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

(limited to '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 

