diff options
| -rw-r--r-- | Makefile | 3 | ||||
| -rw-r--r-- | execsumm/ExecutiveSummaryDraft.tex | 41 | ||||
| -rw-r--r-- | graph.py | 14 |
3 files changed, 12 insertions, 46 deletions
@@ -1,5 +1,6 @@ .POSIX: .SUFFIXES: .tex .pdf +PYTHON = python all: progreport/document.pdf execsumm/document.pdf progreport/document.pdf: progreport/document.tex format.tex com.tex execsumm/document.pdf: execsumm/document.tex format.tex com.tex @@ -9,4 +10,6 @@ clean: find . -regex ".*\.\(pdf\|log\)" -exec rm -f {} + view: open -a Preview execsumm/document.pdf +plot.png: graph.py + ${PYTHON} graph.py img .PHONY: all clean diff --git a/execsumm/ExecutiveSummaryDraft.tex b/execsumm/ExecutiveSummaryDraft.tex deleted file mode 100644 index 498aac9..0000000 --- a/execsumm/ExecutiveSummaryDraft.tex +++ /dev/null @@ -1,41 +0,0 @@ -\documentclass{article} -\usepackage{hyperref} -\usepackage[scr]{rsfso} -\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}} -$$\x' = -{1\over R_1C_1C_2\rload} -\pmatrix{0&-C_2\rload &0 \cr - 0&-C_2(R_1+\rload)&C_1R_1\cr - 0&C_2R_1 &-C_1R_1} \x + -{1\over R_1} -\pmatrix{\omega\cos(\omega t)\cr - \omega\cos(\omega t)\cr - 0} -.$$ -\section{Application of Laplace Transformation} -We can apply the Laplace Transformation in order to solve this system of differential equations. -We have the three equations for $x'$, $y'$, and $z'$, and we can take the Laplace Transform of each of these equations" -$$\mathscr{L}\{x' = \frac{-y}{C_1R_1} + \frac{\omega\cos(\omega t)}{R_1}\} \Rightarrow sX(s) - x(0) = \frac{Y(s)}{C_1R_1} + \frac{\omega s}{R_1(s^2 + \omega^2)}$$ -$$\mathscr{L}\{y' = y\frac{-R_1 - \rload}{R_1C_1\rload} + \frac{z}{C_2\rload} + \frac{\omega\cos(\omega t)}{R_1} \} \Rightarrow sY(s) - y(0) = Y(s)(\frac{-R_1-\rload}{R_1C_1\rload}) + \frac{Z(s)}{C_2\rload} + \frac{\omega s}{R_1(s^2 + \omega^2)}$$ -$$\mathscr{L}\{z' = \frac{y}{C_1\rload} - \frac{z}{C_2\rload} \} \Rightarrow sZ(s) - z(0) = \frac{Y(s)}{C_1\rload} - \frac{Z(s)}{C_2\rload}$$ - -The last two equations we get can be used to solve for $Z(s)$, which we find to be $$Z(s) = \frac{\omega s(C_1C_2\rload^2)}{(s^2 + \omega^2)(s^2 + s(C_1R_1\rload + R_1C_2\rload + \rload^2C_2) + \rload)}$$ - -We can now find the partial fraction decomposition of this: -$$Z(s) = \frac{\omega s(C_1C_2\rload^2)}{(s^2 + \omega^2)(s^2 + s(C_1R_1\rload + R_1C_2\rload + \rload^2C_2) + \rload)} = $$ $$\frac{As + B}{s^2 + \omega^2} + \frac{Cs + D}{(s^2 + \omega^2)(s^2 + s(C_1R_1\rload + R_1C_2\rload + \rload^2C_2) + \rload)} = \omega sC_1C_2\rload^2$$ -To simplify notation, let $b = (s^2 + \omega^2)(s^2 + s(C_1R_1\rload + R_1C_2\rload + \rload^2C_2) + \rload)$ -We find that $$A = \frac{C_1C_2\rload^2\omega (\rload - \omega^2)}{b^2\omega^2 + \rload^2 - 2\rload - 2\rload\omega^2 + \omega^2}$$ -$$B = \frac{bC_1C_2\rload^2\omega^3}{b^2\omega^2 + \rload^2 - 2\rload\omega^2 + \omega^4}$$ -$$C = \frac{-C_1C_2\rload^2\omega(\rload - \omega^2)}{b^2\omega^2 + \rload^2 - 2\rload\omega^2 + \omega^4}$$ -\end{document}
\ No newline at end of file @@ -3,9 +3,10 @@ import matplotlib.pyplot as plt import tkinter from math import sin,cos,e +from sys import argv def f(x): - return G*sin(x) + B*cos(x) + 2*e**(rer1 * x) * (reE * cos(imr1 * x) - imE * sin(imr1 * x)) + return G*sin(A*x) + B*cos(B*x) + 2*e**(rer1 * x) * (reE * cos(imr1 * x) - imE * sin(imr1 * x)) for w in range(100,10000,100): c1 = 2.5*10**(-6) c2 = 1*10**(-6) @@ -26,10 +27,10 @@ for w in range(100,10000,100): D = - b*c1*c2*rload**3*w/(b**2*w**2 + rload**2 - 2*rload*w**2 + w**4) # from https://www.wolframalpha.com/input/?i=solve+for+x1%2Cx2%2Cx3%2Cx4+in+%7B%7B1%2C0%2C1%2C0%7D%2C+%7Bb%2C1%2C0%2C1%7D%2C+%7BR%2C+b%2C+w%5E2%2C+0%7D%2C+%7B0%2C+R%2C+0%2C+w%5E2%7D%7D*%7Bx1%2Cx2%2Cx3%2Cx4%7D+%3D+%7B0%2C0%2Cw*c_1*c_2*R%5E2%2C0%7D and an insane partial fraction decomposition E = (D-C*r1)/(r2-r1) F = C - E # Another PFD of Cs-D/(as^2+bs+c) - G = B/w + G = A/w # Final solution should be - # Gsin(theta) + Bcos(theta) + Ee^(r1 t) + Fe^(r2 t) + # Gsin(A * theta) + Bcos(B * theta) + Ee^(r1 t) + Fe^(r2 t) # But E, F, r1, and r2 are complex. Luckily, conjugates make it that # = 2e^(Re(r1) t) ( Re(E)cos(Im(r1)t) - Im(E)sin(Im(r1)t) ) rer1 = r2.real @@ -37,7 +38,7 @@ for w in range(100,10000,100): imr1 = r2.imag imE = F.imag # switched because positive ones were needed - print("%.2E * sin(t) + %.2E * cos(t) + 2e^(%.2f t) (%.2E cos(%.2f t) - %.2E sin(%.2f t))" % (G, B, rer1, reE, imr1, imE, imr1)) + print("%.2E * sin(%.2E t) + %.2E * cos(%.2E t) + 2e^(%.2f t) (%.2E cos(%.2f t) - %.2E sin(%.2f t))" % (G, A, B, B, rer1, reE, imr1, imE, imr1)) x = [] y = [] @@ -46,4 +47,7 @@ for w in range(100,10000,100): y.append(f(i/100)) plt.plot(x,y) -plt.show() +if len(argv) > 1 and argv[1] == 'img': + plt.savefig('plot.png') +else: + plt.show() |