The Problem of Absolute Stability

Motivation
v0 +
_
u
G(s)
Plant + compensator
  u
yR
Actuator
Assume that the actuator is linear
  Ky
and assume that th e closed loop system is stable for K  [ K min , K max ].
K max

K min
y
Hurwitz sector
8-1
Aizerman conjecture
Assume now that the actuator is nonlinear, for instance, a saturator
 ( y)  u
y
Moreover,assume    (t , y), i.e., may change in time.
Would the system be asymptotic
ally stable if  (t , y)  K min , K max ?
This question was posed by M.A. Aizerman in 1940’s.
Aizerman conjecture : If the closed loop system is asymptotic
ally stable
for all K  [0, K * ], it is also asymptotic
ally
stable for all  (t , y )  [0, K * ].
8-2
Kalman conjecture
Kalman conjecture : If the closed loop system is asymptotically stable
for all K  [0, K * ], it is also stable
for all  (t , y ) such that 0 
 ( y ,t )
y
 K *.
A answer was first proposed by A.I. Lurie. Popov, Kalman, Yakubovich and
others contributed to the solution. Sometimes this problem is called the Lurie
problem.
N.B : In modern terms, this problem is a problem of robustness:
If we have a nominal feedback  ( y ), that is the class
of nonlinearities defined as a sector, where does theasymptotic
al
stability exist ?
8-3
Problem Formulation
Plant x  Ax  Bu
u  Cx
Controller
x  Rn , u  R p
y Rp
u   (t , y)
 : R  R p  R p
Assumptions
( A, B, C )
A
 (t, y)
controllab le & observable
Hurwitz
piecewise continuous in t 
 and  satisfies the sector conditions
locally Lipschitz in y

For p  1

y 2  y (t , y)  y 2

sector condition

t  R
[a, b] local sector condition
y  
(, ) global sector condition
8-4
Problem (Continued)
u
y
y
y
Generaliza tion for p  1 (if  is decentralized)
We can rewrite the sector condition  as
[ (t , y)  y][ (t , y)  y]  0, t  R
 [ a, b]

where y   or
[, ]

8-5
Problem (Continued)
Consider the decentralized feedback
  1 (t , y1 ) 


 (t , y )  


 2 (t , y p ) 
Assume each  i (t , yi ) satisfies the sector conditions with  i ,  i , ai , bi .
Define K min  diag (1 ,, p )
K max  diag ( 1 ,,  p )
and
  { y  R P : ai  yi  bi }
Then p-dim sector condition is
[ (t , y )  K min y ]T [ (t , y )  K max y ]  0 t  R ,
where K max  K min  0

y 
symmetric positive definite diagonal matrix
8-6
Problem (Continued)
- Generalization for centralized case
Assume L  R p p such that  (t , y ) satisfies
 (t , y )  Ly 2   y 2 t  R , y    R p
Introduce K min  L  I
K max  L  I
Then
[ (t , y )  K min y ]T [ (t , y )  K max y ]   (t , y )  Ly 2   2 y 2  0
2
2
where again K max  K min  0
Def : A nonlineari ty  : R  R p  R p is called a sector nonlineari ty
if  K max , K min  R p p such that K max  K min is p.d. and symmetric
and [ (t , y )  K min y ]T [ (t , y )  K max y ]  0, t  R , y   or y  R p .
8-7
Problem (Continued)
Notation   [ K min , K max ]
or if the inequality is strict,   ( K min , K max )
Consider again
x  Ax  Bu , x  R n 

y  Cx,
u, y  R p 

u   (t , y )

  [ K min , K max ] 
or

  ( K min , K max ) 
(1)
Define : System (1) is absolutely stable in the sector [ K min , K max ]
if eq. point 0 is globally uniformly asymptotic ally stable
for any nonlineari ty from the sector.
8-8
Remarks
Remarks
1)
2)
3)
Absolute stability  not another type of stability
Absolute stability gives a measure of robustness
No constructive necessary and sufficient conditions have been
found as yet. The main tool is the Kalman-Yakubovich-Popov Lemma.
8-9
Solution

