208581 Nonlinear Systems in Mechanical Engineering
Lesson 14
We want to design an additional feedback control v such that the
overall control u    t , x   v stabilizes the actual system (1). To do that
1 Lyapunov Redesign
The Lyapunov redesign uses a Lyapunov function of a nominal
system to design an additional control component that makes the design
robust to large matched uncertainties.
Our task is to design a stabilizing controller for a nonlinear
uncertain system
the uncertain term
  t , x,   t , x   v     t , x    0 v , 0   0  1,
x  Rn is the state and u  R p is the control input. The functions
f and G are defined for  t , x, u   0,    D, and the function  is
where D  R
n
(5)
 : 0,    D  R is a nonnegative continuous function. We can
design v with the knowledge of the Lyapunov function V , the function  ,
and the constant  0 . The design of v is called Lyapunov redesign.
(1)
where
 t , x, u   0,    D  R p ,
must satisfy the inequality
where
x  f  t , x   G  t , x  u    t , x, u   ,
defined for

 Example 1 [1]: Show that the perturbed feedback linearizable system fits
into this design framework.
is a domain that
f , G, and  are piecewise
continuous in t and locally Lipschitz in x and u. The functions f and G
contains the origin. We assume that
are known precisely, while the function  is an unknown function that
lumps together various uncertain terms due to model simplification,
parameter uncertainty, disturbance, and so on.
A nominal model of (1) is
x  f  t , x   G t , x  u.
(2)
Suppose a feedback control law u    t , x  uniformly asymptotically
stabilizes the origin of (2) with a known Lyapunov function V  t , x  that
satisfies the inequalities
1  x   V  t , x    2  x  ,
(3)
V V

 f  t , x   G  t , x   t , x     3  x  ,
t x 
(4)

for all  t , x   0,    D, where
1 ,  2 , and  3 are class- K functions.
1
Copyright
2007 by Withit Chatlatanagulchai
208581 Nonlinear Systems in Mechanical Engineering
Lesson 14
Solution
Consider the feedback linearizable system
where
Since the nominal system is feedback linearizable, we can take a
stabilizing feedback control as
x  f  x   G  x  u,
where f : D  R and G : D  R
n
n p
  x, u   1  x   2  x  u.
  x     x    1  x  Kz    x    1  x  KT  x  ,
are smooth functions on a domain
D  R n and there is a diffeomorphism T : D  Rn such that
Dz  T  D  contains the origin and T  x  satisfies the partial differential
where K is chosen so that
z   A  BK  z
T
f  x   AT  x   B  x    x  ,
x
T
G  x   B  x  ,
x
 A, B 
is controllable and
  x
is Hurwitz. A Lyapunov function for
the nominal closed-loop system
equations
where
 A  BK 
can be taken as V  z   z Pz , where P is the solution of the Lyapunov
T
equation
P  A  BK    A  BK  P  I .
T
is nonsingular for all x  D. The
With u    x   v, the uncertain term
change of variables z  T  x  transforms the system into the form
  x, u 
satisfies the
inequality
z  Az  B  x  u    x   .
  x,   x   v   1  x    2  x   x    2  x   1  x  Kz
Consider also the perturbed system
 2  x  v .
x  f  x    f  x   G  x    G  x  u
Thus, to satisfy (5), we need the inequalities
with smooth perturbations that satisfy the conditions
T
 f  x   B  x  1  x  ,  G  x   G  x   2  x 
x
2  x   0  1
(6)
1  x   2  x   x   2  x   1  x  KT  x     x 
(7)
and
on D. The perturbed system can be represented in the form (1), that is,
z  Az  B  x    x   B  x  u    x, u   ,
to hold over a domain that contains the origin.

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Copyright
2007 by Withit Chatlatanagulchai
208581 Nonlinear Systems in Mechanical Engineering
Lesson 14
wT v  wT   wT v  w
1.1 Designing Discontinuous v
Applying the control u    t , x   v
to the actual system (1), we
  1   0  w 2   w
  w 2   w
Choosing a Lyapunov candidate to be the same as that of the nominal
system, its derivative is given by
V V
V

 f  G   G  v   
t x
x
V
  3  x  
G v   
x
  3  x   wT v  wT  ,
1.1.2. When (5) is satisfied with
 t, x 
, for all  t , x   0,    D, and sgn  w is a p 1  0
th
dimensional vector whose i component is sgn  wi  , we obtain
with
w
,
w2
 t, x  
wT v  wT   wT v  w 1 
2

 wT v  w 1    t , x    0 v  
  w 1   w 1   0 w 1
Choosing
v    t , x 

v    t , x  sgn  w ,
V by choosing v so that wT v  wT   0.
1.1.1. When (5) is satisfied with
2
Choosing
wT  V / x  G . Then, we can choose v to cancel the
on
2
 0.
V

