From 82b0e3725b6dd740a00972a7ceb10f9f22ee4b9b Mon Sep 17 00:00:00 2001
From: Holden Rohrer <hr@hrhr.dev>
Date: Fri, 17 Apr 2020 22:44:19 -0400
Subject: lots of document work

---
 Makefile              |   2 +-
 execsumm/document.tex | 123 ++++++++++++++++++++------------------------------
 format.tex            |   4 ++
 3 files changed, 53 insertions(+), 76 deletions(-)

diff --git a/Makefile b/Makefile
index fc28933..18166f2 100644
--- a/Makefile
+++ b/Makefile
@@ -3,7 +3,7 @@
 PYTHON = python
 all: progreport/document.pdf execsumm/document.pdf
 progreport/document.pdf: progreport/document.tex format.tex com.tex
-execsumm/document.pdf: execsumm/document.tex format.tex com.tex
+execsumm/document.pdf: execsumm/document.tex format.tex com.tex plot.png
 .tex.pdf:
 	cd ${@D} && pdftex --jobname ${*F} ${<F}
 clean:
diff --git a/execsumm/document.tex b/execsumm/document.tex
index 1b2627d..42265fa 100644
--- a/execsumm/document.tex
+++ b/execsumm/document.tex
@@ -4,11 +4,11 @@
 
 \input ../com
 
-\section{Matrix Representation and Homogeneous Solution}
+\section{Matrix Representation and Laplace Transform}
 
 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}}
+\def\x{{bf x}}
 $$\x' =
 {1\over R_1C_1C_2\rload}
 \pmatrix{0&-C_2\rload      &0 \cr
@@ -19,81 +19,41 @@ $$\x' =
          \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).$$
-Let $r_1$ and $r_2$ designate these two non-zero roots.
-In the original matrix $A$, this being the transformed matrix $A/c$,
-$Av = cr_1v$ because $A/c * v = r_1.$
-
-The trivial zero eigenvalue corresponds to a unit x-direction vector
-by inspection. The two remaining roots, in the general case of a non-%
-degenerate system, which hasn't been explicitly ruled out, the middle
-row can be ``ignored'' because it is linearly dependent in the
-following system:
-$$
-\pmatrix{-r&-C_2\rload &0\cr
-         * &*          &*\cr
-         0 &C_2R_1     &-r-C_1R_1}
-$$
-With $x = 1$, $\displaystyle y = -{r\over C_2\rload}$ and
-$\displaystyle z = y{C_2R_1\over r+C_1R_1}.$
-
-\def\bu{\par\leavevmode\llap{\hbox to \parindent{\hfil $\bullet$ \hfil}}}
-\noindent The eigenvalues and respective eigenvectors are:
-
-\bu $\lambda_1 = 0, v_1 = \pmatrix{1\cr0\cr0}$
-
-\def\num#1{\pmatrix{\rload\cr
-                    -{r_#1\over C_2}\cr
-                    -{R_1r_#1\over (r_#1+C_1R_1)}}}
-
-\bu $\displaystyle\lambda_2 = {r_1C_1C_2R_1\rload}, v_2 = \num1.$
-\vskip 10pt
-\bu $\displaystyle\lambda_3 = {r_2C_1C_2R_1\rload}, v_3 = \num2.$
-
-This corresponds to a solution of the form $f = Ce^{\lambda t}v,\cdots$
-where $C\in {\bf C},$ $f\in {\bf R}\to{\bf R}$. If the non-trivial
-eigenvectors are complex, %% TRY TO PROVE THIS!!
-their exponential solutions form, in the reals,
-\def\re{{\rm Re}}\def\im{{\rm Im}}
-$g = C_1\cos(\re(\lambda)t)\re(v) + C_2\sin(\re(\lambda)t)\im(v).$
-
-\section{Nonhomogeneous System}
-
-Extending to the nonhomogeneous system will take slightly different
-paths depending on if the system has complex roots or has real roots.
-But in either case, $\cos x*{\rm polynomial}+\sin x*{\rm polynomial}$
-should be a particular solution. This would allow usage of method of
-undetermined coefficients, which may be attempted, but variation of
-parameters will be used so that a general form for both systems with
-both complex and real eigenvalues may be found. 
-Essentially, the solution is ${\bf x} = c_1{\bf\lambda_1}(t) +
-c_2{\bf \lambda_2}(t) + c_3{\bf \lambda_3}(t){\bf\lambda_p}(t)$,
-where the particular solution ${\bf\lambda_p}(t)$ is:
-$${\bf \lambda}(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
-${\bf g}(t) = {1\over R_1}
-\pmatrix{\omega\cos(\omega t)\cr
-         \omega\cos(\omega t)\cr
-         0}.$
 
