:PROPERTIES:
:ID:       6dbe2931-cc18-48fc-8cc1-6c71935a6be3
:END:
#+title: LRC circuit
#+author: Preston Pan
#+html_head: <link rel="stylesheet" type="text/css" href="../style.css" />
#+html_head: <script src="https://polyfill.io/v3/polyfill.min.js?features=es6"></script>
#+html_head: <script id="MathJax-script" async src="https://cdn.jsdelivr.net/npm/mathjax@3/es5/tex-mml-chtml.js"></script>
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* Introduction
LRC circuits are equivalent to mass-spring oscillation systems in terms of the differential equation generated. In other
words, they are an example of a wave generator. First we introduce the LRC circuit without a voltage source. Later,
another circuit diagram will include a possibly variable voltage source.
#+name: LRC Circuit
#+header: :exports results :file lrc_circuit.png 
#+header: :imagemagick yes :iminoptions -density 600 :imoutoptions -geometry 400 
#+header: :fit yes :noweb yes :headers '("\\usepackage{circuitikz}")
#+begin_src latex :exports results :file 
    \documentclass{article}
    \usepackage{circuitikz}
    \begin{document}
    \begin{center}
    \begin{circuitikz} \draw
    (0,0) to[resistor, l=\mbox{$R$}] (0,4)
      to[inductor, l=\mbox{$L$}] (4,4)
      to[capacitor, l=\mbox{$C$}] (4,0)
      (4,0) -- (0,0)
      (2,0) -- (2,-1)
      to (2, -1) node[shape=ground]{};
    \end{circuitikz}
    \end{center}
    \end{document}
#+end_src

#+RESULTS: LRC Circuit
#+begin_export latex
#+end_export

* Mass-Spring Equation Equivalence
We know these relations for the given circuit elements above:
\begin{align}
v(t) = L\frac{di}{dt} \\
i(t) = C\frac{dv}{dt} \\
v = iR
\end{align}
if we analyze the current current signal, Kirchhoff's voltage law tells us that the total voltage
drop throughout this circuit is zero. We use the capacitor equation in integral form and sum the voltages:
\begin{align*}
L\frac{di}{dt} + \frac{1}{C}\int i(t)dt + iR = 0
\end{align*}
We then take a derivative to remove the integral:
\begin{align*}
L\frac{d^{2}i}{dt^{2}} + R\frac{di}{dt} + \frac{1}{C}i = 0 \\
(LD^{2} + RD + \frac{1}{C}) i(t) = 0
\end{align*}
it is clear that the characteristic polynomial of this homogeneous linear differential equation is:
\begin{align*}
L\lambda^{2} + R\lambda + \frac{1}{C} = 0
\end{align*}
which, utilizing the quadratic formula, has the solutions:
\begin{align*}
\lambda_{1} = \frac{-R + \sqrt{R^{2} - \frac{4L}{C}}}{2L},
\lambda_{2} = \frac{-R - \sqrt{R^{2} - \frac{4L}{C}}}{2L}
\end{align*}
which implies the general solution to this differential equation is:
\begin{align*}
i(t) = \sum_{n=0}^{\infty} A_{n}e^{\lambda_{1} t} + B_{n}e^{\lambda_{2} t}
\end{align*}
We can make this nicer by setting $-\frac{R}{2L} = m$, $\frac{\sqrt{R^{2} - \frac{4L}{C}}}{2L} = p$,
then $\lambda_{1} = m + p$, $\lambda_{2} = m - p$. Then:
\begin{align*}
i(t) = \sum_{n=0}^{\infty} A_{n}e^{(m + p) t} + B_{n}e^{(m - p) t} \\
i(t) = e^{m}(\sum_{n=0}^{\infty} A_{n}e^{pt} + B_{n}e^{-pt})
\end{align*}
Then we can just recast our notation for the constants $A_{n}$ and $B_{n}$ to include this $e^{m}$ term:
\begin{align*}
i(t) = \sum_{n=0}^{\infty} A_{n}e^{pt} + B_{n}e^{-pt}
\end{align*}
** Dampened Oscillation
In the case $R^{2} < \frac{4L}{C}$, $p = i\frac{\sqrt{\sigma}}{2L}$ for some $\sigma > 0$. We re-case $\lambda = \frac{\sqrt{\sigma}}{2L}$ so $p = i\lambda$. Then:
\begin{align*}
i(t) = \sum_{n=0}^{\infty} A_{n}e^{i\lambda t} + B_{n}e^{-i\lambda t}
\end{align*}