inverselaplace ((-2s+24))/((s^2-13))

解答

的拉普拉斯逆变换

\frac{( - 2 s + 24 )}{( s ^{2} - 13 )}

- \frac{13 + 12}{13} e^{- 13 t} + \frac{- 13 + 12}{13} e^{13 t}

求解步骤

L^{- 1} {\frac{( - 2 s + 24 )}{( s ^{2} - 13 )}}

将

\frac{- 2 s + 24}{s ^{2} - 13}

用部份分式展开:

- \frac{13 + 12}{13 ( s + 13 )} + \frac{- 13 + 12}{13 ( s - 13 )}

= L^{- 1} {- \frac{13 + 12}{13 ( s + 13 )} + \frac{- 13 + 12}{13 ( s - 13 )}}

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{13 + 12}{13 ( s + 13 )}} + L^{- 1} {\frac{- 13 + 12}{13 ( s - 13 )}}

L^{- 1} {\frac{13 + 12}{13 ( s + 13 )}} : \frac{13 + 12}{13} e^{- 13 t}

L^{- 1} {\frac{- 13 + 12}{13 ( s - 13 )}} : \frac{- 13 + 12}{13} e^{13 t}

= - \frac{13 + 12}{13} e^{- 13 t} + \frac{- 13 + 12}{13} e^{13 t}

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