Lyapunov Based Redesign
• Motivation
Consider x  f (t , x)  G (t , x)u, x  R n , u  R p
f : R  D  R n
G : R  D  R n p
But the real system is
x  f (t , x)  G (t , x)u  G (t , x) (t , x, u )
 f (t , x)  G (t , x)(u   (t , x, u ))
 : R  D  R p  R p
 is unknown but not necessarily small. We assume
it has a known bound.
 (t , x, u)  (t , x, u)
11-1
Problem
Find a state feedback controller so that the closed loop system is stable in a
sufficiently strong sense.
Approach :
u   (t , x)  v(t , x)
(i)
(ii)
(i) chosen so that nominal closed loop system is
asymptotically stable.
(ii) chosen so as to cancel the effect of uncertainty.
11-2
Solution
Assume that u   (t, x) results in the uniformly asymptotically stable
nominal closed loop system,
x  f (t , x)  G(t , x) (t , x)
Suppose also that V (t , x) is a Lyapunov function that proves
the following.
1 (|| x ||)  V (t , x)   2 (|| x ||) : l.p.d. and dec.
V V

 f (t , x)  G(t , x) (t, x)  3 (|| x ||) : l.n.d.
t x
where  i  K , i.e.  i : R   R ,  i (0)  0,  i () is strictly increasing.
11-3
Solution (Continued)
Suppose finally that ||  (t , x, (t , x)  v) ||  (t , x)  k || v ||,
where 0  k  1,  : R  D  R
The actual system with u   (t , x)  v becomes
x  f (t , x)  G (t , x) (t , x)  G (t , x)[v   (t , x, (t , x )  v )]
V V
V

( f  G ) 
G (v   )
t
x
x
V
  3 (|| x ||) 
G (v   )
x
V
Let wT
G
x
Then V   3 (|| x ||)  wT v  wT 
Then V 
11-4
Solution (Continued)
Due to the matching condition, v can wipe out  .
There are two ways at least : when  is bounded in ||  ||2 or in ||  || .
i) ||  ||2
||  (t , x, (t , x)  v) ||2   (t , x)  k || v ||2 , 0  k  1
 T v   T    T v  ||  ||2 ||  ||2   T v  ||  ||2   (t , x)  k || v ||2 
Let  (t , x)   (t , x), (t , x)  R  D
and choose
v
 (t , x) 
1  k ||  ||2
Then  T v    T
T v  T  
 (t , x) 
 (t , x)

||  ||2
1  k ||  ||2
1 k
 (t , x)
||  ||2   ||  ||2  ||  ||2 k
1 k
1
k
  (

) ||  ||2   ||  ||2
1 k 1 k
   ||  ||2   ||  ||2  0
 (t , x)
1 k
11-5
Solution (Continued)
ii) ||  ||
||  (t , x, (t , x)  v) ||   (t , x)  k || v || , 0  k  1

Now    i i  max |  i |  | i |  ||  ||1||  ||
T
i 1
i
 T v   T    T v  ||  ||1||  ||
  T v  ||  ||1   (t , x)  k || v || 
Let  (t , x)   (t , x), (t , x)  R  D
 (t , x)
and choose v 
sgn( )
1 k
k 
  (t , x)


Then  T v   T    T  
sgn( )   ||  ||1   (t , x) 
1  k 
 1 k



k
 
||  ||1   ||  ||1 
||  ||1
1 k
1 k
k 
 1
  

 ||  ||1   ||  ||1
1

k
1

k


   ||  ||1   ||  ||1  0
11-6
Smooth Control
• Smooth Control ( 
2
case)
  (t , x) 
 1  k ||  || , if  (t , x) ||  ||2  

2
v
2
   (t , x)  , if  (t , x) ||  ||  
2

1 k 



where  T 
V
G(t , x)
x
Obviously when  (t , x) ||  ||2    V  0
Analyze what happens when  (t , x) ||  ||2  
11-7
We have
 2 

V   3 (|| x ||2 )    
 
