Nonlinear Control Lecture # 10 Time Varying and Perturbed Systems - PowerPoint PPT Presentation

Nonlinear Control Lecture # 10 Time Varying and Perturbed Systems Nonlinear Control Lecture # 10 Time Varying and Perturbed Systems

Boundedness and Ultimate Boundedness Definition 4.3 The solutions of ˙ x = f ( t, x ) are uniformly bounded if there exists c > 0 , independent of t 0 , and for every a ∈ (0 , c ) , there is β > 0 , dependent on a but independent of t 0 , such that � x ( t 0 ) � ≤ a ⇒ � x ( t ) � ≤ β, ∀ t ≥ t 0 uniformly ultimately bounded with ultimate bound b if there exists a positive constant c , independent of t 0 , and for every a ∈ (0 , c ) , there is T ≥ 0 , dependent on a and b but independent of t 0 , such that � x ( t 0 ) � ≤ a ⇒ � x ( t ) � ≤ b, ∀ t ≥ t 0 + T Nonlinear Control Lecture # 10 Time Varying and Perturbed Systems

Add “Globally” if a can be arbitrarily large Drop “uniformly” if ˙ x = f ( x ) Nonlinear Control Lecture # 10 Time Varying and Perturbed Systems

Lyapunov Analysis: Let V ( x ) be a cont. diff. positive definite function and suppose the sets Ω c = { V ( x ) ≤ c } , Ω ε = { V ( x ) ≤ ε } , Λ = { ε ≤ V ( x ) ≤ c } are compact for some c > ε > 0 Ω c Ω ε Λ Nonlinear Control Lecture # 10 Time Varying and Perturbed Systems

Suppose V ( t, x ) = ∂V ˙ ∂x f ( t, x ) ≤ − W 3 ( x ) , ∀ x ∈ Λ , ∀ t ≥ 0 W 3 ( x ) is continuous and positive definite Ω c and Ω ε are positively invariant k = min x ∈ Λ W 3 ( x ) > 0 ˙ V ( t, x ) ≤ − k, ∀ x ∈ Λ , ∀ t ≥ t 0 ≥ 0 V ( x ( t )) ≤ V ( x ( t 0 )) − k ( t − t 0 ) ≤ c − k ( t − t 0 ) x ( t ) enters the set Ω ε within the interval [ t 0 , t 0 + ( c − ε ) /k ] Nonlinear Control Lecture # 10 Time Varying and Perturbed Systems

Suppose ˙ V ( t, x ) ≤ − W 3 ( x ) , ∀ x ∈ D with � x � ≥ µ, ∀ t ≥ 0 Choose c and ε such that Λ ⊂ D ∩ {� x � ≥ µ } Ω c Ω ε B b B µ Nonlinear Control Lecture # 10 Time Varying and Perturbed Systems

Let α 1 and α 2 be class K functions such that α 1 ( � x � ) ≤ V ( x ) ≤ α 2 ( � x � ) V ( x ) ≤ c ⇒ α 1 ( � x � ) ≤ c ⇔ � x � ≤ α − 1 1 ( c ) If B r ⊂ D, c = α 1 ( r ) ⇒ Ω c ⊂ B r ⊂ D � x � ≤ µ ⇒ V ( x ) ≤ α 2 ( µ ) ε = α 2 ( µ ) ⇒ B µ ⊂ Ω ε What is the ultimate bound? V ( x ) ≤ ε ⇒ α 1 ( � x � ) ≤ ε ⇔ � x � ≤ α − 1 1 ( ε ) = α − 1 1 ( α 2 ( µ )) Nonlinear Control Lecture # 10 Time Varying and Perturbed Systems

Theorem 4.4 Suppose B µ ⊂ D ⊂ R n and α 1 ( � x � ) ≤ V ( x ) ≤ α 2 ( � x � ) ∂V ∂x f ( t, x ) ≤ − W 3 ( x ) , ∀ x ∈ D with � x � ≥ µ, ∀ t ≥ 0 where α 1 and α 2 are class K functions and W 3 ( x ) is a continuous positive definite function. Choose c > 0 such that Ω c = { V ( x ) ≤ c } is compact and contained in D and suppose µ < α − 1 2 ( c ) . Then, Ω c is positively invariant and there exists a class KL function β such that for every x ( t 0 ) ∈ Ω c , � β ( � x ( t 0 ) � , t − t 0 ) , α − 1 � � x ( t ) � ≤ max 1 ( α 2 ( µ )) , ∀ t ≥ t 0 If D = R n and α 1 ∈ K ∞ , the inequality holds ∀ x ( t 0 ) , ∀ µ Nonlinear Control Lecture # 10 Time Varying and Perturbed Systems

