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# Why does Periodogram not reproduce squared magnitude of Fourier output?

Posted 9 years ago
 All, I am relatively new to Mathematica, and attempting to build some data analysis tools for analysis of time series. Right now I am working my way through understanding the Fourier function, and having difficulty understanding the difference between Fourier and Periodogram outputs. I set up a very simple example of a 10 Hz Sin wave and expected to get a very simple spectrograph with a peak at 10 Hz. I can get close to this with the Fourier function by squaring the absolute value of the output. Based on my reading on the Periodogram function, it should do exactly the same, but i can't understand why the two graphs are so different. data = Table[Sin[2*\[Pi]*10 x], {x, 0, 1, .001}] ; ListLinePlot[Abs[Fourier[data]]^2, PlotRange -> {{0, 60}, All}] Periodogram[data]  I'm hoping to get Periodogram to work because it seems much simpler to apply a window using this function. Any advice or insight would be greatly appreciated. Thanks, Mike.
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Posted 9 years ago
 Period gram has the option ScalingFunction Which has the default value "dB". To compare with the absolute value squared of the Fourier transform you will want to make that option be "Absolute"
Posted 9 years ago
 David, That solved it nicely. Thank you for the help! Mike.
Posted 4 years ago
 ScalingFunctions :)