 1 k 

T
  3 (|| x ||2 ) 
  3 (|| x ||2 ) 
2
(1  k )
2
(1  k )
||  ||22   ||  ||2  k ||  ||2 || v ||2
|| 
||22
k 2
  ||  ||2 
||  ||22
(1  k )
2
2
2
  3 (|| x ||2 ) 
||  ||2   ||  ||2    3 (|| x ||2 ) 
||  ||22  ||  ||2



1 2
2z
Here f ( z )   z  z is maximized at

1 2
and max(  z  z ) 


1  z 
2
1 2 


 
 4
2 4

Thus V  3 (|| x ||2 )  , when  (t , x) ||  ||2  
4
On the other hand, when  (t , x) ||  ||2   , V is
V  3 (|| x ||2 )  3 (|| x ||2 ) 
Thus V  3 (|| x ||2 ) 

4

4
irrespective of the value of  (t , x) ||  ||2
11-8
Smooth Control (Continued)
Then take r large, so that
1


(r ) 
3
2
4

i.e. r ( )  31 ( )
2
1
1

Then V ( x)    3 (|| x ||2 )   3 (|| x ||2 ) 
2
2
4
1
   3 (|| x ||2 ),
 || x ||2  r ( )
2
Q


Q   x :  3 (|| x ||)  
2

r ( )  0 as   0
11-9
- When  is chosen small, we can arrive at a sharper result.
n
Assume that D  R
such that 3 (|| x ||2 )   2 ( x)
 (t , x)  0  0,  (t , x)  1 ( x) where  : R n  R .
Then, when  (t , x) ||  ||2   ,
2
V   3 (|| x ||2 ) 
||  ||22   ||  ||2

02
2
  ( x) 
||  ||22  1 ( x) ||  ||2

T
 1    ( x) 
1 2
1   ( x)   1
   ( x)  


2
2 ||  ||2    1 202 /   ||  ||2 
P
where P is positive definite if   202 12 .
202
Thus choosing  
12
we
have V   1  2 ( x ).
,
2
Also V   2 ( x)   1  2 ( x) when  (t , x) ||  ||2   .
2
We conclude V   1  2 ( x) which shows that the origin is uniformly
2
asymptotically stable.
11-10
Example
Ex: x1  x2
x2  a[sin( x1  1 )  sin 1 ]  bx2  cu
Let aˆ and cˆ denote the nominal values of a and c.
u
ˆ u 
Let u  cu
cˆ
Choose u   ( x)  aˆ sin( x1  1 )  sin 1   ( k1x1  k2 x2 )
where k1 & k2 are chosen so that
1 
0
 k k  b  is Hurwitz.
 1 2

Then
x1  x2
x2  k1x1  (k2  b) x2   ( x, u )
ˆ  acˆ 
 ac
 c  cˆ 
sin(
x


)

sin(

)



 k1x1  k2 x2 
1
1
1



ˆ
ˆ
c
c




 ( x, u )  
11-11
Example (Continued)
Hence  ( x, ( x)  v) 
ˆ  acˆ
ac
c  cˆ
c  cˆ
x1 
k1x1  k2 x2 
v
cˆ
cˆ
cˆ
 1 || x ||2  k || v ||2
ˆ  acˆ c  cˆ
aac

|| k ||2 ,
cˆ
cˆ
c  cˆ
We need
 k 1
where 1 
k 
c  cˆ
.
cˆ
cˆ
Calculate the control v. Let V  xT Px be the Liapunov function
for the nominal closed loop system where P is defined by
k1 
1 
0
0
P

P
1 k  b 
 k k  b    I  solve P
2


 1 2

 P11 P12  0 
V
T
T
Then w 
G  2 x PB  2  x1 x2  
 
x
 P21 P22  1 
 2( P12 x1  P22 x2 )
11-12
Example (Continued)
Choose
  (t , x) 
 1  k ||  || , if  ||  ||2  

