Lecture 7
Lyapunov Stability Theory
Theme,Generate parameter adjustment rules
which guarantee stability
1,Introduction
#0F The MIT rule pros and cons
#0F The stabilityproblem
#0F Ordinary Di#0Berential Equations
#0F Dynamical Systems
2,Lyapunov's Stability Theory
#0F The stability concept
#0F Lyapunov's idea
#0F Finding Lyapunov Functions
3,Application to adaptive control
#0F A simple problem
#0F State feedback
#0F Output feedback
4,Conclusions
The Big Picture
Model-ReferenceAdaptive Systems
1,The idea
2,The MIT Rule
3,Lyapunov's stability theory
4,Design of MRAS using Lyapunov theory
5,Applications to adaptive control
6,Bounded-input,bounded-output stability
7,Applications to adaptive control
8,Relations between MRAS and STR
9,Conclusions
Lyapunov Himself
Stability Concepts
#0F Linear systems are very special
#0F Stable systems and stable solutions
#0F An example
#0F Lyapunovs stability concepts
#7B Stability of mechanical systems
#7B A state space approach
#7B The key ideas
c#0D K,J,#C5str#F6m and B,Wittenmark 1
Stable Solutions
Consider the MRAS designed with MIT rule
Model
e
y
Process
u
S
+
–
u
c
q
y
m
-
g
s
kG(s)
k
0
G(s)
P
P
u
c
= sin!t ! = 1#28a#29,2#28b#29 and 3#28c#29
0 5 10 15 20
0
1
0 5 10 15 20
0
1
0 5 10 15 20
0
1
Time
Time
Time
#28a#29
^
#12
#28b#29
^
#12
#28c#29
^
#12
Conclusion?
Preliminaries
Consider the solution x#28t#29=0to
dx
dt
= f#28x#29 f#280#29=0
Existence and uniqueness
kf#28x#29,f#28y#29k#14Lkx,yk L#3E0
#0FSensitivity to perturbations
#0F What perturbations?
#7B Initial conditions
#7B Variations in f#28x#29
Stability concepts!
Lyapunov Stability
Definition 1
The solution x#28t#29=0is stable if for given
"#3E0there exists a number #0E#28"#29 #3E 0 such that
all solutions with initial conditions
kx#280#29k #3C#0E
have the property
kx#28t#29k#3C" for 0 #14t#3C1
The solution is unstable if it is not stable,The
solution is asymptotically stable if it is stable
and #0E can be found such that all solutions with
kx#280#29k #3C#0Ehave the property that kx#28t#29k!0
as t !1.
Notice
#0F Stabilityofaparticular solution
#0F Local concept
#0F How to make it global?
Lyapunov's Idea
Inspiration from mechanics
Stable and unstable equilibria
x
1
x
2
x=0
V(x)=const
dx
dt
Condition
@V
T
@x
f#28x#29 #3C 0
c#0D K,J,#C5str#F6m and B,Wittenmark 2
Formalities
Definition 2
A continuously di#0Berentiable function V,
R
n
! R is called positive de#0Cnite in a region
U #1A R
n
containing the origin if
#0F 1,V #280#29=0
#0F2,V #28x#29 #3E 0; x 2 U and x 6=0
A function is called positive semide#0Cnite if
Condition 2 is replaced by V #28x#29 #15 0.
Theorem 1
If there exists a function V,R
n
! R that is
positive de#0Cnite such that
dV
dt
=
@V
T
@x
dx
dt
=
@V
T
@x
f#28x#29=,W#28x#29
is negative semide#0Cnite,then the solution
x#28t#29=0is stable,If dV=dt is negative de#0Cnite,
then the solution is also asymptotically stable.
