三角関数のラプラス変換

$3$\displaystyle{L\text{\hspace{1}} \{ \sin {\omega}t \} =\frac{\hspace{2}{\omega}\hspace{2}}{\hspace{2} {s}^{2}+{{\omega}}^{2} \hspace{2}}}$

$3$\displaystyle{L\text{\hspace{1}} \{ \cos {\omega}t \} =\frac{\hspace{2}s\hspace{2}}{\hspace{2} {s}^{2}+{{\omega}}^{2} \hspace{2}}}$

■証明：正弦関数

ラプラス変換の定義より

$3$\displaystyle{ L\text{\hspace{1}} \{ \sin {\omega}t \} =\int _{0}^{\infty} {e}^{ - st }\sin {\omega}tdt }$

$3$\displaystyle{={\displaystyle{ \int _{0}^{\infty} $ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}s\hspace{2}}{e}^{ - st } $ }}^{\prime }\sin {\omega}tdt}$

部分積分法を用いて

$3$\displaystyle{ = \[ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}s\hspace{2}}{e}^{ - st }\sin {\omega}t \] _{0}^{ \text{\infty} } }$ $3$\displaystyle{- \displaystyle{ \int _{0}^{\infty} \left({ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}s\hspace{2}}{e}^{ - st } }\right){ \left({ \sin {\omega}t }\right) }^{\prime }dt }}$

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}s\hspace{2}}\displaystyle{ \int _{0}^{\infty} {e}^{ - st } $ {\omega}\cos {\omega}t $ }dt}$

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}s\hspace{2}}\displaystyle{ \int _{0}^{\infty} { $ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}s\hspace{2}}{e}^{ - st } $ }^{\prime }{\omega}\cos {\omega}t }dt}$

もう一度，部分積分法を用いて

$3$\displaystyle{\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}s\hspace{2}} \{ \[ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}s\hspace{2}}{e}^{ - st }{\omega}\cos {\omega}t \] _{0}^{ \text{\infty} } }}$ $3$\displaystyle{- \left.{ \displaystyle{ \int _{0}^{\infty} \left({ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}s\hspace{2}}{e}^{ - st } }\right){ \left({ {\omega}\cos {\omega}t }\right) }^{\prime }dt } }\right\}}$

$3$\displaystyle{\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}s\hspace{2}} \{ \[ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}s\hspace{2}}{e}^{ - st }{\omega}\cos {\omega}t \] _{0}^{ \text{\infty} } }}$ $3$\displaystyle{\displaystyle{ - \displaystyle{\int _{0}^{ \text{\infty} } \frac{\hspace{2}1\hspace{2}}{\hspace{2}s\hspace{2}}{e}^{ - st }{{\omega}}^{2}\sin {\omega}tdt } \} }}$

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}s\hspace{2}} $ \frac{\hspace{2}{\omega}\hspace{2}}{\hspace{2}s\hspace{2}}- \frac{\hspace{2} {{\omega}}^{2} \hspace{2}}{\hspace{2}s\hspace{2}}\displaystyle{ \int _{0}^{\infty} {e}^{ - st }\sin {\omega}tdt } $ }$

ここで

$3$\displaystyle{ L\text{\hspace{1}} \{ \sin {\omega}t \} =\int _{0}^{\infty} {e}^{ - st }\sin {\omega}tdt }$

であることより

$3$\displaystyle{L \{ \sin {\omega}t \} \text{\hspace{1}}}$ $3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}s\hspace{2}} $ \frac{\hspace{2}{\omega}\hspace{2}}{\hspace{2}s\hspace{2}}- \frac{\hspace{2} {{\omega}}^{2} \hspace{2}}{\hspace{2}s\hspace{2}}\text{\hspace{1}}L\text{\hspace{1}} \{ \sin {\omega}t \} $ }$

$3$\displaystyle{L\text{\hspace{1}} \{ \sin {\omega}t \} \text{\hspace{1}}}$ $3$\displaystyle{=\frac{\hspace{2}{\omega}\hspace{2}}{\hspace{2} {s}^{2} \hspace{2}}- \frac{\hspace{2} {{\omega}}^{2} \hspace{2}}{\hspace{2} {s}^{2} \hspace{2}}\text{\hspace{1}}L\text{\hspace{1}} \{ \sin {\omega}t \} }$

この方程式を $3$\displaystyle{L\text{\hspace{1}} \{ \sin {\omega}t \} \text{\hspace{1}}}$ について解くと

$3$\displaystyle{L\text{\hspace{1}} \{ \sin {\omega}t \} $ 1+\frac{\hspace{2} {{\omega}}^{2} \hspace{2}}{\hspace{2} {s}^{2} \hspace{2}} $ =\frac{\hspace{2}{\omega}\hspace{2}}{\hspace{2} {s}^{2} \hspace{2}}}$

$3$\displaystyle{ L\text{\hspace{1}} \{ \sin {\omega}t \} =\frac{\hspace{2}{\omega}\hspace{2}}{\hspace{2} {s}^{2}$1+\frac{\hspace{2} {{\omega}}^{2} \hspace{2}}{\hspace{2} {s}^{2} \hspace{2}}$ \hspace{2}} }$

