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I know that the total harmonic distortion (THD) in a power system with n harmonics is calculated by the below equation:

THD=\$\frac{\sqrt{(V_2^2+...+V_n^2)}}{V_1}\$

but what about any single harmonic? Is it true that calculate harmonic distortion specifically caused by third harmonic (for example) this way?:

\$3^{rd}HD=\frac{V_3}{V_1}\$

if it's incorrect then, how to calculate it?

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THD=\$\sqrt{\frac{(V_2^2+...+V_n^2)}{V_1}}\$

The THD formula you have looks wrong to me so...

The RMS of all harmonics \$\sqrt{V_2^2 + V_3^2 + ... V_N^2}\$

Then this is divided by V1: -

THD = \$\dfrac{\sqrt{V_2^2 + V_3^2 + ... V_N^2}}{V_1}\$

This now makes sense because it's the ratio of two voltages - you had V1 within the sq root area and this is wrong.

So, to answer your question, yes the 3rd harmonic distortion is \$\dfrac{V_3}{V_1}\$

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  • \$\begingroup\$ yes you're right it was a mistake, sorry.And could you let me know why only odd ones are important? And is it true that 9th one is not? If Yes, why? \$\endgroup\$ – Reyhaneh Amouie Jun 21 '14 at 9:34
  • \$\begingroup\$ All harmonics are relevant even if they are a nuisance. I know of nothing particular about the 9th harmonic that sets it out as being different. The ear is less tolerant of odd harmonics that's for sure. \$\endgroup\$ – Andy aka Jun 21 '14 at 9:46
  • \$\begingroup\$ That mistake might as well be mine, I edited the question and that error might have slipped from my keyboard. \$\endgroup\$ – Vladimir Cravero Jun 21 '14 at 10:09

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