inverselaplace 4/(s*(s^2-4s+5))

解答

的拉普拉斯逆变换

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

\frac{4}{5} H (t) - \frac{4}{5} e^{2 t} cos (t) + \frac{8}{5} e^{2 t} sin (t)

求解步骤

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

将

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

用部份分式展开:

\frac{4}{5 s} + \frac{- 4 s + 16}{5 ( s ^{2} - 4 s + 5 )}

= L^{- 1} {\frac{4}{5 s} + \frac{- 4 s + 16}{5 ( s ^{2} - 4 s + 5 )}}

展开

\frac{- 4 s + 16}{5 ( s ^{2} - 4 s + 5 )} : - \frac{4}{5} \cdot \frac{s - 2}{( s - 2 ) ^{2} + 1} + \frac{8}{5} \cdot \frac{1}{( s - 2 ) ^{2} + 1}

= L^{- 1} {\frac{4}{5 s} - \frac{4}{5} \cdot \frac{s - 2}{( s - 2 ) ^{2} + 1} + \frac{8}{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{4}{5 s}} - \frac{4}{5} L^{- 1} {\frac{s - 2}{( s - 2 ) ^{2} + 1}} + \frac{8}{5} L^{- 1} {\frac{1}{( s - 2 ) ^{2} + 1}}

L^{- 1} {\frac{4}{5 s}} : \frac{4}{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{4}{5} H (t) - \frac{4}{5} e^{2 t} cos (t) + \frac{8}{5} e^{2 t} sin (t)

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