laplacetransform te^{-t}cos(t)

解答

拉普拉斯变换

t e^{- t} cos (t)

- \frac{- s ^{2} - 2 s}{( s ^{2} + 2 s + 2 ) ^{2}}

求解步骤

L {e^{- t} t cos (t)}

使用拉普拉斯变换表:

L {t^{k} f (t)} = (- 1)^{k} \frac{d ^{k}}{d s ^{k}} (L {f (t)})

对于

t e^{- t} cos (t) : f (t) = e^{- t} cos (t), k = 1

= (- 1)^{1} \frac{d}{d s} (L {e^{- t} cos (t)})

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

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

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

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

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