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beamerExample
Commits
10acdf5b
Commit
10acdf5b
authored
May 04, 2022
by
Leif Andersson
Browse files
Förkortat beamerExample.tex för att det skall passa med eightbeamer
parent
f7786014
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2
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10acdf5b
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...
@@ -17,9 +17,9 @@
*.xdv
*-converted-to.*
# these rules might exclude image files for figures etc.
#
*.ps
#
*.eps
#
*.pdf
*.ps
*.eps
*.pdf
## Generated if empty string is given at "Please type another file name for output:"
.pdf
...
...
beamerExample.tex
View file @
10acdf5b
...
...
@@ -34,7 +34,7 @@
% ----------------------------------------------------------------------
\begin{frame}
{
\hspace*
{
10pt
}
Alexandr Mihailovich Lyapunov (1857--1918)
}
\begin{frame}
{
Alexandr Mihailovich Lyapunov (1857--1918)
}
\begin{columns}
\begin{column}
{
0.3
\textwidth
}
\includegraphics
[width=\textwidth]
{
figures/lyapunov
_
stamp
}
...
...
@@ -334,270 +334,4 @@ then $x=0$ is unstable.
\end{itemize}
\end{frame}
% ----------------------------------------------------------------------
\begin{frame}
%
\frametitle
{
\large
Proof of (1) in Lyapunov's Linearization Method
}
Lyapunov function candidate
$
V
(
x
)=
x
^
TPx
$
.
$
V
(
0
)=
0
$
,
$
V
(
x
)
>
0
$
for
$
x
\neq
0
$
, and
\begin{align*}
\dot
{
V
}
(x)
&
=x
^
TPf(x)+f
^
T(x)Px
\\
&
=x
^
TP[Ax+g(x)]+[x
^
TA+g
^
T(x)]Px
\\
&
=x
^
T(PA+A
^
TP)x+2x
^
TPg(x)=-x
^
TQx+2x
^
TPg(x)
\end{align*}
%
\begin{equation*}
x
^
TQx
\geq\lambda
_{
\min
}
(Q)
\Vert
x
\Vert
^
2
\end{equation*}
%
and for all
$
\gamma
>
0
$
there exists
$
r>
0
$
such that
%
\begin{equation*}
\Vert
g(x)
\Vert
<
\gamma\Vert
x
\Vert
,
\qquad
\forall\Vert
x
\Vert
<r
\end{equation*}
%
Thus, choosing
$
\gamma
$
sufficiently small gives
%
\begin{equation*}
\dot
{
V
}
(x)
\leq
-
\big
(
\lambda
_{
\min
}
(Q)-2
\gamma\lambda
_{
\max
}
(P)
\big
)
\Vert
x
\Vert
^
2<0
\end{equation*}
%
\end{frame}
% ----------------------------------------------------------------------
\end{document}
\begin{frame}
%
\frametitle
{
Invariant Sets
}
\textbf
{
Definition
}
A set
$
M
$
is called
\textbf
{
invariant
}
if for
the system
$$
\dot
{
x
}
=
f
(
x
)
,
$$
$
x
(
0
)
\in
M
$
implies that
$
x
(
t
)
\in
M
$
for all
$
t
\geq
0
$
.
\medskip
\begin{center}
\psfrag
{
x0
}
[][]
{$
x
(
0
)
$}
\psfrag
{
xt
}
[][]
{$
x
(
t
)
$}
\psfrag
{
M
}
[][]
{$
M
$}
\includegraphics
[width=0.5\hsize]
{
figures/invariant
_
set.eps
}
\end{center}
\end{frame}
% ----------------------------------------------------------------------
\begin{frame}
%
\frametitle
{
Invariant Set Theorem
}
\textbf
{
Theorem
}
Let
$
\Omega\in\mathbf
{
R
}^
n
$
be a bounded and closed set
that is invariant with respect to
$$
\dot
{
x
}
=
f
(
x
)
.
$$
Let
$
V:
\mathbf
{
R
}^
n
\rightarrow\mathbf
{
R
}$
be a radially unbounded
$
C
^
1
$
function such that
$
\dot
{
V
}
(
x
)
\leq
0
$
for
$
x
\in\Omega
$
. Let
$
E
$
be the set of points
in
$
\Omega
$
where
$
\dot
{
V
}
(
x
)=
0
$
. If
$
M
$
is the largest invariant set in
$
E
$
, then every solution with
$
x
(
0
)
\in\Omega
$
approaches
$
M
$
as
$
t
\rightarrow\infty
$
(proof on p.~73)
\begin{center}
\psfrag
{
Om
}
[][]
{$
\Omega
$}
\psfrag
{
E
}
[][]
{$
E
$}
\psfrag
{
M
}
[][]
{$
M
$}
\includegraphics
[width=0.4\hsize]
{
figures/lyap
_
invariant.eps
}
\end{center}
\end{frame}
%----------------------------------------------------------------------
\begin{frame}
%
\frametitle
{
Example---Stable Limit Cycle
}
Show that
$
M
=
\{
x:
\,\Vert
x
\Vert
=
1
\}
$
is a globally stable limit cycle for
\begin{align*}
\dot
{
x
}_
1
&
=x
_
1-x
_
2-x
_
1(x
_
1
^
2+x
_
2
^
2)
\\
\dot
{
x
}_
2
&
=x
_
1+x
_
2-x
_
2(x
_
1
^
2+x
_
2
^
2)
\end{align*}
Let
$
V
(
x
)=(
x
_
1
^
2
+
x
_
2
^
2
-
1
)
^
2
$
.
