laplacetransform 1-e^{-2t}

解答

拉普拉斯变换

1 - e^{- 2 t}

\frac{2}{s ( s + 2 )}

求解步骤

L {1 - e^{- 2 t}}

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)}

= L {1} - L {e^{- 2 t}}

L {1} : \frac{1}{s}

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

= \frac{1}{s} - \frac{1}{s + 2}

化简

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

= \frac{2}{s ( s + 2 )}

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