Stability Analysis of ContinuousTime 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)
1
Overview




Switched systems
Stability
Stability analysis:
 A control-theoretic approach
 A geometric approach
 An integrated approach
Conclusions
2
Switched Systems
Systems that can switch between
several modes of operation.
Mode 1
Mode 2
3
Example 1
Switched power converter
100v
linear filter
50v
4
Example 2
A multi-controller scheme
+
plant
controller1
switching logic
controller2
Switched controllers are “stronger”
than regular controllers.
5
More Examples
Air traffic control
Biological switches
Turbo-decoding
……
For more details, see:
- Introduction to hybrid systems, Branicky
- Basic problems in stability and design of
switched systems, Liberzon & Morse
6
Synthesis of Switched Systems
Driving: use mode 1 (wheels)
Braking: use mode 2 (legs)
The advantage: no compromise
7
Gestalt Principle
“Switched systems are more than the
sum of their subsystems.“
 theoretically interesting
 practically promising
8
Differential Inclusions
x { f ( x), g ( x)},
xR
n
(DI)
A solution is an absolutely continuous
n
function x()  R satisfying (DI) for
almost all t.
x  Ax, Bx
(LDI)
Example:
x(t )  ...exp(t4 A)exp(t3 B)exp(t2 A)exp(t1B) x0
9
Global Asymptotic Stability (GAS)
Definition The differential inclusion
x { f ( x), g ( x)},
xR ,
is called GAS if for any solution
(i) lim x (t )  0,
n
x(t )
t 
(ii)
  0,   0 such that:
| x(0) |   | x(t ) |  .
10
The Challenge
Why is stability analysis difficult?
(i) A DI has an infinite number of
solutions for each initial
condition.
(ii) The gestalt principle.
11
Absolute Stability [Lure, 1944]
u
x  Ax  bu
T
yc x
y
 ( y, t )
ky
Sk  { () : 0  y ( y, t )  ky }
2

y
12
Absolute Stability
The closed-loop system:
x  Ax  b ( c x).
T
(CL)
A is Hurwitz, so CL is asym. stable for
any   S0 .
Absolute Stability Problem
Find k*  min{k :   Sk s.t. CL is not stable}.
For k  k *, CL is asym. stable for any   Sk .
13
Absolute Stability and
Switched Systems
x  Ax  b ( c x)
T
 ( y)  0
x  Ax
 ( y)  ky
x  Ax  kbc x
T
Absolute Stability Problem Find
k*  min{k : x  co{Ax, Ax  kbc x} is unstable}.
T
14
Example
0 1
0
 0
T
A
 , b    , c    , Bk : A  kbc
 2 1
 1
1
x  Ax
1
 0


 2  k 1
x  B10 x
15
A Solution of the Switched System
x(2.85)  e
0.9 A 0.5 B10 0.95 A 0.5 B10
e
e
e
This implies that k *  10.
x0
16
Two Remarks
Although both x  Ax and x  B10 x are
stable, x {Ax, B10 x} is not stable.
Instability requires repeated switching.
17
Optimal Control Approach
Write x {Ax, Bk x} as the bilinear control
system:
x(t )  Ax(t )  u(t )( Bk  A) x(t ), u(t ) [0,1]
x(0)  z.
Fix T
 0. Define: J (u;T , z ) : | x (T ; u, z ) | / 2.
Problem Find a control u maximizing
2
J.
u is the worst-case switching law (WCSL).
Analyze the corresponding trajectory x.
18
Optimal Control Approach
Consider J (u; T , z ) as T   :
k k*
k k*
k k*
z
J (u)  0
J (u)  
19
Optimal Control Approach
Theorem (Pyatnitsky) If k  k * then:
(1) The function
V ( z ) : lim sup J (u; T , z )
T 
is finite, convex, positive, and
homogeneous (i.e. V (cz )  c2V ( z )).
(2) For every initial condition z , there
exists a solution x such that
V ( x(t ))  V ( z ).
20
Solving Optimal Control Problems
| x (T ) |2 is a functional:
x (T )  F (u(t), t [0, T ])
Two approaches:
1. Hamilton-Jacobi-Bellman (HJB)
equation.
2. Maximum Principle.
21
HJB Equation
Find V (, ): R  R  R such that V (T , y ) || y ||2 / 2,
n
d

MAX  V (t , x(t ))   0.

