推移則

${\displaystyle \mathcal{L}\left\{f\left(t\right)\right\}=F\left(s\right)}$ のとき

${\displaystyle \mathcal{L}\left\{f\left(t-a\right)u\left(t-a\right)\right\}=e^{-as}F\left(s\right)}$

ただし， ${\displaystyle u\left(t\right)}$  は， ${\displaystyle t\geqq 0}$  で ${\displaystyle u\left(t\right)=1}$，${\displaystyle t<0}$  で ${\displaystyle u\left(t\right)=0}$  となる関数である．

${\displaystyle \mathcal{L}\left\{e^{-at}f\left(t\right)\right\}=F\left(s+a\right)}$　

■証明

${\displaystyle t>0}$  のとき， ${\displaystyle f\left(t-a\right)u\left(t-a\right)}$  のラプラス変換は定義式より

${\displaystyle \mathcal{L}\left\{f\left(t-a\right)u\left(t-a\right)\right\}}$ ${\displaystyle ={{\displaystyle \int }}_0^{\infty }{e}^{-st}f\left(t-a\right)u\left(t-a\right)dt}$

${\displaystyle t-a=\tau }$  とおくと， ${\displaystyle dt=d\tau }$，${\displaystyle t:0\to \infty }$ のとき${\displaystyle \tau :-a\to \infty }$ となるので

${\displaystyle {{\displaystyle =\int }}_{-a}^{\infty }{e}^{-s\left(\tau +a\right)}f\left(\tau \right)u\left(\tau \right)d\tau }$

${\displaystyle {{\displaystyle =\int }}_{-a}^0{e}^{-s\left(\tau +a\right)}f\left(\tau \right)u\left(\tau \right)d\tau }$${\displaystyle +{{\displaystyle \int }}_0^{\infty }{e}^{-s\left(\tau +a\right)}f\left(\tau \right)u\left(\tau \right)d\tau }$

${\displaystyle u\left(\tau \right)}$  は， ${\displaystyle \tau \geqq 0}$  で ${\displaystyle u\left(\tau \right)=1}$ ，${\displaystyle \tau <0}$  で ${\displaystyle u\left(\tau \right)=0}$より

${\displaystyle ={\int }_0^{\infty }{e}^{-s\left(\tau +a\right)}f\left(\tau \right)d\tau }$

${\displaystyle ={\int }_0^{\infty }{e}^{-sa}·e^{-s\tau }f\left(\tau \right)d\tau }$

${\displaystyle =e^{-sa}{\int }_0^{\infty }{e}^{-s\tau }f\left(\tau \right)d\tau }$

${\displaystyle =e^{-sa}F\left(s\right)}$

■証明

${\displaystyle \mathcal{L}\left\{e^{-at}f\left(t\right)\right\}}$

${\displaystyle ={\int }_0^{\infty }{e}^{-st}\left\{e^{-at}f\left(t\right)\right\}dt}$

${\displaystyle ={\int }_0^{\infty }{e}^{-st}·e^{-at}f\left(t\right)dt}$

${\displaystyle ={\int }_0^{\infty }{e}^{-\left(s+a\right)t}f\left(t\right)dt}$

ラプラス変換の定義式  ${\displaystyle \mathcal{L}\left\{f\left(t\right)\right\}=F\left(s\right)={\int }_0^{\infty }{e}^{-st}f\left(t\right)dt}$  と比較して

${\displaystyle \mathcal{L}\left\{e^{-at}f\left(t\right)\right\}=F\left(s+a\right)}$

　

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最終更新日： 2023年6月6日