FindingLyapunov Functions
"Transforming one di#0Ecult problem to an-
other"
Let the linear system
dx
dt
= Ax
be stable,Pick Q positive de#0Cnite,The
equation
A
T
P + PA=,Q
has always a unique solution with P positive
de#0Cnite and the funtion
V #28x#29=x
T
Px
is a Lyapunov function
Example
System matrix
A =
#12
a
1
a
2
a
3
a
4
#13
Q =
#12
q
1
0
0 q
2
#13
where q
1
#3E 0 and q
2
#3E 0 are positive,Assume
that the matrix P has the form
P =
#12
p
1
p
2
p
2
p
3
#13
The Lyapunov equation becomes
0
@
2a
1
2a
3
0
a
2
a
1
+ a
4
a
3
0 2a
2
2a
4
1
A
0
@
p
1
p
2
p
3
1
A
=
0
@
,q
1
0
,q
2
1
A
Conditions for existence,When is P#3E0?
Proofs are given in the mathematics course on
matrices!
Time-Varying Systems
dx
dt
= f#28x;t#29
Definition 3
The solution x#28t#29=0is uniformly stable if
for "#3E0there exists a number #0E#28"#29 #3E 0,
independent of t
0
,such that
kx#28t
0
#29k #3C#0E#29kx#28t#29k#3C" 8t#15t
0
#150
The solution is uniformly asymptotically stable
if it is uniformly stable and there is c#3E0,
independent of t
0
,such that x#28t#29 ! 0 as
t !1,uniformly in t
0
,for all kx#28t
0
#29k #3Cc.
Definition 4
A continuous function #0B,#5B0;a#29 ! #5B0;1#29
is said to belong to class K if it is strictly
increasing and #0B#280#29 = 0,It is said to belong
to class K
1
if a = 1 and #0B#28r#29 !1as
r !1.
c#0D K,J,#C5str#F6m and B,Wittenmark 3
Lyapunov Theorem
Theorem 2
Let x = 0 be an equilibrium point and
D = fx 2 R
n
jkxk #3Crg,Let V be a
continuously di#0Berentiable function such that
#0B
1
#28kxk#29 #14V #28x;t#29#14 #0B
2
#28kxk#29
dV
dt
=
@V
@t
+
@V
@x
f#28x;t#29 #14,#0B
3
#28kxk#29
for 8t #15 0,where #0B
1
,#0B
2
,and #0B
3
are class
K functions,Then x = 0 is uniformly
asymptotically stable.
Design of MRAS using Lyapunov
Theorem
#0F The idea
#7B Determine a controller structure
#7B Derive the Error Equation
#7B Find a Lyapunov function
#7B Determine an adaptation law
#0F A #0Crst order system
#0F State feedback
#0F Output feedback
Adaptation of Feedforward Gain
Process model
dy
dt
=,ay + ku
Desired response
dy
m
dt
=,y
m
+ k
o
u
c
Controller
u = #12u
c
Introduce the error e = y,y
m
and the error
equation becomes
de
dt
=,ae+#28k#12,k
o
#29u
c
Candidate Lyapunov function
V #28e;#12#29=
#0D
2
e
2
+
k
2
#28#12,
k
0
k
#29
2
Time derivative
dV
dt
= #0De#28,ae + k#28#12,
k
0
k
#29u
c
#29+b#28#12,
k
0
k
#29
d#12
dt
=,#0Dae
2
+#28k#12,k
0
#29
,
d#12
dt
+ #0Du
c
e
#01
Finding the Adjustment Law
System model
dy
dt
=,ay +ku
d#12
dt
=
Desired equilibrium
e =0
#12=#12
0
=
b
0
b
Lyapunov function
V #28e;#12#29=
#0D
2
e
2
+
k
2
#28#12,#12
0
#29
2
dV
dt
=,#0Dae
2
+#28b#12,b
0
#29
,
d#12
dt
+#0Du
c
e
#01
Choosing the adjustment rule
d#12
dt
=,#0Du
c
e
Gives
dV
dt
=,#0Dae
2
c#0D K,J,#C5str#F6m and B,Wittenmark 4
Adaptation of Feedforward Gain
Lyapunov rule
d#12
dt
=,#0Du
c
e
MIT Rule
d#12
dt
=,#0Dye
q
S
–
Model
Process
+
S
–
Model
Process
+
y
y
e
e q
u
c
u
c
kG(s)
kG(s)
k
0
G(s)
k
0
G(s)
y
m
y
m
-
g
s
-
g
s
P
P
P
P
(a)
(b)
Is there really magic in this world?
