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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] A=Fourier[F]; ListPlot[Abs[A]] 3 Replies
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Posted 9 years ago
 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 9 years ago
 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 9 years ago
 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]] 