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Time resolution of Redbook (16/44) PCM

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DonH56

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You're right that this is, strictly speaking, a function of signal rather than sampling frequency. But mansr's model is looking at the best case resolution which is, clearly, a function of sampling rate in a band-limited digital system. I should have specified that though. It's perhaps more important to note that this is a also a function of signal amplitude, but since such discussions focus on full-scale impulse or step functions this falls away.

Even with these caveats the timing accuracy of 16/44.1 digital is at least an order of magnitude greater than even the most accurate analog system.

My bad, I did not follow the link, was working from memory. And as JJ pointed out it should really relate to signal bandwidth though again the equation's the same. For amplitude, you could stick "*A" in the denominator to match the usual form but by convention we assume full-scale signals to "b" bits. There are (much) more complex versions but this is more than good enough to prove your point, that all of us agree with, regarding timing resolution. Seems "they" always manage to get that wrong...
 

voodooless

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You're right that this is, strictly speaking, a function of signal rather than sampling frequency. But mansr's model is looking at the best case resolution which is, clearly, a function of sampling rate in a band-limited digital system.

That is rather deceptive, isn’t it. That means that your best case time resolution of a 96 kHz sampled track only happens at 48 kHz (and at full scale).. it won’t do you any practical good, will it?
 

DonH56

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voodooless

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That's the time resolution of the system.

Just asserting it doesn’t really make it any more convincing ;). The time resolution of a 1 kHz sine or a 10 kHz sine (equal level) in the same system is not the same.
 
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j_j

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Just asserting it doesn’t really make it any more convincing ;). The time resolution of a 1 kHz sine or a 10 kHz sine (equal level) in the same system is not the same.

What's the actual bandwidth of the signal. Both 1kHz and 10kHz sine waves have zero bandwidth, after all.
 

voodooless

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What's the actual bandwidth of the signal. Both 1kHz and 10kHz sine waves have zero bandwidth, after all.
So? Am I wrong or not? If so, what did I not understand about the linked article?
 
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j_j

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So? Am I wrong or not? If so, what did I not understand about the linked article?

System bandwidth affects how you can change both of those frequencies equally. Bandwidth, not "highest frequency". Until you CHANGE a signal there is no "difference" to be detected.
 

dc655321

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So? Am I wrong or not? If so, what did I not understand about the linked article?

I think the operative word here is system - the frequency variable in the equation at @mansr's page is the bandwidth of the system.

So, a system bandwidth of ~20kHz (i.e RBCD; your 1kHz and 10kHz sine waves are encompassed here) has a timing resolution of:
Best case (full-scale) ~ 0.1ns
Worst case (LSB) ~ 8us
 

audio2design

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Timing "resolution", if we use that term ("accuracy" is not what we are measuring here), is a function of the signal frequency and converter resolution (bit depth). Sampling rate does not matter, and SNR is not an explicit part of the equation. @charleski provided the correct equation except it is a function of signal frequency, not sampling frequency (under certain defined conditions blah blah blah but this is "the" equation generally used).

If you have timing errors on the order of tmin, like jitter, then you will reduce the SNR, but that is something different. Dither typically reduces the SNR (you are adding noise, after all) but allows you to resolve signals that are less than 1 lsb. Think of dither as providing a little noise so now the signal, instead of being always below 1 lsb, flirts with the lsb level enough that you can reconstruct (processing using DSP or your ears and brain) the signal from the noise.


If I can reconstruct the signal, then I can extract timing information more accurately. Dither shapes the noise, so if I know the nature of the signal I am capturing and it is narrow band, I can extract timing information that was lost just as the signal was lost.


The limitation on timing resolution will always be SNR and bandwidth, regardless of bit depth, whether it is part of an "equation" or not. Timing is information, just like any other information in a digital signal, and it will be limited by Shannon-Hartley even if you are extracting a timing event instead of bits. There is a lot of work on this in time-of-arrival in sonar, radar, etc.


At the end of the day, it is just a thought exercise as we know if timing was "off", then SNR would be poor (it works in reverse), and we know that modern digital systems have SNR way beyond any analog audio systems, so must have better timing accuracy, and when you consider the important timing relationship between the two channels, digital is so much better than analog it is not worth even mentioning.
 

DonH56

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If I can reconstruct the signal, then I can extract timing information more accurately. Dither shapes the noise, so if I know the nature of the signal I am capturing and it is narrow band, I can extract timing information that was lost just as the signal was lost.

The limitation on timing resolution will always be SNR and bandwidth, regardless of bit depth, whether it is part of an "equation" or not. Timing is information, just like any other information in a digital signal, and it will be limited by Shannon-Hartley even if you are extracting a timing event instead of bits. There is a lot of work on this in time-of-arrival in sonar, radar, etc.

At the end of the day, it is just a thought exercise as we know if timing was "off", then SNR would be poor (it works in reverse), and we know that modern digital systems have SNR way beyond any analog audio systems, so must have better timing accuracy, and when you consider the important timing relationship between the two channels, digital is so much better than analog it is not worth even mentioning.

Not sure where the argument? I was responding to this:

audio2design said:
Don't hold me to this but it has been a while since I went through the math, but if I am not mistaken, then the timing accuracy is a factor of SNR, not bit depth (though related) as dither can increase SNR which increases timing resolution.

