inverselaplace 1/(s^2+6s+25)

解答

的拉普拉斯逆变换

\frac{1}{s ^{2} + 6 s + 25}

e^{- 3 t} \frac{1}{4} sin (4 t)

求解步骤

L^{- 1} {\frac{1}{s ^{2} + 6 s + 25}}

\frac{1}{s ^{2} + 6 s + 25} = \frac{1}{( s + 3 ) ^{2} + 16}

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

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 + 3 ) ^{2} + 16} : a = - 3, F (s) = \frac{1}{s ^{2} + 16}

= e^{- 3 t} L^{- 1} {\frac{1}{s ^{2} + 16}}

L^{- 1} {\frac{1}{s ^{2} + 16}} : \frac{1}{4} sin (4 t)

= e^{- 3 t} \frac{1}{4} sin (4 t)

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