attempted (but wrong) start of solution

1 files changed, 1 insertions, 33 deletions

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}