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1 Example System: Predator Prey Model https://www.math.duke.edu/education/webfeatsII/Word2HTML/Predator-prey.doc

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2 Example System: Predator Prey Model Populations Oscillate Without predators the prey grows unbounded Without prey the predators become extinct

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3 Chapter 3 Lyapunov Stability – Autonomous Systems System of differential equations Question: Is this system “well behaved”? Follow Up Question: What does well behaved mean? Do the states go to fixed values? Do the states stay bounded ? Do we know limits on the size of the states? Bottom line in this chapter is that we want to know if a differential equation, which we can’t solve, is “well behaved”? If we could “solve” the system then theses questions may be easy to answer. We are assuming we can’t solve our systems of interest.

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4 Preview x1x1 x2x2 x(0) System stops moving, “stable” How can we know which way our system behaves? x1x1 x2x2 x(0) System state grows, “unstable”

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5 No explicit time dependence. The solution will evolve with time, i.e. x(t) Open or closed-loop system We will work to quantify “well behaved” Can have multiple equilibrium points Three main issues: System is nonlinear f is a vector Can’t find a solution

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6 Evolution of the state Khalil calls this the “Challenge and answer form to demonstrate stability” Challenger proposes an ε bound for the final state The answerer has to produce a bound on the initial condition so that the state always stays in the ε bound. Answerer has to provide an answer for every ε proposed Bottom line: If we start close enough to x e we stay close to x e x1x1 x2x2 Starting time (typically 0)

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7 Pendulum without friction. Is (0,0) a stable equilibrium point in the sense of Definition 2? Yes

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8 A equilibrium point could be Stable and not Convergent An equilibrium point could be Convergent and Not Stable x (t) goes to x e as t goes to

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9 Pendulum without friction. Is (0,0) convergent? Is (0,0) asymptotically stable? No

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10 Pendulum with friction. Is (0,0) a stable equilibrium point in the sense of Definition 2? Is (0,0) convergent? Is (0,0) asymptotically stable? Yes

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11 Includes a sense of “how fast” the system converges. Exponential is “stricter” than asymptotic

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12 Can perform this shift to any equilibrium point of interest (a system may have multiple equilibrium points)

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13 V is a scalar function x1x1 V(x 1 ) Could be x 1 (t) Could be V(x 1 (t)) x1x1 V(x 1 ) x1x1 x1x1 Example: Are each of these PSD, PD, ND, or NSD? PSD PD None ND Notation: PSD write as V 0 PD write as V>0 NSD write as V 0 ND write as V<0

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14 x1x1 V(x 1 ) Example:

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15 Don’t loose generality by restricting Q to be symmetric in quadratic form i.e., Q symmetric See chapter 2 i.e. only the symmetric part contributes to the quadratic

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17 V is a general function of x 1 and x 2

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18 Where is this going? x1x1 V(x) Let’s say that x 1 is the state of our system x 1 2 could be an abstraction of the energy stored in the system Analogy would work for potential energy of a spring or charge on a capacitor How could that happen? Could be negative t V(x 1 (t)) V(x 1 (t)) will not increase System equations

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19 Eventually we will perform the control design to make this true Can take as many derivatives as you need Recall that we are considering the equilibrium point at the origin; thus, we already know 0

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20 Every convergent sequence has a convergent subsequence in B r B r is a ball about the origin. Our Theorem only applies at the origin

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21 Closeness of x to x=0 implies closeness of V(x) to V(0) Definition of stability

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24 mg l Not required to be able to derive these equations for this class.

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25 Example 3 (cont) mg l h l-h v

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26 Example 3 (cont) Note: open interval that does not include 2 Note: Stable implies that the system (pendulum position and velocity) remains bounded Does not mean that the system has stopped moving The energy is constant (V=E) but the system continuously moves exchanging kinetic and potential energy Limit Cycle)

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27 Limit Cycles Oscillation leads to a closed path in the phase plane, called periodic orbits. For example, the pendulum has a continuum of closed paths If there is a single, isolated periodic orbit such that all trajectories tend to that periodic orbit then it is called a stable limit cycle. Can also have unstable limit cycles. Khalil, Nonlinear Systems 3 rd Ed, p59

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28 Example: LC Oscillator Khalil, Nonlinear Systems 3 rd Ed, p59 1

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29 Example: Oscillator Khalil, Nonlinear Systems 3 rd Ed, p59 i R1 i R2 i=h(v) v v sys vDvD

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30 Example: Oscillator

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31 ND? NSD? We would expect that the friction will take energy out of the system, thus the system will stop moving. friction

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32 Example 4 (cont) Physically, we know that the system will stop because of the friction but our analysis does not show this (we only show it is Stable but can’t show Asymptotic Stability). We know the system is asymptotically stable even if that is not illustrated by this specific Lyapunov analysis. Note: We will return to this later and fix using the Invariant Set Theorem.