2
v
2
   (t , x)  , if  ||  ||  
2

1 k 
where  
202
12
Then the control u is
u 1
 ( ( x)  v)
cˆ cˆ
 stabilize the origin globally!!
u
11-13
Backstepping
Consider a system
  f ( )  g ( )
  u
1
2
where   R n , u  R and f () with f (0)  0 and g () are smooth in D  R n .
Design a state feedback controller to stabilize the system at (  0,   0)
u


g ( )

+
f ( )

f ()
11-14
Backstepping (Continued)
Suppose (1) is stabilized by    ( ) with  (0)  0, i.e., the origin of
  f ( )  g ( ) ( ) is asymptotic
ally stable. Furthermore, suppose that
we knowa positive Lyapunov function V ( ) where
V ( )

[ f ( )  g ( ) ( )]   W ( ),   D
A
where W ( ) is positive definite.
T hen, from (1),(2)
  [ f ( )  g ( ) ( )]  g ( )[   ( )]
  u
u


+
  ( )
g ( )
+


f ()  g () ()
11-15
Backstepping (Continued)
Let denote z     ( ), then
  [ f ( )  g ( ) ( )]  g ( ) z
z  u  
u
+

z
g ( )
+


f ()  g () ()
 
backstepping   ( ) through the integrator
where    [ f ( )  g ( ) ]
11-16
Backstepping (Continued)
Let v  u  , then
  [ f ( )  g ( ) ( )]  g ( ) z
z  v
3
4
which is similar to the original system but  has an asymptotically
stable origin when the input is 0.
Let Va ( ,  )  V ( )  12 z 2 , then
Va  V [ f ( )  g ( ) ( )]  V g ( ) z  zv   W ( )  V g ( ) z  zv



T hus choosing v   V g ( )  Kz , K  0, then
Va   W ( )  Kz 2
which shows that th
e origin (  0, z  0) is asymptotic
ally stable.
Since  (0)  0, the origin (  0,   0) is asymptotic
ally stable
( z     ( )).
Then the state feedback control law
u   [ f ( )  g ( ) ]  V g ( )  K [   ( )]
( u    v)
11-17
Lemma & Example
Lemma: Consider the system defined by (1), (2). Let  ( )
be a stabilizing state feedback control for (1)
with  (0)  0 and V ( ) be a Lyapunov function
satisfying A . T hen theabove state feedback control
u stabilizes the origin of (1), (2) with
V ( )  12 [   ( )]2 as a Lyapunov function.
Ex:
x1  x12  x13  x2
x2  u
x1  x12  x13  x2
x1   x1  x13
let x2   ( x1 )   x12  x1
Then V ( x1 )  12 x12
 V   x12  x14   x12 , x1  R
Thus
11-18
Example (Continued)
u  x1 ( x12  x13  x2 )  xV1  [ x2   ( x1 )]
 (2 x1  1)( x12  x13  x2 )  x1  ( x2  x12  x1 )
Va ( x)  12 x12  12 ( x2  ( x12  x1 )) 2
 12 x12  12 ( x2  x12  x1 ) 2
Let’s consider
  f ( )  g ( )
  f a ( , )  g a ( , )u
If g a ( , )  0 over the domain of interest, choose
u
1
g a ( , )
[ua  f a ( , )]
T hen   ua . T hus from the previous lemma
u  a (,  ) 


1
g a ( , ) 

[ f ( )  g ( ) ]  V g ( )  K [   ( )]  f a (,  )
with Va (,  )  V ( )  12 [   ( )]2
as a stabilizing state feedback control and a Lyapunov function.
11-19
Recursive Backstepping
Consider the following strict feedback system
x  f 0 ( x)  g 0 ( x) z1
z1  f1 ( x, z1 )  g1 ( x, z1 ) z2
z2  f 2 ( x, z1 , z2 )  g 2 ( x, z1 , z2 ) z3