-To do this in general is extremely time-consuming, and we'd like to
-enlist the aid of a computer algebra system like Maxima but we have been
-unable to do this as of yet.
+\section{Application of Laplace Transformation}
+\def\L{{\it L}}
+\def\frac#1#2{{#1\over #2}}
+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"
+$$\L\{x' = {-y \over C_1R_1} + \frac{\omega\cos(\omega t)}{R_1}\} \Rightarrow sX(s) - x(0) = \frac{Y(s)}{C_1R_1} + \frac{\omega s}{R_1(s^2 + \omega^2)}$$
+$$\L\{y' = y\frac{-R_1 - \rload}{R_1C_1\rload} + \frac{z}{C_2\rload} + \frac{\omega\cos(\omega t)}{R_1} \}$$
+$$\Rightarrow sY(s) - y(0) = Y(s)(\frac{-R_1-\rload}{R_1C_1\rload}) + \frac{Z(s)}{C_2\rload} + \frac{\omega s}{R_1(s^2 + \omega^2)}$$
+$$\L\{z' = \frac{y}{C_1\rload} - \frac{z}{C_2\rload}  \} \Rightarrow sZ(s) - z(0) = \frac{Y(s)}{C_1\rload} - \frac{Z(s)}{C_2\rload}$$
+
+The last two equations we get can be used to solve for $Z(s)$, which we find to be $$Z(s) = \frac{\omega s(C_1C_2\rload^2)}{(s^2 + \omega^2)(s^2 + sb + \rload)}$$
+with $b=C_1R_1\rload + R_1C_2\rload + \rload^2C_2$ to simplify notation.
+
+We can now find the partial fraction decomposition of this%
+\footnote{1}{\link{Wolfram Alpha}{https://www.wolframalpha.com/input/?i%
+=solve+for+x1\%2Cx2\%2Cx3\%2Cx4+in+\%7B\%7B1\%2C0\%2C1\%2C0\%7D\%2C+\%7%
+Bb\%2C1\%2C0\%2C1\%7D\%2C+\%7BR\%2C+b\%2C+w\%5E2\%2C+0\%7D\%2C+\%7B0\%2%
+C+R\%2C+0\%2C+w\%5E2\%7D\%7D*\%7Bx1\%2Cx2\%2Cx3\%2Cx4\%7D+\%3D+\%7B0\%2%
+C0\%2Cw*c_1*c_2*R\%5E2\%2C0\%7D}}: 
+$$Z(s) = \frac{\omega s(C_1C_2\rload^2)}{(s^2 + \omega^2)(s^2 + sb + \rload)} = $$ $$\frac{As + B}{s^2 + \omega^2} + \frac{Cs + D}{s^2 + sb + \rload} = \omega sC_1C_2\rload^2$$
+We find that $$A = \frac{C_1C_2\rload^2\omega (\rload - \omega^2)}{b^2\omega^2 + \rload^2 - 2\rload - 2\rload\omega^2 + \omega^2}$$
+$$B = \frac{bC_1C_2\rload^2\omega^3}{b^2\omega^2 + \rload^2 - 2\rload\omega^2 + \omega^4}$$
+$$C = \frac{-C_1C_2\rload^2\omega(\rload - \omega^2)}{b^2\omega^2 + \rload^2 - 2\rload\omega^2 + \omega^4}$$
+$$D = {-bC_1C_1\rload^3 \omega \over b^2w^2 + \rload^2 - 2\rload\omega^2 + \omega^4}$$
+
+The second term of $Z(t)$ can be further simplified into (with $r_1$
+and $r_2$ as the roots of $s^2 + sb + \rload$, as can be found through
+the quadratic equation---note that they are complex, but this is handled
+later) the form $${E\over s-r_1} + {F\over s-r_2}.$$
+$$E = {D - Cr_1 \over r_2 - r_1}$$
+$$F = \overline{E} = C - E.$$
 
-\iffalse
 \section{Particular Case}
+\def\bu{\pre{$\bullet$}}
 
 There is a particular case which is intended to be investigated:
 \bu $C_1 = 2.5\times 10^{-6} F$
@@ -104,7 +64,20 @@ There is a particular case which is intended to be investigated:
 \bu $R_1 = 200\ohm$
 
 \bu $\rload = 1000\ohm$
-\fi
+
+Using Python with matplotlib (\link{Git}{https://git.hrhr.dev/diffeq-pr%
+oj/tree/graph.py}), we found that the system converges rapidly towards
+a steady state with some nearly undetectable oscillation:
+
+\centerline{\pdfximage{../plot.png}\pdfrefximage\pdflastximage}
+
+This doesn't match up with physical intuition that these varying curves
+(representing $Z(t)$ with various frequencies of the electromotive force
+driving the system). However, it does match that because the capacitors
+absorb some of the variance in current from the source, $Z(t)$ is
+smaller with smaller values (blue is the smallest frequency at $100Hz$).
+
+However, the initial oscillation in every curve makes a lot of sense
 
 \section{Possible Generalization}
 
diff --git a/format.tex b/format.tex
index 061366c..c185fc0 100644
--- a/format.tex
+++ b/format.tex
@@ -51,3 +51,7 @@ Mayer
 
 %% Bullet Points and Numbering
 \def\pre#1{\par\leavevmode\llap{\hbox to \parindent{\hfil #1 \hfil}}}
+
+%% Margins
+\hsize=5.5in
+\hoffset=.5in
-- 
cgit