From c67d6e8264b928c3e3d1bd3b353c787f42793be5 Mon Sep 17 00:00:00 2001
From: SpaceOddity404 <40719093+SpaceOddity404@users.noreply.github.com>
Date: Tue, 14 Apr 2020 22:31:18 -0400
Subject: Add files via upload

---
 execsumm/ExecutiveSummaryDraft.tex | 87 ++++++++++++++++++++++++++++++++++++++
 1 file changed, 87 insertions(+)
 create mode 100644 execsumm/ExecutiveSummaryDraft.tex

diff --git a/execsumm/ExecutiveSummaryDraft.tex b/execsumm/ExecutiveSummaryDraft.tex
new file mode 100644
index 0000000..a9c133b
--- /dev/null
+++ b/execsumm/ExecutiveSummaryDraft.tex
@@ -0,0 +1,87 @@
+\documentclass{article}
+\usepackage{hyperref}
+\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}}
+$$\x' =
+{1\over R_1C_1C_2\rload}
+\pmatrix{0&-C_2\rload      &0 \cr
+         0&-C_2(R_1+\rload)&C_1R_1\cr
+         0&C_2R_1          &-C_1R_1}   \x +
+{1\over R_1}
+\pmatrix{\omega\cos(\omega t)\cr
+         \omega\cos(\omega t)\cr
+         0}
+.$$
+The characteristic polynomial is
+$$-\lambda( (-\lambda-C_2(R_1+\rload))(-\lambda-C_1R_1)
+- C_2C_1R_1^2).$$
+Expanded,
+$$-\lambda( \lambda^2 + \lambda(C_2(R_1+\rload)+C_1R_1)
++ C_1C_2R_1\rload).$$
+In terms of its roots (with $b=C_2(R_1+\rload)+C_1R_1$ and
+$c = C_1C_2R_1\rload,$
+$$-\lambda{(\lambda-{-b+\sqrt{b^2-4c}\over2})
+(\lambda-{-b-\sqrt{b^2-4c}\over 2})}$$
+%For reference,
+%$$b^2-4c = C_2^2(R_1+\rload)^2 + C_1^2\rload^2
+%           - 2C_1C_2\rload(3R_1+\rload).$$
+Let $r_1$ and $r_2$ designate these two non-zero roots.
+In the original matrix $A$, this being the transformed matrix $A/c$,
+$Av = cr_1$ because $A/c * v = r_1.$
+
+The trivial zero eigenvalue corresponds to a unit x-direction vector
+by inspection. The two remaining roots, in the general case of a non-%
+degenerate system, which hasn't been explicitly ruled out, the middle
+row can be ``ignored'' because it is linearly independent in the
+following system:
+$$
+\pmatrix{-r&-C_2\rload &0\cr
+         * &*          &*\cr
+         0 &C_2R_1     &-r-C_1R_1}
+$$
+With $x = 1$, $\displaystyle y = -{r\over C_2\rload}$ and
+$\displaystyle z = y{C_2R_1\over r+C_1R_1}.$
+
+\def\bu{\par\leavevmode\llap{\hbox to \parindent{\hfil $\bullet$ \hfil}}}
+\noindent The eigenvalues and respective eigenvectors are:
+
+\bu $\lambda_1 = 0, v_1 = \pmatrix{1\cr0\cr0}$
+
+\def\num#1{\pmatrix{1\cr
+                    -{r_#1\over C_2\rload}\cr
+                    -{R_1r_#1\over \rload(r_#1+C_1R_1)}}}
+
+\bu $\displaystyle\lambda_2 = {r_1C_1C_2R_1\rload}, v_2 = \num1.$
+
+\bu $\displaystyle\lambda_3 = {r_2C_1C_2R_1\rload}, v_3 = \num2.$
+
+This corresponds to a solution of the form $f = Ce^{\lambda t}v,\cdots$
+where $C\in {\bf C},$ $f\in {\bf R}\to{\bf R}$. If the non-trivial
+eigenvectors are complex, %% TRY TO PROVE THIS!!
+\def\re{{\rm Re}}\def\im{{\rm Im}}
+their exponential solutions form, in the reals,
+$g = C_1\cos{\re(\lambda)t}\re v + C_2\sin{\re(\lambda)t}\im v.$ %% DOUBLE CHECK.
+
+\section{Nonhomogeneous System}
+
+Extending to the nonhomogeneous system will take slightly different
+paths depending on if the system has complex roots or has real roots.
+But in either case, $\cos x*{\rm polynomial}+\sin x*{\rm polynomial}$
+should be a particular solution. 
+
+We can apply the method of Variation of Parameters to this. Essentially, the solution is $\bf{x} = c_1\bf{\lambda_1}(t) + c_2\bf{\lambda_2}(t) + c_3\bf{\lambda_3}(t)\bf{\lambda_p}(t)$, where the particular solution $\bf{\lambda_p}(t)$ is:
+$$\bf{\lambda_p}(t) = \bf{X}(t)\int\bf{X^{-1}}(t)\bf{g}(t)dt,$$ where $\bf{X}(t)$ is the Fundamental matrix for the equation and $\bf{g}(t) = {1\over R_1}
+\pmatrix{\omega\cos(\omega t)\cr
+         \omega\cos(\omega t)\cr
+         0}.$
+\end{document}
\ No newline at end of file
-- 
cgit