RMS values in FFT-plot

Joacim Rådstam

Posted on 07.04.2011 12:51
How is the total RMS calculation defined for the FFT-plot in version 7.0.1? I understand that the first value is calculated for the entire frequency range and the value enclosed by () is calculated for the frequency range currently showing in the plot.
Dejan Cencelj

Posted on 12.04.2011 08:46
[img]http://sk-zalec.org/images/stories/image001.png[/img]
Joacim Rådstam

Posted on 14.04.2011 08:25
I'm unfamiliar with this formulation so I have some additional comments. It would be nice if you could give an online reference for this formulation. Why is the first and last value doubled, x[0] and x[n-1]? The window correction factor for the Hanning window is correct (1/sqrt(1.5)) for the special case of a frequency resolution of 1 Hz. But that's rarely the case, I believe it should be dependent on the resolution df = (sampling frequency)/(window length). The factor for the Flat top window is incorrect. Should be 1/sqrt(3.77) = 0.515 for df = 1. (Using the window defenition in ISO 18431-1.) I really recommend these articles I used for theoretical background information: http://www.bksv.com/doc/bv0031.pdf http://www.bksv.com/doc/bv0032.pdf
DEWESoft Administrator
Latin America Regional Manager
Posted on 18.04.2011 08:42
The first and the last line in FFT contain only half of the power, because FFT is symetric. That is why we need to multiply them by two. The df is also compensated. There is a simple test to check the window: create an formula with (for an example) sinewave of 100 Hz, choose sample rate of 1024 and calculate the FFT. The sine wave result will fall exactly on the frequency line of the FFT. The correction factors compensates the window characteristic for all windows that the amplitude is always correct (1 in our case). If you enable RMS value, also the RMS value will be correct all the time. When the frequency is set to fall in between the lines, the amplitude will be wrong depending on window side lobe characteristic, but the RMS will still be perfect. With this simple test we prove the correctness of the factors.
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