Nonlinear Control Lecture # 21 Special nonlinear Forms Nonlinear - PowerPoint PPT Presentation

Nonlinear Control Lecture # 21 Special nonlinear Forms Nonlinear Control Lecture # 21 Special nonlinear Forms

Controller Form Definition A nonlinear system is in the controller form if x = Ax + B [ ψ ( x ) + γ ( x ) u ] ˙ where ( A, B ) is controllable and γ ( x ) is a nonsingular matrix for all x in the domain of interest u = γ − 1 ( x )[ − ψ ( x ) + v ] ⇒ x = Ax + Bv ˙ Any system that can be represented in the controller form is said to be feedback linearizable Nonlinear Control Lecture # 21 Special nonlinear Forms

Example 8.7 ( m -link robot) M ( q )¨ q + C ( q, ˙ q ) ˙ q + D ˙ q + g ( q ) = u q is an m -dimensional vector of joint positions and M ( q ) is a nonsingular inertial matrix � 0 � 0 � q � � � I m x = , A = , B = q ˙ 0 0 I m ψ = − M − 1 ( C ˙ γ = M − 1 q + D ˙ q + g ) , Nonlinear Control Lecture # 21 Special nonlinear Forms

An n -dimensional single-input system x = f ( x ) + g ( x ) u ˙ is transformable into the controller form if and only if ∃ h ( x ) such that x = f ( x ) + g ( x ) u, ˙ y = h ( x ) has relative degree n Search for a smooth function h ( x ) such that L g L i − 1 and L g L n − 1 h ( x ) = 0 , i = 1 , 2 , . . . , n − 1 , h ( x ) � = 0 f f � h ( x ) , L n − 1 h ( x ) � T ( x ) = col L f h ( x ) , · · · f Nonlinear Control Lecture # 21 Special nonlinear Forms

The Lie Bracket: For two vector fields f and g , the Lie bracket [ f, g ] is a third vector field defined by [ f, g ]( x ) = ∂g ∂xf ( x ) − ∂f ∂xg ( x ) Notation: ad 0 f g ( x ) = g ( x ) , ad f g ( x ) = [ f, g ]( x ) ad k f g ( x ) = [ f, ad k − 1 g ]( x ) , k ≥ 1 f Properties: [ f, g ] = − [ g, f ] For constant vector fields f and g , [ f, g ] = 0 Nonlinear Control Lecture # 21 Special nonlinear Forms

Example 8.8 � 0 � � � x 2 f = , g = − sin x 1 − x 2 x 1 � 0 � � 0 � � � � � 0 x 2 0 1 [ f, g ] = − 1 0 − sin x 1 − x 2 − cos x 1 − 1 x 1 � � − x 1 ad f g = [ f, g ] = x 1 + x 2 Nonlinear Control Lecture # 21 Special nonlinear Forms

� � � � x 2 − x 1 f = , ad f g = − sin x 1 − x 2 x 1 + x 2 ad 2 f g = [ f, ad f g ] � − 1 � � � 0 x 2 = 1 1 − sin x 1 − x 2 � � � � 0 1 − x 1 − − cos x 1 − 1 x 1 + x 2 � � − x 1 − 2 x 2 = x 1 + x 2 − sin x 1 − x 1 cos x 1 Nonlinear Control Lecture # 21 Special nonlinear Forms

Example 8.9 f ( x ) = Ax , g is a constant vector field ad f g = [ f, g ] = − Ag, ad 2 f g = [ f, ad f g ] = − A ( − Ag ) = A 2 g ad k f g = ( − 1) k A k g Distribution: For vector fields f 1 , f 2 , . . . , f k on D ⊂ R n , let ∆( x ) = span { f 1 ( x ) , f 2 ( x ) , . . . , f k ( x ) } The collection of all vector spaces ∆( x ) for x ∈ D is called a distribution and referred to by ∆ = span { f 1 , f 2 , . . . , f k } Nonlinear Control Lecture # 21 Special nonlinear Forms

If dim(∆( x )) = k for all x ∈ D , we say that ∆ is a nonsingular distribution on D , generated by f 1 , . . . , f k A distribution ∆ is involutive if g 1 ∈ ∆ and g 2 ∈ ∆ ⇒ [ g 1 , g 2 ] ∈ ∆ If ∆ is a nonsingular distribution, generated by f 1 , . . . , f k , then it is involutive if and only if [ f i , f j ] ∈ ∆ , ∀ 1 ≤ i, j ≤ k Nonlinear Control Lecture # 21 Special nonlinear Forms

