Problem 예제.

Views 537 • Comments 0 • Last Updated at 2 months ago • Full screen

$\mathcal{L}[\sin t] = \dfrac{1}{s^2+1}$ 을 이용하여 ① $\mathcal{L}[\cos \omega t]$ ② $\mathcal{L}[\sin at]$ 를 구하면?

Subscribe to see the correct answer.

① $\dfrac{1}{s+a}$ ② $\dfrac{1}{s-\omega}$

① $\dfrac{1}{s^2-a^2}$ ② $\dfrac{1}{s^2-\omega^2}$

① $\dfrac{1}{s+a}$ ② $\dfrac{s}{s+\omega}$

① $\dfrac{s}{s^2+\omega^2}$ ② $\dfrac{a}{s^2+ a^2}$

Solution

$\mathcal{L}(\cos \omega t) = \dfrac{s}{s^2 + \omega^2}$

$\mathcal{L}(\sin a t) = \dfrac{a}{s^2 + a^2}$

Correct answer: 4