ラプラス変換

mathematics

定義と基本的性質

定義と例

F(s) = \int_{0}^{\infty} e^{-s t} f(t) dt

を $f(t)$ のラプラス変換という。$F(s)$ を像関数、$f(t)$ は原関数と呼ぶ。ここで、$f(t)$ は $t>0$ で与えられているとする ($s$ は一般に複素数)。

また

L\left[f(t) \right] = F(s)

と記述することもある。

例題 1

$f(t) = 1$ のラプラス変換を求めよ。

\begin{aligned}
F(s)
	&= \int_{0}^{\infty} e^{-s t} 1 dt \\
	&= \left[ \frac{e^{-s \infty} - 1}{-s} \right] \\
	&= \frac{1}{s}
\end{aligned}

ここで

\begin{aligned}
e^{-s t}
	&= e^{-(\Re s + i \Im s) t} \\
	&= e^{-\Re s t} e^{-i \Im s t} \\
	&= e^{-\Re s t} \left\{\cos(\Im s t) - i \sin(\Im s t) \right\}
\end{aligned}

として、$\Re s > 0$ のとき

\lim_{t \rightarrow \infty} e^{-s t} = 0

となることを利用した。

例題 2

$f(t) = t$ のラプラス変換を求めよ.

\begin{aligned}
F(s)
	&= \int_{0}^{\infty} e^{-s t} t dt \\
	&= \left[\frac{t e^{-s t}}{(-s)} \right]_{0}^{\infty} - \int_{0}^{\infty} \frac{e^{-s t}}{(-s)} 1 dt \\
	&= 0 - \left[ \frac{e^{-s t}}{s^{2}} \right]_{0}^{\infty} \\
	&= \frac{1}{s^{2}}
\end{aligned}

ラプラス変換の性質

線形性

\begin{aligned}
L[c_{1} f_{1}(t) + c_{2} f_{2}(t)]
 &= \int_{0}^{\infty} e^{-s t} \left\{c_{1} f_{1}(t) + c_{2} f_{2}(t)\right\} dt \\
 &= c_{1} \int_{0}^{\infty} e^{-s t} f_{1}(t) dt +
    c_{2} \int_{0}^{\infty} e^{-s t} f_{2}(t) dt \\
 &= c_{1} L[f_{1}(t)] + c_{2} L[f_{2}(t)]
\end{aligned}

これを線形性という.

例題 3

$\sin t$ のラプラス変換を求めよ。

\sin t = \frac{e^{i t} - e^{-i t}}{2 i}

より

\begin{aligned}
F(s)
 &= \int_{0}^{\infty} e^{-s t} \sin t dt \\
 &= \int_{0}^{\infty} e^{-s t} \frac{e^{i t} - e^{-i t}}{2 i} dt \\
 &= \frac{1}{2 i} \int_{0}^{\infty} (e^{-(s - i) t} - e^{-(s + i) t}) dt \\
 &= \frac{1}{2 i} \left[
  \frac{e^{-(s - i) t}}{-(s - i)} -
  \frac{e^{-(s + i) t}}{-(s + i)}
 \right]_{0}^{\infty} \\

 &= \frac{1}{2 i} \left[
  \frac{1}{s - i} -
  \frac{1}{s + i}
 \right]_{0}^{\infty} \\

 &= \frac{1}{2 i} \frac{s + i - s + i}{s^{2} - i^{2}} \\
 &= \frac{1}{s^{2} + 1}
\end{aligned}

相似性

$a>0$ のとき

L[f(a t)] = \int_{0}^{\infty} e^{-s t} f(a t) dt

$a t = x$ とおくと

\begin{aligned}
a t &= x \\
  t &= \frac{ x}{a} \\
 dt &= \frac{dx}{a} \\
t:0 &\rightarrow \infty \\
x:0 &\rightarrow \infty (a>0)
\end{aligned}

より

\begin{aligned}
L[f(a t)]
 &= \int_{0}^{\infty} e^{-s t} f(a t) dt \\
 &= \int_{0}^{\infty} e^{-s \frac{x}{a}} f(x) \frac{dx}{a} \\
 &= \frac{1}{a} \int_{0}^{\infty} e^{-s \frac{x}{a}} f(x) dx \\
 &= \frac{1}{a} F\left(\frac{s}{a}\right)
\end{aligned}

これを相似性という。

例題 4

$\sin(\omega t)$ のラプラス変換を求めよ.

