相似定理

$3$\displaystyle{L\text{\hspace{1}} \{ f $ at $ \} =\frac{\hspace{2}1\hspace{2}}{\hspace{2}a\hspace{2}}F $ \frac{\hspace{2}s\hspace{2}}{\hspace{2}a\hspace{2}} $ }$

$3$\displaystyle{L\text{\hspace{1}} \{ \frac{\hspace{2}1\hspace{2}}{\hspace{2}a\hspace{2}}f $ \frac{\hspace{2}t\hspace{2}}{\hspace{2}a\hspace{2}} $ \} =F $ as $ }$

■証明

$3$\displaystyle{L\text{\hspace{1}} \{ f $ at $ \} = \int _{0}^{\infty} {e}^{ - st }f $ at $ dt }$

ここで， $3$\displaystyle{{\tau}=at}$ とおくと， $3$\displaystyle{dt=\frac{\hspace{2}1\hspace{2}}{\hspace{2}a\hspace{2}}d{\tau}}$ , $3$\displaystyle{f $ at $ =f ${\tau}$ }$ となり

$3$\displaystyle{L\text{\hspace{1}} \{ f ${\tau}$ \} = \int _{0}^{\infty} {e}^{ - \frac{\hspace{2}s\hspace{2}}{\hspace{2}a\hspace{2}}{\tau} }f ${\tau}$ \text{\hspace{1}}\cdot \text{\hspace{1}}\frac{\hspace{2}1\hspace{2}}{\hspace{2}a\hspace{2}}d{\tau} }$

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}a\hspace{2}} \int _{0}^{\infty} {e}^{ - \frac{\hspace{2}s\hspace{2}}{\hspace{2}a\hspace{2}}{\tau} }f ${\tau}$ d{\tau} }$

$3$\displaystyle{=\frac{\hspace{2}1\hspace{2}}{\hspace{2}a\hspace{2}}F $ \frac{\hspace{2}s\hspace{2}}{\hspace{2}a\hspace{2}} $ }$

$3$\displaystyle{{\tau}=at}$ の関係から

$3$\displaystyle{L\text{\hspace{1}} \{ f $ at $ \} =\frac{\hspace{2}1\hspace{2}}{\hspace{2}a\hspace{2}}F $ \frac{\hspace{2}s\hspace{2}}{\hspace{2}a\hspace{2}} $ }$

■証明

$3$\displaystyle{L\text{\hspace{1}} \{ f $ \frac{\hspace{2}t\hspace{2}}{\hspace{2}a\hspace{2}} $ \} = \int _{0}^{\infty} {e}^{ - st }f $ \frac{\hspace{2}t\hspace{2}}{\hspace{2}a\hspace{2}} $ dt }$

ここで， $3$\displaystyle{{\tau}=\frac{\hspace{2}t\hspace{2}}{\hspace{2}a\hspace{2}}}$ とおくと， $3$\displaystyle{dt=ad{\tau}}$ , $3$\displaystyle{f $ at $ =f ${\tau}$ }$ となり

$3$\displaystyle{L\text{\hspace{1}} \{ f ${\tau}$ \} = \int _{0}^{\infty} {e}^{ - sa{\tau} }f ${\tau}$ \text{\hspace{1}}\cdot \text{\hspace{1}}ad{\tau} }$

$3$\displaystyle{=a \int _{0}^{\infty} {e}^{ - sa{\tau} }f ${\tau}$ d{\tau} }$

$3$\displaystyle{=aF $ as $ }$

$3$\displaystyle{{\tau}=\frac{\hspace{2}t\hspace{2}}{\hspace{2}a\hspace{2}}}$ の関係から

$3$\displaystyle{L\text{\hspace{1}} \{ f $ \frac{\hspace{2}t\hspace{2}}{\hspace{2}a\hspace{2}} $ \} \text{\hspace{1}}}$ $3$\displaystyle{=aF $ as $ }$

ラプラス変換の線形性より

$3$\displaystyle{\frac{\hspace{2}1\hspace{2}}{\hspace{2}a\hspace{2}}\text{\hspace{1}}L\text{\hspace{1}} \{ f $ \frac{\hspace{2}t\hspace{2}}{\hspace{2}a\hspace{2}} $ \} =L\text{\hspace{1}} \{ \frac{\hspace{2}1\hspace{2}}{\hspace{2}a\hspace{2}}f $ \frac{\hspace{2}t\hspace{2}}{\hspace{2}a\hspace{2}} $ \} =F $ as $ }$

ホーム>>カテゴリー分類>>微分>>ラプラス変換>>相似定理

学生スタッフ作成

　最終更新日： 2023年6月6日