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Fourier of stepfunction results in "non-injective" spectrum

Posted 9 years ago

I create a stepfunction and calculate the Discrete Fourier Transformation with "Fourier". The result does not seem injective. Is that right?

F = Table[ Piecewise[{{1, Abs[i] < 5}}], {i, -10, 10, .001}]; ListPlot[F]

StepFunction

A=Fourier[F]; ListPlot[Abs[A]]

Fourier Transform

POSTED BY: Holger Schlüter
3 Replies

This likely has to do with folding of the results due to the Nyquist behavior (http://en.wikipedia.org/wiki/Nyquist_frequency)

You can see it occur in a Manipulate like this

Manipulate[F = Table[Piecewise[{{1, Abs[i] < x}}], {i, -10, 10, 1}];
 A = Fourier[F]; GraphicsRow[{ListPlot[Abs[F]], ListPlot[Abs[A]]}],
 {x, 1, 10, 1}]

or this

Manipulate[
 F = Table[Piecewise[{{1, i < x}}], {i, -10, 10, 1}];
 A = Fourier[F]; GraphicsRow[{ListPlot[Abs[F]], ListPlot[Abs[A]]}],
 {x, 1, 10, 1}]
POSTED BY: David Reiss

Thank you David, but still I am wondering isn't the result odd? Why would this curve alternate between two values? Shouldn't the fourier transform of a step function be a Sync function?

POSTED BY: Holger Schlüter

Note that since you've chose a very large number of points it appears that function is multivlaued. However this is just an illusion because the point values are so closely spaced and they are alternating between the two "curves". You can see this by choosing a much smaller number of points as in

F = Table[Piecewise[{{1, Abs[i] < 5}}], {i, -10, 10, 1}];
A = Fourier[F]; ListPlot[Abs[A]]

enter image description here

POSTED BY: David Reiss
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