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Old 30th January 2014, 01:52 AM   #1
BigE is offline BigE  Canada
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Default CD vs MP3 as THD

Is it possible to measure the difference between the referenced CD signal and the reconstructed MP3 from different bit-rates and report the difference as THD?
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Old 30th January 2014, 01:55 AM   #2
jcx is offline jcx  United States
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it would be a mistake - the whole point of psychoaoustic lossy compression is to use a model of human hearing and throw out the parts of the music signal that would be masked - these are usually the lower harmonics - so you would expect "High THD" from a properly working lossy CODEC with a complex, harmonically dense source like most music

for single or few widely spaced sine tones the compression will likely preserve the spectrum, most psychoustic model based compression will not introduce new distortion harmonics above the numerical precision of the decoder/possible widowing artifacts of the DFT

Schematics needed for 3V 1KHz Sine wave generator

Last edited by jcx; 30th January 2014 at 02:08 AM.
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Old 30th January 2014, 02:02 AM   #3
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Seems to me an altogether better way would be to subtract the mp3 from the original and report the difference signal. In other words, a null test. THD can only be measured on single tone signals as it relies in a very narrow notch filter.
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Old 30th January 2014, 02:10 AM   #4
jcx is offline jcx  United States
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again ignores what the lossy compression is trying to do - it is designed to leave out (= add 100% distortion to cancel to 0) lots of the signal - that you wouldn't perceive in the presence of the parts that it keeps

the compression is working if there is a big (visible) signal difference - that you can't hear - when compared to the original

the only way to judge the quality of psychoacoustic compression is to compare the playback - by ear - to the original


technical measurements make no sense unless you understand the Codec algorithms, psychoaoustic models and want to measure the implementation accuracy to the model

Last edited by jcx; 30th January 2014 at 02:23 AM.
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Old 30th January 2014, 02:16 AM   #5
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Yep - so how does what I said 'ignore what lossy compression is trying to do' please?

<edit> Delete the leading 'yep' as jcx has added some irrelevancies to his original post. Question still stands.
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Old 30th January 2014, 02:18 AM   #6
sreten is offline sreten  United Kingdom
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Hi,

THD+N is a brute force subtractive number that is easy to
measure but tells you next to nothing, unless its very low.

Its totally meaningless applied to non-linear systems like codecs,
as its based on a sliding single tone test, and its fairly obvious if
applied MP3 would do fine until it runs out of bits at the top end.

rgds, sreten.
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Last edited by sreten; 30th January 2014 at 02:25 AM.
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Old 30th January 2014, 02:53 AM   #7
BigE is offline BigE  Canada
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I see measurement of THD can't be made without using test tones.

Not even if FFTS are employed. The distortion components in THD differ from codec information removal.

Last edited by BigE; 30th January 2014 at 02:58 AM.
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Old 30th January 2014, 03:04 AM   #8
sreten is offline sreten  United Kingdom
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Hi,

THD in simple terms cannot be measured on its own.
Its always effectively THD+N in a bridged nulling circuit.

I don't see what FFT's have to do with it.

rgds, sreten.
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Last edited by sreten; 30th January 2014 at 03:06 AM.
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Old 30th January 2014, 03:18 AM   #9
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If MP3 is compressed music, perhaps the THD+N is also compressed (less prominent)?
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Old 30th January 2014, 03:32 AM   #10
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Quote:
Originally Posted by BigE View Post
I see measurement of THD can't be made without using test tones.
If you use a plurality of tones (more than one) then it becomes an IMD measurement, no longer THD.

Quote:
Not even if FFTS are employed.
The FFT might well be useful to analyse the output - if it shows the original tones they can be recognized and set to zero, Its still IMD though and not THD.
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