Lecture # 22 Special nonlinear Forms
∗ 1 0 . . . 0 ∗ 0 1 0 . . . ∂ ˜ . . f ... . . . . ∂z = . . . 0 1 ∗ 0 0 . . . . . . By integration ˜ f ( z ) = A c z + φ ( z 1 ) Nonlinear Control Lecture # 22 Special nonlinear Forms
∂ ˜ ∂T − 1 ∂z = ∂h h ˜ h ( z ) = h ( T − 1 ( z )) , ∂x ∂z ∂T − 1 � τ 1 � = · · · τ 2 τ n x = T − 1 ( z ) ∂z ∂ ˜ h � � ( − 1) n − 1 L ad n − 1 ( − 1) n − 2 L ad n − 2 τ h, τ h, · · · L τ h ∂z = f f ∂ ˜ � 1 , h ˜ 0 � ∂z = 0 , · · · ⇒ h = z 1 Nonlinear Control Lecture # 22 Special nonlinear Forms
Theorem 8.3 An n -dimensional single-output (SO) systems x = f ( x ) , ˙ y = h ( x ) is transformable into the observer form if and only if there is a domain D 0 such that ∀ x ∈ D 0 � ∂ Φ � � h, L n − 1 h � rank ∂x ( x ) = n, Φ = col L f h, · · · f and the unique vector field solution τ of � 0 , ∂ Φ · · · 0 , 1 � ∂x τ = b, b = col [ ad i f τ, ad j satisfies f τ ] = 0 , 0 ≤ i, j ≤ n − 1 Nonlinear Control Lecture # 22 Special nonlinear Forms
m � x = f ( x ) + ˙ g i ( x ) u i , y = h ( x ) i =1 � g i ( z ) = ∂T � When will ˜ ∂x g i ( x ) be independent of z 2 to z n ? � � x = T − 1 ( z ) g i , ( − 1) n − k − 1 e k +1 ] = ( − 1) n − k ∂ ˜ ∂T g i ∂x [ g i , ad n − k − 1 τ ] = [˜ f ∂z k +1 ∂ ˜ g i = 0 ⇔ [ g i , ad n − k − 1 τ ] = 0 f ∂z k +1 Nonlinear Control Lecture # 22 Special nonlinear Forms
Corollary 8.1 Suppose the assumptions of Theorem 8.3 are satisfied. Then, the change of variables z = T ( x ) transforms the system into the observer form if and only if [ g i , ad k f τ ] = 0 , , for 0 ≤ k ≤ n − 2 and 1 ≤ i ≤ m Moreover, if for some i the foregoing condition is strengthened to [ g i , ad k f τ ] = 0 , , for 0 ≤ k ≤ n − 1 then the vector field γ i is constant Nonlinear Control Lecture # 22 Special nonlinear Forms
Example 8.17 � � � � β 1 ( x 1 ) + x 2 b 1 x = ˙ + y = x 1 u, f 2 ( x ) b 2 � � � � h ( x ) x 1 Φ( x ) = = L f h ( x ) β 1 ( x 1 ) + x 2 � � � ∂ Φ � ∂ Φ 1 0 ∂x = ; rank ∂x ( x ) = 2 , ∀ x ∂β 1 1 ∂x 1 � 0 � 0 ∂ Φ � � ∂x τ = ⇒ τ = 1 1 Nonlinear Control Lecture # 22 Special nonlinear Forms
� ∗ � � 0 � � � ad f τ = [ f, τ ] = − ∂f 1 1 ∂xτ = − = − ∂f 2 ∂f 2 ∗ 1 ∂x 2 ∂x 2 � � � 0 0 � [ τ, ad f τ ] = ∂ ( ad f τ ) 0 τ = − ∂ 2 f 2 ∂ 2 f 2 1 ∂x ∂x 2 ∂x 1 ∂x 2 2 [ τ, ad f τ ] = 0 ⇔ ∂ 2 f 2 = 0 ⇔ f 2 ( x ) = β 2 ( x 1 ) + x 2 β 3 ( x 1 ) ∂x 2 2 [ g, τ ] = 0 ( g and τ are constant vector fields) � � b 1 � 0 0 � = 0 if ∂β 3 [ g, ad f τ ] = = 0 or b 1 = 0 − ∂β 3 0 b 2 ∂x 1 ∂x 1 Nonlinear Control Lecture # 22 Special nonlinear Forms
� � 1 τ 1 = ( − 1) 1 ad 1 f τ = − ad f τ = β 3 ( x 1 ) � � 0 τ 2 = ( − 1) 0 ad 0 f τ = τ = 1 � τ 1 , ∂T � τ 2 = I ∂x ∂T 1 ∂T 1 � � � � ∂x 1 ∂x 2 1 0 1 0 = β 3 ( x 1 ) 1 0 1 ∂T 2 ∂T 2 ∂x 1 ∂x 2 ∂T 1 = 0 and ∂T 1 = 1 ⇒ T 1 = x 1 ∂x 2 ∂x 1 Nonlinear Control Lecture # 22 Special nonlinear Forms
∂T 2 = 1 and ∂T 2 + β 3 ( x 1 ) = 0 ∂x 2 ∂x 1 � x 1 ⇒ T 2 ( x ) = x 2 − β 3 ( σ ) dσ 0 z = Az + φ ( y ) + γ ( y ) u, ˙ y = Cz � 0 � 1 � � A = C = 1 0 , 0 0 � β 1 ( y ) + � y � � � 0 β 3 ( σ ) dσ b 1 φ = , γ = β 2 ( y ) − β 1 ( y ) β 3 ( y ) b 2 − b 1 β 3 ( y ) Nonlinear Control Lecture # 22 Special nonlinear Forms
Special Case: SISO system x = f ( x ) + g ( x ) u, ˙ y = h ( x ) Suppose the assumptions of Corollary 8.1 hold with [ g, ad k f τ ] = 0 , , for 0 ≤ k ≤ n − 1 z = T ( x ) → ˙ z = A c z + φ ( y ) + γu, y = C c z � T , γ ρ � = 0 � Rel deg = ρ ⇔ γ = 0 , 0 , . . . , γ ρ , . . . , γ n Minimum Phase ⇔ γ ρ s n − ρ + · · · + γ n − 1 s + γ n Hurwitz Nonlinear Control Lecture # 22 Special nonlinear Forms
Recommend
More recommend
