Part 1
Stability Analysis of Linear
Switched Systems:
An Optimal Control Approach
Michael Margaliot
School of Elec. Eng.
Tel Aviv University, Israel
Joint work with: Gideon Langholz (TAU),
Daniel Liberzon (UIUC), Michael S. Branicky
(CWRU), Joao Hespanha (UCSB).
1
Overview





Switched systems
Global asymptotic stability
The edge of stability
Stability analysis:
 An optimal control approach
 A geometric approach
 An integrated approach
Conclusions
2
Switched Systems
Systems that can switch between
several possible 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 
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
Linear Systems
x  Ax.
Solution:
x(t )  exp( At ) x(0).
Definition: The system is globally asymptotically
x(0).
stable if lim x(t )  0,
t 
Theorem:
stability  Re( λ)  0, λ  eig( A).
A is called a Hurwitz matrix.
9
Linear Switched Systems
Two (or more) linear systems:
x(t )  A1 x(t ),
x(t )  A2 x(t ).
A system that can switch between them:
x(t )  Aσ (t ) x(t ),
σ (t )
2
1
σ : R  {1,2}.
...
t1
t1  t2
t
x(T )  ...exp(t4 A1 )exp(t3 A2 )exp(t2 A1 )exp(t1 A2 ) x0 .
10
10
Stability
Linear switched system: x(t )  Aσ (t ) x(t ),
σ : R  {1,2}.
Definition: Globally uniformly asymptotically
x(t )  0,
x(0), σ .
stable (GUAS):
In other words,
lim x(t )  lim(...exp(t3 A2 ) exp(t2 A1 ) exp(t1 A2 ) x0 )  0
t 
t 
for any t1 , t2 ,....  0
AKA, “stability under arbitrary switching”.
11
11
A Necessary Condition for GUAS
The switching law σ (t )  1 yields x(t )  A1 x(t ).
Thus, a necessary condition for GUAS
is that both A1 , A2 are Hurwitz.
Then instability can only arise due to
repeated switching.
12
12
Why is the GUAS problem
difficult?
Answer 1:
The number of possible switching laws
is huge.
13
13
Why is the GUAS problem difficult?
Answer 2: Even if each linear subsystem is
stable, the switched system may not be GUAS.
0 1
x
x
 2 1
14
 0 1
x
x
 12 1
14
Why is the GUAS problem difficult?
Answer 2: Even if each linear subsystem is
stable, the switched system may not be GUAS.
15
15
Stability of Each Subsystem is
Not Enough
A multi-controller scheme
+
plant
controller1
switching logic
controller2
Even when each closed-loop is stable,
the switched system may not be GUAS.
16
Easy Case #1
A trajectory of the switched system:
x(T )  ...exp(t4 A1 )exp(t3 A2 )exp(t2 A1 )exp(t1 A2 ) x0 .
Suppose that the matrices commute:
A1 A2  A2 A1.
Then x(T )  exp((T  s) A1 )exp(sA2 ) x0 ,
and since both matrices are Hurwitz, the
switched system is GUAS.
17
17
Easy Case #2
Suppose that both matrices are upper
triangular:
 1 1 
x
 x,
 0 2 
 3 7 
x
 x.
 0 2.5 
Then x  2 x , x  2.5x , so
2
Now
2 2
2
| x (t ) | exp(2t ) | x (0) |.
2
2
x   x  x , x  3x  7 x so x (t )  0.
1
1
1 2 1
1 2
This proves GUAS.
18
18
Optimal Control Approach
Pioneered by E. S. Pyatnitsky (1970s).
