Download is Lyapunov stable

Eigenvalues, Zeros and Poles
x  Ax  Bu
y  Cx  Du ,
x(0)  x0
x  Rn , y  Rr ,
(1)
u  Rm
Definition: Roots of the characteristic polynomial of a system are
called eigenvalues of the system.
characteristic
p( )  det( I  A)  (  1 )(  2 )...(  n )
polynomial
eigenvalues
i , i  1,2,...,n .
nxn
adj ( sI  A)
Ri
X (s)  [sI  A] x0 
x0  
x0
det[ sI  A]
i 1 ( s  i )
n
1
n
x(t )    i eit ,
i 1
i 
ˆ Ri x0
nx1
Transfer function of the system (1):
Y (s)
H (s) 
 C ( sI  A) 1 B  D
U (s)
C[ adj ( sI  A)]B  D[det( sI  A)]

[det( sI  A)]
In general H (s ) is a matrix. H ij (s ) is then the transfer function from the
input U j (s ) to the output Yi (s ) .
After cancelling the common terms in nominator and denominator,
one gets
W (s)
H ij (s ) 
( s)
Definition:
Poles of H ij (s ) (or the system (1)) are the roots of  (s ).
Zeros of H ij (s ) (or the system (1)) are the roots of W (s ).
Result: Poles of a system are a subset of system eigenvalues.
Stability
x (t )  Ax(t )  Bu (t )
y (t )  Cx(t )  Du (t )
Zero input stability:
x (t )  Ax(t ), x(t0 )  x0
(2)
Definition: (Equilibrium) A constant solution x (t )  xd , t  t 0 of the
system (2) is called an equilibrium of the system x  f (x) .
How to find
xd ?
What is xd for
linear systems?
Solve 0  f ( xd ) !
If A is invertible then xd is zero. Otherwise, there
are infinitely many equilibria. Solve 0  Axd .
Definition (Norm): Let V be a vector space (a space with appropriate
addition and multiplication by scalars operations). Norm of a vector is
defined as a positive valued function that satisfies.
x 0  x0
. :V  R
x   x
x  y  x  y , x, y V
Generalization of
the notion ...........
Definition: (Lyapunov stability)
Let xd be an equilibrium of the system x (t )  f ( x (t ))
The equilibrium is called Lyapunov stable if for every arbitrary small
  0 , there exists a  ( ) such that
x(t0 )  xd   ( , t0 )

x(t )  xd   , t  t0
Definition: (Asymptotic stability)
Let xd be a Lyapunov stable equilibrium of the system x (t )  f
Then xd is asymptotically stable if there exists a   0 such that
x(t0 ) - xd < d

( x(t ))
lim x(t )  xd  0, t  t0
t 
Theorem:
The zero equilibrium of the system x (t )  Ax(t ) is
Lyapunov stable if and only if
 (t , t0 )  c, t  t0
Proof:  (if)
Solution:
x(t )  (t , t0 ) x0 , t  t0
x(t )  (t , t0 ) x0  (t , t0 ) x0
Norm property
(t , t0 )  c, t  t0 
x(t )  (t , t0 ) x0  (t , t0 ) x0  c x0
 ( , t0 ) ̂
 (only if)


c
x(t )  
Let xd  0 be Lyapunov stable but  (t , t 0 ) not bounded.
Then there exist a  ji (t , t 0 ) which is not bounded. Choose
x0  0
... 0 1 0
i.
... 0
T
Then,
 x(t )   (t , t0 ) x0   ji (t , t0 )   ji (t , t0 )  c  x(t )  
This result is contraditory. Therefore,
(t , t0 )  c
Theorem:
The zero equilibrium of the system x (t )  Ax(t ) is
Lyapunov stable if and only if lim  (t , t 0 )  0, t  t 0
t 
Proof: Similar to the previous proof.
Theorem:
1)
For the system
let the
x (t )  Ax(t ) ,
eigenvalues of the system be i , i  1,2,... n . Then,
x (t )  Ax(t ) is Lyapunov stable
 Re i  0, i  1,2,...., n
and eigenvalues with Re i  0 have
multiplicity one.
2)
x (t )  Ax(t ) is asymptotically stable  Re i  0, i  1,2,...., n
x (t )  Ax(t )  Bu (t ), x(t0 )  0
y (t )  Cx (t )  Du (t )
Definition: (BIBO sability) The system
is bounded input
bounded output (BIBO) stable if and only if output of the system is
bounded for all bounded inputs.
Theorem:
is BIBO stable
 All poles of
have negative real part.
Theorem:
is asymptotically stable

is BIBO stable.
Example:
Find the equilibria and analyze the stability for the following system!
é
2
-x
x
+
x
2 1
1 -1
ê
x=
ê
-x2 x1
ë
ù
ú
ú
û
Example:
Find the transfer function
for the following system!
Is the system
a) Lyapunov stable?
b) asymptotically stable?
c) BIBO stable?
 2
x t    0
 0
y t   1 1
2 1
1
 3 1  xt   1  u t 
0
0 1
1 xt 