f i and gi depend
only on x, z1 ,, zi
zk 1  f k 1 ( x, z1 ,, zk 1 )  g k 1 ( x, z1 ,, zk 1 ) zk
zk  f k ( x, z1 ,, zk )  g k ( x, z1 ,, zk )u
where x  Rn , z1 to z k are scalars and f0 to f k vanish at the origin.
We assume gi ( x, z1 ,, zi )  0, for 1  i  K over the domain of interest.
11-20
Recursive procedure
Recursive procedure
x  f 0 ( x)  g 0 ( x) z1
 Consider a stabilizin g feedback control z1  0 ( x) with 0 (0)  0 and
V0 ( x) such that
V0
x
[ f 0 ( x)  g 0 ( x)0 ( x)]   W ( x)
p.d.f
 Consider
x  f 0 ( x)  g0 ( x) z1
z1  f1 ( x, z1 )  g1 ( x, z1 ) z2
Then using the previous result, obtain
1 ( x, z1 )  g11 [ x0 ( f 0  g 0 z1 )  Vx0 g 0  K1 ( z1  0 )  f1 ] , K1  0
V1 ( x, z1 )  V0 ( x)  12 [ z1  0 ( x)]2
11-21
Recursive procedure (Continued)
x  f 0 ( x)  g 0 ( x) z1
Next consider
z1  f1 ( x, z1 )  g1 ( x, z1 ) z2
z2  f 2 ( x, z1 , z2 )  g 2 ( x, z1 , z2 ) z3
Then we recognize that
 x    z2
   ,
,
 z1  u  z3
 f  g0 z1 
 0  fa  f2
f  0
,
g

,



f1 

 g1  g a  g 2
Thus, similarly, obtain the state feedback control
2 ( x, z1 , z2 ) 

1 1
g 2 x
( f 0  g 0 z1 ) 
1
z1
( f1  g1 z2 ) 
V1
z1
g1  K 2 ( z2  1 )  f 2

and
V2 ( x, z1, z2 )  V1 ( x, z1 )  12 [ z2  1 ( x, z1 )]2
11-22
Extended Linearization (Gain scheduling method)

Motivation







Plant – nonlinear
Controller – linear
Design method – classical linearization
Assumption – no single linear controller satisfies the
performance specification
Idea – design a set of controllers, each good at a particular
operating point, and switch (schedule) the gains of the controllers
accordingly
Problem – now we have a nonlinear (piecewise linear) system
with time dependent jump
Solution – no good tool but some theory is being developed
mostly simulation in the past
11-23
Structure & Examples
gains
Structure
Gain
Scheduler
Controller
Plant
-Examples

y
Linearized model at an operating point,
Tank system
qin  qin0 , h  h 0
qin
h
a
qout
 A( x)dx  qi  a 2 gh
d h
dt 0
Operating
point
G p (s) 

s 
where  
1
A(h 0 )
a 2 gh 0
qin0


2 A(h 0 )h 0 2 A(h 0 )h 0
where A(h) is the cross section of the tank at height h,
a is the cross section of the outlet pipe.
11-24
Control Goal
Control goal : h  h0
h0 +
Gc
-
Gp
Use P I controller: Gc ( s )  K (1 
1
)
Ti s
Choose K and Ti so that th
e closed loop system has the natural frequency
 and relative damping  . T hen K and Ti should be chosen as
qin0
K  2A(h )  0
2h
0
 K
2  

2
qin0
2  
Ti 


T

i
 2 A(h 0 )h 0 2
2
T hus for a desired h  h 0 , we schedule the gain appropriately.
P roblem: How will the system behave with swit ches?
11-25
Nonlinear Actuator

