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How many bits, how fast, just how much resolution is enough?


BlueSkyy

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re: it's a fact that a 44.1KHz sample rate is mathematically adequate to reproduce all audible frequencies.

 

I would say (based on a post somewhere up above) that it's a fact that a 44.1KHz sample rate is mathematically adequate to reproduce all audible [sine wave] frequencies.

 

i.e. out to 20 kHz …

The problem may be the relative timing of tones. You can distinguish e.g. multiple HF tones based on their relative timing or relative phase. The smallest audible timing differences are measured in microseconds and simply cannot be encoded using a 44.1KHz sample rate. Some analog recordings might do better..

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Please..not this bull carp again. Redbook can encode timing to within about 55 picoseconds accuracy. It is not limited by the time between samples.

Of course it is. Imagine two equivalent 10 KHz pure analog sinusoidal tones, one started 100 microseconds before the other. A digital (PCM) 44.1/48 KHz sampled recording of this will sound different from a higher resolution digital recording..

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Think about those numbers you picked for a moment please. 10,000 hz tones offset from each other by 100 microseconds.

 

It makes no difference really, but are you talking one tone in each channel or both in the same channel?

 

Hint:___this isn't going to work out the way you think it will.

Ah, yes, wrong example. :) Make that two 10KHz tones with a 1 (or 99) microsecond relative time delay. Would 48K - and (e.g.) 96K samples/second PCM recordings be indistinguishable in the analog domain?

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I would say yes. Instead let me ask, what would convince you?

 

I can send you files as you described and let you listen to see if they sound different. They won't.

 

There are two things you have in mind which are making you think they would be different which are misconceptions about how digitally sampled audio works.

 

First that you can't discriminate in time between sample points. You are missing that if a wave starts at a slightly different time, even in between samples, it generates a different value in all subsequent samples than if it started right on a sample. The reconstructed wave is continuous and will reconstruct the waves offset by the correct amount of time even between samples.

 

The other is that higher sample rates somehow are more accurate on below 20 khz signals because they have more sample points. That isn't how it works. The higher sample rate gets you more bandwidth. Both 48 khz and 96 khz will recreate a 10 khz signal accurately enough you won't see an analog difference.

 

So what would convince you?

Thanks, I was hoping for mathematical proof :) I'll consider your arguments..

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Shannon-Nyquist theorem. That is the mathematical proof.

 

View the video linked in my signature. You can skip to the 20 min 50 sec. mark, and watch about two minutes. Using analog sources and analog monitoring gear with AD/DA in between he shows you can move a band-limited square wave thru various amounts of delay between sample points and see the wave shape you get on the analog o-scope is exactly the same other than moving in time relative to a second squarewave. What more proof could you want? You have the theorem predicting something, and an analog monitoring system showing the theorem works as advertised.

Thanks again. So this theorem proves that there can be no audible difference between redbook and higher resolution recordings other than different audible artifacts of DA-conversion. I suppose that's possible..

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