Basic idea:
(1) A relaxation: linear switched
system → bilinear control system
(2) characterize the “most
destabilizing” control u *
(3) the switched system is GUAS iff
x * (t )  0
19
19
Optimal Control Approach
Relaxation: the switched system:
x  Aσ x,
σ : R  {1,2},
→ a bilinear control system:
x  ( A1  ( A2  A1 )u ) x,
u U ,
where U is the set of measurable functions
taking values in [0,1].
u0
x  A1 x
u  1/ 2
u 1
x  (1/ 2)( A1  A2 ) x x  A2 x
20
Optimal Control Approach
The bilinear control system (BCS)
x  ( A1  ( A2  A1 )u ) x, u U ,
is globally asymptotically stable (GAS) if:
x(t )  0,
 x(0),  u U .
Theorem The BCS is GAS if and only if the
linear switched system is GUAS.
21
Optimal Control Approach
The most destabilizing control:
x  ( A1  ( A2  A1 )u ) x, u U ,
x(0)  x0 .
Fix a final time t f  0.. Let J (u) | x(t f ; u) |.
Optimal control problem: find a control
that maximizes J (u).
u*U
Intuition: maximize the distance to the origin.
22
Optimal Control Approach
and Stability
Theorem The BCS is GAS iff
x *(t f )  0.
23
Edge of Stability
The BCS: x  ( A1  Bu) x,
B  A2  A1.
Consider x  ( A1  Bk u ) x,
Bk  ( A2  A1 )k .
k 0
x  ( A1  0u ) x
GAS
k 1
k ε0
x  ( A1  Bεu ) x,
GAS
x  ( A1  B1u ) x,
original BCS
24
Edge of Stability
The BCS: x  ( A1  Bu) x,
B  A2  A1.
Consider x  ( A1  Bk u ) x,
Bk  ( A2  A1 )k .
k 0
x  ( A1  0u ) x
GAS
k 1
k ε0
x  ( A1  Bεu ) x,
GAS
x  ( A1  B1u ) x,
original BCS
Definition: k* is the minimal value of k>0
such that GAS is lost.
25
Edge of Stability
The BCS: x  ( A1  Bu) x,
B  A2  A1.
Consider x  ( A1  Bk u ) x,
Bk  ( A2  A1 )k .
Definition: k* is the minimal value of k>0
such that GAS is lost.
The system x  ( A1  Bk*u) x is said to be on
the edge of stability.
26
Edge of Stability
The BCS: x  ( A1  Bu) x,
B  A2  A1.
Consider x  ( A1  Bk u ) x,
Bk  ( A2  A1 )k .
Definition: k* is the minimal value of k>0
such that GAS is lost.
0
k* 1
k
0
1 k*
k
Proposition: our original BCS is GAS iff
k*>1.
27
Edge of Stability
The BCS: x  ( A1  Bu) x,
B  A2  A1.
Consider x  ( A1  Bk u ) x,
Bk  ( A2  A1 )k .
Proposition: our original BCS is GAS iff
k*>1.
→ we can always reduce the problem of
analyzing GUAS to the problem of
determining the edge of stability.
28
Edge of Stability When n=2
Consider
x  ( A1  Bk u ) x.
The trajectory x* corresponding to u*:
k k*
k k*
k k*
x0
x0
A closed
periodic trajectory
29
Solving Optimal Control Problems
2
| x(t f ; u ) | is a functional:
x(t f ; u)  F (u(t), t [0, t f ])
Two approaches:
1. The Hamilton-Jacobi-Bellman (HJB)
equation.
2. The Maximum Principle.
30
Solving Optimal Control Problems
1. The HJB equation.
Intuition: there exists a function V (, ) : R  R
n

R
V (t , x *(t ))  const,
and V can only decrease on any other
trajectory of the system.
31
The HJB Equation
Find V (, ) : R  R
n

R such that V (t f , y) || y ||2 / 2,
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 (HJB).
32
The HJB for a BCS:
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 is difficult.
Note: u* depends on Vx only.
33
The Maximum Principle
 2
Let  (t ) : Vx (t , x *(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
dt
Vx
 Vx (uA  (1- u ) B)
    (uA  (1- u ) B)
A differential equation for  (t ), with a
boundary condition at t f .
34
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.
