Discontinuous Systems

Transcription

1 Discontinuous Systems Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University

2 Outline Simple Examples Bouncing ball Heating system Gearbox/cruise control Simulating Hybrid Systems Simulink with Stateflow Stateflow/Simulink Bouncing ball Power conditioning system Inverted pendulum Brocket s Necessary Condition Some systems cannot be stabilized by smooth state feedback Extensions to BNC Solutions to Discontinuous Differential Equations Various notions of solution may be appropriate

3 SIMPLE EXAMPLES

4 Bouncing Ball Free fall: Collision: y = g + ( ) = y( t) = 0 + ( ) = ( ), [ 0,1] y t y t cy t c

5 Bouncing Ball v = v gt 1 2v x = v0t gt = 0 t = 2 g v0 2v0 2v0 t1 =, t2 = c,, ti = c g g g N N N 2v0 i 1 2v0 N i 1 2v 0 1 c TN = t 1 i = c c i= = i= 1 = i= 1 g g g 1 c 2v0 1 For 0 c< 1, lim TN = N g 1 c Zeno phenomenon transitions in finite time i 1 Time for N transitions

6 Heater continuous state: x R, x + 50 q = off x = f( xq, ), f( xq, ) = x q = on discrete state: q,, { on off } on q = off, x 73 off q = off, x > 73 qk ( + 1 ) = ϕ( xqk, ( )), ϕ( xq, ) = off q = on, x 77 on q = on, x < 77

7 Automobile Gearbox Control Problem 1: Automated gearbox - coordinate automatic gearshift with throttle command Problem 2: Cruise control automate throttle and gearbox to maintain speed Background R. W. Brockett, "Hybrid Models for Motion Control," in Perspectives in Control, H. L. Trentelman and J. C. Willems, Ed. Boston: Birkhauser, 1993, pp S. Hedlund and A. Rantzer, "Optimal Control of Hybrid Systems," presented at Conference on Decision and Control, Phoenix, AZ, pp , F. D. Torrisi and A. Bemporad, "HYSDEL-A Tool for Generating Computational Hybrid Models for Analysis and Synthesis Problems," IEEE Transactions on Control Systems Technology, vol. 12, pp , 2004.

8 Transmission [ ] ( ω) u { q q q q q } ( ω) sin( α) i M q v f i eng u Fb cv M vehg u ω f eng q wheel 0,1 throttle position engine speed engine torque-speed characteristic,,,, transmission state Rq, i = 1,,5 corresponding gear ratios i R rear gear ratio r fin wheel F b Rq Rfin = r wheel radius brake force

9 Cruise Control Continuous control - throttle u and brake F are chosen so that b Rq R i fin f 2 eng ( ω) u F b = cv + M veh g sin ( α) + M q ( ) ( ) i k P v v + k I v v d r wheel - a standard feedback linearized design with PI controller. - notice that control depends on the discrete state. feng ( ω ) Discrete control - ad hoc gear shift strategy. ω ω u ω ω u ω ω u ω ω u ω ω u ωl ωu ω q 0, neutral q, gear1 1 v+ kv + kv= kv P I I Rq R 1 ω = r wheel fin v q, gear2 2 v+ kv + kv= kv P I I Rq R 2 ω = r wheel fin v q, gear3 3 v+ kv + kv= kv P I I Rq R 3 ω = r wheel fin v q, gear4 4 v+ kv + kv= kv P I I Rq R 4 ω = r wheel fin v q, gear5 5 v+ kv + kv= kv P I I Rq R 5 ω = r wheel fin v ω < ω stall ω < ω l ω < ω l ω ω l < ω < l ω

10 Cruise Control Issues Choice of shift thresholds Wide spread implies large speed deviation before shift Narrow spread opens possibility of excessive shifting, even chattering Does not explicitly consider throttle and brake limits It must be verified that the engine does not stall or exceed red line

11 SIMULATING HYBRID SYSTEMS WITH STATEFLOW/SIMULINK

12 Stateflow Stateflow is a Simulink toolbox Provides a graphical means to incorporate discrete event process into Simulink Based on the concept of statecharts Harel, D., Statecharts: A Visual Formalism for Complex Systems. Science of Computer Programming, : p Has evolved to represent an implementation of UML

13 Simulating Hybrid Systems in Stateflow/SIMULINK Stateflow Interface Continuous Outputs Finite State Machine Event Generator Switched Dynamical System Discrete Inputs Continuous Inputs Discrete Disturbances Mode Selector Continuous Disturbances Discrete Outputs Simulink

14 Stateflow: Action Language Categories State creator tool Stateflow coder Default state Event trigger Transition action Merge tool State entry action State exit action State on event_name action State during action Event trigger Condition Condition action

15 Bouncing Ball Free fall: Collision: y = g + ( ) = y( t) = 0 + ( ) = ( ), [ 0,1] y t y t cy t c

16 Power Conditioning System i L R o E + - R pre C vo R bleed High power drives in vehicle applications Startup (precharge) Normal (current regulation) Shutdown (bleed) Background (Boost converters) M. Senesky, G. Eirea, and T. J. Koo, "Hybrid Modeling and Control of Power Electronics," in Hybrid Systems: Computation and Control, vol. 2623, Lecture Notes in Computer Science. New York: Springer-Verlag, 2003, pp P. Gupta and A. Patra, "Hybrid Sliding Mode Control of DC-DC Power Converters," presented at IEEE Tencon 2003, Bangalore, 2003.

