ラプラス変換
$$\begin{eqnarray}
\mathfrak{L}\left[ {f\left( t \right)} \right]
&=&\int_0^\infty {f\left( t \right){e^{–st}}}\mathrm{d}t
\end{eqnarray}$$
\( e^{-at} \)のラプラス変換
$$\begin{eqnarray}
f\left( t \right)&=&e^{at}
\\\mathfrak{L}\left[ {f\left( t \right)} \right]
&=& \int_0^\infty {f\left( t \right){e^{ –st}}}\mathrm{d}t
\\&=& \int_0^\infty {e^{at}e^{ –st}}\mathrm{d}t
\\&=& \int_0^\infty {e^{-(s-a)t}}\mathrm{d}t
\\&&\;\cdots\;u=-(s-a)t,\;t:0\rightarrow \infty,\;u:0\rightarrow -\infty
\\&&\;\cdots\;\frac{\mathrm{d}u}{\mathrm{d}t}=-(s-a),\;\mathrm{d}t=\frac{-1}{s-a}\mathrm{d}u
\\&=&\int_0^{-\infty} {e^{u}} \frac{-1}{s-a} \mathrm{d}u
\\&=&\frac{-1}{s-a}\int_0^{-\infty} {e^{u}} \mathrm{d}u
\\&=&\frac{-1}{s-a}\left[ {e^{u}} \right]_0^{-\infty}
\\&=&\frac{-1}{s-a}\left[ {e^{-\infty}} - {e^{0}}\right]
\\&=&\frac{-1}{s-a}\left[ 0 - 1\right]
\\&=&\frac{-1}{s-a}\left(-1\right)
\\&=&\frac{1}{s-a}
\end{eqnarray}$$
0 件のコメント:
コメントを投稿