Example 8.10 D = R 3 ; ∆ = span { f 1 , f 2 }     2 x 2 1  ,  , f 1 = 1 f 2 = 0 dim(∆( x )) = 2 , ∀ x ∈ D   0 x 2   0 [ f 1 , f 2 ] = ∂f 2 ∂x f 1 − ∂f 1 ∂x f 2 = 0   1 rank [ f 1 ( x ) , f 2 ( x ) , [ f 1 , f 2 ]( x )] =   2 x 2 1 0  = 3 , rank 1 0 0 ∀ x ∈ D  0 x 2 1 ∆ is not involutive Nonlinear Control Lecture # 21 Special nonlinear Forms

Example 8.11 D = { x ∈ R 3 | x 2 1 + x 2 3 � = 0 } ; ∆ = span { f 1 , f 2 }     2 x 3 − x 1  , f 2 =  , dim(∆( x )) = 2 , ∀ x ∈ D f 1 = − 1 − 2 x 2   0 x 3   − 4 x 3 [ f 1 , f 2 ] = ∂f 2 ∂x f 1 − ∂f 1 ∂x f 2 = 2   0   2 x 3 − x 1 − 4 x 3  = 2 , rank − 1 − 2 x 2 2 ∀ x ∈ D  0 x 3 0 ∆ is involutive Nonlinear Control Lecture # 21 Special nonlinear Forms

Theorem 8.2 The n -dimensional single-input system x = f ( x ) + g ( x ) u ˙ is feedback linearizable in a neighborhood of x 0 ∈ D if and only if there is a domain D x ⊂ D , with x 0 ∈ D x , such that 1 the matrix G ( x ) = [ g ( x ) , ad f g ( x ) , . . . , ad n − 1 g ( x )] has f rank n for all x ∈ D x ; 2 the distribution D = span { g, ad f g, . . . , ad n − 2 g } is f involutive in D x . Nonlinear Control Lecture # 21 Special nonlinear Forms

Example 8.12 � a sin x 2 � 0 � � x = ˙ + u − x 2 1 1 � − a cos x 2 ad f g = [ f, g ] = − ∂f � ∂xg = 0 � 0 � − a cos x 2 [ g ( x ) , ad f g ( x )] = 1 0 rank[ g ( x ) , ad f g ( x )] = 2 , ∀ x such that cos x 2 � = 0 span { g } is involutive Find h such that L g h ( x ) = 0 , and L g L f h ( x ) � = 0 Nonlinear Control Lecture # 21 Special nonlinear Forms

∂h ∂xg = ∂h = 0 ⇒ h is independent of x 2 ∂x 2 L f h ( x ) = ∂h a sin x 2 ∂x 1 L g L f h ( x ) = ∂ ( L f h ) g = ∂ ( L f h ) = ∂h a cos x 2 ∂x ∂x 2 ∂x 1 ∂h L g L f h ( x ) � = 0 in D 0 = { x ∈ R 2 | cos x 2 � = 0 } if ∂x 1 � = 0 � � � � h x 1 Take h ( x ) = x 1 ⇒ T ( x ) = = L f h a sin x 2 Nonlinear Control Lecture # 21 Special nonlinear Forms

Example 8.13 (A single link manipulator with flexible joints)     x 2 0 − a sin x 1 − b ( x 1 − x 3 ) 0     f ( x ) =  , g =     x 4 0    c ( x 1 − x 3 ) d   0 ad f g = [ f, g ] = − ∂f 0   ∂xg =   − d   0 Nonlinear Control Lecture # 21 Special nonlinear Forms

  0 f g = [ f, ad f g ] = − ∂f bd   ad 2 ∂xad f g =   0   − cd   − bd f g ] = − ∂f 0 ad 3 f g = [ f, ad 2 ∂xad 2   f g =   cd   0   0 0 0 − bd 0 0 bd 0   rank  = 4   0 − d 0 cd  d 0 − cd 0 Nonlinear Control Lecture # 21 Special nonlinear Forms

span( g, ad f g, ad 2 f g ) is involutive Why? The system is feedback linearizable. Find h ( x ) such that ∂ ( L i − 1 ∂ ( L 3 h ) f h ) f g = 0 , i = 1 , 2 , 3 , g � = 0 , h (0) = 0 ∂x ∂x ∂h ∂h ∂xg = 0 ⇒ = 0 ∂x 4 L f h ( x ) = ∂h x 2 + ∂h [ − a sin x 1 − b ( x 1 − x 3 )] + ∂h x 4 ∂x 1 ∂x 2 ∂x 3 ∂ ( L f h ) g = 0 ⇒ ∂ ( L f h ) = 0 ⇒ ∂h = 0 ∂x ∂x 4 ∂x 3 Nonlinear Control Lecture # 21 Special nonlinear Forms