Simulation
MIT rule
0 5 10 15 20
0
1
0 5 10 15 20
0
1
0 5 10 15 20
0
1
Time
Time
Time
#28a#29
^
#12
#28b#29
^
#12
#28c#29
^
#12
Lyapunov rule
0 5 10 15 20
0
1
0 5 10 15 20
0
1
0 5 10 15 20
0
1
Time
Time
Time
#28a#29
^
#12
#28b#29
^
#12
#28c#29
^
#12
A First Order System
Process model
dy
dt
=,ay + bu
Desired response
dy
m
dt
=,a
m
y
m
+ b
m
u
c
Controller
u = #12
1
u
c
,#12
2
y
The error
e = y,y
m
de
dt
=,a
m
e,#28b#12
2
+ a,a
m
#29y +#28b#12
1
,b
m
#29u
c
Candidate forLyapunov function
V #28e;#12
1;#12
2
#29=
1
2
#12
e
2
+
1
b#0D
#28b#12
2
+ a,a
m
#29
2
+
1
b#0D
#28b#12
1
,b
m
#29
2
#13
A First Order System,cont
Derivative of Lyapunov function
dV
dt
=
e
de
dt
+
1
#0D
#28b#12
2
+a,a
m
#29
d#12
2
dt
+
1
#0D
#28b#12
1
,b
m
#29
d#12
1
dt
=,a
m
e
2
+
1
#0D
#28b#12
2
+ a,a
m
#29
#12
d#12
2
dt
,#0Dye
#13
+
1
#0D
#28b#12
1
,b
m
#29
#12
d#12
1
dt
+ #0Du
c
e
#13
Adaptation law
d#12
1
dt
=,#0Du
c
e
d#12
2
dt
= #0Dye
c#0D K,J,#C5str#F6m and B,Wittenmark 5
A First Order System,cont
The Lyapunov function
V #28e;#12
1;#12
2
#29=
1
2
#12
e
2
+
1
b#0D
#28b#12
2
+ a,a
m
#29
2
+
1
b#0D
#28b#12
1
,b
m
#29
2
#13
Its derivative
dV
dt
=,a
m
e
2
Negative semi-de#0Cnite but not negative de#0Cnite
#7F
V =,2a
m
e
de
dt
=
,2a
m
e#28,a
m
e,#28b#12
2
+a,a
m
#29y +#28b#12
1
,b
m
#29u
c
#29
Error goes to zero but parameters do not
necessarily go to their correct values!
Comparison with MIT rule
Lyapunov
-
S
P
+
e
u y
S
P
P
P
-
+
u
c
G
m
(s)
G(s)
q
1
q
2
g
s
-
g
s
MIT rule
-
S
P
+
e
u y
S
P
P
P
-
+
u
c
G
m
(s)
G(s)
q
1
q
2
g
s
-
g
s
a
m
s + a
m
a
m
s + a
m
Simulation
Process inputs and outputs
0 2040608010
1
1
0 2040608010
5
0
5
Time
Time
#28a#29
y
m
y
#28b#29
u
Parameters
0 2040608010
2
4
0 2040608010
1
1
Time
Time
#12
1
#12
2
#0D =5
#0D=1
#0D=0:2
#0D=5
#0D=1
#0D=0:2
State Feedback
Same idea
#0F Find a controller structure
#0F Derive error equation
#0F Find Lyapunov equation
c#0D K,J,#C5str#F6m and B,Wittenmark 6
State Feedback
Process model
dx
dt
= Ax+Bu
Desired response to command signals
dx
m
dt
= A
m
x
m
+ B
m
u
c
Control law
u = Mu
c
,Lx
The closed-loop system
dx
dt
=#28A,BL#29x+BMu
c
= A
c
#28#12#29x+B
c
#28#12#29u
c
Parametrization
A
c
#28#12
0
#29=A
m
B
c
#28#12
0
#29=B
m
