Nonlinear Control Theory Lecture 9 Periodic Perturbations Averaging Singular Perturbations Khalil Chapter (9, 10) 10.3-10.6, 11
Today: Two Time-scales Averaging ẋ = ǫf(t,x,ǫ) The state x moves slowly compared to f. Singular perturbations ẋ = f(t,x,z,ǫ) ǫż = (t,x,z,ǫ) The state x moves slowly compared to z.
Example: Vibrating Pendulum I m θ l asinωt x=lsin(θ), y=lcos(θ) asin(ωt)
Newton s law in tangential direction m(l θ aω 2 sinωtsinθ) = m sinθ k(l θ+aωcosωtsinθ) (incl. viscous friction in joint) Let ǫ=a/l,τ= ωt, α= ω 0 l/ωa, and β=k/mω 0 x 1 = θ x 2 = ǫ 1 (dθ/dτ)+cosτsinθ f 1 (τ,x) = x 2 cosτsinx 1 f 2 (τ,x) = α βx 2 α 2 sinx 1 the state equation is given by +x 2 cosτcosx 1 cos 2 τsinx 1 cosx 1 dx dτ = ǫf(τ,x)
Averaging Assumptions Consider the system ẋ = ǫf(t,x,ǫ), x(0)=x 0 wherefand its derivatives up to second order are continuous and bounded. Letx av be defined by the equations ẋ av = ǫf av (x av ), x av (0)=x 0 1 f av (x) = lim T T T 0 f(τ,x,0)dτ
Example: Vibrating Pendulum II The averaged system ẋ = ǫf av (x) [ ] x = ǫ 2 α βx 2 α 2 sinx 1 1 4 sin2x 1 has f av x (π,0) = [ ] 0 1 α 2 0.5 α β which is Hurwitz for0< α<1/ 2, β>0. Can this be used for rigorous conclusions?
Periodic Averaging Theorem Letfbe periodic intwith periodt. Let x = 0 be an exponentially stable equilibrium of ẋ av = ǫf(x av ). If x 0 is sufficiently small, then x(t,ǫ) = x av (t,ǫ)+o(ǫ) for allt [0, ] Furthermore, for sufficiently small ǫ>0, the equation ẋ = ǫ f(t, x, ǫ) has a unique exponentially stable periodic solution of period T in an O(ǫ) neighborhood of x = 0.
General Averaging Theorem Under certain conditions on the convergence of 1 T f av (x) = lim f(τ,x,0)dτ T T 0 there exists a C > 0 such that for sufficiently small ǫ > 0 for allt [0,1/ǫ]. x(t,ǫ) x av (t,ǫ) < Cǫ
Example: Vibrating Pendulum III The Jacobian of the averaged system is Hurwitz for 0< α<1/ 2, β>0. For a/l sufficiently small and ω > 2ω 0 l/a the unstable pendulum equilibrium(θ, θ)=(π,0) is therefore stabilized by the vibrations.
Periodic Perturbation Theorem Consider ẋ = f(x)+ǫ (t,x,ǫ) wheref,, f/ x and / x are continuous and bounded. Let be periodic intwith periodt. Let x = 0 be an exponentially stable equilibrium point for ǫ = 0. Then, for sufficiently small ǫ>0, there is a unique periodic solution which is exponentially stable. x(t, ǫ) = O(ǫ)
Proof ideas of Periodic Perturbation Theorem Let φ(t,x 0,ǫ) be the solution of ẋ = f(x)+ǫ (t,x,ǫ), x(0)=x 0 Exponential stability of x = 0 for ǫ = 0, plus bounds on the magnitude of, shows existence of a bounded solution x for small ǫ>0. The implicit function theorem shows solvability of x = φ(t,0,x,ǫ) for small ǫ. This gives periodicity of x. Putz=x x. Exponential stability ofx=0for ǫ=0gives exponential stability ofz=0for small ǫ>0.
Proof idea of Averaging Theorem For small ǫ>0defineuandy by Then u(t,x) = t 0 x = y+ ǫu(t,y) [f(τ,x,0) f av (x)]dτ ẋ=ẏ+ ǫ u(t,y) + ǫ u(t,y) ẏ t y [ I+ ǫ u ] ẏ = ǫf(t,y+ ǫu,ǫ) ǫ u y t (t,y) = ǫf av (y)+ǫ 2 p(t,y,ǫ) Withs=ǫt, dy ( s ) = f av (y)+ǫq ds ǫ,y,ǫ which has a unique and exponentially stable periodic solution for small ǫ. This gives the desired result.
Application: Second Order Oscillators For the second order system introduce y = rsin φ ÿ+ω 2 y = ǫ (y,ẏ) (1) ẏ/ω = rcosφ (rsin φ,ωrcosφ)cos φ f(φ,r,ǫ) = ω 2 (ǫ/r) (rsin φ,ωrcosφ)sin φ f av (r) = 1 2π = Then (1) is equivalent to 2π 1 2πω 2 dr dφ 0 2π 0 f(φ,r,0)dφ (rsin φ,ωrcosφ)cos φdφ = ǫf(φ,r,ǫ) and the periodic averaging theorem may be applied.
Illustration: Van der Pol Oscillator I C i C L i L i + V resistive element linear osc-part For an ordinary resistance we will get a damped oscillation. For a negative resistance/admittance chosen as we get i=h(v) = ( V+ 1 3 V3 ) }{{} ivesvanderpol eq. i C +i L +i=0, i=h(v) CL d2 V dt 2 +V+Lh (V) dv dt =0
Example: Van der Pol Oscillator I The vacuum tube circuit equation (a k a thevanderpolequation) gives f av (r) = 1 2π The averaged system ÿ+y = ǫẏ(1 y 2 ) 2π 0 = 1 2 r 1 8 r3 dr dφ rcos φ(1 r 2 sin 2 φ)cos φdφ ( ) 1 = ǫ 2 r 1 8 r3 has equilibriar=0,r=2with df av dr = 1 r=2 so small ǫ give a stable limit cycle, which is close to circular with radiusr=2.
