Transcript Lesson 9 - nuu.edu.tw

Lecture 07 Analysis (III) -Stability
7.1
7.2
7.3
7.4
Bounded-Input Bounded-Output (BIBO) Stability
Asymptotic Stability
Lyapunov Stability
Linear Approximation of a Nonlinear System
Modern Control Systems
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Bounded-Input Bounded-Output (BIBO) stablility
Definition: For any constant N, M >0
Any bounded input yields bounded output, i.e.
u(t )  N    y(t )  M  
For linear systems:
T ( s) 
p( s )
 C ( sI  A) 1 B
q( s )
BIBO Stability ⇔All the poles of the transfer function lie in the LHP.
q( s)  0
Solve for poles of the transfer function T(s)
Characteristic Equation
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Asymptotic stablility
Whenu(t )  0, i. e. thesystem x  Ax
x(t )  0 as t  
For linear systems:
x  Ax  Bu
y  Cx
Asymptotically stable ⇔ All the eigenvalues of the A matrix
have negative real parts
(i.e. in the LHP)
T ( s) 
p( s )
C adj[ sI  A]B
 C ( sI  A)1 B 
q( s)
sI  A
sI  A  0
Solve for the eigenvalues for A matrix
Note: Asy. Stability is indepedent
of B and C Matrix
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Asy. Stability from Model Decomposition
nn
Suppose that all the eigenvalues of A are distinct. A  R
Let vi the eigenvector of matrix A with respect to eigenvalue i
i.e. i , satisfyingAvi  i vi , i  1,,n
Coordinate Matrix T  [v1, v2 , ,vn ]


A  T 1 AT
  T 1 AT
1  1
 
2    0
 
  
n   0
0
2
0
0
B  T 1 B
0 0  1 
0
0  2 
 

   
 
 n  n 
1t
2t
C  CT
z  T 1 ATz  T 1 Bu
y  CTz  Du
nt
1
ξ
(
0
)

T
x(0)
x(t )  T (t )  v1e ξ1(0)  v2e ξ2 (0)   vne ξn (0),
Hence, system Asy. Stable ⇔ all the eigenvales of A at lie in the LHP
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Asymptotic Stablility versus BIBO Stability
In the absence of pole-zero cancellations, transfer function poles are
identical to the system eigenvalues. Hence BIBO stability is
equivalent to asymptotical stability.
Conclusion: If the system is both controllable and observable, then
BIBO Stability ⇔ Asymptotical Stability
Methods for Testing Stability
• Asymptotically stable
• All the eigenvalues of A lie in the LHP
• BIBO stable
• Routh-Hurwitz criterion
• Root locus method
• Nyquist criterion
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• ....etc.
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Lyapunov Stablility
A state xe of an autonomoussystemis called an equilibrium state,
if startingat thatstate thesystem will not move
fromit in theabsence of theforcinginput.
In other words, consider the system x  f ( x(t ), u(t ))
equilibrium statexe must satisfy f ( xe ,0)  0,
Example:
x2
Equilibrium point
1  1
0
x  
x   u (t )

 2  3 1
Set
t  t0
x1
u(t )  0 ,
1   x1e 
 x1e  0
0
we get 
0   



 2  3  x2 e 
 x 2 e   0
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Definition: An equilibrium state xe of an autonomous system is
stable in the sense of Lyapunov if for every   0 , exist a  ( )  0
such that x0  xe    x(t, x0 )  xe   for t  t0
x2

x(t )

x1
xe
x0
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Definition: An equilibrium state xe of an autonomous system is
asymptotically stable if
(i) it is stable
(ii) there exist a  e  0 such that
x0  xe  e  x(t )  xe  0, as t  
x2

e
xe

x1
x0
x(t )
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Lyapunov Theorem
x  f (x )
Consider the system
Eq. State:
xe  0
(6.1)
 f (0)  0
A function V(x) is called a Lapunov fuction V(x) if
(1) V ( x)  0, x  0
(2) V (0)  0 for x  0
dV ( x ) dV ( x )
(3)

f ( x)  0
dt
dx
Then eq. state of the system (6.1) is stable.
Moreover, if the Lyapunov function satisfies
dV ( x)
 0, x  0
dt
and
dV ( x )
0  x0
dt
Then eq. state of the system (6.1) is asy. stable.
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Explanation of the Lyapunov Stability Theorem
1. The derivative of the Lyapunov function along the trajectory is negative.
2. The Lyapunov function may be consider as an energy function of the system.
x2
V ( x(t ))
0
x(t )
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x1
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Lyapunov’s method for Linear system: x  Ax where
The eq. state
A 0
x  0 is asymptotically stable.
⇔
For any p.d. matrix Q , there exists a p.d. solution of the
Lyapunov equation
T
A P  PA  Q
Proof: Choose
V ( x)  xT Px
T
T


