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受欢迎的微积分 >

inverselaplace 1/(s^2+2pis+pi^2)

解答

的拉普拉斯逆变换

\frac{1}{s ^{2} + 2 \cdot π \cdot s + π ^{2}}

解答

e^{- π t} t

求解步骤

L^{- 1} {\frac{1}{s ^{2} + 2 π s + π ^{2}}}

\frac{1}{s ^{2} + 2 π s + π ^{2}} = \frac{1}{( s + π ) ^{2}}

= L^{- 1} {\frac{1}{( s + π ) ^{2}}}

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

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

L^{- 1} {\frac{1}{s ^{2}}} : t

= e^{- π t} t

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