laplacetransform e^{-2((t-1))}*1((t-1))

解答

拉普拉斯变换

e^{- 2 ((t - 1))} \cdot 1 ((t - 1))

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

求解步骤

L {1 \cdot e^{- 2 ((t - 1))} ((t - 1))}

展开

e^{- 2 (t - 1)} 1 (t - 1) : e^{- 2 t + 2} t - e^{- 2 t + 2}

= L {e^{- 2 t + 2} t - e^{- 2 t + 2}}

利用拉普拉斯变换的线性特性:

对于函数

f (t), g (t)

和常数

a, b : L {a \cdot f (t) + b \cdot g (t)} = a \cdot L {f (t)} + b \cdot L {g (t)}

= e^{2} L {e^{- 2 t} t} - e^{2} L {e^{- 2 t}}

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

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

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

整理

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

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

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