Compatibility conditions
A,A
m
= BL
B
m
= BM
The Error Equation
Process
dx
dt
= Ax+Bu
Desired response
dx
m
dt
= A
m
x
m
+ B
m
u
c
Control law
u = Mu
c
,Lx
Error
e = x,x
m
de
dt
=
dx
dt
,
dx
m
dt
= Ax+ Bu,A
m
x
m
,B
m
u
c
Hence
de
dt
= A
m
e +#28A,A
m
,BL#29x+#28BM,B
m
#29u
c
=A
m
e+#28A
c
#28#12#29,A
m
#29x+#28B
c
#28#12#29,B
m
#29u
c
=A
m
e+#09
,
#12,#12
0
#01
The Lyapunov Function
The error equation
de
dt
= A
m
e+#09
,
#12,#12
0
#01
Try
V #28e;#12#29=
1
2
,
#0De
T
Pe+#28#12,#12
0
#29
T
#28#12,#12
0
#29
#01
Hence
dV
dt
=,
#0D
2
e
T
Qe+ #0D#28#12,#12
0
#29#09
T
Pe+#28#12,#12
0
#29
T
d#12
dt
=,
#0D
2
e
T
Qe+#28#12,#12
0
#29
T
#12
d#12
dt
+#0D#09
T
Pe
#13
where Q positive de#0Cnite and
A
T
m
P + PA
m
=,Q
Adaptation law
d#12
dt
=,#0D#09
T
Pe
gives
dV
dt
=,
#0D
2
e
T
Qe
Discuss!
Adaptation of Feedforward Gain
Error
e =#28kG#28p#29#12,k
0
G#28p#29#29u
c
= kG#28p#29#28#12,#12
0
#29u
c
Introduce a realization of G#28s#29
dx
dt
= Ax+ B#28#12,#12
0
#29u
c
e = Cx
Candidate forLyapunov function
V =
1
2
,
#0Dx
T
Px+#28#12,#12
0
#29
2
#01
where
A
T
P + PA=,Q
c#0D K,J,#C5str#F6m and B,Wittenmark 7
Adaptation of Feedforward Gain
Derivative of V
dV
dt
=
#0D
2
#12
dx
T
dt
Px+x
T
P
dx
dt
#13
+#28#12,#12
0
#29
d#12
dt
=
#0D
2
,
Ax+ Bu
c
#28#12,#12
0
#29
#01
T
Px
+
#0D
2
x
T
P
,
Ax+Bu
c
#28#12,#12
0
#29
#01
+#28#12,#12
0
#29
d#12
dt
=,
#0D
2
x
T
Qx+#28#12,#12
0
#29
#12
d#12
dt
+ #0Du
c
B
T
Px
#13
Adaptation law
d#12
dt
=,#0Du
c
B
T
Px
gives
dV
dt
=,
#0D
2
x
T
Qx
Discuss
Output Feedback
Process
dx
dt
= Ax+ B#28#12,#12
0
#29u
c
e = Cx
Adaptation law
d#12
dt
=,#0Du
c
B
T
Px
Can we #0Cnd P such that
B
T
P = C
The adaptation law then becomes
d#12
dt
=,#0Du
c
e
Kalman-YakubovichLemma
Definition 5
A rational transfer function G with real
coe#0Ecients is positive real #28PR#29 if
ReG#28s#29 #15 0 for Res #150
A transfer function G is strictly positive real
#28SPR#29 if G#28s,"#29 is positive real for some real
"#3E0.
Lemma 1
The transfer function
G#28s#29=C#28sI,A#29
,1
B
is strictly positive real if and only if there exist
positive de#0Cnite matrices P and Q such that
A
T
P + PA=,Q
and
B
T
P = C
Summary
#0F Lyapunov Stability Theory
#7B Stability concept
#7B Lyapunovs theorem
#7B How to use it?
