Problem 예제.

Multiple Choice • Views 662 • Comments 0 • Last Updated at 1 month ago

$f(t) = \mathcal{L}^{-1} \left[ \dfrac{1}{s^2+6s+10} \right]$ 의 값은 얼마인가?

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$e^{-t} \sin 5t$

$e^{-3t} \sin 5 \omega t$

$e^{-3t} \sin t$

$e^{-3t} \cos t$

Solution

$\mathcal{L} (f(t) e^{-at} ) = F(s+a)$

$\mathcal{L}^{-1} \left[ \dfrac{1}{s^2+6s+10} \right] = \mathcal{L}^{-1} \left[ \dfrac{1}{(s+3)^2+ 1} \right] = \sin (t) e^{-3t}$

Correct answer: 3