inverselaplace (e^{-6s})/(s^2+2s-8)

解答

的拉普拉斯逆变换

\frac{e ^{- 6 s}}{s ^{2} + 2 s - 8}

H (t - 6) (\frac{1}{6} e^{2 (t - 6)} - \frac{1}{6} e^{- 4 (t - 6)})

求解步骤

L^{- 1} {\frac{e ^{- 6 s}}{s ^{2} + 2 s - 8}}

应用逆变换法则: if

L^{- 1} {F (s)} = f (t)

则

L^{- 1} {e^{- a s} F (s)} = H (t - a) f (t - a)

其中

H (t)

是赫维赛德阶跃函数

= H (t - 6) L^{- 1} {\frac{1}{s ^{2} + 2 s - 8}} (t - 6)

L^{- 1} {\frac{1}{s ^{2} + 2 s - 8}} : \frac{1}{6} e^{2 t} - \frac{1}{6} e^{- 4 t}

= H (t - 6) (\frac{1}{6} e^{2 (t - 6)} - \frac{1}{6} e^{- 4 (t - 6)})

\int \frac{4 x + x ^{2}}{x ^{2}} d x

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