#0F Adaptive laws with Guaranteed Stability
#0F Simple design procedure
#7B Find control law
#7B Derive Error Equation
#7B Find Lyapunov Function
#7B Choose adjustment law so that
dV=dt #14 0
#0F Remark
#7B Strong similarities with MIT rule
#7B Actually simpler
#7B No normalization
c#0D K,J,#C5str#F6m and B,Wittenmark 8
Lyapunov Stability Theory
Theme,Generate parameter adjustment rules
which guarantee stability
1,Introduction
#0F The MIT rule pros and cons
#0F The stabilityproblem
#0F Ordinary Di#0Berential Equations
#0F Dynamical Systems
2,Lyapunov's Stability Theory
#0F The stability concept
#0F Lyapunov's idea
#0F Finding Lyapunov Functions
3,Application to adaptive control
#0F A simple problem
#0F State feedback
#0F Output feedback
4,Conclusions
The Big Picture
Model-ReferenceAdaptive Systems
1,The idea
2,The MIT Rule
3,Lyapunov's stability theory
4,Design of MRAS using Lyapunov theory
5,Applications to adaptive control
6,Bounded-input,bounded-output stability
7,Applications to adaptive control
8,Relations between MRAS and STR
9,Conclusions
Lyapunov Himself
Stability Concepts
#0F Linear systems are very special
#0F Stable systems and stable solutions
#0F An example
#0F Lyapunovs stability concepts
#7B Stability of mechanical systems
#7B A state space approach
#7B The key ideas
c#0D K,J,#C5str#F6m and B,Wittenmark 1
Stable Solutions
Consider the MRAS designed with MIT rule
Model
e
y
Process
u
S
+
–
u
c
q
y
m
-
g
s
kG(s)
k
0
G(s)
P
P
u
c
= sin!t ! = 1#28a#29,2#28b#29 and 3#28c#29
0 5 10 15 20
0
1
0 5 10 15 20
0
1
0 5 10 15 20
0
1
Time
Time
Time
#28a#29
^
#12
#28b#29
^
#12
#28c#29
^
#12
Conclusion?
Preliminaries
Consider the solution x#28t#29=0to
dx
dt
= f#28x#29 f#280#29=0
Existence and uniqueness
kf#28x#29,f#28y#29k#14Lkx,yk L#3E0
#0FSensitivity to perturbations
#0F What perturbations?
#7B Initial conditions
#7B Variations in f#28x#29
Stability concepts!
Lyapunov Stability
Definition 1
The solution x#28t#29=0is stable if for given
"#3E0there exists a number #0E#28"#29 #3E 0 such that
all solutions with initial conditions
kx#280#29k #3C#0E
have the property
kx#28t#29k#3C" for 0 #14t#3C1
The solution is unstable if it is not stable,The
solution is asymptotically stable if it is stable
and #0E can be found such that all solutions with
kx#280#29k #3C#0Ehave the property that kx#28t#29k!0
as t !1.
Notice
#0F Stabilityofaparticular solution
#0F Local concept
#0F How to make it global?
Lyapunov's Idea
Inspiration from mechanics
Stable and unstable equilibria
x
1
x
2
x=0
V(x)=const
dx
dt
Condition
@V
T
@x
f#28x#29 #3C 0
c#0D K,J,#C5str#F6m and B,Wittenmark 2
Formalities
Definition 2
A continuously di#0Berentiable function V,
R
n
! R is called positive de#0Cnite in a region
U #1A R
n
containing the origin if
#0F 1,V #280#29=0
#0F2,V #28x#29 #3E 0; x 2 U and x 6=0
A function is called positive semide#0Cnite if
Condition 2 is replaced by V #28x#29 #15 0.
Theorem 1
If there exists a function V,R
n
! R that is
positive de#0Cnite such that
dV
dt
=
@V
T
@x
dx
dt
=
@V
T
@x
f#28x#29=,W#28x#29
is negative semide#0Cnite,then the solution
x#28t#29=0is stable,If dV=dt is negative de#0Cnite,
then the solution is also asymptotically stable.
