I start from a signal, I do discrete Fourier transform of the signal, and I want to get back this same signal doing a discrete inverse Fourier transform. However, when I do so, I do not obtain the same signal back. What am I missing?

Here I give the code of a simple example:

signal[i_] := 
 UnitStep[(i*samplingPeriode) + 0.5] - 
  UnitStep[(i*samplingPeriode) - 0.5]

initialTime = -5;
finalTime = 5;
signalDuration = finalTime - initialTime;

bandwidth = 50;
nyquistRate = 2*bandwidth;
nyquistSamplingPeriode = 1/nyquistRate;

samplingFrequency = 50*nyquistRate;
samplingPeriode = 1/samplingFrequency;
numberOfSamples = signalDuration/samplingPeriode;

yValuesSampledSignal = 
  Table[signal[i], {i, initialTime/samplingPeriode, 
xValuesSampledSignal = 
  Table[i*samplingPeriode, {i, initialTime/samplingPeriode, 
sampledSignal = 
  Partition[Riffle[xValuesSampledSignal, yValuesSampledSignal], 2];

yValuesDiscreteFourierTransform = 
   Chop[Fourier[yValuesSampledSignal, FourierParameters -> {1, -1}]];
xValuesDiscreteFourierTransform = 
  Table[i*(1/signalDuration), {i, 0, numberOfSamples}];
discreteFourierTransform = 
    yValuesDiscreteFourierTransform], 2];

absYValuesDiscreteFourierTransform = 
    Fourier[yValuesSampledSignal, FourierParameters -> {1, -1}]];
absDiscreteFourierTransform = 
    absYValuesDiscreteFourierTransform], 2];

argYValuesDiscreteFourierTransform = 
    Fourier[yValuesSampledSignal, FourierParameters -> {1, -1}]];
argDiscreteFourierTransform = 
    argYValuesDiscreteFourierTransform], 2];

yValuesRecontructedSignal = 
  InverseFourier[yValuesSampledSignal, FourierParameters -> {1, -1}];
xValuesRecontructedSignal = xValuesSampledSignal;
reconstructedSignal = 
   Riffle[xValuesRecontructedSignal, yValuesRecontructedSignal], 2];

ListLinePlot[sampledSignal, PlotRange -> All]
ListPlot[Take[absDiscreteFourierTransform, 50], Filling -> Axis, 
 PlotRange -> All]
ListPlot[Take[argDiscreteFourierTransform, 50], Filling -> Axis, 
 PlotRange -> All]
ListLinePlot[Abs[reconstructedSignal], PlotRange -> All]

enter image description here


closed as off-topic by Daniel Lichtblau, user9660, Yves Klett, Öskå, dr.blochwave Dec 21 '15 at 18:41

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question arises due to a simple mistake such as a trivial syntax error, incorrect capitalization, spelling mistake, or other typographical error and is unlikely to help any future visitors, or else it is easily found in the documentation." – Daniel Lichtblau, Community, Yves Klett, Öskå, dr.blochwave
If this question can be reworded to fit the rules in the help center, please edit the question.

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  • $\begingroup$ As I remember, the reconstructed signal have to be created from Fourier image of initial one: recsig=InverseFourier@Fourier@sig but in your sample of code I've seen the recsig=InverseFourier[sig] which is wrong. $\endgroup$ – Rom38 Dec 21 '15 at 7:08
  • $\begingroup$ This is nicely set up for a first question to the forum, but, as it seems to arise from a small programming error, I'm voting to close. $\endgroup$ – Daniel Lichtblau Dec 21 '15 at 16:28

Rom38 is correct, you did have a typo where you were applying the inverse transform on the original data, not on the transformed data. Just change that line to read

yValuesRecontructedSignal = (1/samplingPeriode) InverseFourier[
    yValuesDiscreteFourierTransform, FourierParameters -> {1, -1}];

Also, I find it easier and more clear to use Transpose, as in

`Partition[Riffle[ list1, list2], 2] = Transpose @ {list1, list2}`

But really, you don't have to keep the data as a 2D list of {x, y} values. You run into trouble when you run your last plotting command, ListLinePlot[Abs[reconstructedSignal]] - since it takes the absolute values of both axes (meaning your negative time values are turned into positive time values).

I find it easier to just specify the DataRange of the ListLinePlot, as in

 DataRange -> {initialTime, finalTime}]
 DataRange -> {initialTime, finalTime}]

enter image description here

where I'm using the shorthand notation, f[ g[ x] ] = f @ g @ x

  • $\begingroup$ Hi, Thank you all for your replies! It helped a lot! Sorry for the mistake... $\endgroup$ – Gabriel Dec 21 '15 at 23:33

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