Time- and Energy-Based Stability

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1 Time- and Energy-Based Stability Outline Root Locus Lyapunov s methods 1 Copyright c 2015 Roderic Grupen

2 8+ Root Locus ω n 2 Θ( s) Θ( ref s) 2 s + 2 s + ω 2 n ζω n 0 y γ θ ref θ act 128 The homogeneous (unforced) differential equation of motion: 1+GH = s 2 +2ζω n s+ω 2 n = 0 stability: bounded input bounded output (BIBO) Im K 8+ unstable consider: 1+GH = s 2 +2s+K K from 8 K=1 K from 8 Re K=0 K 2 Copyright c 2015 Roderic Grupen

3 Analytic Stability Lyapunov s Direct Method Stability - the origin of the state space is stable if there eists a region, S(r), such that states which start within S(r) remain within S(r). Asymptotic Stability - a system is asymptotically stable in S(r) if as t, the system state approaches the origin of the state space. S(r) 3 Copyright c 2015 Roderic Grupen

4 Analytic Stability - Lyapunov s Second Method Define: an arbitrary scalar function, V(, t), called a Lyapunov function, continuous is all first derivatives, where is the state and t is time, Iff: If the function, V(,t), eists such that: (a) V(0,t) = 0, and (b) V(,t) > 0, for 0 (positive definite), and (c) V/ t < 0 (negative def inite), Then: the system described by V is asymptotically stable in the neighborhood of the origin....if a system is stable, then there eists a suitable Lyapunov function....if, however, a particular Lyapunov function does not satisfy these criteria, it is not necessarily true that this system is unstable. 4 Copyright c 2015 Roderic Grupen

5 EXAMPLE: spring-mass-damper system dynamics: ẍ+ B Mẋ+ K M = 0 K m B E = v 0 (Mv)dv + 0 (K)d = 1 2 Mv K2 = 1 2 Mẋ K2 Lyapunov function: V(,t) = E = Mẋ2 2 (a) V(0,t) = 0, (b) V(,t) > 0, + K2 2 (c) V/ t negative def inite? 5 Copyright c 2015 Roderic Grupen

6 EXAMPLE: spring-mass-damper m K B B>0 B= t=0...the entire state space is asymptotically stable for B > 0. 6 Copyright c 2015 Roderic Grupen

7 EXAMPLE: population dynamics system dynamics: 1 = # males 2 = # females ẋ 1 = 1 +α 1 2 = 1 (α 2 1) ẋ 2 = 2 +β 1 2 = 2 (β 1 1) equilibrium points: ẋ = 0 (a) 1 = 2 = 0 (b) 1 = 1 β, 2 = 1 α Lyapunov function: V(0,t) = 0 V(,t) > 0 } choose V(,t) = V t = 2 1 ẋ ẋ 2 = 2 2 1(α 2 1)+2 2 2(β 1 1) 0 7 Copyright c 2015 Roderic Grupen

8 EXAMPLE: population dynamics 8 Copyright c 2015 Roderic Grupen

9 Putting it All Together - Control Law Synthesis we now have some powerful tools under our belt: tools for generating analytical descriptions of system dynamics Newton/Euler (outward/inward) iterations Lagrange s Equation (from calculus of variations) tools for analyzing linear controls, CLTF tools for eamining stability root locus Lyapunov s Second Method 9 Copyright c 2015 Roderic Grupen

10 Putting it All Together - Balancing the Cart-Pole 1. describe the dynamics of the system The Lagrangian and Lagrange s Equations 2. linearize around the desired equilibrium position, write the dynamics as a set of n first-order equations q = Aq 3. add a(parametric) linear controller and describe the net(plant + compensator) system behavior q = Aq+Bu(t) 4. use the properties of stable systems to find control parameters B that stabilize the system 10 Copyright c 2015 Roderic Grupen

11 Control Law Synthesis Cart/Pole Inverted Pendulum y m θ y M p p u(t) position and velocity of the pendulum p = Lsin(θ) ẋ p = Lcos(θ) θ+ẋ y p = Lcos(θ) ẏ p = Lsin(θ) θ construct the Lagrangian L = T V: T = 1 2 Mẋ m(ẋ2 p+ẏ 2 p) = 1 2 Mẋ m (( L 2 cos 2 (θ) θ 2 2Lcos(θ) θẋ+ẋ 2 )+ (L 2 sin 2 (θ) θ 2 )) = 1 2 (M +m)ẋ m ( L 2 cos 2 (θ) θ 2 +L 2 sin 2 (θ) θ 2 ) mlcos(θ) θẋ = 1 2 (M +m)ẋ ml2 θ2 mlcos(θ) θẋ 11 Copyright c 2015 Roderic Grupen