FindingLyapunov Functions
"Transforming one di#0Ecult problem to an-
other"
Let the linear system
dx
dt
= Ax
be stable,Pick Q positive de#0Cnite,The
equation
A
T
P + PA=,Q
has always a unique solution with P positive
de#0Cnite and the funtion
V #28x#29=x
T
Px
is a Lyapunov function
Example
System matrix
A =
#12
a
1
a
2
a
3
a
4
#13
Q =
#12
q
1
0
0 q
2
#13
where q
1
#3E 0 and q
2
#3E 0 are positive,Assume
that the matrix P has the form
P =
#12
p
1
p
2
p
2
p
3
#13
The Lyapunov equation becomes
0
@
2a
1
2a
3
0
a
2
a
1
+ a
4
a
3
0 2a
2
2a
4
1
A
0
@
p
1
p
2
p
3
1
A
=
0
@
,q
1
0
,q
2
1
A
Conditions for existence,When is P#3E0?
Proofs are given in the mathematics course on
matrices!
Time-Varying Systems
dx
dt
= f#28x;t#29
Definition 3
The solution x#28t#29=0is uniformly stable if
for "#3E0there exists a number #0E#28"#29 #3E 0,
independent of t
0
,such that
kx#28t
0
#29k #3C#0E#29kx#28t#29k#3C" 8t#15t
0
#150
The solution is uniformly asymptotically stable
if it is uniformly stable and there is c#3E0,
independent of t
0
,such that x#28t#29 ! 0 as
t !1,uniformly in t
0
,for all kx#28t
0
#29k #3Cc.
Definition 4
A continuous function #0B,#5B0;a#29 ! #5B0;1#29
is said to belong to class K if it is strictly
increasing and #0B#280#29 = 0,It is said to belong
to class K
1
if a = 1 and #0B#28r#29 !1as
r !1.
c#0D K,J,#C5str#F6m and B,Wittenmark 3
Lyapunov Theorem
Theorem 2
Let x = 0 be an equilibrium point and
D = fx 2 R
n
jkxk #3Crg,Let V be a
continuously di#0Berentiable function such that
#0B
1
#28kxk#29 #14V #28x;t#29#14 #0B
2
#28kxk#29
dV
dt
=
@V
@t
+
@V
@x
f#28x;t#29 #14,#0B
3
#28kxk#29
for 8t #15 0,where #0B
1
,#0B
2
,and #0B
3
are class
K functions,Then x = 0 is uniformly
asymptotically stable.
Design of MRAS using Lyapunov
Theorem
#0F The idea
#7B Determine a controller structure
#7B Derive the Error Equation
#7B Find a Lyapunov function
#7B Determine an adaptation law
#0F A #0Crst order system
#0F State feedback
#0F Output feedback
Adaptation of Feedforward Gain
Process model
dy
dt
=,ay + ku
Desired response
dy
m
dt
=,y
m
+ k
o
u
c
Controller
u = #12u
c
Introduce the error e = y,y
m
and the error
equation becomes
de
dt
=,ae+#28k#12,k
o
#29u
c
Candidate Lyapunov function
V #28e;#12#29=
#0D
2
e
2
+
k
2
#28#12,
k
0
k
#29
2
Time derivative
dV
dt
= #0De#28,ae + k#28#12,
k
0
k
#29u
c
#29+b#28#12,
k
0
k
#29
d#12
dt
=,#0Dae
2
+#28k#12,k
0
#29
,
d#12
dt
+ #0Du
c
e
#01
Finding the Adjustment Law
System model
dy
dt
=,ay +ku
d#12
dt
=
Desired equilibrium
e =0
#12=#12
0
=
b
0
b
Lyapunov function
V #28e;#12#29=
#0D
2
e
2
+
k
2
#28#12,#12
0
#29
2
dV
dt
=,#0Dae
2
+#28b#12,b
0
#29
,
d#12
dt
+#0Du
c
e
#01
Choosing the adjustment rule
d#12
dt
=,#0Du
c
e
Gives
dV
dt
=,#0Dae
2
c#0D K,J,#C5str#F6m and B,Wittenmark 4
Adaptation of Feedforward Gain
Lyapunov rule
d#12
dt
=,#0Du
c
e
MIT Rule
d#12
dt
=,#0Dye
q
S
–
Model
Process
+
S
–
Model
Process
+
y
y
e
e q
u
c
u
c
kG(s)
kG(s)
k
0
G(s)
k
0
G(s)
y
m
y
m
-
g
s
-
g
s
P
P
P
P
(a)
(b)
Is there really magic in this world?
