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Topic: What's the most commonly accepted definition of THD? (Read 18453 times) previous topic - next topic

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  • pdq
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What's the most commonly accepted definition of THD?
Reply #25
I thought of this thread recently when I came across an audio op amp with a spec of 0.00003% THD+noise. My first thought was that this must be a case of using %power instead of %voltage, but a closer examination showed that it really is 0.3 ppm THD+noise as voltage. It is also rated at 0.000009% (0.09 ppm) DC gain linearity error.

My next thought was, why would anyone need this kind of performance in an audio amplifier?

What's the most commonly accepted definition of THD?
Reply #26
According to physical meaning THD should compare energy (intensity) of signals, not their amplitudes so the "power" definition of THD is more correct, I think. Though "amplitude" (RMS voltage) definition is conventionally agreed and commonly used in audio.
keeping audio clear together - soundexpert.org

  • Woodinville
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What's the most commonly accepted definition of THD?
Reply #27
According to physical meaning THD should compare energy (intensity) of signals, not their amplitudes so the "power" definition of THD is more correct, I think. Though "amplitude" (RMS voltage) definition is conventionally agreed and commonly used in audio.


It's not a question of "should" it's a question of how it's been defined.

You can argue that the definition is in fact kind of ridiculous (it is), that it's misnamed (it's really (S+N)/N) where N is all noise sources, including harmonic distortions that we would define by polynomial distortion products, or that it out to be power ratio, BUT


It is what it is.

And like SNR,when  you're working with audio, it's "mostly useless". So it goes.

Yes, yes, if the SNR is 120dB and the peak level is reasonable in terms of human audition, that's enough SNR, but that's not usually the case.
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J. D. (jj) Johnston