FFT from measured data - Scaling y-axis

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Mepe
Mepe on 28 May 2018
Edited: David Goodmanson on 19 Jun 2018
Hi all,
As a data basis, I have measured data the volts were recorded. Matlab will now be used to perform an FFT. I have the following questions:
  • What unit do I have on the ordinate axis after the FFT? Also volts?
  • How is scaled correctly? By hiding the negative frequencies (Nyquist), I would actually have to double the amplitude, right?
  • Do I have to multiply all values of the FFT again with 20 * log10 (FFT) to represent the ordinate in db?
Here the basis Matlab Code:
load('TimeDomain.mat')%loading of the time domain signal
L=2500; %length of the signal
Fs=500000;%sampling frequency
N=2^nextpow2(L);%scale factor
t=(0:L-1)*10^-3;%time domain array
f=linspace(0,Fs/2,length(t));%frequency domain array
FFT=abs(fft(Timedomain,N));
figure(1)
plot(f,FFT(1:2500))
Thank you so much for your support!
Frank

Accepted Answer

Star Strider
Star Strider on 28 May 2018
Edited: Star Strider on 28 May 2018
Your code appears to be correct.
I would change two lines:
FFT=abs(fft(Timedomain,N)/L); % Divide By ‘L’ To Scale For Signal Vector Length
figure(1)
plot(f,FFT(1:2500)*2) % Your Second Point Is Correct: Multiply By ‘2’ Since You Are Plotting A One-Sided Fourier Transform
The units are the same as the original (here Volts) unless you square ‘FFT’ or use the equivalent ‘20*log10(FFT)’ transformation, since the units then become power (Watts).
EDIT Corrected typographical error, clarified explanation.
  7 Comments
David Goodmanson
David Goodmanson on 19 Jun 2018
I don't agree that 20*log10(fft) means that the expression refers to Watts. Systems like this one have linear quantities such as voltage, and squared quantities such as power ~~ V^2. The factor of 20 pertains to a linear quantity such as Volts, not Watts.
dB is always expressed in terms of the log of a ratio, and sometimes the denominator of that ratio is a specified reference level.
In terms of power, dB Watts is 10*log10(P_signal/1W).
In terms of voltage, dB Volts is 20*log10(V_signal/1V).
Suppose the voltage signal V1 is doubled to make voltage signal V2. Then the dB increase is 20*log10(V2/V1) = 6 dB. In terms of power, the dB increase is 10*log10(V2^2/V1^2) = 6 dB. It's always the same increase in dB, whether the calculation is done in terms of voltage or in terms of power.
To convert from voltage to power, you have to know something about the impedance of the system, which is most often 50 ohms. If V is expressed in rms volts, then P = Vrms^2/50.
1 Vrms (0 dBVrms) produces .02 W = 20 mW, and 10*(log10(20mW/1mW) = 13 dBmW.
So 0 dBVrms corresponds to 13 dBmW.
If it were a 1 ohm system then 0 dBVrms would correspond to 0 dBW but that assumption is uncommon.

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