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\def\rload{R_{\rm load}}
\input ../format
\titlesub{Part 3: Executive Summary}{Mystery Circuit}
\input ../com
\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 $A/c$,
$Av = cr_1v$ 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.
\iffalse
\section{Particular Case}
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 $\rload = 1000\ohm$
\fi
\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
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