A different angle – nonlinear actuator
r
Gc ()
actuator
f ()
Plant G p
y
Assume
1
( s  1)3
1
Gc ( s )  K (1  )
Ti s
G p (s) 
f(u)
u
Large gain
f (u )  u 4
(nonlinear value) so u 4 is good enough

only open or close
(operating point is not at 0)
11-26
Step Responses
with K  0.15, Ti  1, step responses at different operating conditions are
Introduce gain scheduling through the inverse
f 1 (u ) : f ( f 1 (u ))  u
Since f () may be unknown or for simplicit y, we may not want
f 1 (u ) but an estimate fˆ 1 (u ) so that
f ( fˆ 1 (u ))  u
11-27
Approximation
Now we use this approximat e inverse for gain scheduling :
r
+
u
Gc (s )
fˆ 1 (u )
f ( fˆ 1 (u )) v
G p (s )
yp
-
Now u  v
Approximat e fˆ by two linear function
f(u)
16
fˆ
f
3
Domain
0.433u
0u3

1
ˆ
Then f (u )  
0.0538u  1.139 3  u  16
1.3
2
Domain
11-28
Results
Resulting behavior for the same plant wit h controller
No measuremen t of external conditions is required.
11-29
Classification
( ) Scheduling on the operating conditions
gain
+
-
scheduler
Operating point
plant
Gc (s )
(  ) Scheduling on the reference signal
gain scheduler
r (t ) +
-
plant
Gc (s )
( ) Scheduling on the plant output
gain scheduler
r (t )
+
-
y
Gc (s )
plant
11-30
Issues
In all cases, nonlinear time varying systems, many become unstable
even if at each frozen ti me all are fine.
The system is time varying
b
with frozen ti me
s( s  a)
Controller : Gc ( s )  K (u (t ))(1  Td (u (t )) s )
G p (s) 
Consider two systems
x  f (t , x), x  R n
(1)
f : R  R n  R n
and x  f (r , x), x  R n , r  R (2)
frozen ti me system
It is easy to show that (1) can be unstable even if (2) is exponentia lly stable.
11-31
Example
Ex: x  A(t ) x
  1  a cos 2 t
1  a sin t cos t 
A
, a0
2 
 1  a sin t cos t  1  a sin t 
det A  2  (2  a )  (2  a )  0
For a  2, Re   0. Thus the frozen system is exponentia lly stable. But
 e( a 1)t cos t e t sin t 
 (t ,0)   ( a 1)t

sin t e t cos t 
 e
So for 1  a, the system is unstable. However if A (t ) is small the above
problem doesn' t occur.
Theorem: Consider x  A(t ) x. Suppose A(t ) is differenti able and Re i ( A(t ))  0,
t  0, i. Then 0 is uniformly asym. stable provided
sup A (t ) is sufficient ly small.
t
Proof: See Ch 5 in Nonlinear System Analysis
11-32
Formalization

A version of scheduling on the output
 y 
x     f ( y, z )  Bu where y  R m , z  R n m , u  R m
 z 
f (0,0)  0 twice conti. differenti able in all variable s.
 ueq ( y ) and zeq ( y ) such that
f ( y, zeq ( y ))  Bu eq ( y )  0


Family of equilibriu m point parameteri zed by y
A linearizat ion at y  y0 is :
 y  y0 
d  y  y0  f

(
y
,
z
(
y
))
 B(u  ueq ( y0 ))
0 eq
0 
 z  z ( y )

dt 
x
eq
0 
 z  zeq ( y0 )

Design a controller
Ac ( y0 ), Bc ( y0 ), Cc ( y0 )
11-33
Block Diagram
Block diagram
gains
r
e
Gc (s )
u
gain scheduler
u
ueq
plant
y
H
In the scheduler
xc  Ac ( y (t )) xc  Bc ( y (t ))e
u  Cc ( y (t )) x

gain scheduling on the current output
11-34
Conditions
The theory gives condition under whi ch y   the system is stable.
These conditions are that y   and nonoutput nonlineari ty approaches zero
 scheduling variable should vary slowly.
 scheduling variable should capture the plant' s nonlineari ties.
Ref : Analysis of Gain Scheduled Control for Nonlinear Plants
by J.S. Shamma, M. Athans
IEEE Tr. on Automatic Contol , vol.35, no.8, pp. 898 - 907, Aug. 1990
11-35