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Stability Analysis of Continuous-Time

Switched Systems: A Variational Approach

Michael Margaliot

School of EE-Systems Tel Aviv University, Israel

Joint work with: Michael S. Branicky (CWRU)

Daniel Liberzon (UIUC)

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Overview Switched systems Stability Stability analysis:

A control-theoretic approach A geometric approach An integrated approach

Conclusions

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Switched Systems Systems that can switch between

several modes of operation.

Mode 1

Mode 2

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Example 1

Switched power converter

100v 50vlinear filter

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Example 2

A multi-controller scheme

plant

controller1

+

switching logiccontroller

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Switched controllers are “stronger” than regular controllers.

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More Examples

Air traffic controlBiological switchesTurbo-decoding……

For more details, see:

- Introduction to hybrid systems, Branicky

- Basic problems in stability and design of

switched systems, Liberzon & Morse

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Synthesis of Switched Systems

Driving: use mode 1 (wheels)

Braking: use mode 2 (legs)

The advantage: no compromise

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Gestalt Principle

“Switched systems are more than the

sum of their subsystems.“

theoretically interesting

practically promising

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Differential Inclusions

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

Example:

{ ( ), ( )}, (DI)nx f x g x x R

( ) nx R

, (LDI)x Ax Bx

4 3 2 1 0( ) ...exp( )exp( )exp( )exp( )x t t A t B t A t B x

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Global Asymptotic Stability (GAS)

Definition The differential inclusion

is called GAS if for any solution

(i)

(ii)

{ ( ), ( )}, ,nx f x g x x R

( )x tlim ( ) 0,t

x t

0, 0 such that:

| (0) | | ( ) | .x x t

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The Challenge

Why is stability analysis difficult?

(i) A DI has an infinite number of solutions for each initial condition.

(ii) The gestalt principle.

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Absolute Stability [Lure, 1944]

x Ax bu Ty c x

( , )y t

u y

2{ ( ) : 0 ( , ) }kS y y t ky y

ky

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Absolute Stability

0.S

The closed-loop system:

( ). (CL)Tx Ax b c x

* min{ : s.t. CL is not stable}.kk k S

Absolute Stability Problem

Find

A is Hurwitz, so CL is asym. stable for

any

For CL is asym. stable for any*,k k .kS

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Absolute Stability and Switched Systems

( )Tx Ax b c x

x Ax

( ) 0y ( )y ky

Tx Ax kbc x

* min{ : { , } is unstable}.Tk k x co Ax Ax kbc x

Absolute Stability Problem Find

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Example0 1 0 0 0 1

, , , :2 1 1 1 2 1

TkA b c B A kbc

k

10x B xx Ax

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A Solution of the Switched System

* 10.k This implies that

10 100.5 0.50.9 0.950(2.85) B BA Ax e e e e x

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Although both and are

stable, is not stable.

Instability requires repeated switching.

x Ax 10x B x10{ , }x Ax B x

Two Remarks

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Optimal Control ApproachWrite as the bilinear control

system:

{ , }kx Ax B x

( ) ( ) ( )( ) ( ), ( ) [0,1]

(0) .kx t Ax t u t B A x t u t

x z

.x

u is the worst-case switching law (WCSL).

u .J Problem Find a control maximizing

Analyze the corresponding trajectory

2( ; , ) : | ( ; , ) | / 2.J u T z x T u zFix Define: 0.T

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Optimal Control Approach

( ; , )J u T zConsider as :T

*k k

( ) 0J u

*k k

( )J u

*k kz

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Optimal Control ApproachTheorem (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

( ) : lim sup ( ; , )T

V z J u T z

*k k

2( ) ( )V cz c V z

,zx

( ( )) ( ).V x t V z

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Solving Optimal Control Problems

is a functional:

Two approaches:

1. Hamilton-Jacobi-Bellman (HJB)

equation.

2. Maximum Principle.

2| ( ) |x T

( ) ( , [0, ])x T F u(t) t T

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HJB Equation

Find such that

Integrating:

or

An upper bound for ,

obtained for the maximizing Eq. (HJB).

( , ( )) 0. (HJB)[0,1]

dMAX V t x t

dtu

( , ( )) (0, (0)) 0V T x T V x

2| ( ) | / 2x T

( , ) : nV R R R 2( , ) || || / 2,V T y y

u

2| ( ) | / 2 (0, (0)).x T V x

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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.

