Amplitude of FFT is not correct

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Meikel Vollmers
Meikel Vollmers 2021-5-17
评论： Meikel Vollmers 2021-5-17
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.
Regards
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Jonas
Jonas 2021-5-17
can you provide the data file?
Meikel Vollmers
Meikel Vollmers 2021-5-17
Sure, data file is attached. The fundamental is 1060Hz. Powergui FFT says the amplitude is 333V, my calculation says 167V.
SteadyState = 350000;

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David Goodmanson
David Goodmanson 2021-5-17
编辑：David Goodmanson 2021-5-17
Hi Meikel,
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
  1 个评论
Meikel Vollmers
Meikel Vollmers 2021-5-17
Hey David, thank you for the answer!
Windowing is important, but i have never read about the reduction by the average value of the windowing function. So i will just multiply my spectrum with another factor of 2. Thank you!

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