inverselaplace 1/(s(s^2+4s+5))

解答

的拉普拉斯逆变换

\frac{1}{s ( s ^{2} + 4 s + 5 )}

\frac{1}{5} H (t) - \frac{1}{5} e^{- 2 t} cos (t) - \frac{2}{5} e^{- 2 t} sin (t)

求解步骤

L^{- 1} {\frac{1}{s ( s ^{2} + 4 s + 5 )}}

将

\frac{1}{s ( s ^{2} + 4 s + 5 )}

用部份分式展开:

\frac{1}{5 s} + \frac{- s - 4}{5 ( s ^{2} + 4 s + 5 )}

= L^{- 1} {\frac{1}{5 s} + \frac{- s - 4}{5 ( s ^{2} + 4 s + 5 )}}

展开

\frac{- s - 4}{5 ( s ^{2} + 4 s + 5 )} : - \frac{1}{5} \cdot \frac{s + 2}{( s + 2 ) ^{2} + 1} - \frac{2}{5} \cdot \frac{1}{( s + 2 ) ^{2} + 1}

= L^{- 1} {\frac{1}{5 s} - \frac{1}{5} \cdot \frac{s + 2}{( s + 2 ) ^{2} + 1} - \frac{2}{5} \cdot \frac{1}{( s + 2 ) ^{2} + 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{1}{5 s}} - \frac{1}{5} L^{- 1} {\frac{s + 2}{( s + 2 ) ^{2} + 1}} - \frac{2}{5} L^{- 1} {\frac{1}{( s + 2 ) ^{2} + 1}}

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

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

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

= \frac{1}{5} H (t) - \frac{1}{5} e^{- 2 t} cos (t) - \frac{2}{5} e^{- 2 t} sin (t)

\int \frac{1}{25 x ^{2} - 16} d x

in v erse l a pl a ce \frac{2 s + 3}{s ^{2} + 7}

d er i v a t i v e y = \frac{x}{x + 1}

\int \frac{1}{x ( 8 x + 9 )} d x

t^{3} \frac{d y}{d t} + 3 t^{2} y = 2 cos (t), y (π) = 0