Amplitude of FFT is not correct

Question

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 values J = fft(g1); % FFT vac_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.

John Michell · Accepted Answer

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 cos^2 function, and since the average value of cos^2 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);
yf = fft(y);
yfH = fft(yH);
ind = 1:N/2;
figure(1)
stem(f(ind),2*abs(yf(ind)))
grid on
figure(2)
stem(f(ind),2*abs(yfH(ind)))
grid on

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