3 rd Tutorial on EG4321/EG7040 Nonlinear Control

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1 3 rd Tutorial on EG4321/EG7040 Nonlinear Control Lyapunov Stability Dr Angeliki Lekka 1 1 Control Systems Research Group Department of Engineering, University of Leicester arch 9, 2017 Dr Angeliki Lekka (al385@leacuk) Lyapunov Stability arch 9, / 17

2 Stability According to Lyapunov Things to remember Lyapunov s 2 nd method 1 Choose a positive Lyapunov function candidate 2 Determine its derivative along the trajectory of the nonlinear system 3 Check for radial unboundedness of the Lyapunov function (automatically satisfied when using quadratic functions) How do we choose a Lyapunov function? Unfortunately, there is no general way of finding a suitable Lyapunov function One must use Experience and/or intuition and/or physical insights Quadratic functions x Px a good first choice, although not always suitable Dr Angeliki Lekka (al385@leacuk) Lyapunov Stability arch 9, / 17

3 Global results Global results (1) Example Consider the nonlinear system ẋ 1 = x 2 x 1 (x1 2 + x2 2 ) ẋ 2 = x 1 x 2 (x1 2 + x2 2 ) with an equilibrium point at the origin of the state-space Let V (x 1, x 2 ) = x1 2 + x 2 2 be a candidate Lyapunov function, then its time derivative is equal to: V (x 1, x 2 ) = V (x 1, t) dv (x 1, t) + V (x 1, t) dt x 1 dt dt dt + V (x 2, t) dv (x 2, t) + V (x 2, t) dt x 2 dt dt dt Dr Angeliki Lekka (al385@leacuk) Lyapunov Stability arch 9, / 17

4 Global results Global results (2) Example Substituting ẋ 1 and ẋ 2 we get: V (x 1, x 2 ) = 2x 1 ẋ x 2 ẋ V (x 1, x 2 ) = 2x 1 (x 2 x1 3 x 1 x2 2 ) + 2x 2 ( x 1 x1 2 x 2 x2 3 ) = 2x 1 x 2 2x1 4 2x1 2 x2 2 2x 1 x 2 2x1 2 x2 2 2x2 4 = 2x 4 1 4x 2 1 x 2 2 2x 4 2 = 2(x x 2 1 x x 4 2 ) = 2(x x 2 2 ) 2 So, V is globally negative definite (assuming the point at the origin is the only equilibrium) Dr Angeliki Lekka (al385@leacuk) Lyapunov Stability arch 9, / 17

5 Local results Local results (1) Example Consider the nonlinear system ẋ 1 = x 1 (x x 2 2 2) 4x 1 x 2 2 ẋ 2 = 4x 2 1 x 2 + x 2 (x x 2 2 2) with an equilibrium point at the origin of the state-space Let V (x 1, x 2 ) = x1 2 + x 2 2 be a candidate Lyapunov function, then its time derivative is equal to: V (x 1, x 2 ) = V (x 1, t) dv (x 1, t) + V (x 1, t) dt x 1 dt dt dt + V (x 2, t) dv (x 2, t) + V (x 2, t) dt x 2 dt dt dt Dr Angeliki Lekka (al385@leacuk) Lyapunov Stability arch 9, / 17

6 Local results Local results (2) Example Substituting ẋ 1 and ẋ 2 we get: V (x 1, x 2 ) = 2x 1 ẋ x 2 ẋ V (x 1, x 2 ) = 2x 1 (x1 3 + x 1 x2 2 2x 1 4x 1 x2 2 ) + 2x 2 (4x1 2 x 2 + x1 2 x 2 + x2 3 2x 2 ) = 2x x1 2 x2 2 4x1 2 8x1 2 x x1 2 x x1 2 x x2 4 4x2 2 = 2x x x 2 1 x 2 2 4x 2 1 4x 2 2 = 2(x x 2 2 ) 2 4(x x 2 2 ) = 2(x x 2 2 )(x x 2 2 2) V < 0 only if x1 2 + x < 0, so V is locally negative definite in the region defined by x1 2 + x 2 2 < 2 Dr Angeliki Lekka (al385@leacuk) Lyapunov Stability arch 9, / 17

7 Young s inequality any nonlinear systems can be represented as a sum of a linear part and a nonlinear part ẋ = f (x) = }{{} Ax + g(x) }{{} linear part nonlinear part assuming A is stable, one can find a P to solve the Lyapunov equation; we know the derivative of the linear part is negative definite, the goal is to bound the nonlinear part, so the linear part becomes dominant and hence, asymptotic stability can be inferred If we ve got 2 vectors z and y, we can always bound them by Young s Inequality for all γ > 0 2z y 1 γ x 2 + γ y 2 Dr Angeliki Lekka (al385@leacuk) Lyapunov Stability arch 9, / 17

8 Young s inequality Example Let us choose a quadratic candidate Lyapunov function V (x) = x Px Its time derivative is then calculated as: V V (x) (x) = f (x) x = 2x P(Ax + g(x)) }{{} 2x Pẋ = x (A P + PA)x + 2x Pg(x) = x Qx + 2x Pg(x) }{{} indefinite term Young s inequality is normally used to handle the indefinite part Dr Angeliki Lekka (al385@leacuk) Lyapunov Stability arch 9, / 17

9 Exams Question Exams 2015, Question 1ciii (1) Show that a sufficient condition for the system described by ẋ 1 = x 2 ẋ 2 = K x 1 B x 2(1 + x 2 2 ) to be asymptotically stable is for there to exist a scalar λ such that the following inequality holds: ( B + 4λK 2 4λB > 2K Hint: Use the Lyapunov function V tot (x) = V 1 (x) + V 2 (x) = x Px + λx 4 2 where λ > 0 ) 2 Dr Angeliki Lekka (al385@leacuk) Lyapunov Stability arch 9, / 17