Approach to the solution
To find out conditions of absolute stability
Find a Lyapunov function good for a continuum of systems – all with
nonlinearities in the sector  Two types of Lyapunov functions are
typically used.
V ( x)  xT Px
P  PT  0
and
V ( x)  xT Px   0Cx y ( )d
P  PT  0,   0
Here the conditions are less conservative (Popov Criterion).
8-10
Solution (Continued)
In both approaches , the proof of V  0 is done using a technica l result
that allows to complete squares. This is the Kalman - Yakubovich - Popov lemma.
It gives conditions under whi ch a certain stable transfer function Z ( s ) is
strictly positive real, i.e., a stable transfer function Z ( s ) is strictly positive
real if Re Z ( jw)  0, w  (, ) or Z ( jw)  Z ( jw)  0, w  (, )
( just positive real   )
If Z ( s) is a ( p  p) matrix, the positive realness is defined as follows :
Each Z ij ( s) is analytic in Re s  0
Z * ( s)  Z ( s* ) for Re s  0, and
Z T ( s* )  Z ( s)  0 for Re s  0.
Further Z ( s) is strictly positive real if Z ( s   ) is positive real
for some   0.
8-11
Kalman – Yakubovich-Popov Lemma
Kalman - Yakubovich - Popov Lemma : Let
Z ( s )  C ( sI  A) 1 B  D
be an ( p  p) transfer function matrix wi th A Hurwitz and ( A, B, C )
observable and controllab le. Then Z ( s ) is strictly positive real if and
only if  P  PT  0, P  R n n , w  R p p and L  R p n and   0 such that
AT P  PA   LT L  P
PB  C T  LT w
wT w  D  DT
8-12
Circle Criterion

Circle Criterion
x  Ax  Bu
x  Rn
y  Cx
u, y  R p
u   (t , y )
At this point, assume A is Hurwitz and   [0, K ], that is
 T (t , y )( (t , y )  Ky)  0, t , y  R p
Choose V ( x)  xT Px
Then V  xT Px  xT Px  [ xT AT   T BT ]Px  xT P[ Ax  B ]
 xT ( AT P  PA) x  2 xT PB
Since  2 T (  Ky)  0, add this to the RHS and obtain an inequality .
Then
V  xT ( AT P  PA) x  2 xT PB  2( T  yT K )
 xT ( AT P  PA) x  2 xT (C T K  PB)  2 T
8-13
Circle Criterion (Continued)
To show when V  0, we use the trick - completion of the squares using
the K - Y - P lemma : Assume  P, L &  such that
AT P  PA   LT L  P
P  PT  0
PB  C T K  2LT
Then
V  xT Px  xT LT Lx  2 2xT LT  2 T
 xT Px  [ Lx  2 ]T [ Lx  2 ]
 xT Px
Now such P, L and  do exist as it follows from the K - Y - P lemma
if and only if
Z ( s)  KC ( sI  A) 1 B  I is strictly positive real
8-14
K – Y – P Lemma
K - Y - P lemma : Z ( s )  C ( sI  A) 1 B  D is strictly positive real iff
A is Hurwitz, ( A, B, C ) controllab le and observable and
AT P  PA   LT L  P
PB  C T  LT w
wT w  D  DT
[C ]  KC
[ w]  2I
Thus D  I
Thus we can show V  0.
8-15
K – Y – P Lemma (Continued)
Lemma : System x  Ax  Bu
y  Cx
u   (t , y )
with A Hurwitz and ( A, B, C ) observable and controllab le
and  T [  Ky]  0, t , y  R p is absolutely stable in the
sector [0, K ] if Z ( s )  I  KG( s ) where G ( s )  C ( sI  A) 1 B
is strictly positive real.
A graphical illustrati on for the case p  1
Im G( jw)
Re{1  KG( jw)}  0, w
1
Re G( jw)
K
sector [0, K ]
G( jw) is to the right of the vertical line - K1 .
8-16
Generalization
Generaliza tion : Remove condition on A asymptotic stability, sectors ( ,  ).