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33 This is a Local stability result because we have limited the range of the state variables for which the result applies.

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34 Have we learned anything useful? Given control design problem: t(sec) x

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35 There is a missing piece in our original proof that prevents us from applying the result globally:

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36 Evolution of the state (green line) Constant V(x 1,x 2 ) contours State x 1 gets large but it is trapped by the constant Lyapunov function

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37 Fixes the problem that the Theorem 2 was local. Add this condition to AS Theorem: Global AS: The states will go to zero as time increases from any finite starting state

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38 scalar function like radially unbounded

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39 V is PD iif there are class K functions that upper and lower bound the Lyapunov function.

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40 Raleigh-Ritz Theorem.

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41 Summary: If the state starts within some ball, then the state remains within some other ball for all times. We will eventually use t here

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43 Reminder of Theorem 2 (need on next slide)

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44 Solve the differential inequality Solve rhs and substitute a new upper bound Solve lhs Find p th root p

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45 Now want to address the problem demonstrated in the pendulum with friction example, couldn’t show function was negative definite using energy arguments. “What happens in M stays in M” Now want to address the problem demonstrated in the pendulum with friction example, couldn’t show function was negative definite using energy arguments.

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46 An oscillator has a limit cycle Slotine and Li, Applied Nonlinear Control

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50 "vanish identically" -> exactly =0 (vs. approaching zero) iv)

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51 i) PD iv) Radially Unbounded ii) NSD Stays at x 2 =0 forever, i.e. doesn’t move iii) Doesn’t vanish along a trajectory except x 1 =0

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52 Different from Lyapunov Theorem because: 1)V does not need to be Positive Definite 2)Applies to multiple equilibrium points and limit cycles

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53 Example of LaSalle’s Theorem System: Equilibrium Points:

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54 Lyapunov Function Candidate: Lyapunov function design is an iterative process! Can we change the Lyapunov function to yield a better derivative? : It would be nice if this was a “1” Example of LaSalle’s Theorem (cont)

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55 Lyapunov Function Candidate: Can we further change the Lyapunov function to yield a better derivative? Example of LaSalle’s Theorem (cont)

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56 Example of LaSalle’s Theorem (cont) We have enough information to conclude the system is “stable” in some region -> set is invariant wrt the system

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57 Lyapunov Function Candidate: Example of LaSalle’s Theorem (cont) M N

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58 Example of LaSalle’s Theorem (cont)

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59 Not negative beyond this point

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61 D However, it does look like there should be a region such that the system converges. Can we define that region?

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62 Define Region of Attraction (RA): Question we want to answer: Where can the system start (Initial Conditions at t=0) so that we know it will move to the equilibrium point?

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63 Original assumption, V is decreasing We are assuming we know there is a bound on x 2 Solve differential inequality 0 k We have now shown there is a bound on x 2

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64 0 1 k=2 Find R A for k=2: x Phase portrait: This is our R A :

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65 Estimate Region of Attraction (RA) Using LaSalle’s Theorem: Doesn’t mean it is the entire RA

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66 Finish Example 14 Could be larger,

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67 Contour lines of V(x) V(x) x1x1 x2x2 Finish Example 14

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70 Proven the “If” part of the Theorem. If Q>0 and found P then AS real part of eigenvalues is >0 Now prove the “only If” part of the Theorem.

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73 Example: Is the following system AS? In MATLAB: >> P=lyap([0 -1;1 -1],[1 0;0 1]) Wouldn’t it be easier just to find the eigenvalues of A? Yes, but having P (and hence a Lyapunov function) will be useful later.

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74 Linearization of a Nonlinear System Assume small = 0 by assumption of equilibrium point Change of variables to make the origin the equilibrium

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76 Example: Linearize System Eigenvalues = 1,1 Origin is unstable Reminder of Jacobian: 2x2

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77 Compare favorably close to the equilibrium point Linearized Original System May compare less favorably further away from equilibrium point Example : Linearize System (cont)

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79 Example System: Predator Prey Model Populations Oscillate Without predators the prey grows unbounded Without prey the predators become extinct

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80 Example System: Predator Prey Model Lost idea of extinction (100,100) Lost idea of oscillating populations (0,0) Linearization at the equilibrium points

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81 Unstable equilibrium In general, proving a system is unstable is not very “constructive” in designing control systems.

