laplacetransform t*e^{-4t}cos(5t)

解答

拉普拉斯变换

t \cdot e^{- 4 t} cos (5 t)

- \frac{- s ^{2} - 8 s + 9}{( s ^{2} + 8 s + 41 ) ^{2}}

求解步骤

L {e^{- 4 t} t cos (5 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^{- 4 t} cos (5 t) : f (t) = e^{- 4 t} cos (5 t), k = 1

= (- 1)^{1} \frac{d}{d s} (L {e^{- 4 t} cos (5 t)})

L {e^{- 4 t} cos (5 t)} : \frac{s + 4}{( s + 4 ) ^{2} + 25}

\frac{d}{d s} (\frac{s + 4}{( s + 4 ) ^{2} + 25}) = \frac{- s ^{2} - 8 s + 9}{( s ^{2} + 8 s + 41 ) ^{2}}

= (- 1)^{1} \frac{- s ^{2} - 8 s + 9}{( s ^{2} + 8 s + 41 ) ^{2}}

= - \frac{- s ^{2} - 8 s + 9}{( s ^{2} + 8 s + 41 ) ^{2}}

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