35
Result #1 (Margaliot & Langholz, 2003)
An explicit solution for the HJB equation,
when n=2, and {A,B} is on the “edge of
stability”.
This yields an easily verifiable necessary
and sufficient condition for stability of
second-order switched linear systems.
36
Basic Idea
The HJB eq. is:
Thus,
0= max{Vx Bx  uVx ( A  B ) x}.
u
u  0  0=Vx Bx
u  1  0=Vx Ax
Let
H A : R2  R
x(t )  Ax(t ),
be a first integral of
that is,
d A
0  H ( x(t ))  H xA Ax.
dt
Then V is a concatenation of two
first integrals H A ( x) and H B ( x).
37
1
1
 0
 0
Example: A    2  1 B    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...
38
Nonlinear Switched Systems
x { f ( x), f ( x)}
1
with
2
(NLDI)
x  f ( x) GAS.
i
Problem: Find a sufficient condition
guaranteeing GAS of (NLDI).
39
Lie-Algebraic Approach
For the sake of simplicity, we present
the approach for LDIs, that is,
x  { Ax , Bx}
and
x(t )  ...exp( Bt2 ) exp( At1 ) x(0).
40
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.
41
Lie Brackets and Geometry
x { Ax,  Ax, Bx,  Bx}.
Consider
x  Ax
x ( 0)
x  Bx
x ( 4 )
x   Bx
x   Ax
A calculation yields:
x(4ε )  x(0)  ε 2[ A, B]( x(0)).
42
Geometry of Car Parking
This is why we can park our car.
The  2 term is the reason it takes
so long.
 Bx
Ax
 Ax
Bx
[ A, B]( x)
43
Nilpotency
We saw that [A,B]=0 implies GAS.
What if [A,[A,B]]=[B,[A,B]]=0?
Definition: k’th order nilpotency all Lie brackets involving k terms
vanish.
[A,B]=0
→ 1st order nil.
[A,[A,B]]=[B,[A,B]]=0 → 2nd order nil.
44
Nilpotency and Stability
We saw that 1st order nilpotency
Implies GAS.
A natural question:
Does k’th order nilpotency
imply GAS?
45
Some Known Results
Switched linear systems:
k=2 implies GAS (Gurvits,1995).
k 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)
46
A Partial Answer
Result #2 (Margaliot & Liberzon, 2004)
3rd order nilpotency implies GAS.
Proof: Consider the WCSL
1,  T (t )( A  B) x(t )  0
u *(t )  
T
0,  (t )( A  B) x(t )  0
Define the switching function
m(t ) :  (t )Cx(t ), C  A  B
T
47
Differentiating m(t) yields
m(t )   (t )Cx(t )   (t )Cx(t )
T
T
  (t )[C, A]x(t ).
T
2nd 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
3rd order nilpotency  m
  0  m(t )  at  b
 up to a single switching in the WCSL. 48
Singular Arcs
If m(t)0, then the Maximum Principle
provides no direct information.
Singularity can be ruled out using
the auxiliary system.
49
Summary
 Parking cars is an underpaid job.
 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 worst-case trajectory.
50
Summary: Optimal Control Approach
Advantages:
 reduction to a single control u *
 leads to necessary and sufficient conditions
for GUAS
 allows the application of powerful tools
(high-order MPs, HJB equation, Liealgebraic ideas,….)
 applicable to nonlinear switched systems
Disadvantages:
 requires characterizing u *
 explicit results for particular cases only
51
More Information
1. Margaliot. “Stability analysis of switched systems using
variational principles: an introduction”, Automatica,
42: 2059-2077, 2006.
2. Sharon & Margaliot. “Third-order nilpotency, nice
reachability and asymptotic stability”, J. Diff. Eqns.,
233: 136-150, 2007.
3. Margaliot & Branicky. “Nice reachability for planar
bilinear control systems with applications to planar
linear switched systems”, IEEE Trans. Automatic
Control, 54: 1430-1435, 2009.
Available online:
52
www.eng.tau.ac.il/~michaelm
52