Förkortat beamerExample.tex för att det skall passa med eightbeamer

.gitignore

@@ -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

@@ -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)}
-
-$$
-m\ddot{x}=-b\dot{x}\vert\dot{x}\vert-k_0x-k_1x^3
-$$
-$$
-V(x,\dot{x})=(2m\dot{x}^2+2k_0x^2+k_1x^4)/4>0, \qquad V(0,0)=0
-$$
-$\dot{V}(x,\dot{x})=-b\vert\dot{x}\vert^3$ 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