2
 w v  w 2    t , x    0 v 2 
  w 2   w 2   0 w 2
x  f  t , x   G  t , x   t , x   G  t , x  v    t , x,   t , x   v  .(8)
destabilizing effect of

T
have the resulting closed-loop system
where
2
(9)
  1   0  w 1   w 1
  w 1   w 1
 t, x 
, for all  t , x   0,    D,
with a nonnegative function   t , x  
1  0
 0.
we obtain
1.2 Designing Continuous v
The discontinuous v causes some theoretical as well as practical
problems as our discussion in sliding-mode control. In this section, we
design a continuous approximation of v in (9).
Consider the feedback control law
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
  t , x   w / w 2  , if   t , x  w 2   ,
v
2

   t , x  w /   , if   t , x  w 2   .
208581 Nonlinear Systems in Mechanical Engineering
Lesson 14
When the uncertainty  vanishes at the origin, we can arrive at a
sharper result as stated in the following corollary.
(10)
Corollary 14.1: Suppose there is a ball Ba 
It is shown in [1] that, with the Lyapunov candidate V  t , x  of the
contains the origin and Br 
x
2
n
2
 a , a  r such that
3  x 2    2  x  ,
nominal system, the control (10) achieves uniform ultimate boundedness
of the solution of the closed-loop system. The following theorem states the
result.
Theorem 14.3: Consider the system (1). Let D  R
x
  t , x   0  0,
be a domain that
  t , x   1  x  ,
 r  D. Let   t , x  be a stabilizing
feedback control law for the nominal system (2) with a Lyapunov function
V  t , x  that satisfies (3) and (4) in 2-norm for all t  0 and x  D, with
for all x  Ba , where
 : Rn  R is a positive definite function of x.
Assume also that all the assumptions of Theorem 14.3 are satisfied. Then,
1 ,  2 , and  3 . Suppose the uncertain term 
satisfies (5) in 2-norm for all t  0 and x  D. Let v be given by (10) and
  2 3  21 1  r    / 1   0  .
choose
Then,
for
any
some class K functions
for all
  min 202 1   0  / 12 , b1  a  ,
the origin of the closed-loop
system (8) is uniformly asymptotically stable. If
 i  r   ki r c , then the
origin is exponentially stable.
Proof See page 585 of [1].
x  t0  2   21 1  r   , there exists a finite time t1 such that the solution
of the closed-loop system (8) satisfies

where

x  t  2   x  t0  2 , t  t0 , t0  t  t1 ,
(11)
x  t  2  b    , t  t1 ,
(12)
 is a class KL function and b is a class K function defined by
 

b    11  2      11  2  31  1   0  / 2  .
If all the assumptions hold globally and
and (12) hold for any initial state x  t0  .
 1 belongs to class K  , then (11)
Proof See page 584 of [1].
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Copyright
2007 by Withit Chatlatanagulchai
208581 Nonlinear Systems in Mechanical Engineering
Lesson 14
 Example 2 [1]: Design a stabilizing controller for the feedback
linearization system in previous example using Lyapunov redesign method.
Solution
Suppose (6) is satisfied in
Solution
We know that the pendulum is feedback linearizable with
T  x   x. A nominal stabilizing feedback control can be taken as
 aˆ 
 
2
. Suppose further that (7) becomes
1  x    2  x   x    2  x  
1
 x  Kz 2  1
z
where â and ĉ are the nominal values of
a and c , and k1 and k 2 are
chosen such that
2
1
 0

A  BK  

 k1   k2  b 
for all z  Br  Dz . We use the continuous control (10) with
 2 1   0 
1 z 2
2r 2min  P  
T
T
  1
, w  2 z PB,   min 
,
.
2
1   0 
1   0  max  P  
 1
is Hurwitz. With u    x   v , the uncertainty term

is given by
ˆ 
 c  cˆ 
1  acˆ  ac
 c  cˆ 
sin x1  
 k1 x1  k2 x2   


 v.
 cˆ 
 cˆ 
  cˆ 
  
cˆ
It can be verified that all the assumptions of Theorem 14.3 and Corollary
14.1 are satisfied with
1  r   min  P  r 2 ,  2  r   max  P  r 2 , 3  r   r 2 ,   z   z 2 ,
Hence,
  1 x 2   0 v , x  R 2 , v  R,
and a  r. Thus, the origin of the perturbed closed-loop system is
exponentially stable.

where
0 
 Example 3 [1]: Design a stabilizing controller for the pendulum equation
x1  x2 ,
ˆ  acˆ c  cˆ
c  cˆ
k
ac
, 1  , and k 

cˆ
cˆ
cˆ
cˆ
k12  k22 .
 0  1 and taking v as in the previous example, we find that
the control law u    x   v makes the origin globally exponentially
Assuming
x2  a sin x1  bx2  cu,
with
1
 
  x      sin x1     k1 x1  k2 x2  ,
cˆ
cˆ
stable.

