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Diffstat (limited to 'execsumm')
| -rw-r--r-- | execsumm/document.tex | 123 |
1 files changed, 48 insertions, 75 deletions
diff --git a/execsumm/document.tex b/execsumm/document.tex index 1b2627d..42265fa 100644 --- a/execsumm/document.tex +++ b/execsumm/document.tex @@ -4,11 +4,11 @@ \input ../com -\section{Matrix Representation and Homogeneous Solution} +\section{Matrix Representation and Laplace Transform} 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}} +\def\x{{bf x}} $$\x' = {1\over R_1C_1C_2\rload} \pmatrix{0&-C_2\rload &0 \cr @@ -19,81 +19,41 @@ $$\x' = \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).$$ -Let $r_1$ and $r_2$ designate these two non-zero roots. -In the original matrix $A$, this being the transformed matrix $A/c$, -$Av = cr_1v$ because $A/c * v = r_1.$ - -The trivial zero eigenvalue corresponds to a unit x-direction vector -by inspection. The two remaining roots, in the general case of a non-% -degenerate system, which hasn't been explicitly ruled out, the middle -row can be ``ignored'' because it is linearly dependent in the -following system: -$$ -\pmatrix{-r&-C_2\rload &0\cr - * &* &*\cr - 0 &C_2R_1 &-r-C_1R_1} -$$ -With $x = 1$, $\displaystyle y = -{r\over C_2\rload}$ and -$\displaystyle z = y{C_2R_1\over r+C_1R_1}.$ - -\def\bu{\par\leavevmode\llap{\hbox to \parindent{\hfil $\bullet$ \hfil}}} -\noindent The eigenvalues and respective eigenvectors are: - -\bu $\lambda_1 = 0, v_1 = \pmatrix{1\cr0\cr0}$ - -\def\num#1{\pmatrix{\rload\cr - -{r_#1\over C_2}\cr - -{R_1r_#1\over (r_#1+C_1R_1)}}} - -\bu $\displaystyle\lambda_2 = {r_1C_1C_2R_1\rload}, v_2 = \num1.$ -\vskip 10pt -\bu $\displaystyle\lambda_3 = {r_2C_1C_2R_1\rload}, v_3 = \num2.$ - -This corresponds to a solution of the form $f = Ce^{\lambda t}v,\cdots$ -where $C\in {\bf C},$ $f\in {\bf R}\to{\bf R}$. If the non-trivial -eigenvectors are complex, %% TRY TO PROVE THIS!! -their exponential solutions form, in the reals, -\def\re{{\rm Re}}\def\im{{\rm Im}} -$g = C_1\cos(\re(\lambda)t)\re(v) + C_2\sin(\re(\lambda)t)\im(v).$ - -\section{Nonhomogeneous System} - -Extending to the nonhomogeneous system will take slightly different -paths depending on if the system has complex roots or has real roots. -But in either case, $\cos x*{\rm polynomial}+\sin x*{\rm polynomial}$ -should be a particular solution. This would allow usage of method of -undetermined coefficients, which may be attempted, but variation of -parameters will be used so that a general form for both systems with -both complex and real eigenvalues may be found. -Essentially, the solution is ${\bf x} = c_1{\bf\lambda_1}(t) + -c_2{\bf \lambda_2}(t) + c_3{\bf \lambda_3}(t){\bf\lambda_p}(t)$, -where the particular solution ${\bf\lambda_p}(t)$ is: -$${\bf \lambda}(t) = {\bf X}(t)\int{\bf X^{-1}}(t){\bf g}(t)dt,$$ -where ${\bf X}(t)$ is the Fundamental matrix for the equation and -${\bf g}(t) = {1\over R_1} -\pmatrix{\omega\cos(\omega t)\cr - \omega\cos(\omega t)\cr - 0}.$ -To do this in general is extremely time-consuming, and we'd like to -enlist the aid of a computer algebra system like Maxima but we have been -unable to do this as of yet. +\section{Application of Laplace Transformation} +\def\L{{\it L}} +\def\frac#1#2{{#1\over #2}} +We can apply the Laplace Transformation in order to solve this system of differential equations. +We have the three equations for $x'$, $y'$, and $z'$, and we can take the Laplace Transform of each of these equations" +$$\L\{x' = {-y \over 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)}$$ +$$\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)}$$ +$$\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 + sb + \rload)}$$ +with $b=C_1R_1\rload + R_1C_2\rload + \rload^2C_2$ to simplify notation. + +We can now find the partial fraction decomposition of this% +\footnote{1}{\link{Wolfram Alpha}{https://www.wolframalpha.com/input/?i% +=solve+for+x1\%2Cx2\%2Cx3\%2Cx4+in+\%7B\%7B1\%2C0\%2C1\%2C0\%7D\%2C+\%7% +Bb\%2C1\%2C0\%2C1\%7D\%2C+\%7BR\%2C+b\%2C+w\%5E2\%2C+0\%7D\%2C+\%7B0\%2% +C+R\%2C+0\%2C+w\%5E2\%7D\%7D*\%7Bx1\%2Cx2\%2Cx3\%2Cx4\%7D+\%3D+\%7B0\%2% +C0\%2Cw*c_1*c_2*R\%5E2\%2C0\%7D}}: +$$Z(s) = \frac{\omega s(C_1C_2\rload^2)}{(s^2 + \omega^2)(s^2 + sb + \rload)} = $$ $$\frac{As + B}{s^2 + \omega^2} + \frac{Cs + D}{s^2 + sb + \rload} = \omega sC_1C_2\rload^2$$ +We find that $$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}$$ +$$D = {-bC_1C_1\rload^3 \omega \over b^2w^2 + \rload^2 - 2\rload\omega^2 + \omega^4}$$ + +The second term of $Z(t)$ can be further simplified into (with $r_1$ +and $r_2$ as the roots of $s^2 + sb + \rload$, as can be found through +the quadratic equation---note that they are complex, but this is handled +later) the form $${E\over s-r_1} + {F\over s-r_2}.$$ +$$E = {D - Cr_1 \over r_2 - r_1}$$ +$$F = \overline{E} = C - E.$$ -\iffalse \section{Particular Case} +\def\bu{\pre{$\bullet$}} There is a particular case which is intended to be investigated: \bu $C_1 = 2.5\times 10^{-6} F$ @@ -104,7 +64,20 @@ There is a particular case which is intended to be investigated: \bu $R_1 = 200\ohm$ \bu $\rload = 1000\ohm$ -\fi + +Using Python with matplotlib (\link{Git}{https://git.hrhr.dev/diffeq-pr% +oj/tree/graph.py}), we found that the system converges rapidly towards +a steady state with some nearly undetectable oscillation: + +\centerline{\pdfximage{../plot.png}\pdfrefximage\pdflastximage} + +This doesn't match up with physical intuition that these varying curves +(representing $Z(t)$ with various frequencies of the electromotive force +driving the system). However, it does match that because the capacitors +absorb some of the variance in current from the source, $Z(t)$ is +smaller with smaller values (blue is the smallest frequency at $100Hz$). + +However, the initial oscillation in every curve makes a lot of sense \section{Possible Generalization} |