From f86de943734e10fa90fe63ed8431088839f5abde Mon Sep 17 00:00:00 2001 From: SpaceOddity404 <40719093+SpaceOddity404@users.noreply.github.com> Date: Fri, 17 Apr 2020 21:40:05 -0400 Subject: Add files via upload --- execsumm/ExecutiveSummaryDraft.tex | 41 ++++++++++++++++++++++++++++++++++++++ 1 file changed, 41 insertions(+) create mode 100644 execsumm/ExecutiveSummaryDraft.tex diff --git a/execsumm/ExecutiveSummaryDraft.tex b/execsumm/ExecutiveSummaryDraft.tex new file mode 100644 index 0000000..498aac9 --- /dev/null +++ b/execsumm/ExecutiveSummaryDraft.tex @@ -0,0 +1,41 @@ +\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