Problem 예제.

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다음과 같은 파형의 라플라스 변환은?

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

$1-2e^{-s}+e^{-2s}$

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

$s(1-2e^{-s}+e^{-2s})$

Solution

$f(t) = t - 2(t-1)u(t-1)+(t-2)u(t-2)$

$\mathcal{L}(t - 2(t-1)u(t-1)+(t-2)u(t-2)) = \dfrac{1}{s^2} - 2e^{-s}\dfrac{1}{s^2}+e^{-2s}\dfrac{1}{s^2}\\= \dfrac{1}{s^2}(1 - 2e^{-s}+e^{-2s})$

Correct answer: 1