u [0,1]  dt
(HJB)
Integrating: V (T , x(T ))  V (0, x(0))  0
2
|
x
(
T
)
|
/ 2  V (0, x(0)).
or
An upper bound for | x(T ) |2 / 2 ,
obtained for the u maximizing Eq. (HJB).
22
The Case n=2
Margaliot & Langholz (2003) derived an
explicit solution for V ( z ) when n=2.
This yields an easily verifiable necessary
and sufficient condition for stability of
second-order switched linear systems.
23
Basic Idea
A
2
H
:
R
 R is a first
The function
d A
of y(t )  Ay(t ), if 0  H ( y(t ))  H yA Ay.
dt
integral
d
We know that V ( x (t ))  V ( z ), so 0  V ( x (t )).
dt
u  0  x (t )  Ax (t )  Vx Ax  0
u  1  x (t )  Bk * x (t )  Vx Bk * x  0.
Thus, V is a concatenation of two first
integrals H A ( x) and H B ( x).
k*
24
1
1
 0
 0
Example: A    2  1 Bk   2  k 1




x  Ax
1
7 x1
2
A
T
H ( x)  x P0 x exp(
arctan(
))
x1  2 x2
7
1
x  Bk x
7  4k x1
2
H ( x)  x Pk x exp(
arctan(
))
x1  2 x2
7  4k
Bk
T
 2  k 1/ 2 
where Pk  
 and
1 
 1/ 2
k *  6.985...
25
H xA Ax  0 H xA Bx  0
1
H xA
x  Bx
x  Ax
W ( x) : 1
1
H xB Bx  0
H xB Ax  0
Thus,
max{Wx Bx  uWx ( Bk  A) x}  0
u
→ an explicit expression for V (and an
explicit solution of the HJB).
26
More on the Planar Case
Theorem For a planar bilinear control system
x * (t1 )
x  Ax
x * (t3 )  cx * (t1 )
x * (t2 )
x  Bx
[Margaliot & Branicky, 2009]
Corollary GAS of 2nd-order positive
linear switched systems.
27
Nonlinear Switched Systems
x { f ( x), f ( x)}
1
where
2
(NLDI)
x  f ( x), x  f ( x) are GAS.
1
2
Problem Find a sufficient condition
guaranteeing GAS of (NLDI).
28
Lie-Algebraic Approach
For simplicity, consider the linear
differential inclusion:
x  { Ax , Bx}
so
x(t)  ...exp( Bt2 ) exp( At1 ) x(0).
29
Commutation Relations and GAS
Suppose that A and B commute, i.e.
AB=BA, then
x(t )  ...exp( At3 )exp( Bt2 )exp( At1 ) x(0)
 exp( A(...  t3  t1 ))exp( B(...  t4  t2 )) x(0)
Definition The Lie bracket of Ax and
Bx is [Ax,Bx]:=ABx-BAx.
Hence, [Ax,Bx]=0 implies GAS.
30
Lie Brackets and Geometry
Consider x { Ax,  Ax, Bx,  Bx}
x ( 0)
x  Ax
x  Bx
x ( 4 )
x   Bx
x   Ax
Then:
 B  A B A
x(4 )  x(0)  e e e e x(0)  x(0)
  [ A, B]x(0)   ...
2
3
31
Geometry of Car Parking
This is why we can park our car.
2

The
term is the reason this takes
so long.
f ( x)
g ( x)
[ f , g]
32
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?
33
Known Results
Linear 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)
34
A Partial Answer
Theorem (Margaliot & Liberzon, 2004)
2nd order nilpotency implies GAS.
Proof By the PMP, the WCSL satisfies
T

1
,

(t )( A  B ) x (t )  0
~
u(t)  
T
0
,

(t )( A  B ) x (t )  0

Let
m(t )  T (t )Cx(t ), C  A  B
35
Then
m(t )   T (t )Cx(t )   T (t )Cx(t )
  T (t )[C , A]x(t )
1st order nilpotency  m  0  m(t )  const
Differentiating again yields:
m
  T [C , A] x  T [C , A] x
  [[C , A], A] x  u [[C , A], B ] x
T
T
  0  m(t )  at  b
2nd order nilpotency  m
 up to a single switch in the WCSL.
36
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).
37
3rd Order Nilpotency
In this case:
m   [[C, A], A]x  u [[C, A], B]x  0
T
T
further differentiation cannot be carried out.
38
3rd Order Nilpotency
Theorem (Sharon & Margaliot, 2007)
3rd order nilpotency implies
R (T ;U , x0 )  R (T ;PC , x0 ).
4
Proof
(1) Hall-Sussmann canonical system;
(2) A second-order MP
(Agrachev&Gamkrelidze).
39
Conclusions
 Switched systems and differential
inclusions are important in various
scientific fields, and pose
interesting theoretical questions.
 Stability analysis is difficult.
A natural and powerful idea is to
consider the “most unstable” trajectory.
40
More info on the variational approach:
“Stability analysis of switched systems
using variational principles: an
introduction”, Automatica 42(12): 20592077, 2006.
Available online:
www.eng.tau.ac.il/~michaelm
41