Nonlinear Control Lecture # 14 Tracking & Regulation. Nonlinear Control

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1 Nonlinear Control Lecture # 14 Tracking & Regulation

2 Normal form: η = f 0 (η,ξ) ξ i = ξ i+1, for 1 i ρ 1 ξ ρ = a(η,ξ)+b(η,ξ)u y = ξ 1 η D η R n ρ, ξ = col(ξ 1,...,ξ ρ ) D ξ R ρ Tracking Problem: Design a feedback controller such that lim [y(t) r(t)] = 0 t while ensuring boundedness of all state variables Regulation Problem: r is constant

3 Assumption 13.1 b(η,ξ) b 0 > 0, η D η, ξ D ξ Assumption 13.2 η = f 0 (η,ξ) is bounded-input bounded-state stable over D η D ξ Assumption 13.2 holds locally if the system is minimum phase and globally if η = f 0 (η,ξ) is ISS Assumption 13.3 r(t) and its derivatives up to r (ρ) (t) are bounded for all t 0 and the ρth derivative r (ρ) (t) is a piecewise continuous function of t. Moreover, R = col(r,ṙ,...,r (ρ 1) ) D ξ for all t 0

4 The reference signal r(t) could be specified as given functions of time, or it could be the output of a reference model Example: For ρ = 2 ω 2 n, ζ > 0, ω s 2 +2ζω n s+ωn 2 n > 0 ẏ 1 = y 2, ẏ 2 = ω 2 n y 1 2ζω n y 2 +ω 2 n u c, r = y 1 ṙ = y 2, r = ẏ 2 Assumption 13.3 is satisfied when u c (t) is piecewise continuous and bounded

5 Change of variables: e 1 = ξ 1 r, e 2 = ξ 2 r (1),..., e ρ = ξ ρ r (ρ 1) η = f 0 (η,ξ) ė i = e i+1, for 1 i ρ 1 ė ρ = a(η,ξ)+b(η,ξ)u r (ρ) Goal: Ensure e = col(e 1,...,e ρ ) = ξ R is bounded for all t 0 and converges to zero as t tends to infinity Assumption 13.4 r, r (1),..., r (ρ) are available to the controller (needed in state feedback control)

6 Feedback controllers for tracking and regulation are classified as in stabilization State versus output feedback Static versus dynamic controllers Region of validity local tracking regional tracking semiglobal tracking global tracking Local tracking is achieved for sufficiently small initial states and sufficiently small R, while global tracking is achieved for any initial state and any bounded R.

7 Practical tracking: The tracking error is ultimately bounded and the ultimate bound can be made arbitrarily small by choice of design parameters local practical tracking regional practical tracking semiglobal practical tracking global practical tracking

8 Tracking [ η = f 0 (η,ξ), ė = A c e+b ] c a(η,ξ)+b(η,ξ)u r (ρ) Feedback linearization: u = [ a(η,ξ)+r (ρ) +v ] /b(η,ξ) η = f 0 (η,ξ), ė = A c e+b c v v = Ke, A c B c K is Hurwitz η = f 0 (η,ξ), ė = (A c B c K)e A c B c K Hurwitz e(t) is bounded and lim t e(t) = 0 ξ = e+r is bounded η is bounded

9 Example 13.1 (Pendulum equation) ẋ 1 = x 2, ẋ 2 = sinx 1 bx 2 +cu, y = x 1 We want the output y to track a reference signal r(t) e 1 = x 1 r, e 2 = x 2 ṙ ė 1 = e 2, ė 2 = sinx 1 bx 2 +cu r u = 1 c [sinx 1 +bx 2 + r k 1 e 1 k 2 e 2 ] K = [k 1,k 2 ] assigns the eigenvalues of A c B c K at desired locations in the open left-half complex plane

10 Simulation r = sin(t/3), x(0) = col(π/2, 0) Nominal: b = 0.03,c = 1 Figures (a) and (b) Perturbed: b = 0.015,c = 0.5 Figure (c) Reference (dashed) Low gain: K = [ 1 1 ], λ = 0.5±j0.5 3, (solid) High gain: K = [ 9 3 ], λ = 1.5±j1.5 3, (dash-dot)

