inverselaplace (3s)/((s-1)(s^2-4))

解答

的拉普拉斯逆变换

\frac{3 s}{( s - 1 ) ( s ^{2} - 4 )}

- e^{t} - \frac{1}{2} e^{- 2 t} + \frac{3}{2} e^{2 t}

求解步骤

L^{- 1} {\frac{3 s}{( s - 1 ) ( s ^{2} - 4 )}}

将

\frac{3 s}{( s - 1 ) ( s ^{2} - 4 )}

用部份分式展开:

- \frac{1}{s - 1} - \frac{1}{2 ( s + 2 )} + \frac{3}{2 ( s - 2 )}

= L^{- 1} {- \frac{1}{s - 1} - \frac{1}{2 ( s + 2 )} + \frac{3}{2 ( s - 2 )}}

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}{s - 1}} - L^{- 1} {\frac{1}{2 ( s + 2 )}} + L^{- 1} {\frac{3}{2 ( s - 2 )}}

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

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

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

= - e^{t} - \frac{1}{2} e^{- 2 t} + \frac{3}{2} e^{2 t}

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