17 Power Conditioning System 0 Clock Display Power Off 1-1 On Synch Memory Memory1 Memory2 Voltage I_av ge Time Modes1 MainBreaker PreBreaker Time_reset MainBreaker {OFF=-1, ON=1} MainBreaker {OFF=-1, ON=1} PreBreaker {OFF=-1, ON=1} PreBreaker {OFF=-1, ON=1} Bleed Bleed Breaker {OFF=-1, ON=1} Bleed Breaker {OFF=-1, ON=1} Av ge Current Reset I_av ge_reset Av ge Current Reset Time Reset Total Load Current Bank Voltage Bank Voltage Step Main Breaker Mains Voltage Mains Voltage Step1 Precharge Mains Current Mains Current Bleed Bank Current Bank Current Scope Power System Gain 1 10 Constant1 1 s Avge Current 1 s Time I_av ge Time

18 Power Conditioning System

19 Power Conditioning System

20 Inverted Pendulum ~ 1 x = v θ = ω + θω = + ω θ 2 2v cos F sin cosθv + ω = sinθ Suppose we choose F to regulate v, so that the pendulum equations are θ = ω ω = sinθ ucosθ where u is treated as a control.

21 Inverted Pendulum ~ 2 Suppose we wish to design a global feedback controller that will steer any initial state to the upright position [ ] with the constraint u 1,1. Feedback linearization will not work in general, choose u so that θ sinθ + u cosθ = 0 θ + 2 θ + θ = 0 2 θ θ sinθ u = within constraints only near θ = 0. cosθ

22 Inverted Pendulum Swing-up Strategy ( ) ( ) θ= ωω = θ θ = ω + θ 1 2, sin ucos Epend cos 1 2 = 0 when ω = 0, θ = 0 [ εε] 1) pump/remove energy into system until E 0, E, pend E = ω sinθ ucosθ sinθω= uωcosθ 2) wait until pendulum is close to upright 3) apply feedback linearizing control E pend pend

23 Inverted Pendulum: Control Strategy Constant energy trajectories, u = 0, in 'wait' state. Switch to 'stabilize' in blue box.

24 Brockett s Necessary Condition

25 Necessary Condition for Asymptotic Stability ( ) x = f xu,, x R, u R, f 0,0 = 0 ( ) ( ) ( ) n m ( ) Theorem: (Brockett) Suppose f is smooth and the origin is stabilized by a smooth state feedback control u x, n n u 0 = 0. Then the mapping F : R R, ( ) ( ) F x = f xu, x maps neighborhoods of the origin into neighborhoods of the origin, i.e. δ > 0 ε > 0 such that B ( m) δ F B ( ) n alternatively, f B R is a neighborhood of 0 R. ε δ

26 Example x2 x 1 x = u = xu 2 1 Points on F2 axis close to F1 axis are outside image x F F 1 1.0

27 Example 2 x = vx cosθ x1 cos xu 3 1 d y = vsinθ x = sin xu θ x dt = ω x 3 u 2 Notice that points on the F axis close to F axis 2 1 are not in the image of the mapping.

28 Notions of Solution for Discontinuous Dynamics

29 Solutions of ODEs Classical Solutions ( ) ( ) (, ), ( 0) 0 xt = f xt t x = x Caratheodory Solutions t ( ) x( t) = x ( ) 0 + f x s, s ds 0 Satisfies the ode almost everywhere on [0,t] Filippov Solutions (differential inclusion a set) ( ) ( ) (, ), ( 0) 0 xt F xt t x = x

30 Classical Solutions (, ) xt = f xt t ( ) ( ) classical solution: xt is continuously differentiable. Example: brick on ramp with stiction. mv = κsgn v cv + mg sinθ ( ) v t Not a classical solution t

31 Caratheodory Solutions ( ) xt = f xt ( ) ( ) [ ] interval t ab,, a> b is satisfied at almost all points on every Stopping solutions for the brick on ramp problems are not Caratheodory solutions. For these solutions the brick is stopped on a finite v t = t ab ( ) 0 on [, ] κsgn 0 + mg sinθ = 0 ( ) = [ ] interval, i.e, v t 0 on t ab,

32 Brick Example try something else mv = κsgn v cv + mg sinθ κ c v = sgn v v+ gsinθ m m κ c v = v+ gsinθ v > 0 m m κ κ v + gsin θ, + gsinθ v 0 m m = κ c v = v+ gsinθ v< 0 m m

33 Filippov Solutions dx() t dt F( xt ( ), t) : = conv f( S( δ, xt ( )) Λ( δ, xt ( )), t) δ > 0 { y R y }} n S( δ, x) : = x < δ Λ( δ, x) : subset of measure zero on which f is not defined

34 Example: nearest neighbor 3 agents moving in square Q Rule: move diametrically away from nearest neighbor Nearest neighbor to p N Action p i i { p q q Q { p1 p2 p3} { p} } = arg min,, \ i i i = p p i i N N i i

35 Example: nearest neighbor, cont d Consider 1 agent - in which case the only obstacles are the walls. The nearest neighbor is easily identified on the nearest wall. The vector field is well defined everywhere except on the diagonals where it is not defined because there are multiple nearest neighbors. p 1 = p p 1 1 q q

36 Example: nearest neighbor, cont d ( 0, 1) ( 1, 0) ( 0,1) ( 1, 0) x < x < x x < x < x x = f ( x1, x2) = x < x < x x < x < x f α + ( 1 α), 0 α 1 1 0

37 Extension of Brockett s Condition ( xu) Assumption on f, : ( ) m n A R convex f xa, R convex ( ) ( ) m A R conv f x, A f x,conv A Definition: Admissible feedback controls u x are piecewise continuous and solutions are defined in the sense of Filippov x F f( xu, ( x) ) and 0 F u ( 0) f( xu) Theorem ( B ) ( ) (Ryan): For, continuous and satisfying assumption, asymptotic stabilization by discontinuous feedback n m n of 0 R, f R is a neighborhood of 0 R. each neighborhood B