laplacetransform e^{-t}((e^t+e^{-t})/2)

解

ラプラス変換

e^{- t} (\frac{e ^{t} + e ^{- t}}{2})

\frac{1}{2 s} + \frac{1}{2 ( s + 2 )}

解答ステップ

L {e^{- t} (\frac{e ^{t} + e ^{- t}}{2})}

変換ルールを適用:

L {f (t)} = F (s)

ならば

L {e^{a t} f (t)} = F (s - a)

e^{- t} \frac{e ^{t} + e ^{- t}}{2}

向け

: f (t) = \frac{e ^{t} + e ^{- t}}{2}, a = - 1

= L {\frac{e ^{t} + e ^{- t}}{2}} (s + 1)

L {\frac{e ^{t} + e ^{- t}}{2}} : \frac{1}{2 ( s - 1 )} + \frac{1}{2 ( s + 1 )}

= \frac{1}{2 ( ( s + 1 ) - 1 )} + \frac{1}{2 ( ( s + 1 ) + 1 )}

= \frac{1}{2 s} + \frac{1}{2 ( s + 2 )}