inverselaplace 1/((s+2)(s-1)^5)

解答

的拉普拉斯逆变换

\frac{1}{( s + 2 ) ( s - 1 ) ^{5}}

- \frac{1}{243} e^{- 2 t} + \frac{1}{243} e^{t} - \frac{e ^{t} t}{81} + \frac{e ^{t} t ^{2}}{54} - \frac{e ^{t} t ^{3}}{54} + \frac{e ^{t} t ^{4}}{72}

求解步骤

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

将

\frac{1}{( s + 2 ) ( s - 1 ) ^{5}}

用部份分式展开:

- \frac{1}{243 ( s + 2 )} + \frac{1}{243 ( s - 1 )} - \frac{1}{81 ( s - 1 ) ^{2}} + \frac{1}{27 ( s - 1 ) ^{3}} - \frac{1}{9 ( s - 1 ) ^{4}} + \frac{1}{3 ( s - 1 ) ^{5}}

= L^{- 1} {- \frac{1}{243 ( s + 2 )} + \frac{1}{243 ( s - 1 )} - \frac{1}{81 ( s - 1 ) ^{2}} + \frac{1}{27 ( s - 1 ) ^{3}} - \frac{1}{9 ( s - 1 ) ^{4}} + \frac{1}{3 ( s - 1 ) ^{5}}}

利用逆拉普拉斯变换的线性特性:

对于函数

f (s), g (s)

和常数

a, b : L^{- 1} {a \cdot f (s) + b \cdot g (s)} = a \cdot L^{- 1} {f (s)} + b \cdot L^{- 1} {g (s)}

= - L^{- 1} {\frac{1}{243 ( s + 2 )}} + L^{- 1} {\frac{1}{243 ( s - 1 )}} - L^{- 1} {\frac{1}{81 ( s - 1 ) ^{2}}} + L^{- 1} {\frac{1}{27 ( s - 1 ) ^{3}}} - L^{- 1} {\frac{1}{9 ( s - 1 ) ^{4}}} + L^{- 1} {\frac{1}{3 ( s - 1 ) ^{5}}}

L^{- 1} {\frac{1}{243 ( s + 2 )}} : \frac{1}{243} e^{- 2 t}

L^{- 1} {\frac{1}{243 ( s - 1 )}} : \frac{1}{243} e^{t}

L^{- 1} {\frac{1}{81 ( s - 1 ) ^{2}}} : \frac{e ^{t} t}{81}

L^{- 1} {\frac{1}{27 ( s - 1 ) ^{3}}} : \frac{e ^{t} t ^{2}}{54}

L^{- 1} {\frac{1}{9 ( s - 1 ) ^{4}}} : \frac{e ^{t} t ^{3}}{54}

L^{- 1} {\frac{1}{3 ( s - 1 ) ^{5}}} : \frac{e ^{t} t ^{4}}{72}

= - \frac{1}{243} e^{- 2 t} + \frac{1}{243} e^{t} - \frac{e ^{t} t}{81} + \frac{e ^{t} t ^{2}}{54} - \frac{e ^{t} t ^{3}}{54} + \frac{e ^{t} t ^{4}}{72}

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