... ... @@ -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\Vert0, \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