[SS] Fourier Transform of Pulse Signal

 Pulse Signal

다음은 pulse signal의 정의임.

$$ x(t)=\left\{\begin{matrix}1, & |t|\le\frac{\tau}{2} \\ 0, & \text{otherwise} \end{matrix}\right.$$

 

2023.07.05 - [.../Signals and Systems] - [SS] Pulse Signal : Rect function

[SS] Pulse Signal : Rect function

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FT of Pulse

Fourier Transform은 다음과 같음.

$$\begin{aligned} X(\Omega)&=\int^\infty_{-\infty} x(t) e^{-j\Omega t} dt \\ &=\int^{\frac{\tau}{2}}_{-\frac{\tau}{2}} 1 e^{-j \Omega t} dt \\ &=\left. \frac{e^{-j\Omega t}}{-j\Omega} \right| ^{\frac{\tau}{2}} _{-\frac{\tau}{2}} \\ &= \frac{1}{-j\Omega} \left[ e^{-j\Omega \frac{\tau}{2}} - e^{-j\Omega \frac{-\tau}{2}}\right] \\ &= \frac{1}{ j\Omega} \left[ -e^{-j\Omega \frac{\tau}{2}} + e^{-j\Omega \frac{-\tau}{2}}\right] \\ &= \frac{1}{ j\Omega} \left[ -e^{-j\Omega \frac{\tau}{2}} + e^{j\Omega \frac{\tau}{2}}\right] \\ &= \frac{1}{ j\Omega} \left[ e^{j\Omega \frac{\tau}{2}} -e^{-j\Omega \frac{\tau}{2}} \right] \\ &= \frac{2}{ \Omega} \frac{\left[ e^{j\Omega \frac{\tau}{2}} -e^{-j\Omega \frac{\tau}{2}} \right]}{2j} \\ &= \frac{2}{ \Omega} \sin \left(\Omega \frac{\tau}{2} \right)\\ &= \frac {\sin \left(\Omega \frac{\tau}{2} \right)} {\frac{\Omega}{2} } \\ &= \tau \frac {\sin \left(\Omega \frac{\tau}{2} \right)} {\frac{\Omega \tau}{2 } } \\ &= \tau \text{sinc} \left(  \frac{\Omega}{2\pi}\tau \right)\\ \end{aligned}$$

 

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FS of Pulse

위의 pulse signal이 $T$로 반복된다고 가정하고, Fourier Seriese를 계산하면 다음과 같음.

$$ X_k = \frac{\tau}{T} \text{sinc} \left( \frac{k}{T}\tau \right)=\color{red}{\frac{1}{T}} \tau\text{sinc} \left( \color{red}{\frac{k\Omega_0}{2\pi}}\tau \right)$$

 

FT와 FS의 차이를 확인하고, $T$가 어떻게 되는지를 살펴보면 이해가 보다 깊어질 수 있음.

• Fourier Series Coef. : $X_k=\frac{1}{T}\int^{\frac{T}{2}}_{-\frac{T}{2}} \tilde{x}(t) e^{-jk\Omega_0 t} dt=\frac{\Omega_0}{2\pi} \int^{\frac{T}{2}}_{-\frac{T}{2}} \tilde{x}(t) e^{-jk\Omega_0 t}dt$
• Fourier Transform : $X(\Omega)=\int^\infty_{-\infty} x(t) e^{-j\Omega t} dt$

참고로 $\underset{T \to \infty}{\lim}k\Omega_0=\Omega$가 성립함.

즉 주기 $T$를 무한대로 보내면, discrete였던 harmonic $k\Omega_0$가 연속변수 $\Omega$가 된다.

hamonic간의 간격에 해당하던 $\Omega_0$가 $\frac{2\pi}{T}$이기 때문에 $T$가 무한대가 되면 $\Omega_0$는 0에 가까워지게 되기 때문임.

 

2022.09.27 - [Programming/DIP] - [SS] from CTFS to CTFT

[SS] from CTFS to CTFT

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같이보면 좋은 자료들  

2023.10.13 - [.../Signals and Systems] - [SS] Fourier Transform Table

[SS] Fourier Transform Table