\begin{align*}
\frac
{
dV
}{
dt
}&
=2(x
_
1
^
2+x
_
2
^
2-1)
\frac
{
d
}{
dt
}
(x
_
1
^
2+x
_
2
^
2-1)
\\
&
=-2(x
_
1
^
2+x
_
2
^
2-1)
^
2(x
_
1
^
2+x
_
2
^
2)
\leq
0
\quad\text
{
for
}
x
\in\Omega\\
\end{align*}
$
\Omega
=
\{
0
<
\Vert
x
\Vert\leq
R
\}
$
is invariant for
$
R
=
1
$
.
\end{frame}
%----------------------------------------------------------------------
\begin{frame}
%
\frametitle
{
Example---Stable Limit Cycle
}
\begin{equation*}
E=
\{
x
\in\Omega
:
\,\dot
{
V
}
(x)=0
\}
=
\{
x:
\,\Vert
x
\Vert
= 1
\}
\end{equation*}
$
M
=
E
$
is an invariant set, because
\begin{equation*}
\frac
{
d
}{
dt
}
V=-2(x
_
1
^
2+x
_
2
^
2-1)(x
_
1
^
2+x
_
2
^
2)=0
\quad\text
{
for
}
x
\in
M
\end{equation*}
We have shown that
$
M
$
is a stable limit cycle (globally stable in
$
R
-
\{
0
\}
$
)
\begin{center}
\includegraphics
[height=0.3\hsize]
{
figures/guck
}
\end{center}
\end{frame}
% ----------------------------------------------------------------------
\begin{frame}
%
\frametitle
{
A Motivating Example (cont'd)
}
$$
m
\ddot
{
x
}
=-
b
\dot
{
x
}
\vert\dot
{
x
}
\vert
-
k
_
0
x
-
k
_
1
x
^
3
$$
$$
V
(
x,
\dot
{
x
}
)=(
2
m
\dot
{
x
}^
2
+
2
k
_
0
x
^
2
+
k
_
1
x
^
4
)/
4
>
0
,
\qquad
V
(
0
,
0
)=
0
$$
$
\dot
{
V
}
(
x,
\dot
{
x
}
)=-
b
\vert\dot
{
x
}
\vert
^
3
$
gives
$
E
=
\{
(
x,
\dot
{
x
}
)
:
\,\dot
{
x
}
=
0
\}
$
.
Assume there exists
$
(
\bar
{
x
}
,
\dot
{
\bar
{
x
}}
)
\in
M
$
such that
$
\bar
{
x
}
(
t
_
0
)
\neq
0
$
. Then
$$
m
\ddot
{
\bar
{
x
}}
(
t
_
0
)=-
k
_
0
\bar
{
x
}
(
t
_
0
)-
k
_
1
\bar
{
x
}^
3
(
t
_
0
)
\neq
0
$$
so
$
\dot
{
\bar
{
x
}}
(
t
_
0
+)
\neq
0
$
so the trajectory will immediately
leave
$
M
$
. A contradiction to that
$
M
$
is invariant.
Hence,
$
M
=
\{
(
0
,
0
)
\}
$
so the origin is asymptotically stable.
\end{frame}
%----------------------------------------------------------------------
\begin{frame}
%
\frametitle
{
Adaptive Noise Cancellation by Lyapunov Design
}
\begin{center}
\psfrag
{
u
}
[][]
{$
u
$}
\psfrag
{
g1
}
[][][1.4]
{$
\frac
{
b
}{
s
+
a
}$}
\psfrag
{
g2
}
[][][1.4]
{$
\frac
{
\widehat
b
}{
s
+
\widehat
a
}$}
\psfrag
{
x
}
[][]
{$
x
$}
\psfrag
{
xh
}
[][]
{$
\widehat
x
$}
\psfrag
{
xt
}
[][]
{$
\widetilde
x
$}
\psfrag
{
+
}
[][]
{$
+
$}
\psfrag
{
-
}
[][]
{$
-
$}
\includegraphics
[width=0.4\hsize]
{
figures/noise.eps
}
\end{center}
\begin{align*}
\dot
x +a x
&
= bu
\\
\dot
{
\widehat
x
}
+
\widehat
a
\widehat
x
&
=
\widehat
b u
\end{align*}
Introduce
$
\widetilde
x
=
x
-
\widehat
x,
\enskip\widetilde
a
=
a
-
\widehat
a,
\enskip
\widetilde
b
=
b
-
\widehat
b
$
.