0
To eliminate the restriction on A to be Hurwitz
 loop transformation (pole shifting)
v0

G(s)
v0 



G(s)
K min
 (t , y)
K min


y
adding the signal  K min y
adding the signal  K min y
 (t , y)
8-17
Generalization (Continued)
So the scheme does not change. Choose K min so that
( A  BK minC )
is Hurwitz. In this new system, the linear part is
GT ( s )  G ( s )[ I  K minG ( s )]1
or in the state space representa tion
x  ( A  BK minC ) x  Bu
y  Cx
 T    K min y
Obviously
[  K min y ]T [  K max y ]   TT [  K max y  K min y  K min y ]
  TT [ T  ( K max  K min ) y ]   TT [ T  Ky]  0  like before
Apply the above lemma : The system is absolutely stable in [ K min , K max ] if
ZT ( s )  I  KGT ( s ),
K  K max  K min
is strictly positive real and
GT ( s )  G ( s )[ I  K minG ( s )]1 is asymptotic ally stable.
8-18
Generalization : Case 1
A very convenient representa tion can be given to this condition
if p  1 and K min   , K max   .
In this case we have
G (s)
GT ( s ) 
1  G ( s )
G (s)
1  G ( s )
ZT ( s )  1  (    )

1  G ( s ) 1  G ( s )
To give a graphical interpreta tion, consider three cases seperately :
  0,   0,   0
1)     0
Want to find a graphical condition such that GT ( s) is Hurwitz
and ZT ( s) strictly positive real. Begin wit h the latter.
8-19
Generalization : Case 1
 1  G ( j ) 
Re 
 0,   R

 1  G ( j ) 
or
1 /   G ( j ) 
Re 
  0,   R
1
/


G
(
j

)


j Im G( j )
p
1


21
1

Re G( j )
1  G ( j ) is represente d (for point p ) by the line { 1 , p}


1  G ( j ) is represente d (for point p ) by the line { 1 , p}


8-20
Generalization : Case 1
The real part of the ratio of the two complex numbers is positive
when the angle difference is less than  / 2
1   2   2
1  G ( j ) 
Re 
  0,   R
1


G
(
j

)


G ( j )  u  jv

Then  becomes
(1  u )(1  u )  v 2
 0,   R
2
2
(1  u )  v
Since the denominato r is positive,
(1  u )(1  u )  v 2  0,   R
1
1
(u  )(u  )  v 2  0,   R



8-21
Generalization : Case 1
Im G( j )
G ( j *)
v
1

1
Re G( j )

u
This condition can be given a simple geometric interpreta tion
a :v  v :b
 v 2  ab
v
a
b
Inequality  states that Nyquist plot of G( j ) never enter the disk.
8-22
Generalization : Case 1
Let D( ,  ) be a disk with the center on the real axis and diameter
1

 1
D ( ,  )
1

1

Then the above condition is met if G ( j ) does not enter D( ,  ).
Further, GT ( s ) is Hurwitz if the point ( 1 , j 0) is encircled by G ( j )
r times in the counterclo ckwise direction where r is the # of poles of
G ( s ) in the ORHP.
Thus, with p  1, the system is absolutely stable in [ ,  ] if G ( s ) doesn' t
enter the disk D( ,  ) and encircles it r times in the counterclo ckwise
direction.
8-23
Generalization : Case 2
2)   0.
Then, GT ( s )  G ( s ),
T 
ZT ( s )  G ( s )  1
We need G ( s ) asym. stable and
Re(1  G ( j ))  0, R
i.e., Re G ( j )   1 
Im G( j )

1
Re G( j )

Thus G( j ) should be right of the vertical line passing through ( 1  , j 0)
8-24
Generalization : Case 3
3)   0  
For this case, GT ( s )
1  G ( s)
1  G ( s )
G ( s )  Hurwitz(?)
ZT ( s ) 
Since


