inverselaplace 1/(s^2+40s+2000)

解答

的拉普拉斯逆变换

\frac{1}{s ^{2} + 40 s + 2000}

e^{- 20 t} \frac{1}{40} sin (40 t)

求解步骤

L^{- 1} {\frac{1}{s ^{2} + 40 s + 2000}}

\frac{1}{s ^{2} + 40 s + 2000} = \frac{1}{( s + 20 ) ^{2} + 1600}

= L^{- 1} {\frac{1}{( s + 20 ) ^{2} + 1600}}

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 + 20 ) ^{2} + 1600} : a = - 20, F (s) = \frac{1}{s ^{2} + 1600}

= e^{- 20 t} L^{- 1} {\frac{1}{s ^{2} + 1600}}

L^{- 1} {\frac{1}{s ^{2} + 1600}} : \frac{1}{40} sin (40 t)

= e^{- 20 t} \frac{1}{40} sin (40 t)

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