Problem 예제.

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$f(t) = \dfrac{s+2}{(s+1)^2}$ 의 라플라스 역변환은?

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$1+te^{-t}$

$e^{-t}+te^{-t}$

$1-te^{-t}$

$e^{-t}-te^{-t}$

Solution

$\dfrac{s+2}{(s+1)^2} = \dfrac{k_1}{(s+1)^2} + \dfrac{k_2}{s+1}$

$k_1 = \lim_{s \to -1} \left[ (s+1)^2\dfrac{s+2}{(s+1)^2} \right] = 1$

$k_2 = \lim_{s \to -1} \dfrac{d}{ds}\left[ (s+1)^2\dfrac{s+2}{(s+1)^2} \right] = 1$

$\mathcal{L}^{-1} \left[ \dfrac{1}{(s+1)^2} + \dfrac{1}{s+1} \right] = te^{-t} + e^{-t}$

Correct answer: 2