inverselaplace (e^{-2s})/(s(s+1.1))

解答

的拉普拉斯逆变换

\frac{e ^{- 2 s}}{s ( s + 1.1 )}

H (t - 2) (\frac{10}{11} - 0.90909 \dots e^{- 1.1 (t - 2)})

求解步骤

L^{- 1} {\frac{e ^{- 2 s}}{s ( s + 1.1 )}}

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{1}{s ( s + 1.1 )}} (t - 2)

L^{- 1} {\frac{1}{s ( s + 1.1 )}} : \frac{10}{11} H (t) - 0.90909 \dots e^{- 1.1 t}

= H (t - 2) (\frac{10}{11} H (t - 2) - 0.90909 \dots e^{- 1.1 (t - 2)})

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

= {0 \frac{10}{11} - 0.90909 \dots e^{- 1.1 (t - 2)} t < 2 t \geq 2

= H (t - 2) (\frac{10}{11} - 0.90909 \dots e^{- 1.1 (t - 2)})

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