L[\sin t] = \frac{1}{s^{2} + 1} = F(s)

である.

$\omega > 0$ のとき,

\begin{aligned}
L[\sin(\omega t)]
	&= \frac{1}{\omega F\left(\frac{s}{\omega}\right)} \\
	&= \frac{1}{\omega} \frac{1}{\left(\frac{s}{\omega}\right)^{2} + 1} \\
	&=\frac{\omega}{s^{2}+\omega^{2}}
\end{aligned}

$\omega < 0$ のとき,

\sin(\omega t) = -\sin(-\omega t)

より,

\begin{aligned}
L[\sin(\omega t)]
	&= L[-\sin(-\omega t)] \\
	&= -L[\sin(-\omega t)] \\
	&= \frac{-1}{-\omega F\left(\frac{s}{-\omega}\right)} \\
	&= \frac{1}{\omega} \frac{1}{\left(\frac{s}{-\omega}\right)^{2} + 1} \\
	&= \frac{\omega}{s^{2}+\omega ^{2}}
\end{aligned}

よって,

L[\sin(\omega t)] = \frac{\omega}{s^{2}+\omega^{2}}

像関数の移動法則

\begin{aligned}
L[e^{\alpha t} f(t)]
	&= \int_{0}^{\infty} e^{-s t} e^{\alpha t} f(t) dt \\
	&= \int_{0}^{\infty} e^{-(s-\alpha ) t} f(t) dt \\
	&= F(s-\alpha )
\end{aligned}
\therefore L[e^{\alpha t} f(t)] = F(s-\alpha )

これを像関数の移動法則という.

(例) $e^{\alpha t} \sin(\beta t)$ のラプラス変換を求めよ.

\begin{aligned}
L[e^{\alpha t} \sin(\beta t)]
	&= \frac{\beta}{(s-\alpha)^{2}+\beta^{2}} \\
	&= F(s-\alpha )
\end{aligned}

原関数の移動法則

$\mu>0$ の時,

L[f(t-\mu )] = \int_{0}^{\infty} e^{-s t} f(t-\mu) dt

$x=t-\mu$ と置くと,

\begin{aligned}
x &= t-\mu  \\
t &= x+\mu  \\
dt &= dx \\
t &: 0  \rightarrow \infty \\
x &: -\mu \rightarrow \infty
\end{aligned}

より,

\begin{aligned}
L[f(t-\mu)]
	&= \int_{0}^{\infty} e^{-s t} f(t-\mu) dt \\
	&= \int_{-\mu}^{\infty} e^{-s(x+\mu)} f(x) dx \\
	&= e^{-s \mu} \int_{-\mu }^{\infty} e^{-s x} f(x) dx \\
	&= e^{-s \mu} F(s)
\end{aligned}
\therefore L[f(t-\mu)] = e^{-s \mu} F(s)

これを原関数の移動法則という.

原関数の微分法則

\begin{aligned}
L[f'(t)]
	&= \int_{0}^{\infty} e^{-s t} f'(t) dt \\
	&= [e^{-s t} f(t)]_{0}^{\infty} - \int_{0}^{\infty} (-s) e^{-s t} f(t) dt \\
	&= 0 - f(0) + s \int_{0}^{\infty} e^{-s t} f(t) dt \\
	&= s \int_{0}^{\infty} e^{-s t} f(t) dt - f(0) \\
	&= s L[f(t)] - f(0)
\end{aligned}
\therefore L[f'(t)] = s L[f(t)] - f(0)

これを原関数の微分法則という.