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83 Example Not conclusive, doesn’t mean it is stable

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84 Summary x=0 stable x=0 exponentially stable x=0 asymptotically stable convergent x=0 stable time X(t) time X(t) time X(t) x=0 asymptotically stablex=0 exponentially stableconvergent time X(t)

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85 x=0 stable x=0 asymptotically stable Locally, NSD dV/dt Locally, state could grow while V shrinks x=0 asymptotically stable globally 1 2 3 x=0 exponentially stable global, local depend on conditions 7 V is class K 8 x=0 asymptotically stable local Global, asymptotically stable Conditions true in entire state space 9 10 M is asymptotically stable Local, global Summary

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86 Summary Linearization 11 x=0 exponentially stable Inherently local result since it is an approximation of a nonlinear system at a point systemLinear approximation

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87 f 2 (x) Summary f(x) u u g(x)

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88 Homework Set 3.A: Pendulum without friction Set 3.B: Equilibrium points, quadratic Lyapunov function candidate to determine stability Set 3.C: Book problems 3.1, 3.3, 3.6, 3.7 Set 3.D: Book problems 3.4,3.5,3.13, 3.14 Set 3.E Pendulum with friction AS

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89 Homework #3-A “Pend w/out fric” - Find all of the equilibrium points for the pendulum without friction. – Plot the phase portrait from -3

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90 Homework 3-A Pendulum spins Pendulum swings Pendulum doesn’t move (stable EQ point) Pendulum doesn’t move at exactly that point (Unstable EQ point) Pend w/out fric

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91 Homework 3-A Change of variable Shifted system No Is the EQ point stable or unstable? No conclusion can be made Same Lyapunov function candidate that we used in notes Pend w/out fric (cont)

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92 Homework 3-A Change of variable Shifted system Yes Is the EQ point stable or unstable? Stable Same Lyapunov function candidate that we used in notes Pend w/out fric (cont)

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93 Homework #3-B Find the equilibrium points for each of the following and use a quadratic Lyapunov function candidate to determine stability of each equilibrium point. Using a quadratic Lyapunov function, find the conditions on a,b,c,d such that the system is stable at x=0 Using a quadratic Lyapunov function, find u=f(x) such that the system is stable at x=0

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94 Homework 3-B

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95 Homework 3-B

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96 Homework 3-B

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97 Homework 3-B

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98 Homework 3-B

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99 Homework 3-B We just designed a feedback control to stabilize the system We used the Lyapunov function to design the control.

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100 Homework 3-B

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101 Homework #3.C Chapter 3 - Problems 3.1, 3.3, 3.6, 3.7

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102 Homework 3.C System doesn’t move, i.e. derivatives are zero Notation means the second eq point, not the square of the eq point See that (0,0) which corresponds to (1,0) in the original system is an equilibrium point

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103 Homework 3.C See that (0,0) which corresponds to (-1,0) in the original system is an equilibrium point (-1,0)

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104 Homework 3.C Solve for conditions on v (note that this is the voltage on the coil not a Lyapunov function candidate) so that x 1 =y o and the system is at rest, i.e. solve v so that there is an equilibrium point at x 1 =y o. Approach: set derivatives to zero, x 1 to the constant y o solve for v.

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105 Homework 3.C Emphasizing that we are looking for a constant x 3 x 1 =y o

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106 Homework 3.3 (sol) This is why linearization can be so powerful tool - you get to use all of the linear analysis tools.

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107 Homework 3.C a)Verify that the origin is an equilibrium Substitute (0,0)

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108 Homework 3.C

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109 Homework 3.C From pplane8:

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110 Homework 3.C Test this first V can be zero other than at the origin

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111 Homework 3.C

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112 Homework 3.D Chapter 3 - Problems 3.4,3.5,3.13, 3.14

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113 Homework 3.D (Sol)

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114 Homework 3.D (Sol) For small x 2 Because we assumed small x 2

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115 Homework 3.D (Sol)

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116 Homework 3.D (Sol)

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117 Homework 3.D (Sol)

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118 Homework 3.D (Sol) This is a clever way to add a degree of freedom to a quadratic Lyapunov function. a,b don’t change the basic nature of the quadratic function, ie still PD and radially unbounded. a,b provide the opportunity to cancel these cross terms which might have otherwise stopped our analysis (ie had we chosen a=b=1)

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119 Homework 3.D (Sol) Why choose this? Because someone tried a lot of other V’s that did not work.

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120 Homework 4.E Pendulum with friction AS

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