\def\rload{R_{\rm load}}
\input ../format
\titlesub{Part 3: Executive Summary}{Mystery Circuit}

\input ../com

\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}}
$$\x' =
{1\over R_1C_1C_2\rload}
\pmatrix{0&-C_2\rload      &0 \cr
         0&-C_2(R_1+\rload)&C_1R_1\cr
         0&C_2R_1          &-C_1R_1}   \x +
{1\over R_1}
\pmatrix{\omega\cos(\omega t)\cr
         \omega\cos(\omega t)\cr
         0}
.$$

\section{Application of Laplace Transformation}
\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 $${E\over s-r_1} + {F\over s-r_2}.$$
$$E = {D - Cr_1 \over r_2 - r_1}$$
$$F = \overline{E} = C - E.$$

\section{Particular Case}
\def\bu{\pre{$\bullet$}}

There is a particular case which is intended to be investigated:
\bu $C_1 = 2.5\times 10^{-6} F$

\bu $C_2 = 1.0\times 10^{-6} F$

\let\ohm\Omega
\bu $R_1 = 200\ohm$

\bu $\rload = 1000\ohm$

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}

This solution is general to any formulation of the original problem,
but gain may look different for a square or triangular wave, for
example. It is expected that these would exhibit similar behavior
to the sine wave because, because they would input similar amounts of
energy on a similar time scale, but the sensitivity to waveform type
could be investigated in the same way that this paper did, with possible
usage of the Laplace transform to handle discontinuities, but because
there is a general form for a sine wave and the output is proportional
to the input, a Fourier transform could be used to either approximate
or analytically obtain a solution for these types of waves.

\bye