$3$\displaystyle{=\frac{\hspace{2}{\omega}\hspace{2}}{\hspace{2} {s}^{2}+{{\omega}}^{2} \hspace{2}}}$

■証明：余弦関数

ラプラス変換の定義より

$3$\displaystyle{L\text{\hspace{1}} \{ \cos {\omega}t \} =\displaystyle{ \int _{0}^{\infty} {e}^{ - st }\cos {\omega}tdt }}$

$3$\displaystyle{={\displaystyle{ \int _{0}^{\infty} $ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}s\hspace{2}}{e}^{ - st } $ }}^{\prime }\cos {\omega}tdt}$

部分積分法を用いて

$3$\displaystyle{= \[ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}s\hspace{2}}{e}^{ - st }\cos {\omega}t \] _{0}^{\infty}- \displaystyle{ \int _{0}^{\infty} $ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}s\hspace{2}}{e}^{ - st } $ }{ $ \cos {\omega}t $ }^{\prime }dt}$

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}s\hspace{2}}- \frac{\hspace{2}{\omega}\hspace{2}}{\hspace{2}s\hspace{2}}\displaystyle{ \int _{0}^{\infty} {e}^{ - st } }\sin {\omega}tdt}$

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}s\hspace{2}}- \frac{\hspace{2}{\omega}\hspace{2}}{\hspace{2}s\hspace{2}}{\displaystyle{ \int _{0}^{\infty} $ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}s\hspace{2}}{e}^{ - st } $ }}^{\prime }\sin {\omega}tdt}$

もう一度，部分積分法を用いて

$3$\displaystyle{ =\frac{\hspace{2}1\hspace{2}}{\hspace{2}s\hspace{2}}- \frac{\hspace{2}{\omega}\hspace{2}}{\hspace{2}s\hspace{2}} \{ \[ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}s\hspace{2}}{e}^{ - st }\sin {\omega}t \] _{0}^{\infty} }$ $3$\displaystyle{ - \displaystyle{ {\int }\limits_{0}^{\infty} $ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}s\hspace{2}}{e}^{ - st } $ }{ $ \sin {\omega}t $ }^{\prime }dt \} }$

$3$\displaystyle{ =\frac{\hspace{2}1\hspace{2}}{\hspace{2}s\hspace{2}}- \frac{\hspace{2}{\omega}\hspace{2}}{\hspace{2}s\hspace{2}} \{ \[ - \frac{\hspace{2}1\hspace{2}}{\hspace{2}s\hspace{2}}{e}^{ - st }\sin {\omega}t \] _{0}^{\infty} }$ $3$\displaystyle{ +\displaystyle{ {\int }\limits_{0}^{\infty} \frac{\hspace{2}1\hspace{2}}{\hspace{2}s\hspace{2}}{e}^{ - st } }{\omega}\cos {\omega}tdt \} }$

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}s\hspace{2}}- \frac{\hspace{2} {{\omega}}^{2} \hspace{2}}{\hspace{2} {s}^{2} \hspace{2}}\displaystyle{ \int _{0}^{\infty} {e}^{ - st } }\cos {\omega}tdt}$

ここで

$3$\displaystyle{ L\text{\hspace{1}} \{ \cos {\omega}t \} =\int _{0}^{\infty} {e}^{ - st }\cos {\omega}tdt }$

であことより

$3$\displaystyle{L\text{\hspace{1}} \{ \cos {\omega}t \} =\frac{\hspace{2}1\hspace{2}}{\hspace{2}s\hspace{2}}- \frac{\hspace{2} {{\omega}}^{2} \hspace{2}}{\hspace{2} {s}^{2} \hspace{2}}\text{\hspace{1}}L\text{\hspace{1}} \{ \cos {\omega}t \} }$

この方程式を $3$\displaystyle{L\text{\hspace{1}} \{ \cos {\omega}t \} \text{\hspace{1}}}$ について解くと

$3$\displaystyle{L\text{\hspace{1}} \{ \cos {\omega}t \} $ 1+\frac{\hspace{2} {{\omega}}^{2} \hspace{2}}{\hspace{2} {s}^{2} \hspace{2}} $ }$ $3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}s\hspace{2}}}$

$3$\displaystyle{L\text{\hspace{1}} \{ \cos {\omega}t \} \text{\hspace{1}}=\frac{\hspace{2}1\hspace{2}}{\hspace{2} s $ 1+\frac{\hspace{2} {{\omega}}^{2} \hspace{2}}{\hspace{2} {s}^{2} \hspace{2}} $ \hspace{2}}}$

$3$\displaystyle{=\frac{\hspace{2}s\hspace{2}}{\hspace{2} {s}^{2}+{{\omega}}^{2} \hspace{2}}}$

ホーム>>カテゴリー分類>>微分>>ラプラス変換>>三角関数のラプラス変換

学生スタッフ作成

　最終更新日： 2023年6月6日