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Math Help - Fourier Transform

  1. #1
    Member
    Joined
    Jan 2010
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    133

    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
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  2. #2
    Junior Member
    Joined
    Jan 2009
    Posts
    56

    Re: Fourier Transform

    delete, I was wrong.
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