1
$\begingroup$

I'm trying to estimate a frequency spectrum of a given discrete function. I have a file which is filled of values of Sine-function $sin(t)$, where $t$ runs from $0.0$ to $100.0$ and a discrete step is $dt = 0.01$.

Here's a plot of my file

ListPlot[list, Joined -> True, DataRange -> {0, 100}]

enter image description here

Then I do Fourier

With[{number = 10000}, 
ListPlot[(Sqrt[2 Pi]/Sqrt[number]) Abs[Fourier[list]], 
Joined -> True, PlotRange -> {{0, 2}, All}, DataRange -> {0, 100}]]

enter image description here

and I see that the frequency maximum is shifted - it's not $\omega = 1$ like it should be.

So I think, that the problem lies in the DataRange options, but I don't know how to evaluate maximum DataRange correctly. I thought that it should be the range of time $t$ for my case, but it doesn't work (range of $t$ equals $100$).

$\endgroup$
1
$\begingroup$

I think you are mixing up angular frequency $\omega$ and linear frequency $f$. The frequency of your Sin signal is $f=1/2\pi$. Change to DataRange->{0,2 Pi 10000/100} and the spectrum ListPlot has a peak at 1.0. The discrete FT pairs time $t$ with $f$, not with $\omega$. See the first line in Details and Options in Fourier.

$\endgroup$
0
$\begingroup$

Details of the frequency and axis and how Fourier is scaled may be found here.

This could be what you want. I will work with explicit abscissa rather than using DataRange. I start by generating your data but include the time values.

dt = 0.01;
list = Table[{t, Cos[t]}, {t, 0, 100 - dt, dt}];
ListLinePlot[list, Joined -> True]

Mathematica graphics

I now take the Fourier transform and generate the corresponding frequency values. I use the scaling in FourierParameters of {-1,-1} which means that the mean square value of the time history equals the total square value of the spectrum. Also note that the frequency axis goes from 0 to the sample rate less one point. Here I give the frequency axis in radians per second. (Miss out the 2 Pi if you wish to have it in Hz).

ft = Fourier[list[[All, 2]], FourierParameters -> {-1, -1}];
ff = Table[(n - 1) (2 \[Pi])/(dt Length[ft]), {n, Length[ft]}];
ListLinePlot[Transpose[{ff, Abs[ft]}][[1 ;; 200]], 
 PlotRange -> {{0, 2}, All}]

Mathematica graphics

The ordinate of the peak is 0.5. Note there is another peak in the spectrum corresponding to negative frequency values. Hope that helps.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.