また,

\begin{aligned}
L[f''(t)]
	&= s L[f'(t)] - f'(0) \\
	&= s (s L[f(t)] - f(0)) - f'(0) \\
	&= s^{2} L[f(t)] - s f(0) - f'(0)
\end{aligned}
\therefore L[f''(t)] = s^{2} L[f(t)] - s f(0) - f'(0)

同様に,

\begin{aligned}
L[f^{(3)}(t)]
	&= s L[f''(t)] - f''(0) \\
	&= s (s^{2} L[f(t)] - s f(0) - f'(0)) - f''(0) \\
	&= s^{3} L[f(t)] - s^{2} f(0) - s f'(0) - f''(0)
\end{aligned}
\therefore L[f^{(3)}(t)] = s^{3} L[f(t)] - s^{2} f(0) - s f'(0) - f''(0)

一般に,

L[f^{(n)}(t)] = s^{n} L[f(t)] - s^{n-1} f(0) - \cdots - s f^{(n-2)} - f^{(n-1)}

(例) $\sin^{2}(t)$ のラプラス変換を求めよ.

\begin{aligned}
f(t) &= \sin^{2}(t) \\
f'(t)
	&= 2 \sin t \cos t \\
	&= \sin(2t)
\end{aligned}
\begin{aligned}
L[f'(t)] &= s L[f(t)] - f(0) \\
L[\sin(2t)] &= s L[\sin^{2} t] - \sin^{2} 0 \\
L[\sin(2t)] &= s L[\sin^{2} t] \\
L[\sin^{2}(t)]
	&= L[\sin(2t)] \frac{1}{s} \\
	&= \frac{2}{s^{2}+4} \frac{1}{s} \\
	&= \frac{2}{s(s^{2}+4)}
\end{aligned}
\therefore L[\sin^{2}(t)] = \frac{2}{s(s^{2}+4)}

(例) $t^{n}$ のラプラス変換を求めよ.

\begin{aligned}
f(t) &= t^{n} \\
f'(t) &= n t^{n-1} \\
\cdots \\
f^{(n)}(t) &= n!
\end{aligned}
\begin{aligned}
L[n!] &= s^{n} L[t^{n}] - 0 s^{n-1} - 0 s^{n-2} - \cdots 0 \\
n! L[1] &= s^{n} L[t^{n}] \\
n! \frac{1}{s} &= s^{n} L[t^{n}] \\
L[t^{n}] &= \frac{n!}{s^{n+1}}
\end{aligned}
\therefore L[t^{n}] = \frac{n!}{s^{n+1}}

原関数の積分法則

\begin{aligned}
L\left[\int_{0}^{t} f(x) dx\right]
	&= \int_{0}^{\infty} e^{-s t} \left(\int_{0}^{t} f(x) dx \right) dt \\
	&= \left[\frac{e^{-s t}}{-s} \int_{0}^{t} f(x) dx \right]_{0}^{\infty} - \int_{0}^{\infty} \frac{e^{-s t}}{-s} f(t) dt \\
	&= \frac{1}{s} \int_{0}^{\infty} e^{-s t} f(t) dt \\
	&= \frac{1}{s} L[f(x)]
\end{aligned}
\therefore L\left[\int_{0}^{t} f(x) dx\right] = \frac{1}{s} L[f(x)]

これを原関数の積分法則という.

像関数の微分法則

\begin{aligned}
F'(s)
	&= \int_{0}^{\infty} e^{-s t}' f(t) dt \\
	&= \int_{0}^{\infty} (-t) e^{-s t} f(t) dt \\
	&= -\int_{0}^{\infty} e^{-s t} t f(t) dt \\
	&= -L[t f(t)]
\end{aligned}
\therefore L[t f(t)] = -F'(s)

これを像関数の微分法則という.

一般に,

\begin{aligned}
F^{(n)}(s)
    &= \int_{0}^{\infty} \left(e^{-s t}\right)^{(n)} f(t) dt \\
    &= \int_{0}^{\infty} (-t)^{n} e^{-s t} f(t) dt \\
    &= (-1)^{n} \int_{0}^{\infty} e^{-s t} (t)^{n} f(t) dt \\
    &= (-1)^{n} L[t^{n} f(t)]
\end{aligned}
\therefore L[t^{n} f(t)] = (-1)^{n} F^{(n)}(s)

(例) $t \sin(\omega t)$ のラプラス変換を求めよ.

