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Wrong frequencies in Periodogram?

Vladimir Ivanov

Vladimir Ivanov, Pulkovo observatory

Posted 1 year ago

POSTED BY: Vladimir Ivanov

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Mariusz Jankowski

Mariusz Jankowski, University of Southern Maine

Posted 1 year ago

Hi Vladimir, yes that is a bug. It is due to an incorrect setting of the DataRange option. To correct please see the attached notebook. Attachments: For Vladimir- Pe...nb

POSTED BY: Mariusz Jankowski

Vladimir Ivanov

Vladimir Ivanov, Pulkovo observatory

Posted 1 year ago

Mariusz, thank you.

POSTED BY: Vladimir Ivanov

Henrik Schachner

Henrik Schachner, Radiation Therapy Center, Weilheim, Germany

Posted 1 year ago

Well - ok! I do admit that I was simply not willig to even consider a bug in a function which belongs to the prominent Fourier-family ... At least `PeriodogramArray[]` seems to give a correct result: n = 100; data = Cos[2 \[Pi] Range[0, n - 1]*0.25] // N; pdg = Periodogram[data, PlotRange -> All, ScalingFunctions -> {"Linear", "Linear"}, PlotStyle -> Dotted]; pdata = MapIndexed[{(First[#2] - 1)/n, #1} &, PeriodogramArray[data]]; Show[ListLinePlot[pdata[[;; 50]], PlotRange -> All, GridLines -> {{.25}, None}], pdg]

Well - ok! I do admit that I was simply not willig to even consider a bug in a function which belongs to the prominent Fourier-family ...

At least PeriodogramArray[] seems to give a correct result:

n = 100;
data = Cos[2 \[Pi] Range[0, n - 1]*0.25] // N;
pdg = Periodogram[data, PlotRange -> All, ScalingFunctions -> {"Linear", "Linear"}, PlotStyle -> Dotted];
pdata = MapIndexed[{(First[#2] - 1)/n, #1} &, PeriodogramArray[data]];
Show[ListLinePlot[pdata[[;; 50]], PlotRange -> All, GridLines -> {{.25}, None}], pdg]

enter image description here

POSTED BY: Henrik Schachner

Daniel Lichtblau

Daniel Lichtblau, Wolfram Research

Posted 1 year ago

POSTED BY: Daniel Lichtblau

Henrik Schachner

Henrik Schachner, Radiation Therapy Center, Weilheim, Germany

Posted 1 year ago

I guess this is because your data are not perfectly periodic - in the sense that the first value is not identical to the last one. To assure this you just have to include one more point, e.g.: n = 100; data = Cos[2 \[Pi] Range[0, n]*0.25] // N; The length of the data list is now 101, and the periodogram is probably like you are expecting it.

POSTED BY: Henrik Schachner

Vladimir Ivanov

Vladimir Ivanov, Pulkovo observatory

Posted 1 year ago

Henrik, the discrete Fourier transform works for any signal, not only for ones with the first and last elements equal. We can check it formally, making sure that the explicit inverse Fourier transform Table[Chop[1/Sqrt[n] (Fourier[data] . Table[Exp[-2 \[Pi] I t s / n ], {s, 0, n-1}])], {t, 0, n-1}] brings us back to the initial data. And this expression, in the case of my signal, includes Exp[-2 [Pi] I t * 0.25 ] rather than Exp[-2 [Pi] I t * 0.255102 ], so the true frequency must be 0.25.