From 609afd96f4069c69945fedcdfa7787ae4f7de967 Mon Sep 17 00:00:00 2001
From: Holden Rohrer <hr@hrhr.dev>
Date: Sat, 4 Apr 2020 02:21:04 -0400
Subject: attempted (but wrong) start of solution

---
 execsumm/document.tex | 39 +++++++++++++++++++++++++++++++++++++++
 1 file changed, 39 insertions(+)

(limited to 'execsumm')

diff --git a/execsumm/document.tex b/execsumm/document.tex
index 4db9e4a..b5dfdc7 100644
--- a/execsumm/document.tex
+++ b/execsumm/document.tex
@@ -1,4 +1,43 @@
+\def\rload{R_{\rm load}}
+\def\opt#1{\vskip0pt plus #1\vskip 0pt plus -#1}
 \input ../format
 \titlesub{Part 3: Executive Summary}{Mystery Circuit}
 
+\input ../com
+
+\section{Matrix Representation}
+
+To determine the relevant properties of the linear system, matrix form
+is useful:
+\def\x{{\bf x}}
+$$\x' =
+{1\over R_1C_1C_2\rload}\left(
+\pmatrix{0&-C_2\rload&0 \cr
+         *&*         &* \cr
+         0&C_2R_1    &-C_2\rload}   \x +
+\pmatrix{\omega\cos(\omega t)\cr
+         \omega\cos(\omega t)\cr
+         0}
+\right).
+$$
+
+{\it *Linearly dependent, meaning a trivial eigenvalue of 0}
+
+\def\bu{\par\leavevmode\llap{\hbox to \parindent{\hfil $\bullet$ \hfil}}}
+\opt{.15fil}
+\noindent The eigenvalues and respective eigenvectors are:
+
+\bu $\lambda_1 = 0, v_1 = \pmatrix{1\cr0\cr0}$
+
+\def\num{\pmatrix{0\cr C_2\rload\cr C_2R_1}}
+\bu $\lambda_2 = -C_2\rload, v_2 = \num$
+
+\bu $\lambda_3 = -C_2\rload, v_2 = \pmatrix{0\cr0\cr1}$
+
+This gives the solution to the homogenous system
+$$D_1\pmatrix{1\cr0\cr0} + D_2e^{-C_2\rload t}\num
++ D_3te^{-C_2\rload t}\pmatrix{0\cr0\cr1}.$$
+
+Extending to the nonhomogenous system, 
+
 \bye
-- 
cgit