My response was to your statement that timing accuracy was a factor of SNR and not bit depth, whereas for a quantized system the aperture time is directly (OK, inversely) related to bit depth. That is for quantization noise, one component in SNR, and the equation is valid and has been proven theoretically and empirically many times over. Dither is generally used to decorrelate the quantization (and some other) noise; noise shaping is something different. Narrowband or colored dither is often used (or was when I was designing them) in radar and other (e.g., sonar, telcom, EW/ELINT) systems, which was the focus of most of my design career (not audio converters). But this is really beyond the scope of just proving time resolution is better than 1/fs...
 

charleski

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Just asserting it doesn’t really make it any more convincing ;). The time resolution of a 1 kHz sine or a 10 kHz sine (equal level) in the same system is not the same.
I think people are losing sight of the actual question that’s relevant here.

Atonal transients are a common feature of musical signals, and thus it’s reasonable to ask how well their timing can be reproduced in a digital system. These features can be modelled as impulse or step functions, as seen in the video posted earlier. Such functions consist of an infinite series of components and the time resolution of the function as a whole is dependent on the resolution of the highest component that the system can reproduce. This, in turn, is dependent on the sampling rate (in an ideal digital system sampling rate and bandwidth are simply interchangeable terms, and well-implemented modern digital systems come very close to the ideal). The best case is the only case that matters, and this is dependent on sampling rate. In the worst case, any digital system has a time resolution of around 8msec, but this is utterly irrelevant and anyone worrying about that needs to go away and have a think about their life-choices :p.
 

voodooless

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I think people are losing sight of the actual question that’s relevant here.

That might be very true ;)

Atonal transients are a common feature of musical signals, and thus it’s reasonable to ask how well their timing can be reproduced in a digital system. These features can be modelled as impulse or step functions, as seen in the video posted earlier. Such functions consist of an infinite series of components and the time resolution of the function as a whole is dependent on the resolution of the highest component that the system can reproduce. This, in turn, is dependent on the sampling rate (in an ideal digital system sampling rate and bandwidth are simply interchangeable terms, and well-implemented modern digital systems come very close to the ideal). The best case is the only case that matters, and this is dependent on sampling rate.

I get that you can get steeper transients with a higher sample rate. But I find it a bit strange why that would be the only relevant measure? You cannot hear the higher components of the transient above 20 kHz anyway. So why would timing resolution only be relevant in transients?

Am I correct in assuming that the timing resolution of a digital system, measured in degrees of phase is constant over frequency?

In the worst case, any digital system has a time resolution of around 8msec, but this is utterly irrelevant and anyone worrying about that needs to go away and have a think about their life-choices :p.

Oh, I'm not worried at all, just want to understand where I'm wrong in my understanding.
 
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Atonal transients are a common feature of musical signals,.

"Atonal Transients" are nothing but the time-limited sum of a bunch of sine waves.

THAT is what you need to understand.

I can take your "atonal transients", analyze them, and show you the spectrum of your "atonal transients".

All of this is about bandwidth, and nothing but bandwidth. You need to go digest some fourier analysis before you keep heading in this direction.
 

voodooless

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"Atonal Transients" are nothing but the time-limited sum of a bunch of sine waves.

THAT is what you need to understand.

I can take your "atonal transients", analyze them, and show you the spectrum of your "atonal transients".

All of this is about bandwidth, and nothing but bandwidth. You need to go digest some fourier analysis

I get all this. I get that a step response models the whole system, so in that sense, you could say that this dictates the timing resolution. So far I'm with you. But that still doesn't mean that I can encode a 1 kHz sine with the same phase resolution (in time) as a 10 kHz sine. Does it matter, probably not, but that is still correct, isn't it? Let's say we have a stereo track in a 16 bit system, both tracks contain a 1 kHz sine. Now on one of them we shift the sine by 300ps. You cannot encode this difference. Now, do the same for a 10 kHz sine: you can encode that difference. Am I wrong here?

before you keep heading in this direction.

I'm not even sure what direction that is :facepalm:
 

audio2design

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Let's say we have a stereo track in a 16 bit system, both tracks contain a 1 kHz sine. Now on one of them we shift the sine by 300ps. You cannot encode this difference. Now, do the same for a 10 kHz sine: you can encode that difference. Am I wrong here?


Ignoring the value picked of 300ps and whether that fits the bandwidth and SNR of the system, conceptually you are wrong. It does not matter whether it is 1KHz or 10Khz. If you shift one by 300psec, you will make a small change in the output bit stream in either case.
 

voodooless

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Ignoring the value picked of 300ps and whether that fits the bandwidth and SNR of the system, conceptually you are wrong. It does not matter whether it is 1KHz or 10Khz. If you shift one by 300psec, you will make a small change in the output bit stream in either case.

Isn't it the point that the change is too small to make bit's flip? Otherwise, that means the calculations in the link are wrong?
 

audio2design

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Isn't it the point that the change is too small to make bit's flip? Otherwise, that means the calculations in the link are wrong?

Actually a 300psec timing shift will cause a difference in a bit stream at 1K (or 10KHz) with a perfect 16 bit A/D. Even a 50 psec shift will cause a change. The problem is that bit shift is below the quantization noise, so you can't say for sure what is causing those bit flips.


However, back to my one of my points, the formula assumes no noise shaping, but uniform quantization noise, and the real limitation is SNR, not bit depth, as bit depth does not give the full story w.r.t. SNR.


On modern CDs, we dither and noise shape, which changes the structure of the SNR, which makes the actual result more complex, but still way way better than 1/FS and better than any analog format.
 
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