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Problem 예제.

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라플라스 변환

라플라스 변환

$f(t) = u(t-a) - u(t-b)$ 식으로 표시되는 4각파의 라플라스는?

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

$\dfrac{1}{s^2}(e^{as}+e^{bs})$

$\dfrac{1}{s^2}(e^{-as}-e^{-bs})$

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

Solution

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

Correct answer: 1

예제.

Problem 예제.

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