Remarks The ultimate bound is independent of the initial state The ultimate bound is a class K function of µ ; hence, the smaller the value of µ , the smaller the ultimate bound. As µ → 0 , the ultimate bound approaches zero Nonlinear Control Lecture # 10 Time Varying and Perturbed Systems

Example 4.8 x 2 = − (1 + x 2 x 1 = x 2 , ˙ ˙ 1 ) x 1 − x 2 + M cos ωt, M ≥ 0 x 2 = − (1 + x 2 With M = 0 , ˙ 1 ) x 1 − x 2 = − h ( x 1 ) − x 2  1 1  � x 1 2 2 V ( x ) = x T  x + 2 ( y + y 3 ) dy (Example 3.7)  1 0 1 2 3 1   2 2 def V ( x ) = x T  x + 1 2 x 4 = x T Px + 1 2 x 4  1 1 1 1 2 Nonlinear Control Lecture # 10 Time Varying and Perturbed Systems

λ min ( P ) � x � 2 ≤ V ( x ) ≤ λ max ( P ) � x � 2 + 1 2 � x � 4 α 2 ( r ) = λ max ( P ) r 2 + 1 α 1 ( r ) = λ min ( P ) r 2 , 2 r 4 ˙ − x 2 1 − x 4 1 − x 2 V = 2 + ( x 1 + 2 x 2 ) M cos ωt √ −� x � 2 − x 4 ≤ 1 + M 5 � x � √ − (1 − θ ) � x � 2 − x 4 1 − θ � x � 2 + M = 5 � x � (0 < θ < 1) √ − (1 − θ ) � x � 2 − x 4 def ≤ 1 , ∀ � x � ≥ M 5 /θ = µ The solutions are GUUB by � λ max ( P ) µ 2 + µ 4 / 2 b = α − 1 1 ( α 2 ( µ )) = λ min ( P ) Nonlinear Control Lecture # 10 Time Varying and Perturbed Systems

Theorem 4.5 Suppose c 1 � x � 2 ≤ V ( x ) ≤ c 2 � x � 2 ∂V ∂x f ( t, x ) ≤ − c 3 � x � 2 , ∀ x ∈ D with � x � ≥ µ, ∀ t ≥ 0 � for some positive constants c 1 to c 3 , and µ < c/c 2 . Then, Ω c = { V ( x ) ≤ c } is positively invariant and ∀ x ( t 0 ) ∈ Ω c V ( x ( t 0 )) e − ( c 3 /c 2 )( t − t 0 ) , c 2 µ 2 � � V ( x ( t )) ≤ max , ∀ t ≥ t 0 � � x ( t 0 ) � e − ( c 3 /c 2 )( t − t 0 ) / 2 , µ � � � x ( t ) � ≤ c 2 /c 1 max , ∀ t ≥ t 0 If D = R n , the inequalities hold ∀ x ( t 0 ) , ∀ µ Nonlinear Control Lecture # 10 Time Varying and Perturbed Systems

Example 4.9 h ( x 1 ) = x 1 − 1 3 x 3 x 1 = x 2 , ˙ x 2 = − h ( x 1 ) − x 2 + u ( t ) , ˙ 1 | u ( t ) | ≤ d � x 1 � � k k V ( x ) = 1 2 x T x + h ( y ) dy, 0 < k < 1 k 1 0 � x 1 2 3 x 2 1 ≤ x 1 h ( x 1 ) ≤ x 2 12 x 2 5 h ( y ) dy ≤ 1 2 x 2 1 , 1 ≤ 1 , ∀ | x 1 | ≤ 1 0 λ min ( P 1 ) � x � 2 ≤ x T P 1 x ≤ V ( x ) ≤ x T P 2 x ≤ λ max ( P 2 ) � x � 2 � k + 5 � � k + 1 � k k P 1 = 1 P 2 = 1 6 , 2 k 1 2 k 1 Nonlinear Control Lecture # 10 Time Varying and Perturbed Systems

