inverselaplace (-5)/(s(6(5s+1)))

解答

的拉普拉斯逆变换

\frac{- 5}{s ( 6 ( 5 s + 1 ) )}

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

求解步骤

L^{- 1} {\frac{- 5}{s ( 6 ( 5 s + 1 ) )}}

将

\frac{- 5}{s 6 ( 5 s + 1 )}

用部份分式展开:

- \frac{5}{6 s} + \frac{25}{6 ( 5 s + 1 )}

= L^{- 1} {- \frac{5}{6 s} + \frac{25}{6 ( 5 s + 1 )}}

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{5}{6 s}} + L^{- 1} {\frac{25}{6 ( 5 s + 1 )}}

L^{- 1} {\frac{5}{6 s}} : \frac{5}{6} H (t)

L^{- 1} {\frac{25}{6 ( 5 s + 1 )}} : \frac{5}{6} e^{- \frac{t}{5}}

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

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