Extremely high THD in FFT Toolbox

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Marta Grzybek
Marta Grzybek 2020 年 4 月 23 日
コメント済み: Marta Grzybek 2020 年 4 月 24 日
I want to calculate THD of the waveform from my three-phase rectifier in Simulink. I am using the FFT toolbox for this, however when I analyse DC voltage and currents of the rectifier. My fundamental frequency is 400Hz, however as seen in the picture, the spectrum does not show this frequency at all. I am also confused about the very high magnitude of harmonic at 0, which I don't understant. I would appreciate your help.
  3 件のコメント
David Goodmanson
David Goodmanson 2020 年 4 月 23 日
Hi LS,
The large value at zero frequency is because your signal has a large DC offset. If you subtract the mean of the signal from the signal then the DC offset is zero and the somewhat meaningless peak at f=0 goes away.
Marta Grzybek
Marta Grzybek 2020 年 4 月 23 日
編集済み: Marta Grzybek 2020 年 4 月 23 日
Hi Michael,
The fundamental frequency is defined in Hz. When I display THD for the current on AC side on retifier, it correctly shows the 400Hz funadmental harmonic and shows reasonable values of THD. I am very confused about why it displays 2400Hz as fundamental.
Best wishes,
Marta

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David Goodmanson
David Goodmanson 2020 年 4 月 24 日
編集済み: David Goodmanson 2020 年 4 月 24 日
Hi LS,
The signal shows exactly six oscillations in .0025 sec, so it indeed consists almost entirely of the sixth harmonic (plus the DC offset that gives the peak at f=0).
The following plot shows one oscillation of a 400 Hz three phase signal, and since this is a rectifier I tossed in 'abs'. If you follow along the top of the entire envelope, there are six oscillations. So whatever else is going on later in the process, it would not be at all surprising to see a very large peak at the sixth harmonic. Also, note that the envelope has basically no 400 Hz component (which roughly speaking would require the half of the plot to be at a different average height than the other half), so the fundamental basically goes away.
t = 0:1e-5:1/400;
s1 = abs(cos(2*pi*400*t));
s2 = abs(cos(2*pi*400*t + 2*pi/3));
s3 = abs(cos(2*pi*400*t + 4*pi/3));
plot(t,[s1;s2;s3])
  1 件のコメント
Marta Grzybek
Marta Grzybek 2020 年 4 月 24 日
That explains it, thanks a lot!

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