Amplitude of FFT is not correct

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

[Jonas]

are you sure that your amplitude is still too small? you did your normalization correctly by length of fft and you correctly multiplied the amplitude by two for the one sided spectrum

it is not necessary that the length is dividable by 2

[Meikel Vollmers]

Yeah, i.e. for one setpoint the amplitude of the fundamental is 248V when i use the FFT analysis tool.

My calculation says that the amplitude is only about 122V, so theres always a factor 2 missing.

[Jonas]

can you provide the data file?

[Meikel Vollmers]

Sure, data file is attached. The fundamental is 1060Hz. Powergui FFT says the amplitude is 333V, my calculation says 167V.

SteadyState = 350000;

Answer 1

[David Goodmanson]

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

[Meikel Vollmers]

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!