1   .
5
Copyright
2007 by Withit Chatlatanagulchai
208581 Nonlinear Systems in Mechanical Engineering
Lesson 14
state x  t0  that the solution of the closed-loop system is uniformly
1.3 Nonlinear Damping
Consider again the system (1) with
bounded. The following lemma summarizes the result.
  t , x, u     t , x   0  t , x, u  ,
Lemma 14.1: Consider the system (13) and let
be a stabilizing
feedback control for the nominal system (2) with a Lyapunov function
V  t , x  that satisfies (3) and (4) for all t  0 and all x  R n , with some
that is,
x  f  t , x   G  t , x  u    t , x   0  t , x, u  .
We assume that
 0  t , x, u 
1 ,  2 , and  3 . Suppose the uncertain term  0 is
n
p
uniformly bounded for  t , x, u   0,    R  R . Let v be given by (14)
class K functions
(13)
is the only uncertain term. The functions
and take u    t , x   v. Then, for any x  t0   R , the solution of the
n
f , G, and  are precisely known. Even when the upper bound on  0 is
closed-loop system is uniformly bounded.
unknown, using the so called nonlinear damping control
v  kw   t , x  2 , k  0,
2
 Example 4 [1]: Design a controller for the scalar system
(14)
x  x 2  u  x 0  t  ,
we can achieve uniformly boundedness of the solution. Recall that, letting
wT  V / x G, the derivative of V along the trajectories of the closed-

 t, x 

where
V V
V

 f  G   G  v   0 
t x
x
T
  3  x   w  v   0 
V
  3 
2
2
2 w
2
2
Solution
With
We can see that since
  x    x 2  x, the Lyapunov function V  x   x 2 satisfies
(3) and (4) globally with
 2 k0
1  r    2  r   3  r   r 2 . The nonlinear
damping component (14), with
loop system
k02
x 
,
4k
where k0 is an unknown upper bound on
is a bounded function of t , using the Lyapunov redesign
method and nonlinear damping.
loop system satisfies
  3  x   k w
0 t 
k  1, is given by v  2 x3 . The closed-
x   x  2 x3  x 0  t 
0 .
 3 is class K  , it is always true that V is
has a bounded solution no matter how large the bounded disturbance
negative outside some ball. It follows from Theorem 4.18 for any initial
0
is because of the nonlinear damping term 2 x .

3
6
Copyright
2007 by Withit Chatlatanagulchai
208581 Nonlinear Systems in Mechanical Engineering
Lesson 14
2 Backstepping
where W   is positive definite. Since f , g , and
Backstepping is a recursive procedure that interlaces the choice of
a Lyapunov function with the design of feedback control. It breaks a
design problem for the full system into a sequence of design problems for
lower order subsystems. Backstepping can be used to relax the matching
condition, a condition where uncertainty can be seen by control input.
derivative
 are known, the
 can be computed by using the expression


 f    g     .
 
(17)
Step 2: Let
2.1 Integrator Backstepping
Consider the system
z       .
  f    g   ,
(15)
  u,
(16)
(18)
(15) and (16) become
   f    g        g   z ,
z  u  .
T
T
n 1
where  ,    R
is the state and u  R is the control input. The
f : D  Rn and g : D  Rn are smooth in a domain D  R n
that contains   0 and f  0   0.
functions
Using a Lyapunov candidate
Vc  , z   V   
We want to design a state feedback control law to stabilize the
origin   0,   0 . We assume that all functions are known precisely.
1 2
z ,
2
we obtain
2.1.1. Controller Design
V
 f    g       g   z   z u  
 
V
 W   
g   z  z u   .

The controller design can be divided into two steps.
Step 1: Consider the first subsystem (15) alone, having  as input.
Assume that (15) can be stabilized by a smooth state feedback control law
     , with   0  0, that is, the origin of

Vc 

  f    g    


Choosing
is asymptotically stable with a Lyapunov function V   that satisfies
u
V
 f    g       W   ,   D,
 
V
g    kz   ,

(19)
we have
7
Copyright
2007 by Withit Chatlatanagulchai
208581 Nonlinear Systems in Mechanical Engineering
Lesson 14
which is shown in Figure 1(c), and can be viewed as “backstepping”
   through the integrator.
Vc  W    kz 2 ,
  0, z  0 is asymptotically stable. Since
that the origin   0,   0 is asymptotically
which shows that the origin
  0  0,
we conclude
u
stable. Substituting (17) and (18) into (19), we have the state-feedback
control law
u
V

g    k      
 f    g     .