 0, the strict positive realness of ZT ( s ) implies
1 /   G ( j ) 
Re 
 0, 

1 /   G ( j ) 
This means that G ( j ) is inside the disk D( ,  ).
Im G( j )
Thus G ( j ) can' t
encircle ( 1 , j 0)
 1

1
Re G( j )

8-25
Summarizing
Summarizin g :
Theorem : Consider a system
x  Ax  Bu , x  R n
y  cx
, u, y  R
u   (t , y ), t , y
( A, B, C ) observable & controllab le
(i) It is absolutely stable in the sector [ ,  ],     0, if G ( j ) does not
enter D ( ,  ) and encircles it r times in the counterclo ckwise direction,
where r is the # of G ( s ) poles in ORHP.
(ii) It is absolutely stable in the sector [0,  ] if G ( s ) lies to the right of
the vertical passing through (- 1  , j 0) and G ( s ) is Hurwitz.
(iii) It is absolutely stable in the sector [ ,  ],   0   , if G ( s ) is inside
of D ( ,  ) and G ( s ) is Hurwitz.
8-26
Summarizing
    0  replace G by  G and apply (i), (ii)
 by  
 by  
If the sector condition is satisfied locally, we have absolute stability
with finite domain.
(1) absolutely stable in [0,  2 ]
(2) absolutely stable in [1 , 1 ]
1
1
1
2
1
1
whi ch is one is bigger?

1
2
whi ch is better for applicatio n?
(3) absolutely stable in [ 2 ,  2 ]
8-27
Popov Criterion
Need for Popov' s
1
1
1
2
1
Popov

2

1
1
  2  1
Popov Criterion
x  Ax  Bu , x  R n
y  Cx
, u, y  R
u   ( y ) ,  note that th e feedback is time invariant
Also  ( y ) is a decetraliz ed nonlineari ty
8-28
Popov Criterion (Continued)
  1 ( y1 ) 


 ( y)    
 p ( y p )


Each  i ( yi ) is in t he sect or [0,  i ] i  1,  , p
i.e.,
0   i ( yi ) yi   i yi2 ,
or  T ( y )( ( y )  Ky )  0,
yi (locally or globally)
K  diag[ 1 ,  ,  p ]
At t his st age, we assume also t hat A is Hurwit z. T his condit ion can
be removed by loop t ransformat ions. As it was point ed out before,
t he main difference bet ween circle & P opov crit erion is t he Lyapunov
funct ion. In t he P opov crit erion, we choose Lyapunov funct ion as follows:
v( x)  x T Px  2 
y  cx
0
 T ( ) K d where P  0, P  P T and   0
are t o be chosen.
Let F ( y)  2 0y Cx T ( ) Kd  0.
8-29
Popov Criterion (Continued)
For V ( x), we obtain :
d
V ( x)  x T Px  xT Px  F ( y )
dt
F ( y ) dy dx
 x T Px  xT Px 
y dx dt
 x T Px  xT Px  2 T ( y ) KCx
 xT ( AT P  PA) x  2 xT PB ( y )
 2 T ( y ) KC Ax  B ( y )
as for quadratic V ( x)
additional term
Obviously if   0, the additional term is 0 and the two cases are
the same. Adding again
 2 T (  Ky )  0
We obtain
V ( x)  xT ( AT P  PA) x  2 xT PB  2 T KC Ax  B   2 T (  Ky )
 xT ( AT P  PA) x  2 xT ( PB  AT C T K  C T K )  T [2 I  2KCB ]
8-30
Popov Criterion (Continued)
Since  TKCB is a scalar, V can be rewritten as
V ( x)  xT ( AT P  PA) x  2 xT ( PB  AT C T K  C T K )
  T [2 I  KCB  BT C T K ]
Choose  small enough so that
2 I  KCB  BT C T K  0
Now let 2 I  KCB  BT C T K  W TW
Assume that  P  PT , P  R n n , L  R p  n and   0 such that
AT P  PA   LT L  P
PB  C T K  AT C T K  LTW
8-31
Popov Criterion (Continued)
Then
V  xT Px  xT LT Lx  2 xT (C T K  AT C T K  LTW  AT C T K  C T K )   TW TW
 xT Px  xT LT Lx   TW TW  xT LT L   TW T Lx
 xT Px  [ Lx  W ]T [ Lx  W ]  xT Px  0
When P, L &  exist? Recall the K - Y - P lemma
Z ( s )  C ( sI  A) 1 B  D is strictly positive real
iff ( A, B, C ) controllab le & observable , A is Hurwitz
and  P, L,W &  such that
AT P  PA   LT L  P
PB  C T  LTW
W T W  D  DT
So
C T  C T K  AT C T K
D  I  KCB
8-32
Popov Criterion (Continued)
Thus
Z ( s )  ( KC  KCA)( sI  A) 1 B  I  KCB
 I  KC ( sI  A) 1 B  KC[ A( sI  A) 1  I ]B
 I  KG( s )  KC[( sI  A)( sI  A) 1  A( sI  A) 1 ]B