Want to design adaptation law so that
$
\widetilde
x
\to
0
$
\end{frame}
\begin{frame}
Let us try the Lyapunov function
\begin{align*}
V
&
=
\frac
{
1
}{
2
}
(
\widetilde
x
^
2+
\gamma
_
a
\widetilde
a
^
2+
\gamma
_
b
\widetilde
b
^
2)
\\
\dot
V
&
=
\widetilde
x
\dot
{
\widetilde
x
}
+
\gamma
_
a
\widetilde
a
\dot
{
\widetilde
a
}
+
\gamma
_
b
\widetilde
b
\dot
{
\widetilde
b
}
=
\\
&
=
\widetilde
x(-a
\widetilde
x-
\widetilde
a
\widehat
x +
\widetilde
b
u) +
\gamma
_
a
\widetilde
a
\dot
{
\widetilde
a
}
+
\gamma
_
b
\widetilde
b
\dot
{
\widetilde
b
}
= -a
\widetilde
x
^
2
\end{align*}
where the last equality follows if we choose
\begin{align*}
\dot
{
\widetilde
a
}
= -
\dot
{
\widehat
a
}
=
\frac
{
1
}{
\gamma
_
a
}
\widetilde
{
x
}
\widehat
x
\qquad
\dot
{
\widetilde
b
}
= -
\dot
{
\widehat
b
}
= -
\frac
{
1
}{
\gamma
_
b
}
\widetilde
{
x
}
u
\end{align*}
Invariant set:
$
\widetilde
x
=
0
$
.
This proves that
$
\widetilde
x
\to
0
$
.
(The parameters
$
\widetilde
a
$
and
$
\widetilde
b
$
do not necessarily converge:
$
u
\equiv
0
$
.)
\fbox
{
Demonstration if time permits
}
\end{frame}
%%------------------------------
\begin{frame}
\frametitle
{
Results
}
\begin{center}
\psfrag
{
p1
}
[][][1.3]
{$
\hat
{
a
}$}
\psfrag
{
p2
}
[][][1.3]
{$
\hat
{
b
}$}
\includegraphics
[width=0.7\hsize]
{
musik/parameters.eps
}
\\
Estimation of parameters starts at t=10 s.
\end{center}
\end{frame}
%----------------------------------------------------------------------
\begin{frame}
\frametitle
{
Results
}
\begin{center}
\psfrag
{
x-xhat
}
[][][1.3]
{$
x
-
\hat
{
x
}$}
\includegraphics
[width=0.6\hsize,angle=0]
{
musik/adap.ps
}
\\
\includegraphics
[width=0.3\hsize]
{
musik/xt.eps
}
\end{center}
\begin{center}
\begin{small}
Estimation of parameters starts at t=10 s.
\end{small}
\end{center}
\end{frame}
%----------------------------------------------------------------------
\begin{frame}
%
\frametitle
{
Next Lecture
}
\begin{center}
Stability analysis using input-output (frequency) methods
\end{center}
\begin{itemize}
item Stability analysis using input-output (frequency) methods
\end{itemize}
\vfill
\end{frame}
%----------------------------------------------------------------------
\begin{frame}
\frametitle
{
title
}
Let us try the Lyapunov function
\begin{align*}
V
&
=
\frac
{
1
}{
2
}
(
\widetilde
x
^
2+
\gamma
_
a
\widetilde
a
^
2+
\gamma
_
b
\widetilde
b
^
2)
\\
\dot
V
&
=
\widetilde
x
\dot
{
\widetilde
x
}
+
\gamma
_
a
\widetilde
a
\dot
{
\widetilde
a
}
+
\gamma
_
b
\widetilde
b
\dot
{
\widetilde
b
}
=
\\
&
=
\widetilde
x(-a
\widetilde
x-
\widetilde
a
\widehat
x +
\widetilde
b
u) +
\gamma
_
a
\widetilde
a
\dot
{
\widetilde
a
}
+
\gamma
_
b
\widetilde
b
\dot
{
\widetilde
b
}
= -a
\widetilde
x
^
2
\end{align*}
where the last equality follows if we choose
\begin{align*}
\dot
{
\widetilde
a
}
= -
\dot
{
\widehat
a
}
=
\frac
{
1
}{
\gamma
_
a
}
\widetilde
{
x
}
\widehat
x
\qquad
\dot
{
\widetilde
b
}
= -
\dot
{
\widehat
b
}
= -
\frac
{
1
}{
\gamma
_
b
}
\widetilde
{
x
}
u
\end{align*}
Invariant set:
$
\widetilde
x
=
0
$
.
This proves that
$
\widetilde
x
\to
0
$
.
(The parameters
$
\widetilde
a
$
and
$
\widetilde
b
$
do not necessarily converge:
$
u
\equiv
0
$
.)
\fbox
{
Demonstration if time permits
}
\end{frame}
\ No newline at end of file
\end{document}
\ No newline at end of file
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