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diff --git a/execsumm/ExecutiveSummaryDraft.tex b/execsumm/ExecutiveSummaryDraft.tex new file mode 100644 index 0000000..a9c133b --- /dev/null +++ b/execsumm/ExecutiveSummaryDraft.tex @@ -0,0 +1,87 @@ +\documentclass{article} +\usepackage{hyperref} +\def\rload{R_{\rm load}} +\date{} +\begin{document} +\title{Project Executive Summary} +\author{Holden Rohrer and Nithya Jayakumar} + +\maketitle +\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 $A/c$, +$Av = cr_1$ 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 independent 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{1\cr + -{r_#1\over C_2\rload}\cr + -{R_1r_#1\over \rload(r_#1+C_1R_1)}}} + +\bu $\displaystyle\lambda_2 = {r_1C_1C_2R_1\rload}, v_2 = \num1.$ + +\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!! +\def\re{{\rm Re}}\def\im{{\rm Im}} +their exponential solutions form, in the reals, +$g = C_1\cos{\re(\lambda)t}\re v + C_2\sin{\re(\lambda)t}\im v.$ %% DOUBLE CHECK. + +\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. + +We can apply the method of Variation of Parameters to this. 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: +$$\bf{\lambda_p}(t) = \bf{X}(t)\int\bf{X^{-1}}(t)\bf{g}(t)dt,$$ where $\bf{X}(t)$ is the Fundamental matrix for the equation and $\bf{g}(t) = {1\over R_1} +\pmatrix{\omega\cos(\omega t)\cr + \omega\cos(\omega t)\cr + 0}.$ +\end{document}
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