Re: Re: Re: FourierTransform of Sinc Function

• To: mathgroup at smc.vnet.net
• Subject: [mg44405] Re: [mg44359] Re: [mg44311] Re: FourierTransform of Sinc Function
• From: Sseziwa Mukasa <mukasa at jeol.com>
• Date: Sat, 8 Nov 2003 04:50:49 -0500 (EST)
• Sender: owner-wri-mathgroup at wolfram.com

On Friday, November 7, 2003, at 05:16 AM, Kieng wrote:

> I am sorry my question is apparently not so clear.
> I generated a list of data from a sinc function.  I am not doing the
> transform from the sinc function.
> If I use Fourier[list of the data], and plot the result, her is where
> I am
> kind of not sure why I don't get a top hat function, even though I
> used very
> fine step for my sinc function data.

First of all, how are you sampling your sinc function?  Fourier assumes
that the first point in a sampled f(n*tau) where tau is the sample
period corresponds to n = 0.  Sampling sinc(x) for values of x < 0 will
result in a linear phase shift in the resulting spectrum.

Secondly, are you aware that the DC component of the Fourier Transform
is returned as the first element of the list not N/2 where N is the
number of samples?  To get a traditional plot you need to rotate the
data by N/2.

Here is a function to sample a sinc function at n points in the
interval [a,b].

sampledsinc[n_Integer,
a_Real, b_Real] := Table[With[{x = a + (b - a) i/(n - 1)}, If[x ==
0, 1,
Sin[x]/x]], {i, 0, n - 1}] /; a < b

And here is a function to plot the Fourier Transform

plotsampledsinc[n_Integer, a_Real, b_Real] := ListPlot[Re[RotateRight[
Fourier[sampledsinc[n,
a, b]], Quotient[
n, 2]] Table[Exp[2 I Pi a (i(
n - 1)/(n (b - a)) - (n - 1)/(2 (b - a)))], {i, 0, n - 1}]],
PlotJoined -> True, PlotRange -> All] /; a < b

Using a finer step, ie increasing n, only increases the bandwidth
plotted, you need to increase n and the sampled region to increase
resolution.

Regards,

Ssezi

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