Quantization Grid, Split from Topic ID #55966 |
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Quantization Grid, Split from Topic ID #55966 |
Mar 10 2012, 11:58
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#1
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Group: Members Posts: 63 Joined: 21-January 09 From: UK Member No.: 65825 |
I have thought that this phenomana may be real for years . I call it the "quantiztion grid". An analog waveform can change direction or cross the zero line at an infinite number of locations. Sampling and Quantization create a grid to which features of the waveform are statistically constrained. Increasing the sample rate at a fixed bandwidth inreases the quantization grid density. Whether it is perceivable is what is in question. This experiment may have proved it. I will give some thought to a simillar experiment that forum members can do for themselves and start a new topic on it.
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Mar 10 2012, 17:28
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#2
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![]() Group: Members Posts: 715 Joined: 22-April 03 From: /dev/null Member No.: 6130 |
There's no "quantization grid", it's a myth. PCM can and will reproduce audio with sub-sample delay, assuming the signal is band-limited to your sample rate.
Dead horse flogging alert: http://www.stevehoffman.tv/forums/showthread.php?t=85436 http://www.hydrogenaudio.org/forums/index....49043&st=25 Before you ask, it is trivial to band-limit the signal, either using right analog microphones or with a brickwall filter. This post has been edited by AstralStorm: Mar 10 2012, 17:32 -------------------- ruxvilti'a
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Mar 10 2012, 21:30
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#3
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Group: Members Posts: 63 Joined: 21-January 09 From: UK Member No.: 65825 |
The quantization grid is not caused by sampling it is caused by the interaction of quantization levels with sampling points. It may not be perceptable but there is no doubt that a waveform created from a digital file must be formed from a selection of points that are selected from a finite number of pre- determined regularly spaced co-ordinates. The digital file is formed from regularly spaced sampling points and reguarly spaced quantization levels therefore the waveform derived from it must have a coresponding regularity.
This post has been edited by KMD: Mar 10 2012, 21:37 |
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Mar 10 2012, 21:43
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#4
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![]() Group: Developer Posts: 2980 Joined: 2-December 07 Member No.: 49183 |
The digital file is formed from regularly spaced sampling points This means that the reconstructed signal doesn't have frequencies >= Fs/2. and reguarly spaced quantization levels. This adds constant white noise to the signal (simplest case: TPDF dither, no noise shaping). IOW: there is no interaction between sampling and quantisation This post has been edited by lvqcl: Mar 10 2012, 22:17 |
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Mar 10 2012, 22:05
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#5
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Group: Members Posts: 63 Joined: 21-January 09 From: UK Member No.: 65825 |
The output of the D to A is derived from a regularly spaced set of coordinates so it therefore has that reqularity.
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Mar 10 2012, 22:16
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#6
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Group: Members Posts: 128 Joined: 12-December 07 From: Cleveland, OH Member No.: 49500 |
About a year or so ago, j.j. wrote:
In fact, we can resolve, at 20kHz, for a "Redbook CD", a time resolution proportional to, and very near 1/(2*pi*20000*216) in a mechanical fashion. That is a very, very large number of time points per second. It's about 27143360. This post has been edited by Speedskater: Mar 10 2012, 22:18 -------------------- Kevin
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Mar 10 2012, 22:21
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#7
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![]() Group: Developer Posts: 2980 Joined: 2-December 07 Member No.: 49183 |
The output of the D to A is derived from a regularly spaced set of coordinates so it therefore has that reqularity. Here is the graph of the CD noise floor with a standard TPDF dither: ![]() Show me a regularity in this signal please |
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Mar 10 2012, 22:21
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#8
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Group: Members Posts: 63 Joined: 21-January 09 From: UK Member No.: 65825 |
Just for clarify when I say co-ordinate I mean x and y. The cordinates would be more than just 2^16 *44100 per second. I'm thinking twice that or more. Any Idea how jj got that figure.
This post has been edited by KMD: Mar 10 2012, 22:31 |
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Mar 10 2012, 22:24
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#9
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Group: Members Posts: 63 Joined: 21-January 09 From: UK Member No.: 65825 |
ivqci - I am talking about regularity in the shape of the waveform
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Mar 10 2012, 22:31
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#10
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![]() Group: Developer Posts: 2980 Joined: 2-December 07 Member No.: 49183 |
But we cannot hear the shape.
