[SS] Laplace Transform : $\sin^2 \Omega_0 t u(t)$

$$\sin^2 t = \frac{1-\cos 2t}{2}$$ 를 이용한다.

 

$\sin^2 \Omega_0 t$는 다음과 같이 전개가 가능함.

$$\begin{aligned}\sin^2(\Omega_0 t)&=\frac{1-\cos 2\Omega_0t}{2}\\ &=\frac{1}{2}-\frac{1}{2}\cos 2\Omega_0t\end{aligned}$$

 

이를 (Unilateral )Laplace transform하면 다음과 같음.

 

$$\begin{aligned}\mathscr{L}\left[\sin^2\Omega_0 t\right]&=\mathscr{L}\left[\frac{1}{2}\right]-\mathscr{L}\left[\frac{1}{2}\cos 2\Omega_0t\right]\\ &=\frac{1}{2}\mathscr{L}\left[1\right]-\frac{1}{2}\mathscr{L}\left[\cos 2\Omega_0t\right]\\ &=\frac{1}{2}\frac{1}{s}- \frac{1}{2} \frac{s}{s^2+(2\Omega_0)^2}\\ &=\frac{1}{2s}- \frac{s}{2s^2+2(2\Omega_0)^2}\\ &=\frac{\left\{2s^2+2(2\Omega_0)^2\right\}-2s^2}{(2s)\left\{2s^2+2(2\Omega_0)^2\right\}}\\ &=\frac{\left\{s^2-(2\Omega_0)^2\right\}-s^2}{(s)\left\{2s^2+2(2\Omega_0)^2\right\}}\\ &=\frac{4\Omega_0^2}{2s\left(s^2+4\Omega_0^2\right)}\\ &=\frac{2\Omega_0^2}{s\left(s^2+4\Omega_0^2\right)} \end{aligned}$$