Simulation
MIT rule
0 5 10 15 20
0
1
0 5 10 15 20
0
1
0 5 10 15 20
0
1
Time
Time
Time
#28a#29
^
#12
#28b#29
^
#12
#28c#29
^
#12
Lyapunov rule
0 5 10 15 20
0
1
0 5 10 15 20
0
1
0 5 10 15 20
0
1
Time
Time
Time
#28a#29
^
#12
#28b#29
^
#12
#28c#29
^
#12
A First Order System
Process model
dy
dt
=,ay + bu
Desired response
dy
m
dt
=,a
m
y
m
+ b
m
u
c
Controller
u = #12
1
u
c
,#12
2
y
The error
e = y,y
m
de
dt
=,a
m
e,#28b#12
2
+ a,a
m
#29y +#28b#12
1
,b
m
#29u
c
Candidate forLyapunov function
V #28e;#12
1;#12
2
#29=
1
2
#12
e
2
+
1
b#0D
#28b#12
2
+ a,a
m
#29
2
+
1
b#0D
#28b#12
1
,b
m
#29
2
#13
A First Order System,cont
Derivative of Lyapunov function
dV
dt
=
e
de
dt
+
1
#0D
#28b#12
2
+a,a
m
#29
d#12
2
dt
+
1
#0D
#28b#12
1
,b
m
#29
d#12
1
dt
=,a
m
e
2
+
1
#0D
#28b#12
2
+ a,a
m
#29
#12
d#12
2
dt
,#0Dye
#13
+
1
#0D
#28b#12
1
,b
m
#29
#12
d#12
1
dt
+ #0Du
c
e
#13
Adaptation law
d#12
1
dt
=,#0Du
c
e
d#12
2
dt
= #0Dye
c#0D K,J,#C5str#F6m and B,Wittenmark 5
A First Order System,cont
The Lyapunov function
V #28e;#12
1;#12
2
#29=
1
2
#12
e
2
+
1
b#0D
#28b#12
2
+ a,a
m
#29
2
+
1
b#0D
#28b#12
1
,b
m
#29
2
#13
Its derivative
dV
dt
=,a
m
e
2
Negative semi-de#0Cnite but not negative de#0Cnite
#7F
V =,2a
m
e
de
dt
=
,2a
m
e#28,a
m
e,#28b#12
2
+a,a
m
#29y +#28b#12
1
,b
m
#29u
c
#29
Error goes to zero but parameters do not
necessarily go to their correct values!
Comparison with MIT rule
Lyapunov
-
S
P
+
e
u y
S
P
P
P
-
+
u
c
G
m
(s)
G(s)
q
1
q
2
g
s
-
g
s
MIT rule
-
S
P
+
e
u y
S
P
P
P
-
+
u
c
G
m
(s)
G(s)
q
1
q
2
g
s
-
g
s
a
m
s + a
m
a
m
s + a
m
Simulation
Process inputs and outputs
0 2040608010
1
1
0 2040608010
5
0
5
Time
Time
#28a#29
y
m
y
#28b#29
u
Parameters
0 2040608010
2
4
0 2040608010
1
1
Time
Time
#12
1
#12
2
#0D =5
#0D=1
#0D=0:2
#0D=5
#0D=1
#0D=0:2
State Feedback
Same idea
#0F Find a controller structure
#0F Derive error equation
#0F Find Lyapunov equation
c#0D K,J,#C5str#F6m and B,Wittenmark 6
State Feedback
Process model
dx
dt
= Ax+Bu
Desired response to command signals
dx
m
dt
= A
m
x
m
+ B
m
u
c
Control law
u = Mu
c
,Lx
The closed-loop system
dx
dt
=#28A,BL#29x+BMu
c
= A
c
#28#12#29x+B
c
#28#12#29u
c
Parametrization
A
c
#28#12
0
#29=A
m
B
c
#28#12
0
#29=B
m
Compatibility conditions
A,A
m
= BL
B
m
= BM
The Error Equation
Process
dx
dt
= Ax+Bu
Desired response
dx
m
dt
= A
m
x
m
+ B
m
u
c
Control law
u = Mu
c
,Lx
Error
e = x,x
m