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Mar 10 2012, 22:39
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#11
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Group: Members Posts: 63 Joined: 21-January 09 From: UK Member No.: 65825 |
lvqcl- But whatever we do hear must be derived from the regularised data set wouuld you concur
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Mar 11 2012, 12:50
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#12
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Group: Members Posts: 698 Joined: 6-March 10 Member No.: 78779 |
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Mar 12 2012, 01:30
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#13
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![]() Group: Developer Posts: 191 Joined: 8-July 03 Member No.: 7653 |
This regularity is effectively removed by dithering. Not just "wave your arms"/good enough "effectively". But it's removed completely and replaced with your-choice-shaped additive noise by proper dithering&noise shaping. Like the sampling theorem itself this isn't particular intuitive though it becomes clear if you actually step through the mathematics. The first couple of sections of this thesis explain it pretty well. |
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Mar 12 2012, 15:58
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#14
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Group: Members Posts: 63 Joined: 21-January 09 From: UK Member No.: 65825 |
The movement of the quantization levels by dithering averages to zero over a number of points. Statistically the dithered quantization levels still fall on the original quantization points plus a deviation ether side with the statistical distribution shape determined by the noise used. I think it would be useful to be able to statistically analyse the shape of a piece of audio to asertain what process recorded it.
This post has been edited by KMD: Mar 12 2012, 16:07 |
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Mar 12 2012, 17:08
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#15
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![]() ReplayGain developer Group: Developer Posts: 4584 Joined: 5-November 01 From: Yorkshire, UK Member No.: 409 |
ivqci - I am talking about regularity in the shape of the waveform Yes, and what you say is true - easily visible in the waveform, and sometimes visible on a scope (depends on the DAC).However, there's no mechanism in the ear to pick out that regularity. Once you do any kind of filtering (and the ear contains a continuum of bandpass filters), it disappears. You're just left with what looks like random noise. Also, the error signal (original minus quantised = dither + quantisation error) contains remnants of the original signal. The classic Lipschitz and Vanderkooy dither paper from 1984 (I think!) showed graphs of error signals for various orders of dither. rectangular = first order, triangular = second order. More orders add more noise, but make the error even less correlated with the original. As the ear seems insensitive to higher orders of correlation, there's no need to go further. This is all ignoring the real noise and signal levels in most recordings which make this (and even dither itself) academic. Cheers, David. |
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Mar 12 2012, 17:23
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#16
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Group: Members Posts: 63 Joined: 21-January 09 From: UK Member No.: 65825 |
when you say can be seen on a scope I guess you mean before the reconstruction filter. If so the reconstruction filter forms a waveform from a product of those samples and quantization points and so there is literally trillions of points in just a second to define the smoothed output
This post has been edited by KMD: Mar 12 2012, 17:29 |
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Mar 12 2012, 18:23
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#17
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![]() ReplayGain developer Group: Developer Posts: 4584 Joined: 5-November 01 From: Yorkshire, UK Member No.: 409 |
when you say can be seen on a scope I guess you mean before the reconstruction filter. If so the reconstruction filter forms a waveform from a product of those samples and quantization points and so there is literally trillions of points in just a second to define the smoothed output No, with the right test waveform the effect can be quite visible after the reconstruction filter. Mathematically you'll get different values after the oversampling/reconstruction filter, but visually you can see remnants of the original quantisation steps.Cheers, David. |
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Mar 19 2012, 11:14
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#18
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Group: Members Posts: 63 Joined: 21-January 09 From: UK Member No.: 65825 |
Thanks for the discussion.