L f h ( x ) = ∂h x 2 + ∂h [ − a sin x 1 − b ( x 1 − x 3 )] ∂x 1 ∂x 2 f h ( x ) = ∂ ( L f h ) x 2 + ∂ ( L f h ) [ − a sin x 1 − b ( x 1 − x 3 )]+ ∂ ( L f h ) L 2 x 4 ∂x 1 ∂x 2 ∂x 3 ∂ ( L 2 f h ) = 0 ⇒ ∂ ( L f h ) = 0 ⇒ ∂h = 0 ∂x 4 ∂x 3 ∂x 2 ∂ ( L 3 f h ) ∂h g � = 0 ⇔ � = 0 ∂x ∂x 1   x 1 x 2   h ( x ) = x 1 , T ( x ) =   − a sin x 1 − b ( x 1 − x 3 )   − ax 2 cos x 1 − b ( x 2 − x 4 ) Nonlinear Control Lecture # 21 Special nonlinear Forms

Example 8.14 ( Field-Controlled DC Motor)     d 1 ( − x 1 − x 2 x 3 + V a ) 0  , f e ∈ C 2 for x 2 ∈ J  , g = − d 2 f e ( x 2 ) d 2 f =   d 3 ( x 1 x 2 − bx 3 ) 0   d 1 d 2 x 3 d 2 ad f g = 2 f ′ e ( x 2 )   − d 2 d 3 x 1   d 1 d 2 x 3 ( d 1 + d 2 f ′ e ( x 2 ) − bd 3 ) e ( x 2 )) 2 − d 3 ad 2 d 3 f g = 2 ( f ′ 2 f 2 ( x 2 ) f ′′ e ( x 2 )   d 1 d 2 d 3 ( x 1 − V a ) − d 2 e ( x 2 ) − bd 2 d 2 2 d 3 x 1 f ′ 3 x 1 Nonlinear Control Lecture # 21 Special nonlinear Forms

det G = − 2 d 2 1 d 3 2 d 3 x 3 ( x 1 − a )(1 − bd 3 /d 1 ) a = 1 2 V a / (1 − bd 3 /d 1 ) > 0 x 1 � = a and x 3 � = 0 ⇒ rank ( G ) = 3 [ g, ad f g ] = d 2 2 f ′′ e ( x 2 ) g ⇒ span { g, ad f g } is involutive The conditions of Theorem 8.2 are satisfied in the domain D x = { x 1 > a, x 2 ∈ J, x 3 > 0 } Find h ( x ) such that ∂ ( L 2 f h ) ∂h ∂ ( L f h ) ∂xg = 0; g = 0; g � = 0 ∂x ∂x Nonlinear Control Lecture # 21 Special nonlinear Forms

∂h ∂h ∂xg = 0 ⇒ = 0 ∂x 2 L f h ( x ) = ∂h d 1 ( − x 1 − x 2 x 3 + V a ) + ∂h d 3 ( x 1 x 2 − bx 3 ) ∂x 1 ∂x 3 ∂ ( L f h ) g = 0 ⇒ ∂ ( L f h ) ∂h ∂h = 0 ⇒ − d 1 x 3 + d 3 x 1 = 0 ∂x ∂x 2 ∂x 1 ∂x 3 Take h = d 3 x 2 1 + d 1 x 2 3 + c L f h ( x ) = 2 d 1 d 3 x 1 ( V a − x 1 ) − 2 bd 1 d 3 x 2 3 L 2 f h ( x ) = 2 d 2 1 d 3 ( V a − 2 x 1 )( − x 1 − x 2 x 3 + V a ) − 4 bd 1 d 2 3 x 3 ( x 1 x 2 − bx 3 ) ∂ ( L 2 ∂ ( L 2 f h ) f h ) = 4 d 2 g = d 2 1 d 2 d 3 (1 − bd 3 /d 1 ) x 3 ( x 1 − a ) � = 0 ∂x ∂x 2 Nonlinear Control Lecture # 21 Special nonlinear Forms