stability analysis of switched systems: a variational approach
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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 - PowerPoint PPT PresentationTRANSCRIPT
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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)
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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
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
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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.
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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