inverselaplace 1/(s^2+4s+12)

解答

的拉普拉斯逆变换

\frac{1}{s ^{2} + 4 s + 12}

e^{- 2 t} \frac{1}{2 2} sin (22 t)

求解步骤

L^{- 1} {\frac{1}{s ^{2} + 4 s + 12}}

\frac{1}{s ^{2} + 4 s + 12} = \frac{1}{( s + 2 ) ^{2} + 8}

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

应用逆变换法则: if

L^{- 1} {F (s)} = f (t)

则

L^{- 1} {F (s - a)} = e^{a t} f (t)

对于

\frac{1}{( s + 2 ) ^{2} + 8} : a = - 2, F (s) = \frac{1}{s ^{2} + 8}

= e^{- 2 t} L^{- 1} {\frac{1}{s ^{2} + 8}}

L^{- 1} {\frac{1}{s ^{2} + 8}} : \frac{1}{2 2} sin (22 t)

= e^{- 2 t} \frac{1}{2 2} sin (22 t)

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