# Fourier Transform

• Oct 6th 2011, 01:17 AM
rebghb
Fourier Transform
Hello Everyone!

I'm trying to find the Fourier transform of a sinc function without using the duality property. Here's what I got so far:\

$\mathcal{F} \{\frac{sin(\pi t)}{\pi t}\} = \int _R \frac{1}{\pi t} \cdot \frac{e^{i\pi t} - e^{-i \pi t}}{2j} \cdot e^{-i2\pi ft} dt$

Now let's not condier the complex exponentials out of the sine for now, include them later and make use fo the shifting property of the complex exponential.

$I= \int _R \frac{1}{\pi t} \cdot e^{-i2\pi f t}dt = \int _R \frac{\cos (2\pi ft)}{\pi t}dt -i \int _R \frac{\sin (2\pi ft)}{\pi t}dt$. Letting $x = 2\pi f t$ and eliminating the odd integral of $x$ we get $I = -i\frac{1}{\pi} \int _R \frac{\sin(x)}{x} dx = -i$.

So the integral $\int _R \frac{1}{2j} \cdot \frac{1}{\pi t} \cdot e^{-i2\pi ft}dt = -1/2$. When adding the complex exponentials, we get this -1/2 shifted to the right and to the left by 1/2. What does that mean? Am I doing something wrong?

NB: I know the FT of a sinc is a rect
• Oct 9th 2011, 12:24 PM
InvisibleMan
Re: Fourier Transform
delete, I was wrong.