attempted (but wrong) start of solution

4 files changed, 78 insertions, 35 deletions

diff --git a/Makefile b/Makefile
index 934cbc4..6c54cba 100644
--- a/Makefile
+++ b/Makefile
@@ -1,10 +1,12 @@
 .POSIX:
 .SUFFIXES: .tex .pdf
 all: progreport/document.pdf execsumm/document.pdf
-progreport/document.pdf: progreport/document.tex format.tex
-execsumm/document.pdf: execsumm/document.tex format.tex
+progreport/document.pdf: progreport/document.tex format.tex com.tex
+execsumm/document.pdf: execsumm/document.tex format.tex com.tex
 .tex.pdf:
 	cd ${@D} && pdftex --jobname ${*F} ${<F}
 clean:
 	find -E . -regex ".*\.(pdf|log)" -exec rm -f {} +
+view:
+	open -a Preview execsumm/document.pdf
 .PHONY: all clean
diff --git a/com.tex b/com.tex
new file mode 100644
index 0000000..c119bd4
--- /dev/null
+++ b/com.tex
@@ -0,0 +1,34 @@
+\section{Project Topic}
+
+Our group will be working on the \link{Mystery Circuit Modelling
+Scenario from SIMIODE}{https://simiode.org/resources/3187/download/4-23%
+-S-MysteryCircuit-StudentVersion.pdf}. This applies Kirchhoff's Voltage
+and Current Laws to the given circuit, which describe, respectively,
+that the sum of all voltages in a closed loop is zero and the sum of all
+currents at a node is zero. The circuit we're examining is an RLC
+(resistor, inductor, capacitor) circuit, with zeroed initial conditions.
+The specific circuit has two linked loops of resistors and capacitors,
+in which ``gain,'' the ratio between chosen voltage differentials in the
+circuit can be modeled mathematically. Because there are two connected
+loops, there are three different currents. There is the current coming
+off of the battery $x(t)$, the current split at the middle node becoming
+$y(t)$ and $z(t)$. We are examining ${E(t)\over z(t)\rload}$ as the
+``gain'' in the system. The first part uses $\omega = 100$ and the
+entire problem uses $E(t) = \sin(\omega t)$.
+
+\section{Progress}
+
+From Kirchhoff's Voltage law over the first (xy) loop,
+$$E(t) = \sin(\omega t) = x(t)R_1 + {1\over C_1}\int y(t)dt.$$
+Kirchhoff's Voltage law also applies to the second (yz) loop:
+$${1\over C_1}\int y(t)dt = {1\over C_2}\int z(t)dt + z(t)\rload.$$
+Differentiating and rearranging gives:
+$$x'(t) = -{y(t) \over R_1C_1} + {\omega\cos(\omega t) \over R_1},$$
+$$z'(t) = {y(t) \over C_1\rload} - {z(t) \over C_2\rload}$$
+
+Kirchhoff's current law tells us that $y(t) + z(t) = x(t)$, so
+$$y'(t) = x'(t) - z'(t) = -{y(t)\over R_1C_1} +
+{\omega\cos(\omega t) \over R_1} - {y(t)\over C_1\rload}
++ {z(t) \over C_2\rload},$$
+giving a system of differential equations to solve.
+
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
diff --git a/progreport/document.tex b/progreport/document.tex
index 7b34439..af7da22 100644
--- a/progreport/document.tex
+++ b/progreport/document.tex
@@ -2,39 +2,7 @@
 \input ../format
 \titlesub{Part 2: Progress Report}{Topic: Mystery Circuit}
 
-\section{Project Topic}
-
-Our group will be working on the \link{Mystery Circuit Modelling
-Scenario from SIMIODE}{https://simiode.org/resources/3187/download/4-23%
--S-MysteryCircuit-StudentVersion.pdf}. This applies Kirchhoff's Voltage
-and Current Laws to the given circuit, which describe, respectively,
-that the sum of all voltages in a closed loop is zero and the sum of all
-currents at a node is zero. The circuit we're examining is an RLC
-(resistor, inductor, capacitor) circuit, with zeroed initial conditions.
-The specific circuit has two linked loops of resistors and capacitors,
-in which ``gain,'' the ratio between chosen voltage differentials in the
-circuit can be modeled mathematically. Because there are two connected
-loops, there are three different currents. There is the current coming
-off of the battery $x(t)$, the current split at the middle node becoming
-$y(t)$ and $z(t)$. We are examining ${E(t)\over z(t)\rload}$ as the
-``gain'' in the system. The first part uses $\omega = 100$ and the
-entire problem uses $E(t) = \sin(\omega t)$.
-
-\section{Progress}
-
-From Kirchhoff's Voltage law over the first (xy) loop,
-$$E(t) = \sin(\omega t) = x(t)R_1 + {1\over C_1}\int y(t)dt.$$
-Kirchhoff's Voltage law also applies to the second (yz) loop:
-$${1\over C_1}\int y(t)dt = {1\over C_2}\int z(t)dt + z(t)\rload.$$
-Differentiating and rearranging gives:
-$$x'(t) = -{y(t) \over R_1C_1} + {\omega\cos(\omega t) \over R_1},$$
-$$z'(t) = {y(t) \over C_1\rload} - {z(t) \over C_2\rload}$$
-
-Kirchhoff's current law tells us that $y(t) + z(t) = x(t)$, so
-$$y'(t) = x'(t) - z'(t) = -{y(t)\over R_1C_1} +
-{\omega\cos(\omega t) \over R_1} - {y(t)\over C_1\rload}
-+ {z(t) \over C_2\rload},$$
-giving a system of differential equations to solve.
+\input ../com
 
 \section{Further Exploration}