( )V z

The Case n=2

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The function is a first integral of if

We know that so

* ( ).kBH x( )AH x

V

( ( )) ( ),V x t V z

* *

0 ( ) ( ) 0

1 ( ) ( ) 0.

x

k x k

u x t Ax t V Ax

u x t B x t V B x

0 ( ( )) .A Ay

dH y t H Ay

dt ( ) ( ),y t Ay t

2:AH R R

Basic Idea

0 ( ( )).d

V x tdt

Thus, is a concatenation of two first integrals and

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Example:

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10A

10

1 2

72( ) exp( arctan( ))

27A T x

H x x P xx x

kx B x

Axx 1

1

12/1

2/12 kPkwhere and ...985.6* k

0 1

2 1kBk

1

1 2

7 42( ) exp( arctan( ))

27 4kB T

k

k xH x x P x

x xk

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Thus,

→ an explicit expression for V (and an explicit solution of the HJB).

0AxH Ax 0A

xH Bx

x x max{ ( ) } 0ku

W Bx uW B A x

1

1 0BxH Bx 0B

xH Ax ( ) : 1W x

Axx x Bx

AxH

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More on the Planar Case

[Margaliot & Branicky, 2009]

Corollary GAS of 2nd-order positive linear switched systems.

Theorem For a planar bilinear control system

1*( )x t

2*( )x t3 1*( ) *( )x t cx t

x Ax

x Bx

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Nonlinear Switched Systems

where are GAS.

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

1 2 { ( ), ( )} (NLDI)x f x f x1 2 ( ), ( )x f x x f x

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Lie-Algebraic Approach

For simplicity, consider the linear

differential inclusion:

so

},{ BxAxx

2 1(t) ...exp( )exp( ) (0).x Bt At x

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Commutation Relations and GAS

Suppose that A and B commute, i.e.AB=BA, then

Definition The Lie bracket of Ax and

Bx is [Ax,Bx]:=ABx-BAx.

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

3 2 1

3 1 4 2

( ) ...exp( )exp( )exp( ) (0)

exp( (... )) exp( (... )) (0)

x t At Bt At x

A t t B t t x

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Lie Brackets and Geometry

Consider

Then:

{ , , , }x Ax Ax Bx Bx

x Ax

x Axx Bx

x Bx

)0(x

)4( x

2 3

(4 ) (0) (0) (0)

[ , ] (0) ...

B A B Ax x e e e e x x

A B x

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Geometry of Car Parking

This is why we can park our car.

The term is the reason this takes

so long.

2

)(xf

)(xg

],[ gf

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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?

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Known ResultsLinear switched systems:k = 2 implies GAS (Gurvits,1995).k’th order nilpotency implies GAS

(Liberzon, Hespanha, & Morse, 1999)(Kutepov, 1982)

Nonlinear switched systems:k = 1 implies GAS (Mancilla-Aguilar, 2000).An open problem: higher orders of k?

(Liberzon, 2003)

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A Partial Answer

Theorem (Margaliot & Liberzon, 2004)

2nd order nilpotency implies GAS.

0)())(( ,0

0)())(( ,1(t)u~

txBAt

txBAtT

T

( ) ( ) ( ), Tm t t Cx t C A B

Proof By the PMP, the WCSL satisfies

Let

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( ) ( ) ( ) ( ) ( )

( )[ , ] ( )

T T

T

m t t Cx t t Cx t

t C A x t

xBACuxAAC

xACxACm

TT

TT

]],,[[]],,[[

],[],[

0m ( )m t const

Then

1st order nilpotency

Differentiating again yields:

battm )(0m2nd order nilpotency up to a single switch in the WCSL.

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Handling Singularity

If m(t)0, the Maximum Principle

does not necessarily provide enough

information to characterize the WCSL.

Singularity can be ruled out using

the notion of strong extremality

(Sussmann, 1979).

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[[ , ], ] [[ , ], ] 0T Tm C A A x u C A B x

3rd Order Nilpotency

In this case:

further differentiation cannot be carried out.

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3rd Order Nilpotency Theorem (Sharon & Margaliot, 2007) 3rd order nilpotency implies

Proof(1) Hall-Sussmann canonical system;(2) A second-order MP (Agrachev&Gamkrelidze).

40 0 ( ; , ) ( ;PC , ).R T U x R T x

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Conclusions

Stability analysis is difficult. A natural and powerful idea is to consider the “most unstable” trajectory.

Switched systems and differential inclusions are important in various scientific fields, and pose interesting theoretical questions.

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More info on the variational approach:

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

Available online:

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