Singular Perturbations Consider equations of the form ẋ = f(t,x,z,ǫ), x(0)=x 0 ǫż = (t,x,z,ǫ) z(0)=z 0 For small ǫ>0, the first equation describes theslowdynamics, while the second equation defines thefastdynamics. The main idea will be to approximatexwith the solution of the reduced problem x = f(t,x,h(t, x),0) x(0)=x 0 whereh(t, x) is defined by the equation 0 = (t,x,h(t,x),0)
Example: DC Motor I i R L u EMK=kω ω J J dω dt L di dt = ki = kω Ri+u Withx= ω,z=iand ǫ=lk 2 /JR 2 we get ẋ = z ǫż = x z+u
Linear Singular Perturbation Theorem Let the matrixa 22 have nonzero eigenvalues γ 1,...,γ m and let λ 1,...,λ n be the eigenvalues ofa 0 =A 11 A 12 A 1 22 A 21. Then, δ>0 ǫ 0 >0such that the eigenvalues α 1,...,α n+m of the matrix [ ] A11 A 12 A 21 /ǫ A 22 /ǫ satisfy the bounds for0<ǫ<ǫ 0. λ i α i < δ, i=1,...,n γ i n ǫα i < δ, i=n+1,...,n+m
Proof A 22 is invertible, so it follows from the implicit function theorem that for sufficiently small ǫ the Riccati equation ǫa 11 P ǫ +A 12 ǫp ǫ A 21 P ǫ P ǫ A 22 = 0 has a unique solutionp ǫ =A 12 A 1 22 +O(ǫ). The desired result now follows from the similarity transformation [ ][ ] [ ] I ǫpǫ A11 A 12 I ǫpǫ 0 I A 21 /ǫ A 22 /ǫ 0 I = [ ][ ] I ǫpǫ A11 A 12 + ǫa 11 P ǫ 0 I A 21 /ǫ A 22 /ǫ+a 21 P ǫ [ A0 +O(ǫ) 0 = A 22 /ǫ+o(1) ]
Example: DC Motor II In the example we have ẋ = z ǫż = x z+u [ ] A11 A 12 A 21 A 22 = [ ] 0 1 1 1 A 11 A 12 A 1 22 A 21 = 1 so stability of the DC motor model for small is verified. ǫ = Lk2 JR 2 See Khalil for example where reduced system is stable but fast dynamics unstable.
The Boundary-Layer System For fixed(t, x) the boundary layer system dŷ dτ = (t,x,ŷ+h(t,x),0), ŷ(0)=z 0 h(0,x 0 ) describes the fast dynamics, disregarding variations in the slow variables t, x.
Tikhonov s Theorem Consider a singular perturbation problem with f,,h, / x C 1. Assume that the reduced problem has a unique bounded solution x on[0, T] and that the equilibrium ŷ=0of the boundary layer problem is exponentially stable uniformly in(t,x). Then uniformly fort [0,T]. x(t,ǫ) = x(t)+o(ǫ) z(t,ǫ) = h(t, x(t))+ŷ(t/ǫ)+o(ǫ)
Example: High Gain Feedback u + + ψ( ) k 1 /s u p ẋ p =... y=.. y k 2 Closed loop system Reduced model ẋ p = Ax p +Bu p 1 p = ψ(u u p k 2 Cx p ) k 1 u ẋ p = (A Bk 2 C)x p +Bu
Proof ideas of Tikhonov s Theorem Replacefand withfandgthat are identical for x <r, but nicer for largex. For small ǫ,g(t,x,y,ǫ) is close tog(t,x,y,0). y-bound for G(,,, 0)-equation y-bound forg-equation x,y-bound forf,g-equations For small ǫ > 0, the x, y-solutions of the F, G-equations will satisfy x <r. Hence, they also solve thef, -equations
The Slow Manifold For small ǫ>0, the system ẋ = f(x,z) ǫż = (x,z) has the invariant manifold z = H(x,ǫ) It can often be computed approximately by Taylor expansion H(x,ǫ) = H 0 (x)+ǫh 1 (x)+ǫ 2 H 2 (x)+ whereh 0 satisfies 0 = (x,h 0 )
z z The Fast Manifold ǫ=0.001 x = x + z epsilon = 0.001 z = 1/epsilon atan(1 z x) 2 1 0 1 2 3 4 2 1 0 1 2 3 4 x x = x + z z = 1/epsilon atan(1 z x) ǫ=0.1 epsilon = 0.1 2 1 0 1 2 3 4 2 1 0 1 2 3 4 x
Example: Van der Pol Oscillator III Consider With we have the system with slow manifold d 2 v ds 2 µ(1 v2 ) dv ds +v = 0 x = 1 dv µ ds +v 1 3 v3 z = v t = s/µ ǫ = 1/µ 2 ẋ = z ǫż = x+z 1 3 z3 x = z 1 3 z3
z z Illustration: Van der Pol III Phase plot for van der Pol example ǫ=0.001 x = z z = 1/epsilon ( x + z 1/3 z 3 epsilon = 0.001 ) 4 3 2 1 0 1 2 3 4 4 3 2 1 0 1 2 3 4 x x = z z = 1/epsilon ( x + z 1/3 z 3 ) ǫ=0.1 epsilon = 0.1 4 3 2 1 0 1 2 3 4 4 3 2 1 0 1 2 3 4 x