# tdrhq/cis540

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 \documentclass{article} \usepackage{amsmath} \newcommand{\reducesto}{{\le}_m^p} \newcommand{\argmax}{\text{argmax}} \newcommand{\TWOSAT}{{\lang{2SAT}}} \newcommand{\sC}{\mathcal{C}} % \usepackage{fullpage} \usepackage{graphicx} \usepackage{enumerate} \usepackage{fancyhdr} \pagestyle{fancy} %% L/C/R denote left/center/right header (or footer) elements %% E/O denote even/odd pages %% \leftmark, \rightmark are chapter/section headings generated by the %% book document class \fancyhead[R]{\slshape Ritika Goel, Arnold Noronha} \fancyhead[L]{\slshape CIS-540, Homework 4} \fancyfoot[L]{} \fancyfoot[C]{} \fancyfoot[R]{\slshape Page \thepage \text{ of 4}} %\pagestyle{fancy} \begin{document} \title{CIS540, Homework 4}\author{Ritika Goel, Arnold Noronha } \maketitle \section{} \subsection*{(a)} The state is: $$x(t) = \left[ \begin{array}{cc} \varphi(t) \\ \dot{\varphi}(t) \end{array} \right]$$ The transition functions are, \begin{eqnarray} f_1 (x, u) &=& x_2 \\ f_2 (x, u) &=& \frac{g}{l} \sin(x_1) - u \end{eqnarray} \subsection*{(b)} Since the system is in equilibrium, we have two conditions, \begin{eqnarray} f_1 (x, 0) &=& 0 \\ f_2 (x, 0) &=& 0 \end{eqnarray} The first condition gives $x_2 = \dot{\varphi}(t) = 0$, and the second condition gives, $g\sin x_1 /l = 0$ or $x_1 = \varphi(t) = 0, n\pi$. For the case, $\langle 0, 0\rangle$, if there is a small deviation in $\varphi$ and velocity is zero, then the pendulum can never reach the position $\langle 0,0\rangle$, because this equilibrium position has a higher potential energy than the initial position. For $\langle \pi, 0\rangle$, the energy of the system is 0, however, any epsilon deviation has some potential energy in it, so if it ever reaches $\varphi = \pi$, the velocity cannot be zero at the same time. \subsection*{(c)} The linearized differential equation: $$-ml^2 \ddot {\varphi} (t) + mgl \varphi(t) = u(t).$$ This corresonds to the following update function: $$f_2 (x,u) = \frac{gx_1}{l} - \frac{u}{ml^2}.$$ In other words, we can write the transition as: $$\dot{x}(t) = \left[ \begin{array}{cc} 0 & 1\\ g/l & 0 \end{array} \right] x(t) + \left[\begin{array}{cc} 0 \\ -1/ml^2 \end{array}\right] u(t) .$$ \subsection*{(d)} If we set $u(t) = \alpha \varphi(t) + \beta \dot{\varphi}(t)$, then the transition becomes: $$\dot{x}(t) = \left[ \begin{array}{cc} 0 & 1\\ g/l - \alpha/ml^2 & -\beta/ml^2 \end{array} \right] x(t).$$ The eigenvalues can be of this equation can be computed as the solution of \begin{eqnarray} \det\left| \begin{array}{cc} 0 - \lambda & 1\\ g/l - \alpha/ml^2 & -\beta/ml^2-\lambda \end{array} \right| &=& 0, \\ \lambda^2 + \frac{\beta}{ml^2}\lambda + \left(\frac{\alpha}{ml^2} - \frac{g}{l}\right) &=& 0, \end{eqnarray} From which we get, $$\lambda = \frac{1}{2} \left(\frac{-\beta}{ml^2} \pm \sqrt {\frac{\beta^2}{m^2l^4} - 4 \left(\frac{\alpha}{ml^2} - \frac{g}{l}\right)}\right).$$ For the real parts of both solutions to be negative, clearly $\beta > 0$. Now we have two cases: \begin{enumerate} \item if solutions are real, i.e., $\frac{\beta^2}{m^2l^4} - 4 \left(\frac{\alpha}{ml^2} - \frac{g}{l}\right) \ge 0,$ then we can we just need to test that, $\frac{\beta^2}{m^2l^4} > \frac{\beta^2}{m^2l^4} - 4 \left(\frac{\alpha}{ml^2} - \frac{g}{l}\right)$. From these two inequalities we get $$mgl <\alpha \le mgl + \frac{\beta^2}{4ml^2} .$$ \item if the solutions are complex: the real part is always negative if $\beta > 0$. But this case only if $\alpha > mgl + \frac{\beta^2}{4ml^2}$. \end{enumerate} Thus the system is stable for all $\beta > 0$ and $\alpha > mgl$. \end{document}