[FieldTrip] Amplitude of power (ft_freqanalysis) (Hoffman, Steven)

[Antoine Ducorps]

Dear Steven,

The true reason of the formula Amp^2 / 2 is that the power estimation is the one of a continuous signal giving the same power. For example, the power you get from the alternating power line (rated as 220v in Europe) is the one you would get from a DC voltage of that amplitude (220v) when driving it though your device (eg simple resistor as a heater). As such, the amplitude of the 220v alternating  voltage is not 220v but 311v (220 x square root of 2).
Another way to see it is that the power is the integral of the the signal, i.e. the area « under »  the sinewave, which is clearly not the area you would get under of a DC line of same amplitude (5v in your case).
Hoping this helps,

	Antoine.


>   2. Amplitude of power (ft_freqanalysis) (Hoffman, Steven)
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>   6. Re: Question about loading in ERP data (Julian Keil)
>   7. Re: Question about loading in ERP data (Amna Hyder)
>   8. Re: Question about loading in ERP data (Amna Hyder)
> 
> 
> Message: 2
> Date: Tue, 26 Nov 2019 20:34:30 +0000
> From: "Hoffman, Steven" <steven.hoffman5 at montana.edu>
> To: "fieldtrip at science.ru.nl" <fieldtrip at science.ru.nl>
> Subject: [FieldTrip] Amplitude of power (ft_freqanalysis)
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> 
> To whom may be able to help me out,
> 
> We were trying to recover the actual amplitude units for our power spectra, and I noticed something interesting in the output from FieldTrip. I ran a simulation where I take a 20Hz sine wave of +/- 5 "Volts", and calculate the power on it (Using the mtfft method). What we expected the amplitude to be was 25 (since power is the V^2 /Hz). However FieldTrip gave a value closer to about 12 V for the amplitude.
> 
> I then calculated the power spectra of the simulated +/- 5 "Volt" sine wave using the Matlab built in "periodogram.m" function. This too yielded an amplitude not of 25, but of 12.5 .
> 
> After some digging I found some documentation in the matlab that said that the units of the power spectra were actually (Amp ^2 / 2) . So in this case that makes sense as (5^2/2 = 12.5).
> 
> My question is why the divide by two? I don't think I've ever seen amplitude expressed this way, its always V^2 / Hz ...
> 
> Attached are figures of my matlab simulations (One for Field Trip, and for matlab's periodogram.m), and the snippet from the Matlab documentation for periodogram.m where it talks about amplitude being Amp^2 / 2.
> 
>