diff options
| -rw-r--r-- | Makefile | 6 | ||||
| -rw-r--r-- | com.tex | 34 | ||||
| -rw-r--r-- | execsumm/document.tex | 39 | ||||
| -rw-r--r-- | progreport/document.tex | 34 |
4 files changed, 78 insertions, 35 deletions
@@ -1,10 +1,12 @@ .POSIX: .SUFFIXES: .tex .pdf all: progreport/document.pdf execsumm/document.pdf -progreport/document.pdf: progreport/document.tex format.tex -execsumm/document.pdf: execsumm/document.tex format.tex +progreport/document.pdf: progreport/document.tex format.tex com.tex +execsumm/document.pdf: execsumm/document.tex format.tex com.tex .tex.pdf: cd ${@D} && pdftex --jobname ${*F} ${<F} clean: find -E . -regex ".*\.(pdf|log)" -exec rm -f {} + +view: + open -a Preview execsumm/document.pdf .PHONY: all clean @@ -0,0 +1,34 @@ +\section{Project Topic} + +Our group will be working on the \link{Mystery Circuit Modelling +Scenario from SIMIODE}{https://simiode.org/resources/3187/download/4-23% +-S-MysteryCircuit-StudentVersion.pdf}. This applies Kirchhoff's Voltage +and Current Laws to the given circuit, which describe, respectively, +that the sum of all voltages in a closed loop is zero and the sum of all +currents at a node is zero. The circuit we're examining is an RLC +(resistor, inductor, capacitor) circuit, with zeroed initial conditions. +The specific circuit has two linked loops of resistors and capacitors, +in which ``gain,'' the ratio between chosen voltage differentials in the +circuit can be modeled mathematically. Because there are two connected +loops, there are three different currents. There is the current coming +off of the battery $x(t)$, the current split at the middle node becoming +$y(t)$ and $z(t)$. We are examining ${E(t)\over z(t)\rload}$ as the +``gain'' in the system. The first part uses $\omega = 100$ and the +entire problem uses $E(t) = \sin(\omega t)$. + +\section{Progress} + +From Kirchhoff's Voltage law over the first (xy) loop, +$$E(t) = \sin(\omega t) = x(t)R_1 + {1\over C_1}\int y(t)dt.$$ +Kirchhoff's Voltage law also applies to the second (yz) loop: +$${1\over C_1}\int y(t)dt = {1\over C_2}\int z(t)dt + z(t)\rload.$$ +Differentiating and rearranging gives: +$$x'(t) = -{y(t) \over R_1C_1} + {\omega\cos(\omega t) \over R_1},$$ +$$z'(t) = {y(t) \over C_1\rload} - {z(t) \over C_2\rload}$$ + +Kirchhoff's current law tells us that $y(t) + z(t) = x(t)$, so +$$y'(t) = x'(t) - z'(t) = -{y(t)\over R_1C_1} + +{\omega\cos(\omega t) \over R_1} - {y(t)\over C_1\rload} ++ {z(t) \over C_2\rload},$$ +giving a system of differential equations to solve. + diff --git a/execsumm/document.tex b/execsumm/document.tex index 4db9e4a..b5dfdc7 100644 --- a/execsumm/document.tex +++ b/execsumm/document.tex @@ -1,4 +1,43 @@ +\def\rload{R_{\rm load}} +\def\opt#1{\vskip0pt plus #1\vskip 0pt plus -#1} \input ../format \titlesub{Part 3: Executive Summary}{Mystery Circuit} +\input ../com + +\section{Matrix Representation} + +To determine the relevant properties of the linear system, matrix form +is useful: +\def\x{{\bf x}} +$$\x' = +{1\over R_1C_1C_2\rload}\left( +\pmatrix{0&-C_2\rload&0 \cr + *&* &* \cr + 0&C_2R_1 &-C_2\rload} \x + +\pmatrix{\omega\cos(\omega t)\cr + \omega\cos(\omega t)\cr + 0} +\right). +$$ + +{\it *Linearly dependent, meaning a trivial eigenvalue of 0} + +\def\bu{\par\leavevmode\llap{\hbox to \parindent{\hfil $\bullet$ \hfil}}} +\opt{.15fil} +\noindent The eigenvalues and respective eigenvectors are: + +\bu $\lambda_1 = 0, v_1 = \pmatrix{1\cr0\cr0}$ + +\def\num{\pmatrix{0\cr C_2\rload\cr C_2R_1}} +\bu $\lambda_2 = -C_2\rload, v_2 = \num$ + +\bu $\lambda_3 = -C_2\rload, v_2 = \pmatrix{0\cr0\cr1}$ + +This gives the solution to the homogenous system +$$D_1\pmatrix{1\cr0\cr0} + D_2e^{-C_2\rload t}\num ++ D_3te^{-C_2\rload t}\pmatrix{0\cr0\cr1}.$$ + +Extending to the nonhomogenous system, + \bye diff --git a/progreport/document.tex b/progreport/document.tex index 7b34439..af7da22 100644 --- a/progreport/document.tex +++ b/progreport/document.tex @@ -2,39 +2,7 @@ \input ../format \titlesub{Part 2: Progress Report}{Topic: Mystery Circuit} -\section{Project Topic} - -Our group will be working on the \link{Mystery Circuit Modelling -Scenario from SIMIODE}{https://simiode.org/resources/3187/download/4-23% --S-MysteryCircuit-StudentVersion.pdf}. This applies Kirchhoff's Voltage -and Current Laws to the given circuit, which describe, respectively, -that the sum of all voltages in a closed loop is zero and the sum of all -currents at a node is zero. The circuit we're examining is an RLC -(resistor, inductor, capacitor) circuit, with zeroed initial conditions. -The specific circuit has two linked loops of resistors and capacitors, -in which ``gain,'' the ratio between chosen voltage differentials in the -circuit can be modeled mathematically. Because there are two connected -loops, there are three different currents. There is the current coming -off of the battery $x(t)$, the current split at the middle node becoming -$y(t)$ and $z(t)$. We are examining ${E(t)\over z(t)\rload}$ as the -``gain'' in the system. The first part uses $\omega = 100$ and the -entire problem uses $E(t) = \sin(\omega t)$. - -\section{Progress} - -From Kirchhoff's Voltage law over the first (xy) loop, -$$E(t) = \sin(\omega t) = x(t)R_1 + {1\over C_1}\int y(t)dt.$$ -Kirchhoff's Voltage law also applies to the second (yz) loop: -$${1\over C_1}\int y(t)dt = {1\over C_2}\int z(t)dt + z(t)\rload.$$ -Differentiating and rearranging gives: -$$x'(t) = -{y(t) \over R_1C_1} + {\omega\cos(\omega t) \over R_1},$$ -$$z'(t) = {y(t) \over C_1\rload} - {z(t) \over C_2\rload}$$ - -Kirchhoff's current law tells us that $y(t) + z(t) = x(t)$, so -$$y'(t) = x'(t) - z'(t) = -{y(t)\over R_1C_1} + -{\omega\cos(\omega t) \over R_1} - {y(t)\over C_1\rload} -+ {z(t) \over C_2\rload},$$ -giving a system of differential equations to solve. +\input ../com \section{Further Exploration} |