다음과 같은 Real Exponential Function이 있다고 하자.
$$b e^{-at} u(t), a>0$$
해당 Real Exponential Function의 Fourier Transform은 다음과 같음.
$$\begin {align} \int^\infty_{-\infty}b e^{-at} u(t)e^{-j\Omega t} \text{d}t &= b \int^{\infty}_{-\infty}u(t)e ^{-(a+j\Omega)t}\text{d}t\\ \quad &= b \int^{\infty}_{0}e ^{-(a+j\Omega)t}\text{d}t\\ \quad &= b \left [ \frac{e ^{-(a+j\Omega)t}}{-(a+j\Omega)}\right]^\infty_0\\ \quad &= b \left [ 0-\frac{1}{-(a+j\Omega)} \right ] \\ \quad &= \frac{b}{(a+j\Omega)} \\ \\ \therefore \mathcal{FT}\left[b e^{-at} u(t)\right] &= \frac{b}{(a+j\Omega)} \end {align}$$
같이 보면 좋은 자료
2023.10.13 - [.../Signals and Systems] - [SS] Fourier Transform Table
[SS] Fourier Transform Table
0. Fourier Transform Table $x(t)$$X(\omega)$ 1$e^{-a t}u(t), a>0$$\frac{1}{a+j\omega}$ref.2$e^{a t} u(-t) , a>0$$\frac{1}{a-j\omega}$ 3$e^{-a \vert t\vert}, a>0$$\frac{2a}{a^2+\omega^2}$ 4$te^{-a t}u(t), a>0$$\frac{1}{ (a+j\omega)^2}$ 5$t^ne^{-a t}u(t
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