My conclusion on this is. - There is no quantization grid. And no regimentation effect in amplitude due to quantization when dither is used. As dither scans the input up and down over the quantization steps, it makes quantization benign like sampling. In fact the word Quantization is innapropriate for the finished process as the resulting waveform is not a quantized waveform! It has infinite resolution because the dither noise fills the space between quantization steps, so there is no room for anything to go undetected. This is another reason why dither is important. Cheers Owen. This post has been edited by KMD: Mar 19 2012, 11:46 |
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Mar 19 2012, 13:39
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#19
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![]() Group: Members Posts: 3212 Joined: 29-October 08 From: USA, 48236 Member No.: 61311 |
Thanks for the discussion. My conclusion on this is. - There is no quantization grid. And no regimentation effect in amplitude due to quantization when dither is used. As dither scans the input up and down over the quantization steps, it makes quantization benign like sampling. In fact the word Quantization is innapropriate for the finished process as the resulting waveform is not a quantized waveform! It has infinite resolution because the dither noise fills the space between quantization steps, so there is no room for anything to go undetected. This is another reason why dither is important. I question that *any* real world process whether analog or digital, has infinite resolution. A process with infinite resolution would output a signal that is exactly the input signal. The error that would be calculated by subtracting the input and output signals would always be exactly zero. In a proper digital process with randomized quantization, the resolution is not infinite and the timing and level errors inherent in quantization are randomized by dither. There are no other errors unless they are intentionally introduced. In an analog process the resolution of storage, processing and transmission is also not infinite. There is practically speaking no such thing as infinite resolution in the analog domain. The timing and level errors inherent in real world analog audio are produced by natural effects such as thermal noise other noise, as well as transmission and processing losses (or gains). These noises and errors come from many sources. Even reasonably short transmission lines and their line drivers and receivers cause errors. All known forms of analog signal storage introduce relatively massive errors. Every resistor, capacitor, transducer, transformer, inductor and active or passive device in the analog domain adds noise and amplitude and timing errors unless it is perfect and is operating at absolute zero. That pretty well excludes everything in every real world audio experiment from the realm of things with infinite resolution. So, you are comparing errors and non-infinite bandpass and resolution in the analog domain with errors and non-infinite bandpass and resolution in the digital domain. Nothing in the real world is perfect or perfectible or even close to perfect except relatively abstract things like digital data storage and transmission. At this time the only things that are even close to be being perfect at performing audio tasks in the analog domain are relatively short transmission lines, a small but enlarging group of operational amplifiers, and digital<-> analog converters. At this time it appears that what drives the development of low noise, low distortion operational amplifiers is the development of digital<-> analog converters. It has been found by many experimenters and observers that relatively inexpensive digital recorders and players can be free of audible flaws in normal use. This was never achieved and is probably unachievable with purely analog devices. |
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Mar 19 2012, 14:28
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#20
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Group: Members Posts: 63 Joined: 21-January 09 From: UK Member No.: 65825 |
Hello Arnold
Signal to noise ratio, and resolution, are not the same thing. Owen. This post has been edited by KMD: Mar 19 2012, 14:49 |
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Mar 19 2012, 15:14
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#21
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Group: Members Posts: 121 Joined: 25-January 12 Member No.: 96698 |
Is Arnold not saying that the output signal might replicate the input information perfectly, but will also introduce noise. So actually the output signal is not the same as the input even if all the orginial information is there.
QUOTE A process with infinite resolution would output a signal that is exactly the input signal. The error that would be calculated by subtracting the input and output signals would always be exactly zero. this I agree with |
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Mar 19 2012, 15:19
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#22
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Group: Members Posts: 63 Joined: 21-January 09 From: UK Member No.: 65825 |
Hi icstm
An analogue recording system has infinite resolution but not infinite signal to noise ratio A dithered digital recording system has infinite resolution but not infinite signal to noise ratio Owen. |
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Mar 19 2012, 15:24
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#23
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![]() ReplayGain developer Group: Developer Posts: 4584 Joined: 5-November 01 From: Yorkshire, UK Member No.: 409 |
Until you define the meaning of the term "resolution", such statements are meaningless.
If they include the word "infinite", they're probably wrong, too. Cheers, David. |
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Mar 19 2012, 15:29
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#24
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Group: Members Posts: 63 Joined: 21-January 09 From: UK Member No.: 65825 |
Resolution is the smallest detail a system can record. In an undithered digital system that is a detail with height greater than one quantisation step. In a dithered system it is infinitely small.
This post has been edited by KMD: Mar 19 2012, 15:29 |
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Mar 19 2012, 15:48
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#25
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![]() ReplayGain developer Group: Developer Posts: 4584 Joined: 5-November 01 From: Yorkshire, UK Member No.: 409 |
Resolution is the smallest detail a system can record. In an undithered digital system that is a detail with height greater than one quantisation step. In a dithered system it is infinitely small. It's only infinitely small if it persists for infinitely long, the system is DC coupled, and I'm allowed to average the result over an infinite number of samples.Try again Cheers, David. |
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Lo-Fi Version | Time is now: 19th May 2013 - 22:13 |