I
 I  KG( s )  KC[( sI  A  A)( sI  A) 1 ]B
 I  KG( s )  sKC ( sI  A) 1 B
 I  KG( s )  sKG ( s )  I  (1  s ) KG( s )
Thus, the P, L,  exist if
Z ( s )  I  (1  s ) KG( s ) is strictly positive real.
In addition, the pair ( A, KC  KCA) should be observable .
(pair ( A, B) is controllab le by assumption and A is Hurwitz)
Suppose  is chosen so that 1  i  0, where i is the eigenvalue of A,
i  1,, n. Then - 1  i and the pair ( A, KC  KCA) is observable since

( A, C ) is observable .
8-33
Theorem
Theorem : System
x  Ax  Bu , x  R n , u  R p
y  Cx
, y Rp
  1 ( y1 ) 


u   ( y )     
 p ( y p )
A Hurwitz
( A, B, C ) controllab le & observable
is absolutely stable in the sector [0, K ) where K  diag [ 1 ,,  p ]
if   0 with 
1
 i ( A), i such that

Z ( s )  I  (1  s ) KG( s )
is strictly positive real.
8-34
Theorem(Continued)
For p  1, a simple interpreta tion can be given.
Z ( s ) is strictly positive real if
Re[1  (1  j ) KG( j )]  0,   R
or Re[ 1 k  (1  j )G ( j )]  0
 Re[ 1 k  (1  j )(Re G  j Im G )]
 1 k  Re G ( j )    Im G ( j )  0



x
1
k
y
 x  y  0  y  1 ( x  1k )
Consider the plain (Re G ( j ),  Im G ( j ))
Then the above LHS is a line with th e slope
1

intersecti ng
G ( j ) - axis at  1k .
8-35
Theorem(Continued)
 Im G ( j )
Graphically
a
1

 1k
Re G ( j )
K can be as large as possible
Im G
 1k
Re G
Finite sector
Thus with p  1, the system is absolutely stable in [0, K ] if the
Popov plot is to the right of a line intersecti on with the real
axis at  1 k with a slope 1 ,   0
8-36
Theorem(Continued)
On the question of choosing  so that
2 I  KCB  BT C T K  0
For p  1, this implies
1  KCB  0
Thus CB is the coefficien t for s n 1 in the numerator of G ( s ).
Indeed,
G ( s )  C ( sI  A) 1 B  C
Adj( sI  A)
B
det( sI  A)
b1s n 1    bm
 n
s  a1s n 1    an
Adj( sI  A)  [ Is n 1  ( A  a1I ) s n  2    ( An 1  a1 An  2    an 1I )]
here ai are the coefficien t of the characteri stic polynomial of A.
Thus CB  coefficien t for s n 1.
8-37
Theorem(Continued)
Thus
i) if relative degree of G ( s )  1, CB  0 and   [0, )
ii) if relative degree of G ( s )  1 but b1  0,   [0, )
iii) only when relative degree of G ( s )  1 and b1  0

1
k | b1 |
i.e., we have to make sure that
1  kb1  0
8-38