Signum function은 다음과 같음.
$$ \text{sgn}(t)=\left\{\begin{matrix}1, & \text{ for } t \ge 0 \\ -1, & \text{ for } t<0\end{matrix}\right.$$
2023.07.05 - [.../Signals and Systems] - [SS] Signum function : 부호함수
문제는 sgn은 absolutely integrable하지 않아서 직접적으로 Fourier Transform(FT)를 못 구한다.
때문에 다음과 같이 $e^{-a|t|}$와 unit step function $u(t)$의 도움을 받아 구해야함.
$$x(t)=\text{sgn}(t)=\underset{a\to 0}{\lim} e^{-a|t|}\text{sgn}(t)=\underset{a\to 0}{\lim} e^{-a|t|}[u(t)-u(-t)]$$
위 식을 정리하면 다음과 같음.
$$x(t)=\text{sgn}(t)=\underset{a\to 0}{\lim} [e^{-at}u(t)-e^{at}u(-t)]$$
이를 Fourier Transform하면 다음과 같음.
$$\begin{aligned}X(\Omega)&=\int^\infty_{-\infty}x(t)e^{-j\Omega t}dt \\ &= \int^\infty_{-\infty}\underset{a\to 0}{\lim} [e^{-at}u(t)-e^{at}u(-t)]e^{-j\Omega t}dt \\ &=\underset{a\to 0}{\lim} \left[\int^\infty_{-\infty} [e^{-at}u(t)-e^{at}u(-t)]e^{-j\Omega t}dt \right] \\ &=\underset{a\to 0}{\lim} \left[\int^\infty_{-\infty} e^{-at}u(t)e^{-j\Omega t}dt - \int^\infty_{-\infty} e^{at}u(-t)e^{-j\Omega t}dt \right] \\ &=\underset{a\to 0}{\lim} \left[\int^\infty_{-\infty} u(t)e^{-at-j\Omega t}dt - \int^\infty_{-\infty} u(-t)e^{at-j\Omega t}dt \right] \\ &=\underset{a\to 0}{\lim} \left[\int^\infty_{0} e^{-at-j\Omega t}dt - \int^0_{-\infty} e^{at-j\Omega t}dt \right] \\ &=\underset{a\to 0}{\lim} \left[\int^\infty_{0} e^{-(a+j\Omega) t}dt - \int^\infty_{0} e^{-(a-j\Omega) t}dt \right] \\ &=\underset{a\to 0}{\lim} \left[ \left.\frac{e^{-(a+j\Omega)t}}{-(a+j\Omega)}\right|_{0}^{\infty} -\left.\frac{e^{-(a-j\Omega)t}}{-(a-j\Omega)}\right|_{0}^{\infty} \right] \\ &=\underset{a\to 0}{\lim} \left[ \left\{\frac{e^{-\infty}-e^{0}}{-(a+j\Omega)} \right \} - \left\{ \frac{e^{-\infty}-e^{0}}{-(a-j\Omega)} \right\} \right] \\ &=\underset{a\to 0}{\lim} \left[ \frac{1}{(a+j\Omega)}-\frac{1}{(a-j\Omega)}\right] \\ &=\frac{1}{(j\Omega)}-\frac{1}{(-j\Omega)} \\&=\frac{2}{j\Omega} \end{aligned}$$
즉, 다음과 같이 표기가능함.
$$\mathcal{F}[\text{sgn}(t)]=\frac{2}{j\Omega}$$
2022.09.28 - [.../Signals and Systems] - [SS] Fourier Transform of Unit Step Function
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