 


g  


f  
a

f  
2.1.2. Why do we name this method “Backstepping?”
The original system (15)-(16) can be viewed as a cascade
connection of two components as shown in Figure 1(a). The first
component is (15), with  as input, and the second component is the
integrator (16).
By adding and subtracting g     on the right hand side of
u
(15), we obtain an equivalent representation


b
   f    g        g         ,
  u,
u
which is shown in Figure 1(b).
The change of variables
z      

g  

  

z

f    g       
g  

results in the system





f    g       
c
   f    g        g   z ,
Figure 1: (a) Block diagram of the system.
(b) Introducing    . (c) Backstepping
z  u  ,
   through the integrator.
8
Copyright
2007 by Withit Chatlatanagulchai
208581 Nonlinear Systems in Mechanical Engineering
Lesson 14
2.2 Strict-Feedback System
over the domain of interest for some positive definite function W  x  .
To show the recursive nature of the backstepping method, we
consider stabilization problem of the strict-feedback system
Step 2: For convenience, we drop functions’ arguments where
appropriate. Let
x  f 0  x   g 0  x  z1 ,
e1  z1  0  x  .
z1  f1  x, z1   g1  x, z1  z2 ,
The first two subsystems become
z2  f 2  x, z1 , z2   g 2  x, z1 , z2  z3 ,
zk 1  f k 1  x, z1 ,
zk  f k  x, z1 ,
, zk 1   g k 1  x, z1 ,
, zk   g k  x, z1 ,
x   f 0  g 00   g 0e1 ,
e1  f1  g1 z2  0 .
, zk 1  zk ,
, zk  u ,
Using a Lyapunov candidate
where x  R , z1 to z k are scalars, and f 0 to f k vanish at the origin. The
1
V1  x, z1   V0  e12 ,
2
n
reason for the name “strict-feedback” is that the zi equation depends only
on x, z1 ,
, zi , that is, on the state variables that are “fed back.” We
we obtain
assume that
gi  x, z1 ,
V0
 f 0  g 00  g0e1   e1 f1  g1 z2  0
x
V
 W  0 g 0e1  e1 f1  g1 z2  0 .
x

over the domain of interest. The controller design can be divides into the
following steps.
Step 1: Consider the first subsystem
function V0  x  such that

z2  1  x, z1 
where z1 is viewed as input. We assume that there exists a stabilizing
0  0  0,

Choosing
x  f0  x   g0  x  z1 ,
state feedback control law z1  0  x  , with

V1 
, zi   0, 1  i  k

and a Lyapunov
V
1 

0  0 g 0  k1e1  f1  , k1  0,

g1 
x

we have
V0
 f 0  x   g 0  x  0  x    W  x  ,
x 
V1  W  k1e12 ,
9
Copyright
2007 by Withit Chatlatanagulchai
208581 Nonlinear Systems in Mechanical Engineering
Lesson 14
 x  0, e1  0 is asymptotically stable. Since
that the origin  x  0, z1  0 is asymptotically
which shows that the origin  x  0, e1  0, e2  0 is asymptotically stable.
which shows that the origin
0  0  0,
stable.
we conclude
1  0, 0  0,
Since
x   f 0  g 00   g 0e1 ,
The backstepping method inserts virtual control into each
subsystem and is therefore capable of handling the unmatched
uncertainties. For example, the backstepping can be used to stabilize the
system
e2  f 2  g 2 z3  1.
Using a Lyapunov candidate
  f    g       ,   ,
1
V2  V1  e22 ,
2
  f a  ,    g a  ,   u    ,   ,
we obtain
where
V1
V
 f0  g00  g0e1   1  f1  g11  g1e2 
x
z1

 W  k1e12 
  and   are uncertain terms.
Lesson 14 Homework Problems

None
Homework problems are from the required textbook (Nonlinear
V1
g1e2  e2 f 2  g 2 z3  1 .
z1

is
2.3 Relaxing the Matching Condition
e1  f1  g1 z2  0 ,
e2 f 2  g 2 z3  1
 x  0, z1  0, z2  0
asymptotically stable.
The process is repeated until the overall stabilizing state feedback
control law u  k  x, z1 , , zk  is obtained.
Step 3: Let e2  z2  1. The first three subsystems become
V2 
we conclude that the origin

Systems, by Hassan K. Khalil, Prentice Hall, 2002.)
Choosing
References
[1]
z3  2  x, z1 , z2 

1
g2
Nonlinear Systems, by Hassan K. Khalil, Prentice Hall, 2002.


V1


g

k
e

f
1
2 2
2  , k2  0,
 1
z1


we have
V2  W  k1e12  k2e12 ,
10
Copyright
2007 by Withit Chatlatanagulchai