inverselaplace (e^{-5s})/((s-2)^4)

解答

的拉普拉斯逆变换

\frac{e ^{- 5 s}}{( s - 2 ) ^{4}}

\frac{e ^{2 (t - 5)} H ( t - 5 ) ( t - 5 ) ^{3}}{6}

求解步骤

L^{- 1} {\frac{e ^{- 5 s}}{( s - 2 ) ^{4}}}

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 - 5) L^{- 1} {\frac{1}{( s - 2 ) ^{4}}} (t - 5)

L^{- 1} {\frac{1}{( s - 2 ) ^{4}}} : \frac{e ^{2 t} t ^{3}}{6}

= H (t - 5) \frac{e ^{2 (t - 5)} ( t - 5 ) ^{3}}{6}

= \frac{e ^{2 (t - 5)} H ( t - 5 ) ( t - 5 ) ^{3}}{6}

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