For a good example of the conflicting definitions take a look at the following link. This one actually has the two conflicting definitions listed one straight after the other and wrongly claims that they are "equivalent". Apparently that author thinks that a quantity and it's square-root are "equivalent", how brain-dead is that???Link : http://www.answers.com/topic/total-harmonic-distortion
Apparently this was fixed in the original wikipedia article on the 9th of November.
It's been rather awhile since I had any engineering classes... Seems to me the discussion here is not about the definition of THD, but rather how it is expressed. Off hand I cannot recall any time where the context left any ambiguity. I see no real problem here.
...Anyway the main purpose of this thread was to clarify which definition is commonly used (or assumed) when reading audio THD specification. I think we've now established that it is an ampletude ratio and not a power ratio. I think I'll go edit wikipedia again.
..."Most audio Total Harmonic Distortion (THD) measurement systems are in fact Total Harmonic Distortion plus Noise (THD+N) analyzers. They operate by removing the fundamental from the test signal with a sharply tuned band reject or “notch” filter and measuring everything that remains. The amplitude of this “residual” is compared to the amplitude of the fundamental and the result is expressed as a percentage or dB figure."AP might not have the final word on definitions, but they sure know what they are talking about
I don't know how it is used in the audio field, but using the amplitudes makes no sense to me.
btw in some cases the ratio between the amplitudes of two signal is equal to the ratio of the power.(for example when the two signals are sinusoids)
I found this strange definition on http://www.maxim-ic.com/glossary/index.cfm...V/ID/308/Tm/THD"Total Harmonic Distortion (THD): A measure of signal distortion which assesses the energy that occurs on harmonics of the original signal. It is specified as a percentage of the signal amplitude.As an example, if a 12kHz signal is applied to the input, THD would look at energy on the output occurring at 24kHz, 36kHz, 48kHz, etc. and compare it to the energy occurring at 12kHz."
Can you elaborate on why it makes no sense to you?
No, for a resistive load the power ratio is always equal to the square of the amplitude ratio. This is true for sinusoids and other wave shapes as well.
Quote from: uart on 01 January, 2007, 06:13:58 AMCan you elaborate on why it makes no sense to you?QuoteNo, for a resistive load the power ratio is always equal to the square of the amplitude ratio. This is true for sinusoids and other wave shapes as well.Please give me more time to focus the whole question. Also I can check witch definition is used by some popular electronic CAD.btw, when you talk about "resistive load" you are referring to a linear system, while THD is in essence a measure of the non-linearity of a system.
The ratio between the squares of amplitudes of a square wave and a sinusoid it is not equal to the ratio of their power... think what can happens if their amplitude is equal!
This is the reason because I don't understand the sense of using "amplitudes". However, there is a relationship between the square of the amplitudes of the harmonics and the power of the signal, and this can be used to define the THD using the amplitude of the harmonics... but I'm getting confused, I think I need to go over the Parseval theorem.
A "resistive load" is a linear system, but a linear system is not necessarily resisitive.
QuoteThe ratio between the squares of amplitudes of a square wave and a sinusoid it is not equal to the ratio of their power... think what can happens if their amplitude is equal!Actually, it is. Let's assume we have a resistive load to avoid dealing with complex numbers and phase differences. Then P = V^2/R. So (P1)^2 / (P2)^2 = ((V1)^2/R) / ((V2)^2/R) = ((V1)^2)/((V2)^2) = ((V1)/(V2))^2.
What does exactly it mean "the ratio of the rms voltage of the harmonics"? maybe it is not sum(Vnrms), butsqrt(sum(Vnmax^2)) that is the rms of the harmonics, using the mathematical definition of root mean square!
After this research I think I have few doubts about THD ^^. I vote for the square root of the ratio of powers. The first definition at wikipedia is probably wrong.
However, I don't know if I would say the first definition on wikipedia is wrong; I don't see any reason to doubt their assertion that some people use a THD based on power ratio rather than amplitude ratio and that "It is unfortunate that these two conflicting definitions of THD (one as a power ratio and the other as an amplitude ratio) are both in common usage." It's possible that they may overstate the number of people who use the power ratio; I don't really know.
btw, when you talk about "resistive load" you are referring to a linear system, while THD is in essence a measure of the non-linearity of a system.
btw I think I must clarify: it is the square root of a ratio of (average) powers because the relationship that exists between the average power and the amplitude of the harmonics comes from a not so trivial theory.
I simply didn't found any place, but wikipedia, where the definition doesn't use the square root. I found also that the THD% is obliviously defined as 100*THD
This is correct, and it is equal to the definition given by wikipedia that doesn't use the square root, but I don't understand why the need of using RMS values, since we are using harmonics (sinusoids) and their RMS is Vnmax/sqrt(2), so the ratio sum(Vnrms^2)/V1rms is obliviously equal to sum(Vnmax^2)/V1max^2
Due purely to history, it's either expressed in dB, just like SNR usually is, or it's expressed as the amplitude ratio, which can be calculated via 10^(db/20) just like you'd expect.If it's percent, you multiply the amplitude ratio by 100, to make it percentage.So, 1% -> 40dB, and so on.
Look I've got no problems with that, it's how I've always calcuated THD. But I do have a problem with people referring to it as a power ratio (and not including the square-root in the formula) when, as you've pointed out, it (the sqrt) clearly is required to give correct perentage figures.So the Wikipedia definition is wrong as are numerous other texts and sources -YOU NEED THE SQUARE ROOT SIGN IN THE DEFINITION OTHERWISE 1% IS NOT 40dB IT'S ONLY 20dB. CAN'T ANYBODY BUT MYSELF SEE THAT!!!!!
So the Wikipedia definition is wrong as are numerous other texts and sources -
Quote from: uart on 02 January, 2007, 08:55:05 PMSo the Wikipedia definition is wrong as are numerous other texts and sources -Well, with a wiki, you get exactly what you paid for, definition-by-popularity-contest.