From 7d32b4b78455cf2bd0f37d07f976e04e249a8923 Mon Sep 17 00:00:00 2001 From: Holden Rohrer Date: Fri, 17 Apr 2020 22:31:14 -0400 Subject: python makes png --- execsumm/ExecutiveSummaryDraft.tex | 41 -------------------------------------- 1 file changed, 41 deletions(-) delete mode 100644 execsumm/ExecutiveSummaryDraft.tex (limited to 'execsumm') diff --git a/execsumm/ExecutiveSummaryDraft.tex b/execsumm/ExecutiveSummaryDraft.tex deleted file mode 100644 index 498aac9..0000000 --- a/execsumm/ExecutiveSummaryDraft.tex +++ /dev/null @@ -1,41 +0,0 @@ -\documentclass{article} -\usepackage{hyperref} -\usepackage[scr]{rsfso} -\def\rload{R_{\rm load}} -\date{} -\begin{document} -\title{Project Executive Summary} -\author{Holden Rohrer and Nithya Jayakumar} - -\maketitle -\section{Matrix Representation and Homogeneous Solution} - -To determine the relevant properties of the linear system, matrix form -is useful (this form was chosen to reduce fractions' usage): -\def\x{{\bf x}} -$$\x' = -{1\over R_1C_1C_2\rload} -\pmatrix{0&-C_2\rload &0 \cr - 0&-C_2(R_1+\rload)&C_1R_1\cr - 0&C_2R_1 &-C_1R_1} \x + -{1\over R_1} -\pmatrix{\omega\cos(\omega t)\cr - \omega\cos(\omega t)\cr - 0} -.$$ -\section{Application of Laplace Transformation} -We can apply the Laplace Transformation in order to solve this system of differential equations. -We have the three equations for $x'$, $y'$, and $z'$, and we can take the Laplace Transform of each of these equations" -$$\mathscr{L}\{x' = \frac{-y}{C_1R_1} + \frac{\omega\cos(\omega t)}{R_1}\} \Rightarrow sX(s) - x(0) = \frac{Y(s)}{C_1R_1} + \frac{\omega s}{R_1(s^2 + \omega^2)}$$ -$$\mathscr{L}\{y' = y\frac{-R_1 - \rload}{R_1C_1\rload} + \frac{z}{C_2\rload} + \frac{\omega\cos(\omega t)}{R_1} \} \Rightarrow sY(s) - y(0) = Y(s)(\frac{-R_1-\rload}{R_1C_1\rload}) + \frac{Z(s)}{C_2\rload} + \frac{\omega s}{R_1(s^2 + \omega^2)}$$ -$$\mathscr{L}\{z' = \frac{y}{C_1\rload} - \frac{z}{C_2\rload} \} \Rightarrow sZ(s) - z(0) = \frac{Y(s)}{C_1\rload} - \frac{Z(s)}{C_2\rload}$$ - -The last two equations we get can be used to solve for $Z(s)$, which we find to be $$Z(s) = \frac{\omega s(C_1C_2\rload^2)}{(s^2 + \omega^2)(s^2 + s(C_1R_1\rload + R_1C_2\rload + \rload^2C_2) + \rload)}$$ - -We can now find the partial fraction decomposition of this: -$$Z(s) = \frac{\omega s(C_1C_2\rload^2)}{(s^2 + \omega^2)(s^2 + s(C_1R_1\rload + R_1C_2\rload + \rload^2C_2) + \rload)} = $$ $$\frac{As + B}{s^2 + \omega^2} + \frac{Cs + D}{(s^2 + \omega^2)(s^2 + s(C_1R_1\rload + R_1C_2\rload + \rload^2C_2) + \rload)} = \omega sC_1C_2\rload^2$$ -To simplify notation, let $b = (s^2 + \omega^2)(s^2 + s(C_1R_1\rload + R_1C_2\rload + \rload^2C_2) + \rload)$ -We find that $$A = \frac{C_1C_2\rload^2\omega (\rload - \omega^2)}{b^2\omega^2 + \rload^2 - 2\rload - 2\rload\omega^2 + \omega^2}$$ -$$B = \frac{bC_1C_2\rload^2\omega^3}{b^2\omega^2 + \rload^2 - 2\rload\omega^2 + \omega^4}$$ -$$C = \frac{-C_1C_2\rload^2\omega(\rload - \omega^2)}{b^2\omega^2 + \rload^2 - 2\rload\omega^2 + \omega^4}$$ -\end{document} \ No newline at end of file -- cgit