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Stability Analysis of Switched Systems: A Variational Approach. Michael Margaliot School of EE-Systems Tel Aviv University Joint work with Daniel Liberzon (UIUC). Overview. Switched systems Stability Stability analysis: A control-theoretic approach A geometric approach

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stability analysis of switched systems a variational approach

Stability Analysis of Switched Systems: A Variational Approach

Michael Margaliot

School of EE-Systems Tel Aviv University

Joint work with Daniel Liberzon (UIUC)

overview
Overview
  • Switched systems
  • Stability
  • Stability analysis:
    • A control-theoretic approach
    • A geometric approach
    • An integrated approach
  • Conclusions
switched systems
Switched Systems

Systems that can switch between

several modes of operation.

Mode 1

Mode 2

example 1
Example 1

server

example 2

linear filter

Example 2

Switched power converter

50v

100v

example 3
Example 3

A multi-controller scheme

plant

+

controller1

controller2

switching logic

Switched controllers are “stronger” than

regular controllers.

more examples
More Examples
  • Air traffic control
  • Biological switches
  • Turbo-decoding
  • ……
synthesis of switched systems
Synthesis of Switched Systems

Driving: use mode 1 (wheels)

Braking: use mode 2 (legs)

The advantage: no compromise

mathematical modeling with differential inclusions

stronger

MODELING

CAPABILITY

weaker

Mathematical Modeling with Differential Inclusions

easier ANALYSIS harder

the gestalt principle
The Gestalt Principle

“Switched systems are more than the

sum of their subsystems.“

 theoretically interesting

 practically promising

differential inclusions
Differential Inclusions

A solution is an absolutely continuous function satisfying (DI) for all t.

Example:

stability
Stability

The differential inclusion

is called GAS if for any solution

(i)

(ii)

the challenge
The Challenge
  • Why is stability analysis difficult?
  • A DI has an infinite number of solutions for each initial condition.
  • The gestalt principle.
problem of absolute stability
Problem of Absolute Stability

The closed-loop system:

A is Hurwitz, so CL is asym. stable for

any

The Problem of Absolute Stability:

Find

For CL is asym. stable for any

absolute stability and switched systems
Absolute Stability and Switched Systems

The Problem of Absolute Stability: Find

slide19
Although both and are

stable, is not stable.

Instability requires repeated switching.

This presents a serious problem in

multi-controller schemes.

optimal control approach
Optimal Control Approach

Write as a control system:

Fix Define

Problem: Find the control that maximizes

is the worst-case switching law (WCSL).

Analyze the corresponding trajectory

optimal control approach2
Optimal Control Approach

Thm. 1 (Pyatnitsky) If then:

(1) The function

is finite, convex, positive, and homogeneous (i.e., ).

(2) For every initial condition there exists a solution such that

solving optimal control problems
Solving Optimal Control Problems

is a functional:

Two approaches:

  • The Hamilton-Jacobi-Bellman (HJB) equation.
  • The Maximum Principle.
the hjb equation
The HJB Equation

Find such that

Integrating:

or

An upper bound for ,

obtained for the maximizing Eq. (HJB).

slide25

The HJB for a LDI:

Hence,

In general, finding is difficult.

the maximum principle
The Maximum Principle

Let Then,

Differentiating we get

A differential equation for with a

boundary condition at

slide27
Summarizing,

The WCSL is the maximizing

that is,

We can simulate the optimal solution

backwards in time.

the case n 2
The Case n=2

Margaliot & Langholz (2003) derived an

explicit solution for when n=2.

This yields an easily verifiable necessary and sufficient condition for stability of second-order switched linear systems.

slide29

The Basic Idea

The function is a first integral of if

We know that so

Thus, is a concatenation of two first integrals and

example1
Example:

where and

slide31

Thus,

so we have an explicit expression for V (and an explicit solution of HJB).

nonlinear switched systems
Nonlinear Switched Systems

where are GAS.

Problem: Find a sufficient condition guaranteeing GAS of (NLDI).

lie algebraic approach
Lie-Algebraic Approach

For the sake of simplicity, consider

the LDI

so

commutation and gas
Commutation and GAS

Suppose that A and B commute,

AB=BA, then

Definition: The Lie bracket of Ax and Bx is [Ax,Bx]:=ABx-BAx.

Hence, [Ax,Bx]=0 implies GAS.

geometry of car parking
Geometry of Car Parking

This is why we can park our car.

The term is the reason it takes

so long.

nilpotency
Nilpotency

Definition: k’th order nilpotency -

all Lie brackets involving k+1 terms vanish.

1st order nilpotency: [A,B]=0

2nd order nilpotency: [A,[A,B]]=[B,[A,B]]=0

Q: Does k’th order nilpotency imply GAS?

some known results
Some Known Results

Switched linear systems:

  • k = 2 implies GAS (Gurvits,1995).
  • k’th order nilpotency implies GAS (Liberzon, Hespanha, and Morse, 1999).(The proof is based on Lie’s Theorem)

Switched nonlinear systems:

  • k = 1 implies GAS.
  • An open problem: higher orders of k? (Liberzon, 2003)
a partial answer
A Partial Answer

Thm. 1 (Margaliot & Liberzon, 2004)

2nd order nilpotency implies GAS.

Proof: Consider the WCSL

Define the switching function

slide40
Differentiating m(t) yields

1st order nilpotency 

 no switching in the WCSL.

Differentiating again, we get

2nd order nilpotency  

 up to a single switch in the WCSL.

handling singularity
Handling Singularity

If m(t)0, then the Maximum Principle

does not necessarily provide enough

information to characterize the WCSL.

Singularity can be ruled out using

thenotion of strong extermality

(Sussmann, 1979).

slide42

3rd order Nilpotency

In this case:

further differentiation cannot be carried out.

3rd order nilpotency
3rd order Nilpotency

Thm. 2 (Sharon & Margaliot, 2005)

3rd order nilpotency implies

The proof is based on using: (1) the Hall-Sussmann canonical system; and (2) the second-order Agrachev-Gamkrelidze MP.

hall sussmann system
Hall-Sussmann System

Consider the case [A,B]=0.

Guess the solution:

Then

so

and

(HS system)

hall sussmann system1
Hall-Sussmann System

If two controls u, v yield the same values for

then they yield the same

value for

does not depend on u,

Since

we conclude that any

and

measurable control can be replaced with a

bang-bang control with a single switch:

3rd order nilpotency1
3rd order Nilpotency

In this case,

The HS system:

conclusions
Conclusions
  • Switched systems and differential inclusions are important in various scientific fields, and pose interesting theoretical questions.
  • Stability analysis is difficult.

A natural and useful idea is to

consider the most unstable trajectory.

slide48

For more information, see the survey paper:

“Stability analysis of switched systems using variational principles: an introduction”, Automatica 42(12), 2059-2077, 2006.

Available online:

www.eng.tau.ac.il/~michaelm