Problem 예제.

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$\dfrac{1}{s(s+1)}$ 의 라플라스 역변환을 구하면?

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$1+e^{-t}$

$e^{-t} \sin t$

$1-e^{-t}$

$e^{-t} \cos t$

Solution

$\mathcal{L}^{-1} \left[ \dfrac{1}{s(s+1)} \right] = \mathcal{L}^{-1} \left[ \dfrac{k_1}{s} + \dfrac{k_2}{s+1} \right]$

$k_1 = \lim_{s \to 0} s\left[ \dfrac{1}{s(s+1)} \right] = 1$

$k_2 = \lim_{s \to -1} (s+1)\left[ \dfrac{1}{s(s+1)} \right] = -1$

$\mathcal{L}^{-1} \left[ \dfrac{1}{s} - \dfrac{1}{s+1} \right] = 1 - e^{-t}$

Correct answer: 3