inverselaplace 1/((s^2+4)^2*s)

解答

的拉普拉斯逆变换

\frac{1}{( s ^{2} + 4 ) ^{2} \cdot s}

- \frac{1}{16} cos (2 t) - \frac{t sin ( 2 t )}{16} + \frac{1}{16} H (t)

求解步骤

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

将

\frac{1}{( s ^{2} + 4 ) ^{2} s}

用部份分式展开:

- \frac{s}{16 ( s ^{2} + 4 )} - \frac{s}{4 ( s ^{2} + 4 ) ^{2}} + \frac{1}{16 s}

= L^{- 1} {- \frac{s}{16 ( s ^{2} + 4 )} - \frac{s}{4 ( s ^{2} + 4 ) ^{2}} + \frac{1}{16 s}}

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{s}{16 ( s ^{2} + 4 )}} - L^{- 1} {\frac{s}{4 ( s ^{2} + 4 ) ^{2}}} + L^{- 1} {\frac{1}{16 s}}

L^{- 1} {\frac{s}{16 ( s ^{2} + 4 )}} : \frac{1}{16} cos (2 t)

L^{- 1} {\frac{s}{4 ( s ^{2} + 4 ) ^{2}}} : \frac{t sin ( 2 t )}{16}

L^{- 1} {\frac{1}{16 s}} : \frac{1}{16} H (t)

= - \frac{1}{16} cos (2 t) - \frac{t sin ( 2 t )}{16} + \frac{1}{16} H (t)

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