10 Exams Question Exams 2015, Question 1ciii (2) Firstly, let s write the system in state-space form: [ ] [ ] [ ] [ ẋ1 0 1 x1 = ẋ 2 K }{{} B + x 2 }{{}}{{} ẋ A x The time derivative of V 1 (x) is: V 1 (x) = 2x P(Ax + g(x)) }{{} 2x Pẋ = 2 (A P + PA) x + 2x Pg(x) }{{} Q [ = x P11 P Qx + 2[x 1 x 2 ] 12 P 21 P 22 = x 2 2x 1 P 12 B x 3 2 2P 22 B Dr Angeliki Lekka (al385@leacuk) Lyapunov Stability arch 9, / 17 0 B ] x 2 2 } {{ } g(x) ] [ x B ] x 2 2

11 Exams Question Exams 2015, Question 1ciii (3) Similarly, the time derivative of V 2 (x) is: V 2 (x) = 4λx2 3 ẋ ( = 4λx2 3 K x 1 B ) x 2(1 + x2 2 ) So, V tot (x) is equal to: = 4λ K x 1x 3 2 4λ B x 4 2 4λ B x 6 2 V tot (x) = V 1 (x) + V 2 (x) = x 2 B 2x }{{} 1 P 12 x 3 B 2 2P 22 x 2 4 x1 2 x2 2 4λ K x 1x 3 2 4λ B x 4 2 4λ B x 6 2 Dr Angeliki Lekka (al385@leacuk) Lyapunov Stability arch 9, / 17

12 Exams Question Exams 2015, Question 1ciii (4) Collecting terms together yields: V tot (x) = x1 2 x2 2 2 B }{{} x 2 4 (P λ) 4λ B ( B }{{} x 2 6 2x 1 x2 3 }{{} P λ K ) <0 }{{} <0 <0 indefinite Using Young s inequality with z = x 3 2 and y = ϵx 1, the indefinite term can be bound as: 2z y 1 γ x 2 + γ y 2 ( B 2x2 3 P λ K ) x 1 1 γ x γ ϵx 1 2 }{{}}{{} ϵ 2 x 1 2 ϵ 1 γ x γϵ 2 x 2 1 Dr Angeliki Lekka (al385@leacuk) Lyapunov Stability arch 9, / 17

13 Exams Question Exams 2015, Question 1ciii (5) So, now V tot (x) is equal to: V tot (x) = x 2 1 x B x 4 2 (P λ) 4λ B x γ x γϵ 2 x 2 1 and collecting terms together V tot (x) = (1 γϵ 2 )x 2 1 ( 4λ B 1 ) x2 6 x2 2 2(P λ) B γ x 2 4 For V tot (x) to be negative definite the terms (1 γϵ 2 ) and 4λ B 1 γ must be positive Dr Angeliki Lekka (al385@leacuk) Lyapunov Stability arch 9, / 17

14 Exams Question Exams 2015, Question 1ciii (6) 4λ B 1 γ > 0 4λ B > 1 γ γ > 4λB (1) Substituting ϵ in (2) yields: 1 γϵ 2 > 0 1 > γϵ 2 γ < 1 ϵ 2 (2) γ < 1 ϵ 2 γ < 1 ( P12 B + 2λK ) 2 γ < 2 (P 12 B + 2λK) 2 Dr Angeliki Lekka (al385@leacuk) Lyapunov Stability arch 9, / 17

15 Exams Question Exams 2015, Question 1ciii (7) Combining inequalities (1) and (2), we get: 4λB < γ < 2 (P 12 B + 2λK) 2 1 4λB 1 4λB < < γ < (P 12 B + 2λK) 2 (P 12 B + 2λK) 2 4λB > (P 12B + 2λK) 2 4λB > (P 12 B + 2λK) 2 Substituting P 12 = 2K from part ii (see 2nd tutorial or exams solution) yields: 4λB > ( ) B 2 ( B + 4λK 2 2K + 2λK 4λB > 2K ) 2 Dr Angeliki Lekka (al385@leacuk) Lyapunov Stability arch 9, / 17

16 Appendix Appendix Derivative Chain Rule (The Outside-Inside Rule) The derivative of a composite function can be calculated using the chain rule, ie the derivative of a composite function is equal to the derivative of the outside function evaluated at the inside function times the derivative of the inside function, eg V (x) = f (g(x)) V (x) = f (g(x)) g(x) Dr Angeliki Lekka (al385@leacuk) Lyapunov Stability arch 9, / 17

17 Appendix Appendix Example V (x) = (3x 2 + 1) 5 V (x) = 5(3x 2 + 1) 4 ( d dx (3x 2 + 1) ) = 30x(3x 2 + 1) 4 Example V (x) = (sin(x)) 3 V (x) = 3(sin(x)) 2 ( d dx sin(x) ) = 3cos(x)sin 2 (x) Dr Angeliki Lekka (al385@leacuk) Lyapunov Stability arch 9, / 17

EG4321/EG7040. Nonlinear Control. Dr. Matt Turner

EG4321/EG7040. Nonlinear Control. Dr. Matt Turner EG4321/EG7040 Nonlinear Control Dr. Matt Turner EG4321/EG7040 [An introduction to] Nonlinear Control Dr. Matt Turner EG4321/EG7040 [An introduction to] Nonlinear [System Analysis] and Control Dr. Matt