inverselaplace (3(s+1)*e^{-2s})/(s(s+1)(2s+1))

解答

的拉普拉斯逆变换

\frac{3 ( s + 1 ) \cdot e ^{- 2 s}}{s ( s + 1 ) ( 2 s + 1 )}

H (t - 2) (3 - 3 e^{- \frac{t - 2}{2}})

求解步骤

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

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

L^{- 1} {\frac{3 ( s + 1 )}{s ( s + 1 ) ( 2 s + 1 )}} : 3 H (t) - 3 e^{- \frac{t}{2}}

= H (t - 2) (3 H (t - 2) - 3 e^{- \frac{t - 2}{2}})

H (t - 2) = {01 t < 2 t \geq 2

= {0 3 - 3 e^{- \frac{t - 2}{2}} t < 2 t \geq 2

= H (t - 2) (3 - 3 e^{- \frac{t - 2}{2}})

d er i v a t i v e \frac{2 x + 1}{2 x - 1}

d er i v a t i v e f (x) = - x^{3}

\int_{- 3}^{5} 2 y - (y^{2} - 15) d y

\frac{\partial}{\partial x} (3 x arctan (\frac{x}{y}))

d er i v a t i v e 4 e^{x} + x