Stability Analysis of Switched Systems: A Variational Approach

1

Stability Analysis of Switched Systems: A Variational

Approach

Michael Margaliot

School of EE-Systems Tel Aviv University

Joint work with Daniel Liberzon (UIUC)

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Overview Switched systems Stability Stability analysis:

A control-theoretic approach A geometric approach An integrated approach

Conclusions

3

Switched Systems Systems that can switch between

several modes of operation.

Mode 1

Mode 2

4

Example 1

Cta

ta

ta

Cta

x

x

)(

)(,

)(

)(

2

1

2

1

2

1

serve

r

1x 2x

C

)(2 ta)(1 ta

1 1

2 2

( )

( )

x a t

x a t C

1 1

2 2

( )

( )

x a t C

x a t

5

Example 2

Switched power converter

100v 50vlinear filter

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Example 3

A multi-controller scheme

plant

controller1

+

switching logiccontroller

2

Switched controllers are “stronger” than regular controllers.

7

More Examples

Air traffic controlBiological switchesTurbo-decoding……

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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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Mathematical Modeling with Differential Inclusions

easier ANALYSIS harder

MODELING

CAPABILITY

weaker

stronger

Axx

)(xfx

BxAxx ,

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The 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 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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StabilityThe differential inclusion

is called GAS if for any solution

(i)

(ii)

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

( )x tlim ( ) 0tx 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

x Ax bu Ty c x

( , )y t

u y

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

ky

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Problem of Absolute Stability

0.S

The closed-loop system:

( ). (CL)Tx Ax b c x

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

The Problem of Absolute Stability:

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

The Problem of Absolute Stability: 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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Trajectory 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.

This presents a serious problem in

multi-controller schemes.

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

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Optimal Control ApproachWrite as a control system:

Fix Define

Problem: Find the control that maximizes

(t)u~

.~(t)x(t)u~

{ , }kx Ax B x

( ) ( ) ( )( ) ( ), ( ) {0,1}

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

x z

2( ; , ) : | ( ) | / 2.J u T z x T0.T

.J

is the worst-case switching law (WCSL).Analyze the corresponding trajectory

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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 ApproachThm. 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

( ) : lim sup ( ; , )T

V z J u T z

*k k

( ) ( )V cz cV z

,z( )x t

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

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Solving Optimal Control Problems

is a functional:

Two approaches:

1. The Hamilton-Jacobi-Bellman (HJB) equation.

2. The Maximum Principle.

2| ( ) |fx t

( ) ( , [0, ])f fx t F u(t) t t

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The 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)) 0f fV t x t V x

2| ( ) | / 2fx t

( , ) : nV R R R 2( , ) || || / 2,fV t y y

u

2| ( ) | / 2 (0, (0)).fx t V x

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The HJB for a LDI:

Hence,

In general, finding is difficult.

0)(V ?,

0)(V 0,

0)(V 1,~

x

x

x

xBA

xBA

xBA

u

V(t,x)

})({max

)})1(({max

}{max0

xx

x

x

xBAuVBxVV

BxuuAxVV

xVV

tu

tu

tu

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The Maximum Principle

Let Then,

Differentiating we get

A differential equation for with a boundary condition at

,xVV0 xt

u)B)-(1(uA

u)B)-(1(uAV V

u)B)-(1(uAVxVV0

xx

xxxtx

dtd

x( ) : V (t).t 2( ) ( ) / 2 ( ). xf f ft x t x t

.ft( ),t

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Summarizing,

The WCSL is the maximizing

that is,

We can simulate the optimal solution backwards in time.

0

( (1- ) ) , ( ) ( )

( (1- ) ) , (0)

Tf fuA u B t x t

x uA u B x x x

xBuuAT ))1((VxVV txt

1, ( )( ) ( ) 0( )

0, ( )( ) ( ) 0

T

T

t A B x tu t

t A B x t

u~

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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

The Basic Idea

0 ( ( )).dV x t

dt

Thus, is a concatenation of two first integrals and

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Example:

12

10A

10

1 2

72( ) exp( arctan( ))

27A T x

H x x P xx x

Bxx

Axx

1

1 2

7 42( ) exp( arctan( ))

27 4B T

k

k xH x x P x

x xk

1

1

12/1

2/12 kPkwhere and ...985.6* k

0 1

2 1kB k

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Thus,

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

0AxH Ax

1

1

0AxH Bx

0BxH Bx 0B

xH Ax ( ) 1W x

x x max{ ( ) } 0kuW Bx uW B A x

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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 the sake of simplicity, consider

the LDI

so

},{ BxAxx

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

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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.

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 A x A x B x B x

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 it takes

so long.

2

)(xf

)(xg

],[ gf

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NilpotencyDefinition: 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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Some Known ResultsSwitched 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)

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A Partial Answer

Thm. 1 (Margaliot & Liberzon, 2004)

2nd order nilpotency implies GAS.

Proof: Consider the WCSL

Define the switching function

0)())(( ,0

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

txBAt

txBAtT

T

BACtCxttm T ),()(:)(

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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.

( ) ( ) ( ) ( ) ( )

( )[ , ] ( ).

T T

T

m t t Cx t t Cx t

t C A x t

xBACuxAAC

xACxACm

TT

TT

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

],[],[

0m ( )m t const

battm )(0m

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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

the notion of strong extermality

(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

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.

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

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Hall-Sussmann SystemConsider the case [A,B]=0.

( ) ( ) ( ) ( ), ( ) {0,1}.x t Ax t u t Bx t u t

Guess the solution:

1 2 0( ) exp( ( , )) exp( ( , )) .y t Ac t u Bc t u xThen

1 2( ) ,y t c Ay c By so

1

2

1 c

c u

(HS system)

and1 2(0) (0) 0c c

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Hall-Sussmann SystemIf two controls u, v yield the same values for 1 2( ), ( )c t c t then they yield the same value for ( ).x t

measurable control can be replaced with a

Since does not depend on u, 1( )c t

2

0

( ) ( )t

c t u d we conclude that any

bang-bang control with a single switch:1

0 0 ( ; , ) ( ;BB , ).R t U x R t x

and

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3rd order Nilpotency

In this case,

1 2 3

4 5 0

( ) exp( )exp( )exp([ , ] )

exp([ ,[ , ]] ) exp([ ,[ , ]] ) .

x t Ac Bc A B c

A A B c B A B c x

1

2

3 1

24 1

5 1 2

1

1/ 2

c

c u

c c u

c c u

c c c u

The HS system:

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Conclusions

Stability analysis is difficult. A natural and useful 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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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