inverselaplace 3/((s^2+1)(s^2+9))

解答

的拉普拉斯逆变换

\frac{3}{( s ^{2} + 1 ) ( s ^{2} + 9 )}

\frac{3}{8} sin (t) - \frac{1}{8} sin (3 t)

求解步骤

L^{- 1} {\frac{3}{( s ^{2} + 1 ) ( s ^{2} + 9 )}}

将

\frac{3}{( s ^{2} + 1 ) ( s ^{2} + 9 )}

用部份分式展开:

\frac{3}{8 ( s ^{2} + 1 )} - \frac{3}{8 ( s ^{2} + 9 )}

= L^{- 1} {\frac{3}{8 ( s ^{2} + 1 )} - \frac{3}{8 ( s ^{2} + 9 )}}

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{3}{8 ( s ^{2} + 1 )}} - L^{- 1} {\frac{3}{8 ( s ^{2} + 9 )}}

L^{- 1} {\frac{3}{8 ( s ^{2} + 1 )}} : \frac{3}{8} sin (t)

L^{- 1} {\frac{3}{8 ( s ^{2} + 9 )}} : \frac{1}{8} sin (3 t)

= \frac{3}{8} sin (t) - \frac{1}{8} sin (3 t)

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