Transcript Stability Analysis of Switched Systems: A Variational Approach Michael Margaliot School of EE-Systems Tel Aviv University Joint work with Daniel Liberzon (UIUC)

Stability Analysis of Switched
Systems: A Variational Approach
Michael Margaliot
School of EE-Systems
Tel Aviv University
Joint work with 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
 x1   a1 (t )  C 
 

 x2   a2 (t ) 
a1 (t )
a2 ( t )
x1
x2
C
server
 x1   a1 (t ) 
 

 x2   a2 (t )  C 
 x1   a1 (t )  C   a1 (t )  
   
, 

 x2   a2 (t )   a2 (t )  C  
4
Example 2
Switched power converter
100v
linear filter
50v
5
Example 3
A multi-controller scheme
+
plant
controller1
switching logic
controller2
Switched controllers are “stronger” than
regular controllers.
6
More Examples
Air traffic control
Biological switches
Turbo-decoding
……
7
Synthesis of Switched Systems
Driving: use mode 1 (wheels)
Braking: use mode 2 (legs)
The advantage: no compromise
8
Mathematical Modeling with
Differential Inclusions
stronger
x  f (x )
x  Ax, Bx
weaker
x  Ax
easier ANALYSIS
harder
9
The Gestalt Principle
“Switched systems are more than the
sum of their subsystems.“
 theoretically interesting
 practically promising
10
Differential Inclusions
x { f ( x), g ( x)},
xR
n
(DI)
A solution is an absolutely continuous
n
function x()  R satisfying (DI) for all t.
Example: x  Ax, Bx
(LDI)
x(t )  ...exp(t4 A)exp(t3 B)exp(t2 A)exp(t1B) x0
11
Stability
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 ) | 
12
The Challenge
Why is stability analysis difficult?
(i) A DI has an infinite number of
solutions for each initial
condition.
(ii) The gestalt principle.
13
Absolute Stability
u
x  Ax  bu
T
yc x
y
 ( y, t )
ky
Sk  { () : 0  y ( y, t )  ky }
2

y
14
Problem of Absolute Stability
The closed-loop system:
x  Ax  b ( c x).
T
(CL)
A is Hurwitz, so CL is asym. stable for
any   S0 .
The Problem of Absolute Stability:
Find k*  min{k :   Sk s.t. CL is not stable}.
For k  k *, CL is asym. stable for any   Sk .
15
Absolute Stability and
Switched Systems
x  Ax  b ( c x)
T
 ( y)  0
x  Ax
 ( y)  ky
x  Ax  kbc x
T
The Problem of Absolute Stability: Find
k*  min{k : x  co{Ax, Ax  kbc x} is unstable}.
T
16
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
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Trajectory 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
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Although both x  Ax and x  B10 x are
stable, x {Ax, B10 x} is not stable.
Instability requires repeated switching.
This presents a serious problem in
multi-controller schemes.
19
Optimal Control Approach
Write
x {Ax, Bk x}
as a 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 ) |
Problem: Find the control u~(t) that
maximizes J .
2
/ 2.
u~(t) is the worst-case switching law (WCSL).
Analyze the corresponding trajectory ~
x (t).
20
Optimal Control Approach
Consider J (u; T , z ) as T   :
k k*
k k*
k k*
z
J (u)  0
J (u)  
21
Optimal Control Approach
Thm. 1 (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 )  cV ( z ) ).
(2) For every initial condition z , there
exists a solution x (t ) such that
V ( x(t ))  V ( z ).
22
Solving Optimal Control Problems
2
| x(t f ) | is a functional:
x(t f )  F (u(t), t [0, t f ])
Two approaches:
1. The Hamilton-Jacobi-Bellman (HJB)
equation.
2. The Maximum Principle.
23
The HJB Equation
Find V (, ): R  R  R
n
2 / 2,
V
(
t
,
y
)

||
y
||
such that f
d

MAX  V (t , x(t ))   0.

u [0,1]  dt
(HJB)
Integrating: V (t f , x(t f )) V (0, x(0))  0
2
| x(t f ) | / 2  V (0, x(0)).
or
2
An upper bound for | x(t f ) | / 2,
obtained for the u maximizing Eq. (HJB).
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The HJB for a LDI:
0  max{Vt  Vx x}
u
 max{Vt  Vx (uAx  (1  u ) Bx )}
u
 max{Vt  Vx Bx  uVx ( A  B ) x}
u
Hence,
1, Vx ( A  B ) x  0