11 2 (a) 2 (b) Output Output Time (c) Time 5 (d) Output Control Time Time

12 Robust Tracking η = f 0 (η,ξ) ė i = e i+1, 1 i ρ 1 ė ρ = a(η,ξ)+b(η,ξ)u+δ(t,η,ξ,u) r (ρ) (t) Sliding mode control: Design the sliding surface ė i = e i+1, 1 i ρ 1 View e ρ as the control input and design it to stabilize the origin e ρ = (k 1 e 1 + +k ρ 1 e ρ 1 ) λ ρ 1 +k ρ 1 λ ρ 2 + +k 1 is Hurwitz

13 s = (k 1 e 1 + +k ρ 1 e ρ 1 )+e ρ = 0 ρ 1 ṡ = k i e i+1 +a(η,ξ)+b(η,ξ)u+δ(t,η,ξ,u) r (ρ) (t) i=1 u = v or u = 1 ˆb(η,ξ) [ ρ 1 ] k i e i+1 +â(η,ξ) r (ρ) (t) +v i=1 ṡ = b(η,ξ)v+ (t,η,ξ,v) Suppose (t,η,ξ,v) b(η, ξ) (η,ξ)+κ 0 v, 0 κ 0 < 1 ( ) s v = β(η,ξ) sat, β(η,ξ) (η,ξ) µ (1 κ 0 ) +β 0, β 0 > 0

14 sṡ β 0 b 0 (1 κ 0 ) s, s µ ζ = col(e 1,...,e ρ 1 ), ζ = (Ac B c K) ζ +B }{{} c s Hurwitz V 0 = ζ T Pζ, P(A c B c K)+(A c B c K) T P = I V 0 = ζ T ζ+2ζ T PB c s (1 θ) ζ 2, ζ 2 PB c s /θ 0 < θ < 1. For σ µ { ζ 2 PB c σ/θ} {ζ T Pζ λ max (P)(2 PB c /θ) 2 σ 2 } ρ 1 = λ max (P)(2 PB c /θ) 2, c > µ Ω = {ζ T Pζ ρ 1 c 2 } { s c} is positively invariant

15 For all e(0) Ω, e(t) enters the positively invariant set Ω µ = {ζ T Pζ ρ 1 µ 2 } { s µ} Inside Ω µ, e 1 kµ k = LP 1/2 ρ 1, L = [ ]

16 Example 13.2 (Reconsider Example 13.1) ė 1 = e 2, ė 2 = sinx 1 bx 2 +cu r r(t) = sin(t/3), 0 b 0.1, 0.5 c 2 s = e 1 +e 2 ṡ = e 2 sinx 1 bx 2 +cu r = (1 b)e 2 sinx 1 bṙ r (1 b)e 2 sinx 1 bṙ r c e /3+1/9 0.5 ( ) e1 +e 2 u = (2 e 2 +3) sat µ

17 Simulation: µ = 0.1, x(0) = col(π/2,0) b = 0.03, c = 1 (solid) b = 0.015, c = 0.5 (dash-dot) Reference (dashed)

18 Output (a) Time s (b) Time

19 Robust Regulation via Integral Action η = f 0 (η,ξ,w) ξ i = ξ i+1, 1 i ρ 1 ξ ρ = a(η,ξ,w)+b(η,ξ,w)u y = ξ 1 Disturbance w and reference r are constant Equilibrium point: 0 = f 0 ( η, ξ,w) 0 = ξ i+1, 1 i ρ 1 0 = a( η, ξ,w)+b( η, ξ,w)ū r = ξ 1

20 Assumption = f 0 ( η, ξ,w) has a unique solution η = φ η (r,w) ū = a( η, ξ,w) b( η, ξ,w) z = η η, e = def = φ u (r,w) Augment the integrator ė 0 = y r e 1 ξ 1 r e 2 ξ 2. e ρ =. ξ ρ

21 ż = f 0 (z + η,ξ,w) def = f 0 (z,e,r,w) ė i = e i+1, Sliding mode control: for 0 i ρ 1 ė ρ = a(η,ξ,w)+b(η,ξ,w)u s = k 0 e 0 +k 1 e 1 + +k ρ 1 e ρ 1 +e ρ λ ρ +k ρ 1 λ ρ 1 + +k 1 λ+k 0 is Hurwitz ρ 1 ṡ = k i e i+1 +a(η,ξ,w)+b(η,ξ,w)u [ i=0 u = v or u = 1 ρ 1 ] k i e i+1 +â(η,ξ) +v ˆb(η,ξ) i=0