$f(t) = \sin(\omega t)$ とする.

\begin{aligned}
L[f(t)]
	&= L[\sin(\omega t)] \\
	&= \frac{\omega}{s^{2}+\omega^{2}} \\
	&= F(s)
\end{aligned}
\begin{aligned}
L[t f(t)]
	&= -F'(s) \\
	&= \frac{\omega}{(s^{2}+\omega ^{2})^{2}} (s^{2}+\omega ^{2})' \\
	&= \frac{\omega}{(s^{2}+\omega ^{2})^{2}} 2 s \\
	&= \frac{2 \omega s}{(s^{2}+\omega ^{2})^{2}}
\end{aligned}
\therefore L[t \sin(\omega t)] = \frac{2 \omega s}{(s^{2}+\omega ^{2})^{2}}

像関数の積分法則

\begin{aligned}
\int_{s}^{\infty} F(\sigma ) d\sigma 
	&= \int_{s}^{\infty} \left(\int_{0}^{\infty} e^{-\sigma  t} f(t) dt\right) ds \\
	&= \int_{0}^{\infty} \left(\int_{s}^{\infty} e^{-\sigma  t} ds\right) f(t) dt \\
	&= \int_{0}^{\infty} \left[\frac{e^{-\sigma  t}}{-t}\right]_{\sigma =s}^{\sigma =\infty} f(t) dt \\
	&= \int_{0}^{\infty} e^{-s t} \frac{f(t)}{t} dt \\
	&= L\left[\frac{f(t)}{t}\right]
\end{aligned}
\therefore L\left[\frac{f(t)}{t}\right] = \int_{s}^{\infty} F(\sigma) d\sigma 

これを像関数の積分法則という.

(例) $\frac{\sin(\omega t)}{t}$ のラプラス変換を求めよ.

$f(t) = \sin(\omega t)$ とする.

\begin{aligned}
L[f(t)]
	&= \frac{\omega}{s^{2}+\omega^{2}} \\
	&= F(s)
\end{aligned}
L\left[\frac{f(t)}{t}\right] = \int_{s}^{\infty} \frac{\omega}{\sigma^{2}+\omega^{2}} d\sigma 

$\sigma=\omega\tan\theta$ と置くと,

\begin{aligned}
\sigma &= \omega\tan\theta \\
d\sigma &= \omega \frac{d\theta}{\cos^{2}\theta} \\
\sigma^{2}+\omega^{2}
	&= \omega^{2}\tan^{2}\theta+\omega ^{2} \\
	&= \omega^{2}(\tan^{2}\theta+1) \\
	&= \frac{\omega^{2}}{\cos^{2}\theta} \\
\sigma &: s \rightarrow \infty \\
\theta &: \tan^{-1}\left(\frac{s}{\omega}\right) \rightarrow \frac{\pi}{2}
\end{aligned}
\begin{aligned}
L\left[\frac{f(t)}{t}\right]
	&= \int_{s}^{\infty} \frac{\omega}{\sigma^{2}+\omega^{2}} d\sigma  \\
	&= \int_{\tan^{-1}\frac{s}{\omega}}^{\frac{\pi}{2}} \frac{\cos^{2}\theta}{\omega^{2}} \omega \frac{\omega}{\cos^{2}\theta} d\theta  \\
	&= \int_{\tan^{-1}\frac{s}{\omega}}^{\frac{\pi}{2}} d\theta  \\
	&= [\theta ]_{\tan^{-1}\frac{s}{\omega}}^{\frac{\pi}{2}} \\
	&= \frac{\pi}{2} -\tan^{-1}\frac{s}{\omega}
\end{aligned}
\therefore L\left[\frac{\sin(\omega t)}{t}\right] = \frac{\pi}{2} - \tan^{-1}\frac{s}{\omega}

たたみこみのラプラス変換

\begin{aligned}
L[f(t)] L[g(t)]
	&= \int_{0}^{\infty} e^{-s x} f(x) dx \int_{0}^{\infty} e^{-s y} g(y) dy \\
	&= \int_{0}^{\infty} \int_{0}^{\infty} e^{-s(x+y)} f(x) g(y) dx dy
\end{aligned}