V ( x )  x Px  x Px
 xT AT Px  x T PAx
 xT ( AT P  PA) x
  x T Qx  0, for x  0
 AT P  PA  Q
Hence, the eq. state x=0 is asy. stable by Lapunov theorem.
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Asymptotically stable in the large
( globally asymptotically stable)
(1) The system is asymptotically stable for all the initial states x(t0 ) .
(2) The system has only one equilibrium state.
(3) For an LTI system, asymptotically stable and globally
asymptotically stable are equivalent.
Lyapunov Theorem (Asy. Stability in the large)
If the Lyapunov function V(x) further satisfies
(i)  x  ,V ( x)  
(ii) x  ,V ( x)  
Then, the (asy.) stability is global.
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Sylvester’s criterion
A symmetric n n matrix Q is p.d. if and only if all its n
leading principle minors are positive.
Definition
The i-th leading principle minor Qi i  1,2,3,, n of an n n
matrix Q is the determinant of the i  i matrix extracted from
the upper left-hand corner of Q.
Example 6.1:
 q11 q12 q13 
Q  q21 q22 q23 
Q1  q11
q31 q32 q33 
q11 q21
Q2 
Q3  Q
q21 q22
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Remark:
(1) Q1 , Q2 , Qn are all negative
Q is n.d.
(2) All leading principle minors of –Q are positive
Q is n.d.
Example:
V ( x)  2 x12  4 x1 x3  3x12  6 x2 x3  x32
 x1
 x1
x2
x2
2
x3 0
0
2
x3 0
2
0 4  x1 
3 6  x2 
0 1  x3 
0 2  x1 
3 3  x2 
3 1  x3 
Q1  2  0
Q2  6  0
Q3  24  0
Q is not p.d.
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Example:
0 1
x  
x

 1  1
 p11
Let Q  I , Assume P  
 p12
Solve for AT P  PA   I
p12 
p22 
1   1 0 
0  1  p11 p12   p11 p12   0




1  1  p



p
p
p

1

1
0

1

  12
22 
22  
 

 12
 p11 p12  1 3 1
P
 


p
p
1
2
2
22 


 12
p11  3  0
P 50
P is p.d.
System is asymptotically stable
The Lyapunov function is:
V ( x)  xT Px  1 (3x12  2 x1 x2  2 x22 )
2
V ( x)  ( x12  x22 )
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Linear approximation of a function around an operating point xe
Let f (x) be a differentiable function.
Expanding the nonlinear equation into a Taylor series about the
operation point xe , we have
df ( x )
f ( x )  f ( xe ) 
dx
( x  xe ) d 2 f ( x )

1!
dx2
x  xe
( x  xe )2

2!
x  xe
Neglecting all the high order terms, to yield
f ( x)  f ( xe ) 
( x  xe )
 f ( xe )  m  ( x  xe )
1!
x  xe
df ( x)
dx
f ( x)  f ( xe )  m  ( x  xe )
f (x )
f (x )
f ( xe )
where
m
df ( x )
dx x  xe
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xe
x
x
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Multi-dimensional Case:
Let x be a n-dimensional vector, i.e. x  R
f ( x1 ,, xn )
 f ( x1e ,, xne ) 
f
x1
f
 f ( x1e ,, xne ) 
x
x  xe ( x1  x1e ) 
( x  xe ),
x  xe
f
x2
n
x  x e ( x2  x2 e )   
f
where
x
x  xe
 f

 x1
x  xe
f
xn
x  xe
( xn  xne )
f
, ,
xn
x  xe



Let f be a m-dimensional vector function, i.e. f ( x) : Rn  Rm
 f1 ( x1 ,, xn ) 
 f ( x ,, x ) 
n 
f ( x1 ,, xn )   2 1





f
(
x
,

,
x
)
n 
 m 1
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Linear approximation of a function around an operating point xe
Special Case: n=m=2
f ( x)  f ( xe ) 
where x  [ x1, x2 ]
T
f
( x  xe )  A( x  xe )
x x xe
and
f
x
x  xe
 f1

x1 x  x

e

 f
 2
 x1 x  xe
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

x  xe 
A

f 2

x2 x  x 
e 
f1
x2
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Linear approximation of an autonomous nonlinear systems x (t )  f ( x(t ))
Let xe be an equilibrium state, from
x  f ( x(t ))  A( x - xe )
where
A
f
x
x  xe
 f1

x1 x  x

e

 f
 2
 x1 x  xe


x  xe 

f 2

x2 x  x 
e
f1
x2
The linearization of x (t )  f ( x(t )) around the equilibrium state xe is
z  Az
where
z  x - xe and
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z  x - xe  x
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Example : Pendulum oscillator model
From Newton’s Law we have
d 2
J
 MgLsin  0
dt 2
where J is the inertia.
Define
x1   , x2  
x2

 x1  

     MgL
 x2  - J sinx1 
(Reproduced from [1])
We can show that xe  0 is an equilibrium state.
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Example (cont.):
Method 1:
f1 ( x2 )  x2  f1 ( x2 ) - f (0)  ( x2 - 0)  z2
f 2 ( x1 )  sinx1
 f 2 ( x1 ) - f 2 (0)  sinx1 - sin0 
d (sinx1 )
( x1 - 0)  z1
dx1 x 0
1
The linearization around the equilibrium state xe  0 is
z2 
 z1  
 z   - MgL z 
 2   J 1 
where
z  x and z  x
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Example (cont.):
Method 2:
MgL
f1 ( x )  x2 , f 2 ( x )  sinx1
J
f
x
x  xe
 f1

f1



x

x
1 x  xe
2 x  xe 

 f

f 2
2


x2 x  x 
 x1 x  xe
e 
1
 0
  - MgL   A
0
 J

The linearization around the equilibrium state xe  0 is
1  z   z 2 
 0
 z1 
1
MgL

Az



 z 
 z   - MgL z 
0
 J
  2   J 1 
 2
where
z Systems
x
andControl
z  xModern
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