Problem 예제.

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그림과 같은 높이가 1인 펄스의 라플라스 변환은?

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$\dfrac{1}{a-b}\left( \dfrac{e^{-as}+e^{-bs}}{s} \right)$

$\dfrac{1}{s}(e^{-as}+e^{-bs})$

$\dfrac{1}{s}(e^{-as}-e^{-bs})$

$\dfrac{1}{a-b}\left( \dfrac{e^{as}-e^{-bs}}{s} \right)$

Solution

$f(t) = u(t-a) - u(t-b)$

$\mathcal{L}(f(t)) = \mathcal{L}(u(t-a) - u(t-b)) = \dfrac{e^{-as}}{s}- \dfrac{e^{-bs}}{s}=\dfrac{1}{s}(e^{-as}- e^{-bs})$

Correct answer: 3