inverselaplace (e^{-4s})/(s(s^2+16))

解答

的拉普拉斯逆变换

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

H (t - 4) (\frac{1}{16} - \frac{1}{16} cos (4 (t - 4)))

求解步骤

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

应用逆变换法则: if

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

则

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

其中

H (t)

是赫维赛德阶跃函数

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

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

= H (t - 4) (\frac{1}{16} H (t - 4) - \frac{1}{16} cos (4 (t - 4)))

H (t - 4) = {01 t < 4 t \geq 4

= {0 \frac{1}{16} - \frac{1}{16} cos (4 (t - 4)) t < 4 t \geq 4

= H (t - 4) (\frac{1}{16} - \frac{1}{16} cos (4 (t - 4)))

\frac{d y}{d x} = - 3 y

d er i v a t i v e (6 x + 7)^{x}

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