de
dt
=
dx
dt
,
dx
m
dt
= Ax+ Bu,A
m
x
m
,B
m
u
c
Hence
de
dt
= A
m
e +#28A,A
m
,BL#29x+#28BM,B
m
#29u
c
=A
m
e+#28A
c
#28#12#29,A
m
#29x+#28B
c
#28#12#29,B
m
#29u
c
=A
m
e+#09
,
#12,#12
0
#01
The Lyapunov Function
The error equation
de
dt
= A
m
e+#09
,
#12,#12
0
#01
Try
V #28e;#12#29=
1
2
,
#0De
T
Pe+#28#12,#12
0
#29
T
#28#12,#12
0
#29
#01
Hence
dV
dt
=,
#0D
2
e
T
Qe+ #0D#28#12,#12
0
#29#09
T
Pe+#28#12,#12
0
#29
T
d#12
dt
=,
#0D
2
e
T
Qe+#28#12,#12
0
#29
T
#12
d#12
dt
+#0D#09
T
Pe
#13
where Q positive de#0Cnite and
A
T
m
P + PA
m
=,Q
Adaptation law
d#12
dt
=,#0D#09
T
Pe
gives
dV
dt
=,
#0D
2
e
T
Qe
Discuss!
Adaptation of Feedforward Gain
Error
e =#28kG#28p#29#12,k
0
G#28p#29#29u
c
= kG#28p#29#28#12,#12
0
#29u
c
Introduce a realization of G#28s#29
dx
dt
= Ax+ B#28#12,#12
0
#29u
c
e = Cx
Candidate forLyapunov function
V =
1
2
,
#0Dx
T
Px+#28#12,#12
0
#29
2
#01
where
A
T
P + PA=,Q
c#0D K,J,#C5str#F6m and B,Wittenmark 7
Adaptation of Feedforward Gain
Derivative of V
dV
dt
=
#0D
2
#12
dx
T
dt
Px+x
T
P
dx
dt
#13
+#28#12,#12
0
#29
d#12
dt
=
#0D
2
,
Ax+ Bu
c
#28#12,#12
0
#29
#01
T
Px
+
#0D
2
x
T
P
,
Ax+Bu
c
#28#12,#12
0
#29
#01
+#28#12,#12
0
#29
d#12
dt
=,
#0D
2
x
T
Qx+#28#12,#12
0
#29
#12
d#12
dt
+ #0Du
c
B
T
Px
#13
Adaptation law
d#12
dt
=,#0Du
c
B
T
Px
gives
dV
dt
=,
#0D
2
x
T
Qx
Discuss
Output Feedback
Process
dx
dt
= Ax+ B#28#12,#12
0
#29u
c
e = Cx
Adaptation law
d#12
dt
=,#0Du
c
B
T
Px
Can we #0Cnd P such that
B
T
P = C
The adaptation law then becomes
d#12
dt
=,#0Du
c
e
Kalman-YakubovichLemma
Definition 5
A rational transfer function G with real
coe#0Ecients is positive real #28PR#29 if
ReG#28s#29 #15 0 for Res #150
A transfer function G is strictly positive real
#28SPR#29 if G#28s,"#29 is positive real for some real
"#3E0.
Lemma 1
The transfer function
G#28s#29=C#28sI,A#29
,1
B
is strictly positive real if and only if there exist
positive de#0Cnite matrices P and Q such that
A
T
P + PA=,Q
and
B
T
P = C
Summary
#0F Lyapunov Stability Theory
#7B Stability concept
#7B Lyapunovs theorem
#7B How to use it?
#0F Adaptive laws with Guaranteed Stability
#0F Simple design procedure
#7B Find control law
#7B Derive Error Equation
#7B Find Lyapunov Function
#7B Choose adjustment law so that
dV=dt #14 0
#0F Remark
#7B Strong similarities with MIT rule
#7B Actually simpler
#7B No normalization
c#0D K,J,#C5str#F6m and B,Wittenmark 8