22 ṡ = b(η,ξ,w)v+ (η,ξ,r,w) (η,ξ,r,w) b(η,ξ,w) (η,ξ) ( ) s v = β(η,ξ) sat, β(η,ξ) (η,ξ)+β 0, β 0 > 0 µ Assumption 13.6 α 1 ( z ) V 1 (z,r,w) α 2 ( z ) V 1 z f 0 (z,e,r,w) α 3 ( z ), z α 4 ( e ) Assumption 13.7 z = 0 is an exponentially stable equilibrium point of ż = f 0 (z,0,r,w)

23 Theorem 13.1 Under the stated assumptions, there are positive constants c, ρ 1 and ρ 2 and a positive definite matrix P such that the set Ω = {V 1 (z) α 2 (α 4 (cρ 2 )} {ζ T Pζ ρ 1 c 2 } { s c} where ζ = col(e 0,e 1,...,e ρ 1 ), is compact and positively invariant, and for all initial states in Ω lim y(t) r = 0 t Special case: β = k (a constant) and u = v ( ) k0 e 0 +k 1 e 1 + +k ρ 1 e ρ 1 +e ρ u = k sat µ

24 Example 13.4 (Pendulum with horizontal acceleration) ẋ 1 = x 2, ẋ 2 = sinx 1 bx 2 +cu+dcosx 1, y = x 1 d is constant. Regulate y to a constant reference r 0 b 0.1, 0.5 c 2, 0 d 0.5 e 1 = x 1 r, e 2 = x 2 ė 0 = e 1, ė 1 = e 2, ė 2 = sinx 1 bx 2 +cu+dcosx 1 s = e 0 +2e 1 +e 2 ṡ = e 1 +(2 b)e 2 sinx 1 +cu+dcosx 1

25 e 1 +(2 b)e 2 sinx 1 +dcosx 1 c e 1 +2 e ( ) e0 +2e 1 +e 2 u = (2 e 1 +4 e 2 +4) sat µ For comparison, SMC without integrator s = e 1 +e 2, ṡ = (1 b)e 2 sinx 1 +cu+dcosx 1 ( ) e1 +e 2 u = (2 e 2 +4) sat µ Simulation: With integrator (dashed), without (solid) µ = 0.1, x(0) = 0, r = π/2, b = 0.03, c = 1, d = 0.3

26 Output 1 Output Time Time

27 Output Feedback Tracking: Regulation: η = f 0 (η,ξ) ė i = e i+1, 1 i ρ 1 ė ρ = a(η,ξ)+b(η,ξ)u+δ(t,η,ξ,u) r (ρ) (t) η = f 0 (η,ξ,w) ξ i = ξ i+1, 1 i ρ 1 ξ ρ = a(η,ξ,w)+b(η,ξ,w)u y = ξ 1 Design partial state feedback control that uses ξ Use a high-gain observer

28 Tracking sliding mode controller: ( ) k1 e 1 + +k ρ 1 e ρ 1 +e ρ u = β(ξ) sat µ Regulation sliding mode controller: ( ) k0 e 0 +k 1 e 1 + +k ρ 1 e ρ 1 +e ρ u = β(ξ) sat µ ė 0 = e 1 = y r β is allowed to depend only on ξ rather than the full state vector. On compact sets, the η-dependent part of (η,ξ) can be bounded by a constant

29 High-gain observer: ê i = ê i+1 + α i ε i(y r ê 1), 1 i ρ 1 ê ρ = α ρ ε ρ(y r ê 1) λ ρ +α 1 λ ρ 1 + +α ρ 1 λ+α ρ Hurwitz e ê ξ ˆξ = ê+r β(ˆξ) β s (ˆξ) (saturated)

30 Tracking: ( ) k1 ê 1 + +k ρ 1 ê ρ 1 +ê ρ u = β s (ˆξ) sat µ Regulation: ( ) k0 e 0 +k 1 ê 1 + +k ρ 1 ê ρ 1 +ê ρ u = β s (ˆξ) sat µ We can replace ê 1 by e 1 Special case: When β s is constant or function of ê rather than ˆξ, we do not need the derivatives of r, as required by Assumption 13.4

31 The output feedback controllers recover the performance of the partial state feedback controllers for sufficiently small ε. In the regulation case, the regulation error converges to zero Relative degree one systems: No observer ( ) ( ) y r k0 e 0 +y r u = β(y)sat, u = β(y)sat µ µ