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

解答

的拉普拉斯逆变换

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

- \frac{3}{16} e^{- 3 t} + \frac{e ^{- 3 t} t}{4} + \frac{3}{16} e^{t}

求解步骤

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

将

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

用部份分式展开:

- \frac{3}{16 ( s + 3 )} + \frac{1}{4 ( s + 3 ) ^{2}} + \frac{3}{16 ( s - 1 )}

= L^{- 1} {- \frac{3}{16 ( s + 3 )} + \frac{1}{4 ( s + 3 ) ^{2}} + \frac{3}{16 ( s - 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{3}{16 ( s + 3 )}} + L^{- 1} {\frac{1}{4 ( s + 3 ) ^{2}}} + L^{- 1} {\frac{3}{16 ( s - 1 )}}

L^{- 1} {\frac{3}{16 ( s + 3 )}} : \frac{3}{16} e^{- 3 t}

L^{- 1} {\frac{1}{4 ( s + 3 ) ^{2}}} : \frac{e ^{- 3 t} t}{4}

L^{- 1} {\frac{3}{16 ( s - 1 )}} : \frac{3}{16} e^{t}

= - \frac{3}{16} e^{- 3 t} + \frac{e ^{- 3 t} t}{4} + \frac{3}{16} e^{t}

x \to 0 - lim (\frac{3}{1 + e ^{\frac{1}{x}}})

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