メモ: ラプラス変換 - xlogI125’s blog

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ラプラス変換

$t<0$ のとき $y(t)=0$ とする

$\Large \begin{align*} Y(s) &= \mathcal{L}\left[ y(t) \right] \\ &= \int_0^\infty y(t) \mathrm{e}^{-st} \mathrm{d}t \end{align*}$

変数 $s$ の関数として見る
$\Large \begin{align*} Y(s-a) &= \int_0^\infty y(t) \mathrm{e}^{-\left(s - a\right)t} \mathrm{d}t \\ &= \int_0^\infty \left( y(t) \mathrm{e}^{at} \right) \mathrm{e}^{-st} \mathrm{d}t \\ &= \mathcal{L}\left[ \mathrm{e}^{at} y(t) \right] \end{align*}$

微分された関数のラプラス変換

$\Large \begin{align*} \mathcal{L}\left[ \frac{\mathrm{d}y(t)}{\mathrm{d}t} \right] &= s \mathcal{L}\left[ y(t) \right] - y(0) \\ &= s Y(s) - y(0) \end{align*}$

$y(t)=1$ のとき
$\Large \begin{gather*} \mathcal{L}\left[ \frac{\mathrm{d}1}{\mathrm{d}t} \right] = s \mathcal{L}\left[ 1 \right] - 1 \\ \mathcal{L}\left[ 1 \right] = \frac{1}{s} \end{gather*}$

$Y(s-a) = \mathcal{L}\left[ \mathrm{e}^{at} y(t) \right]$ より
$\Large \begin{gather*} \frac{1}{s - a} = \mathcal{L}\left[ \mathrm{e}^{at} 1 \right] \\ \mathcal{L}\left[ \mathrm{e}^{at} \right] = \frac{1}{s - a} \end{gather*}$

機械的に $a = \mathrm{j}\omega$ とする
$\Large \begin{gather*} \mathcal{L}\left[ \mathrm{e}^{\mathrm{j} \omega t} \right] = \frac{1}{s - \mathrm{j}\omega} = \frac{s}{s^2 + \omega^2} + \mathrm{j}\frac{\omega}{s^2 + \omega^2} \end{gather*}$

$Y(s-a) = \mathcal{L}\left[ \mathrm{e}^{at} y(t) \right]$ より
$\Large \begin{gather*} \frac{s-a}{\left(s-a\right)^2 + \omega^2} + \mathrm{j}\frac{\omega}{\left(s-a\right)^2 + \omega^2} = \mathcal{L}\left[ \mathrm{e}^{at} \mathrm{e}^{\mathrm{j} \omega t} \right] \end{gather*}$

導出過程は気にしない
$\Large \begin{gather*} \mathcal{L}\left[ \mathrm{e}^{at} \cos \omega t \right] = \frac{s-a}{\left(s-a\right)^2 + \omega^2} \\ \mathcal{L}\left[ \mathrm{e}^{at} \sin \omega t \right] = \frac{\omega}{\left(s-a\right)^2 + \omega^2} \end{gather*}$

$y(t)=t^n$ のとき
$\Large \begin{gather*} \mathcal{L}\left[ \frac{\mathrm{d}t^n}{\mathrm{d}t} \right] = s \mathcal{L}\left[ t^n \right] - 0 \\ \mathcal{L}\left[ t^n \right] = \frac{n}{s} \mathcal{L}\left[ t^{n-1} \right] \end{gather*}$

$n=1$ のとき
$\Large \begin{gather*} \mathcal{L}\left[ t \right] = \frac{1}{s} \mathcal{L}\left[ 1 \right] = \frac{1}{s^2} \end{gather*}$

$n=2$ のとき
$\Large \begin{gather*} \mathcal{L}\left[ t^2 \right] = \frac{2}{s} \mathcal{L}\left[ t \right] = \frac{2}{s^3} \end{gather*}$

積分された関数のラプラス変換

$t<0$ のとき $f(t)=0$ , $g(t)=0$ とするので $\tau > t$ のとき $g(t-\tau) = 0$
$\Large \begin{align*} \int_0^t f(\tau)g(t-\tau)\mathrm{d}\tau &= \int_0^\infty f(\tau)g(t-\tau)\mathrm{d}\tau - \int_t^\infty f(\tau)g(t-\tau)\mathrm{d}\tau \\ &= \int_0^\infty f(\tau)g(t-\tau)\mathrm{d}\tau - 0 \\ &= \int_0^\infty f(\tau)g(t-\tau)\mathrm{d}\tau \end{align*}$

$\Large \begin{align*} \int_{-\tau_0}^\infty f(\tau)g(t-\tau)\mathrm{d}\tau &= \int_{-\tau_0}^0 f(\tau)g(t-\tau)\mathrm{d}\tau + \int_0^\infty f(\tau)g(t-\tau)\mathrm{d}\tau \\ &= 0 + \int_0^\infty f(\tau)g(t-\tau)\mathrm{d}\tau \\ &= \int_0^\infty f(\tau)g(t-\tau)\mathrm{d}\tau \end{align*}$

