Amplitude of FFT is not correct

Technical Source
2 min readSep 29, 2021

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Hey there,

i want to calculate the power output of a 3 phase inverter. Therefore i have to do a FFT on my voltage Signals, to get the amplitude of the fundamental.

In comparison to the FFT Tool from the Simulink powergui, my amplitude is always lower. I searched for some examples for a correct scaled FFT, but i cant find differences to my code.

f1 = ac_voltage_a.data(SteadyState:1:SteadyState+2^15); % A Timeseries vector is used when system is in steady state
g1 = hanning(length(f1)).*f1; % Using the hanning window
dt=Tsample % Tsample = 1e-6
vac_fenstera_Nfft = length(g1); % Sampled valuesJ = fft(g1); % FFTvac_fenstera_sfft = 2*abs(J)/vac_fenstera_Nfft; % plot(1/dt * (0:(vac_fenstera_Nfft/2-1)) / vac_fenstera_Nfft, (vac_fenstera_sfft(1:vac_fenstera_Nfft/2)));

In comparison to the FFT analysis tool from powergui, the value of my fundamental amplitude is only half the size, even i mutiplied the abs*2.

Where is my fault?

Is it nessesary that the length of g1 is a multiple of 2? I think a DFT could also work out.

ANSWER

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You are multiplying your signal times a window, which reduces the signal amplitude. That has to have a significant effect on the fft. The code below shows the effect, which for the (misnamed) hanning window drops the fundamental by a factor of 2.

For an oscillation at one frequency, the peak amplitude in the frequency domain is reduced by a factor equal to the average value of the window function. Since the Hann window is a cos² function, and since the average value of cos² is 1/2, the frequency domain peak is reduced by that amount.

N = 1000;          % signal length
t = (0:N-1)/N;
f0 = 40;
y = cos(2*pi*f0*t)/N;
yH = (cos(2*pi*f0*t)/N).*hanning(N)';
f = (0:N-1);

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Technical Source
Technical Source

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