˙ − kx 1 h ( x 1 ) − (1 − k ) x 2 V = 2 + ( kx 1 + x 2 ) u ( t ) − 2 3 kx 2 1 − (1 − k ) x 2 ≤ 2 + | kx 1 + x 2 | d k = 3 ⇒ c 1 = λ min ( P 1 ) = 0 . 2894 , c 2 = λ max ( P 2 ) = 0 . 9854 5 � � 3 � 2 � x � d � x � 2 − 0 . 9 × 2 � x � 2 + ˙ − 0 . 1 × 2 V ≤ 1 + 5 5 5 def 0 . 1 × 2 � x � 2 , ≤ ∀ � x � ≥ 3 . 2394 d = µ 5 c = min | x 1 | =1 V ( x ) = 0 . 5367 ⇒ Ω c = { V ( x ) ≤ c } ⊂ {| x 1 | ≤ 1 } � For µ < c/c 2 we need d < 0 . 2278 . Theorem 4.5 holds and � b = µ c 2 /c 1 = 5 . 9775 d Nonlinear Control Lecture # 10 Time Varying and Perturbed Systems

Perturbed Systems: Nonvanishing Perturbation Nominal System: x = f ( x ) , ˙ f (0) = 0 Perturbed System: x = f ( x ) + g ( t, x ) , ˙ g ( t, 0) � = 0 Case 1:(Lemma 4.3) The origin of ˙ x = f ( x ) is exponentially stable Case 2:(Lemma 4.4) The origin of ˙ x = f ( x ) is asymptotically stable Nonlinear Control Lecture # 10 Time Varying and Perturbed Systems

Lemma 4.3 Suppose that ∀ x ∈ B r , ∀ t ≥ 0 c 1 � x � 2 ≤ V ( x ) ≤ c 2 � x � 2 � � ∂V ∂V ∂x f ( x ) ≤ − c 3 � x � 2 , � � � ≤ c 4 � x � � � ∂x � � g ( t, x ) � ≤ δ < c 3 � c 1 θr, 0 < θ < 1 c 4 c 2 Then, for all x ( t 0 ) ∈ { V ( x ) ≤ c 1 r 2 } � x ( t ) � ≤ max { k exp[ − γ ( t − t 0 )] � x ( t 0 ) � , b } , ∀ t ≥ t 0 � c 2 γ = (1 − θ ) c 3 b = δc 4 � c 2 k = , , c 1 2 c 2 θc 3 c 1 Nonlinear Control Lecture # 10 Time Varying and Perturbed Systems

Proof Apply Theorem 4.5 ˙ ∂V ∂x f ( x ) + ∂V V ( t, x ) = ∂x g ( t, x ) − c 3 � x � 2 + � � g ( t, x ) � � � ∂V � ≤ ∂x − c 3 � x � 2 + c 4 δ � x � ≤ − (1 − θ ) c 3 � x � 2 − θc 3 � x � 2 + c 4 δ � x � = def − (1 − θ ) c 3 � x � 2 , ≤ ∀ � x � ≥ δc 4 / ( θc 3 ) = µ x ( t 0 ) ∈ Ω = { V ( x ) ≤ c 1 r 2 } � c 1 ⇔ δ < c 3 � c 1 � c 2 ⇔ b = δc 4 � c 2 µ < r θr, b = µ c 2 c 4 c 2 c 1 θc 3 c 1 Nonlinear Control Lecture # 10 Time Varying and Perturbed Systems

Example 4.10 x 2 = − 4 x 1 − 2 x 2 + βx 3 x 1 = x 2 , ˙ ˙ 2 + d ( t ) β ≥ 0 , | d ( t ) | ≤ δ, ∀ t ≥ 0  3 1  2 8 V ( x ) = x T Px = x T  x (Example 4.5)  1 5 8 16 � 1 −� x � 2 + 2 βx 2 ˙ 8 x 1 x 2 + 5 16 x 2 � V ( t, x ) = 2 2 � 1 8 x 1 + 5 � + 2 d ( t ) 16 x 2 √ √ 29 29 δ −� x � 2 + 2 � x � 2 + 8 βk 2 ≤ � x � 8 Nonlinear Control Lecture # 10 Time Varying and Perturbed Systems

x T P x ≤ c | x 2 | = 1 . 8194 √ c k 2 = max √ 29 k 2 Suppose β ≤ 8(1 − ζ ) / ( 2 ) (0 < ζ < 1) √ − ζ � x � 2 + ˙ 29 δ V ( t, x ) ≤ 8 � x � √ def − (1 − θ ) ζ � x � 2 , ∀ � x � ≥ 29 δ ≤ = µ 8 ζθ (0 < θ < 1) If µ 2 λ max ( P ) < c , then all solutions of the perturbed system, starting in Ω c , are uniformly ultimately bounded by √ � 29 δ λ max ( P ) b = 8 ζθ λ min ( P ) Nonlinear Control Lecture # 10 Time Varying and Perturbed Systems