$x+y=t$,$x=\tau$ と置く,

\begin{aligned}
x+y &= t \\
x &= \tau  \\

x &= \tau  \\
y &= t-x \\

dx dy &= d\tau  dt \\

\int_{0}^{\infty} \int_{0}^{\infty} dx dy &= \int_{0}^{\infty} \int_{0}^{t} d\tau dt
\end{aligned}
\begin{aligned}
L[f(t)] L[g(t)]
	&= \int_{0}^{\infty} \int_{0}^{\infty} e^{-s(x+y)} f(x) g(y) dx dy \\
	&= \int_{0}^{\infty} \left(\int_{0}^{t} e^{-s t} f(t) g(t-\tau ) d\tau\right) dt \\
	&= \int_{0}^{\infty} ((f*g)_{(t)}) dt \\
	&= L[(f*g)_{(t)}]
\end{aligned}
\therefore L[(f*g)_{(t)}] = L[f(t)] L[g(t)]

たたみこみのラプラス変換は,各々の関数をラプラス変換したものの積となる.

(例) $f(t)=\sin t$,$g(t)=\cos t$ の時の $(f*g)_{(t)}$ を求めよ. また, $L[(f*g)_{(t)}]=L[f(t)] L[g(t)]$ を求めよ.

\begin{aligned}
f(\tau) &= \sin\tau \\
g(t-\tau) &= \cos(t-\tau)
\end{aligned}
\begin{aligned}
(f*g)_{(t)}
	&= \int_{0}^{t} f(\tau) g(t-\tau) d\tau  \\
	&= \int_{0}^{t} \sin\tau \cos(t-\tau) d\tau  \\
	&= \int_{0}^{t} \frac{\sin(\tau + (t-\tau)) + \sin(\tau -(t-\tau))}{2} d\tau  \\
	&= \frac{1}{2} \int_{0}^{t} \left(\sin t + \sin(2\tau - t)\right) d\tau  \\
	&= \frac{1}{2} \left[\tau \sin t - \frac{1}{2} \cos(2\tau-t)\right]_{0}^{t} \\
	&= \frac{1}{2} \left(t \sin t - \frac{1}{2} \cos t - \frac{1}{2} \cos(-t)\right) \\
	&= \frac{1}{2} t \sin t
\end{aligned}
\therefore (f*g)_{(t)} = \frac{1}{2} t \sin t
\begin{aligned}
L[(f*g)_{(t)}]
	&= L\left[\frac{1}{2} t \sin t\right] \\
	&= \frac{1}{2} L\left[t \sin t\right] \\
	&= \frac{1}{2} \frac{2s}{(s^{2}+1)^{2}} \\
	&= \frac{s}{(s^{2}+1)^{2}}
\end{aligned}
\begin{aligned}
L[f(t)] L[g(t)]
	&= \frac{1}{s^{2}+1} \frac{s}{s^{2}+1} \\
	&= \frac{s}{(s^{2}+1)^{2}}
\end{aligned}
\therefore L\left[(f*g)_{(t)}\right] = L[f(t)] L[g(t)]

逆ラプラス変換

ラプラス変換表

原関数 像関数
$\delta(t)$ $1$
$1$ $\frac{1}{s}$
$t$ $\frac{1}{s^{2}}$
$t^{n}$ $\frac{n!}{s^{n+1}}$
$e^{\alpha t}$ $\frac{1}{s-\alpha}$
$t e^{\alpha t}$ $\frac{1}{(s-\alpha)^{2}}$
$\sin(\omega t)$ $\frac{\omega}{s^{2}+\omega^{2}}$
$\cos(\omega t)$ $\frac{s}{s^{2}+\omega^{2}}$
$\exp(\alpha t) \sin(\beta t)$ $\frac{\beta}{(s-\alpha)^{2}+\beta^{2}}$
$\exp(\alpha t) \cos(\beta t)$ $\frac{s-\alpha}{(s-\alpha)^{2}+\beta^{2}}$
$t \sin(\omega t)$ $\frac{2\omega s}{(s^{2}+\omega^{2})^{2}}$
$t \cos(\omega t)$ $\frac{s^{2}-\omega^{2}}{(s^{2}+\omega ^{2})^{2}}$

部分分数分解

TODO