12 Control Law Synthesis y m θ y M p p u(t) Cart/Pole Inverted Pendulum V = mgy p = mglcos(θ) so the Lagrangian is: L = 1 2 (M +m)ẋ ml2 θ2 mlcos(θ) θẋ mglcos(θ) and Lagrange s equations are written: d L L = Γ i. dt q i q i 12 Copyright c 2015 Roderic Grupen

13 Cart-Pole Dynamics L = 1 2 (M +m)ẋ ml2 θ2 mlcos(θ) θẋ mglcos(θ) d L L = Γ i dt q i q i for the coordinate of the cart: L ẋ = (M +m)ẋ mlcos(θ) θ d L dt ẋ = (M +m)ẍ+mlsin(θ) θ 2 mlcos(θ) θ L = 0 and u(t) = (M +m)ẍ+mlsin(θ) θ 2 mlcos(θ) θ 13 Copyright c 2015 Roderic Grupen

14 Cart-Pole Dynamics L = 1 2 (M +m)ẋ ml2 θ2 mlcos(θ) θẋ mglcos(θ) d L L = Γ i dt q i q i for the (unactuated) angle of the pendulum: L = ml2 θ mlcos(θ)ẋ θ so d dt L = ml2 θ mlcos(θ)ẍ+mlsin(θ) θẋ θ L θ = mlsin(θ) θẋ+mglsin(θ) 0 = ml 2 θ mlcos(θ)ẍ+mlsin(θ) θẋ mlsin(θ) θẋ mglsin(θ) = ml 2 θ mlcos(θ)ẍ mglsin(θ) 14 Copyright c 2015 Roderic Grupen

15 Linearized Cart-Pole Equation of Motion (M +m)ẍ+mlsin(θ) θ 2 mlcos(θ) θ = u(t) ml 2 θ mlcos(θ)ẍ mglsin(θ) = 0 using small angle assumptions (sin(θ) θ; cos(θ) 1), and assuming small pendulum velocities ( θ 2 0): (M +m)ẍ ml θ = u(t) ml 2 θ mlẍ mglθ = 0 if we define state variables: (q 1 q 2 q 3 q 4 ) = ( ẋ θ θ) then the equations of motion become: (M +m) q 2 ml q 4 = u(t) (a) q 2 + L q 4 = gq 3 (b) if we solve (b) for L q 4 and substitute into (a)... or (M) q 2 mgq 3 = u(t) q 2 = mg M q 3+ u(t) M 15 Copyright c 2015 Roderic Grupen

16 Linearized Cart-Pole Equation of Motion (cont.) similarly, if we solve for q 2 in (a) and substitute into (b), we get q 4 = q 2 = u(t)+ml q 4 M +m (M +m) M g L q 3+ u(t) ML therefore, we can write the cart-pole dynamics in the form of four first order equations (assume that m << M, so that m/m 0): q 1 = q 2 q 2 = 1 M u(t) q 3 = q 4 q 4 = g L q 3+ 1 ML u(t) 16 Copyright c 2015 Roderic Grupen

17 Linearized Cart-Pole Equation of Motion (cont.) q 1 q 2 q 3 q 4 = q = Aq+Bu(t) q 1 q 2 q 3 + q g/l 0 0 1/M 0 1/ML u(t) The characteristic equation of this linear system is det(λi A) = λ 2 (λ 2 g L ) so we get four roots: λ 1 = λ 2 = 0 λ 3 = + g/l λ 4 = g/l (unstable) 17 Copyright c 2015 Roderic Grupen

18 Simplified, Linearized Cart-Pole isolate variables q 3 and q 4 to focus on the unstable roots and to design control inputs that stabilizes the system... [ q3 q 4 ] = [ 0 1 g/l 0 ][ q3 q 4 ] + [ 0 1/ML ] u(t) the characteristic equation of the simplified system verifies that we have isolated the unstable part of the system dynamics det(λi A) = (λ 2 g L ) λ = ± g/l (unstable) 18 Copyright c 2015 Roderic Grupen

19 Simplified, Linearized Cart-Pole Stabilizing Compensator suppose we define a linear control input u(t) = Hq = [h 1 h 2 ] [ q3 q 4 ] [ q3 q 4 ] = = [ ][ ] [ 0 1 q3 0 + g/l 0 q 4 (h 1 q 3 +h 2 q 4 )/ML [ ][ ] 0 1 q3 g/l+h 1 /(ML) h 2 /(ML) q 4 ] pickan(h 1,h 2 )compensatorthatmakeseigenvaluesnegative...e.g., let h 1 = Mg det(λi A) = λ 1 ( 0 (λ h 2 /(ML)) = λ λ h ) 2 ML λ 1,2 = 0, h 2 ML therefore, h 1 = Mg and h 2 < 0, i.e. u(t) = (Mg)θ B θ balances the cart-pole near vertical (θ = θ = 0)! 19 Copyright c 2015 Roderic Grupen