~
u  0, Vx ( A  B ) x  0
 ?, V ( A  B ) x  0

x
In general, finding V(t,x) is difficult.
25
The Maximum Principle
 2
Let  (t ) : Vx (t). Then,  (t f )  x (t f ) / 2  x(t f ).
x
Differentiating 0  Vt  Vx x , we get
0  Vtx  Vxx x  Vx (uA  (1 - u)B)

d V
dt x
 Vx (uA  (1 - u)B)
    (uA  (1 - u)B)
A differential equation for  (t ), with a
boundary condition at t f .
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Summarizing,
T
  (uA  (1- u) B)  ,  (t f )  x(t f )
x  (uA  (1- u) B) x,
x(0)  x0
The WCSL is the ~u maximizing
Vt  Vx x  Vt  T (uA  (1  u) B) x
that is,
1,  T (t )( A  B) x(t )  0
u (t )  
T
0,

(t )( A  B) x(t )  0

We can simulate the optimal solution
backwards in time.
27
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.
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The 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*
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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  Bx
H B ( x)  xT Pk x exp(
2  k 1/ 2 

where Pk  

1 
 1/ 2
7  4k x1
2
arctan(
))
x1  2 x2
7  4k
and k *  6.985...
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1
W ( x)  1
1
H Ax  0
A
x
H xB Bx  0
H xA Bx  0
H xB Ax  0
Thus,
max{Wx Bx  uWx ( Bk  A) x}  0
u
so we have an explicit expression for V
(and an explicit solution of HJB).
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Nonlinear Switched Systems
x { f ( x), f ( x)}
1
where x
2
(NLDI)
 f ( x), x  f ( x) are GAS.
1
2
Problem: Find a sufficient condition
guaranteeing GAS of (NLDI).
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Lie-Algebraic Approach
For the sake of simplicity, consider
the LDI
x  { Ax , Bx}
so
x(t)  ...exp( Bt2 ) exp( At1 ) x(0).
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Commutation and GAS
Suppose that A and B commute,
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.
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Lie Brackets and Geometry
x {A( x),  A( x), B( x), B( x)}
Consider
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
35
Geometry of Car Parking
This is why we can park our car.
2

The
term is the reason it takes
so long.
f ( x)
g ( x)
[ f , g]
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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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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)
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A Partial Answer
Thm. 1 (Margaliot & Liberzon, 2004)
2nd order nilpotency implies GAS.
Proof: Consider the WCSL
T

1
,

(t )( A  B ) x (t )  0
~
u(t)  
T
0
,

(t )( A  B ) x (t )  0

Define the switching function
m(t ) : T (t )Cx(t ), C  A  B
39
Differentiating m(t) yields
m(t )   (t )Cx(t )   (t )Cx(t )
T
T
  (t )[C , A]x(t ).
T
1st order nilpotency  m  0  m(t )  const
 no switching in the WCSL.
Differentiating again, we get
T
T

m


[
C
,
A
]
x


[C , A] x

 T [[C , A], A] x  uT [[C , A], B ] x
2nd order nilpotency  m
  0 m(t )  at  b
 up to a single switch in the WCSL.
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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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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.
42
3rd order Nilpotency
Thm. 2 (Sharon & Margaliot, 2005)
3rd order nilpotency implies
R (t;U , x0 )  R (t;PC , x0 ).
4
The proof is based on using: (1) the HallSussmann canonical system; and (2) the
second-order Agrachev-Gamkrelidze MP.
43
Hall-Sussmann System
Consider the case [A,B]=0.
x(t )  Ax(t )  u(t ) Bx(t ), u(t ) {0,1}.
Guess the solution:
y(t )  exp( Ac1 (t, u))exp(Bc2 (t , u)) x0 .
Then c1 (0)  c2 (0)  0
and y(t )  c1 Ay  c2 By, so
c1  1
c2  u
(HS system)
44
Hall-Sussmann System
If two controls u, v yield the same values
for c1 (t ), c2 (t ) then they yield the same
value for x(t ).
Since c1 (t ) does not depend on u,
t

and c2 (t )  u ( )d we conclude that any
0
measurable control can be replaced with a
bang-bang control with a single switch:
R (t;U , x0 )  R (t;BB , x0 ).
1
45
3rd order Nilpotency
In this case,
x(t )  exp( Ac1 )exp( Bc2 )exp([ A, B]c3 )
exp([ A,[ A, B]]c4 )exp([ B,[ A, B]]c5 ) x0 .
The HS system: c1  1
c2  u
c3  c1u
c4  1/ 2 c u
2
1
c5  c1c2u
46
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.
47
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
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