$\Large \begin{align*} \mathcal{L}\left[ \int_0^t f(\tau)g(t - \tau)\mathrm{d}\tau \right] &= \int_0^\infty \left( \int_0^t f(\tau)g(t-\tau)\mathrm{d}\tau \right) \mathrm{e}^{-st} \mathrm{d}t \\ &= \int_0^\infty \left( \int_0^x f(y)g(x-y)\mathrm{d}y \right) \mathrm{e}^{-sx} \mathrm{d}x \\ &= \int_0^\infty \left( \int_0^\infty f(y)g(x-y)\mathrm{d}y \right) \mathrm{e}^{-sx} \mathrm{d}x \\ &= \int_0^\infty \left( \int_0^\infty f(y)g(x-y) \mathrm{e}^{-sx} \mathrm{d}y \right) \mathrm{d}x \\ &= \int_0^\infty \left( \int_0^\infty f(y) \mathrm{e}^{-sy} g(x-y) \mathrm{e}^{-s\left(x-y\right)} \mathrm{d}y \right) \mathrm{d}x \\ &= \int_0^\infty \left( \int_0^\infty f(y) \mathrm{e}^{-sy} g(x-y) \mathrm{e}^{-s\left(x-y\right)} \mathrm{d}x \right) \mathrm{d}y \\ &= \int_0^\infty f(y) \mathrm{e}^{-sy} \left( \int_0^\infty g(x-y) \mathrm{e}^{-s\left(x-y\right)} \mathrm{d}x \right) \mathrm{d}y \\ &= \int_0^\infty f(y) \mathrm{e}^{-sy} \left( \int_0^\infty g(\tau) \mathrm{e}^{-s\tau} \mathrm{d}\tau \right) \mathrm{d}y \\ &= \left( \int_0^\infty f(y) \mathrm{e}^{-sy} \mathrm{d}y \right) \left( \int_0^\infty g(\tau) \mathrm{e}^{-s\tau} \mathrm{d}\tau \right) \\ &= \left( \int_0^\infty f(t) \mathrm{e}^{-st} \mathrm{d}t \right) \left( \int_0^\infty g(t) \mathrm{e}^{-st} \mathrm{d}t \right) \\ &= \mathcal{L}\left[ f(t) \right] \mathcal{L}\left[ g(t) \right] \\ &= F(s)G(s) \end{align*}$

$g(t)=1$ のとき
$\Large \begin{gather*} \mathcal{L}\left[ \int_0^t f(\tau) \mathrm{d}\tau \right] = F(s) \frac{1}{s} \end{gather*}$

$\Large \begin{align*} \mathcal{L}\left[ \int_0^t y(t) \mathrm{d}t \right] &= \frac{1}{s} \mathcal{L}\left[ y(t) \right] \\ &= \frac{1}{s} Y(s) \end{align*}$

最終値

$\Large \begin{gather*} \left\{ \begin{aligned} &\lim_{s \to 0} \int_0^\infty \left( \frac{\mathrm{d}y(t)}{\mathrm{d}t} \right) \mathrm{e}^{-st} \mathrm{d}t &{} &= \lim_{s \to 0} \mathcal{L}\left[ \frac{\mathrm{d}y(t)}{\mathrm{d}t} \right] &{} &= \lim_{s \to 0} \left( s \mathcal{L}\left[ y(t) \right] - y(0) \right) \\ % &\lim_{s \to 0} \int_0^\infty \left( \frac{\mathrm{d}y(t)}{\mathrm{d}t} \right) \mathrm{e}^{-st} \mathrm{d}t &{} &= \int_0^\infty \left( \frac{\mathrm{d}y(t)}{\mathrm{d}t} \right) \mathrm{d}t &{} &= y(\infty) - y(0) \end{aligned} \right. \end{gather*}$

$\Large \begin{gather*} y(\infty) = \lim_{s \to 0} s \mathcal{L}\left[ y(t) \right] = \lim_{s \to 0} s Y(s) \\ \end{gather*}$

初期値

$\Large \begin{gather*} \left\{ \begin{aligned} &\lim_{s \to \infty} \int_0^\infty \left( \frac{\mathrm{d}y(t)}{\mathrm{d}t} \right) \mathrm{e}^{-st} \mathrm{d}t &{} &= \lim_{s \to \infty} \mathcal{L}\left[ \frac{\mathrm{d}y(t)}{\mathrm{d}t} \right] &{} &= \lim_{s \to \infty} \left( s \mathcal{L}\left[ y(t) \right] - y(0) \right) \\ % &\lim_{s \to \infty} \int_0^\infty \left( \frac{\mathrm{d}y(t)}{\mathrm{d}t} \right) \mathrm{e}^{-st} \mathrm{d}t &{} &= \int_0^\infty 0 \mathrm{d}t &{} &= 0 \end{aligned} \right. \end{gather*}$

$\Large \begin{gather*} y(0) = \lim_{s \to \infty} s \mathcal{L}\left[ y(t) \right] = \lim_{s \to \infty} s Y(s) \\ \end{gather*}$