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Quantization Grid, Split from Topic ID #55966
greynol
post Mar 20 2012, 19:54
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QUOTE (2Bdecided @ Mar 20 2012, 11:43) *
Those original 8-bit quantisation steps are still easily visible...
It wasn't my intention to put this in dispute.

QUOTE (2Bdecided @ Mar 20 2012, 11:52) *
You thanked him a bit early wink.gif
Not really; rather I think the issue is that not everyone is following the claims put forth by the OP (could only be me, but I somehow doubt it).

Thanks for the detour, nonetheless.

This post has been edited by greynol: Mar 20 2012, 19:59


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2Bdecided
post Mar 20 2012, 19:59
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QUOTE (greynol @ Mar 20 2012, 18:54) *
QUOTE (2Bdecided @ Mar 20 2012, 11:43) *
Those original 8-bit quantisation steps are still easily visible...
It wasn't my intention to put this in dispute.

Thanks for the detour, nonetheless.
Then what is your point?

The effect of quantisation is visible beyond the reconstruction filter. That is my point. The "grid" survives, though not perfectly.

btw, I don't think there's any audible relevance to this.

Cheers,
David.
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Arnold B. Kruege...
post Mar 20 2012, 20:02
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QUOTE (2Bdecided @ Mar 20 2012, 14:52) *
QUOTE (greynol @ Mar 20 2012, 18:42) *
QUOTE (Arnold B. Krueger @ Mar 20 2012, 11:36) *
The output of a DAC contains all of the intermediate voltages even if the signal is undithered.

Thank you!

You thanked him a bit early.

I'm not sure why anyone thinks a filter at 22kHz is going to change a square-ish low frequency waveform that much. The example I posted was a low amplitude sine wave at 50Hz; at 8-bits it ends up with square-wave-like transitions at ~250Hz due to quantisation. In this example, you can comfortably fit the first 40 harmonics within the transition band. That's more than you need to make a square wave look something like a square wave.


The quantization filter only changes a low frequency wave enough so that the transition isn't perfectly square. As soon as the wave stops being perfectly square, all of the intermediate voltages are there.
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greynol
post Mar 20 2012, 20:02
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I edited my previous response to include your follow-up post. It should help to present my point, though I think I made it pretty clear earlier by suggesting that people are assuming facts not in evidence about the OPs claims.

This post has been edited by greynol: Mar 20 2012, 20:04
Reason for edit: added link


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2Bdecided
post Mar 20 2012, 20:03
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QUOTE (greynol @ Mar 20 2012, 18:54) *
not everyone is following the claims put forth by the OP
I think the OP has changed their mind throughout this thread.

They've gone from digital = an evil fixed grid that wrecks audio, to digital = infinite resolution. (I'm paraphrasing!). From one extreme to the other. Neither is correct IMO, but the ideas are too vague to critique properly, and I haven't seen a good answer for amplitude "resolution". I know there's a formula for temporal "resolution" in another thread somewhere.

Cheers,
David.

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greynol
post Mar 20 2012, 20:08
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QUOTE (2Bdecided @ Mar 20 2012, 12:03) *
From one extreme to the other.

With such a professorial tone, no less. wink.gif

Thanks for the images. Hopefully people won't come away from this with the wrong idea this time around. I fear that they would without them.

This post has been edited by greynol: Mar 20 2012, 20:30


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2Bdecided
post Mar 20 2012, 20:09
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Ah, maybe we're at cross purposes, and don't even disagree. I said "most of the instantaneous output voltages will be essentially 'on' the quantization steps". Of course there's some smoothing. It just doesn't remove the original "quantisation grid" completely (or much at all, for some waveforms).

Cheers,
David.
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googlebot
post Mar 20 2012, 20:30
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Aren't those pictures just showing that not enough dither has been applied? Not in the sense of audible necessity, but in the sense of optical waveform hygiene.
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drewfx
post Mar 20 2012, 21:13
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QUOTE (2Bdecided @ Mar 20 2012, 13:59) *
The effect of quantisation is visible beyond the reconstruction filter. That is my point. The "grid" survives, though not perfectly.


Isn't this just sort of saying, "there's quantization error/distortion/noise present"?
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pdq
post Mar 20 2012, 21:30
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It's always present. Dither simply replaces much of it with a different kind of noise.
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drewfx
post Mar 20 2012, 21:39
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Understood. The point I'm making is that the "surviving grid" is in fact just the quantization error. Which shouldn't be controversial.
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greynol
post Mar 20 2012, 21:40
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The image of the dithered 50Hz wave is pretty much how I remember the digital signal generators looking when set to minimum amplitude back when I used those things.

...or was it the digital oscilloscopes, I can't remember, though I usually used analog scopes. smile.gif


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greynol
post Mar 20 2012, 21:43
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QUOTE (drewfx @ Mar 20 2012, 13:39) *
The point I'm making is that the "surviving grid" is in fact just the quantization error.

I would say the quantization itself, but maybe this is a distinction without a difference.


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googlebot
post Mar 20 2012, 22:07
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QUOTE (greynol @ Mar 20 2012, 21:43) *
QUOTE (drewfx @ Mar 20 2012, 13:39) *
The point I'm making is that the "surviving grid" is in fact just the quantization error.

I would say the quantization itself, but maybe this is a distinction without a difference.


The "surviving grid" is just correlated quantization error, that has not been decorrelated properly with an amount of noise suited for the input bit depth. It's neither quantization 'itself' nor any other voodoo. 2Bdecided has already explained it all very early in this thread. From there it went down, as if there really was some kind of unavoidable 'grid' residue, different from classical quantization error.

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greynol
post Mar 21 2012, 00:00
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Yet the "error" plainly indicates the available discrete levels.

How is this voodoo?

I used scary quotes around the word error since the actual error is an analog signal that is limited between -1/2 and +1/2 when dither is not applied.

This post has been edited by greynol: Mar 21 2012, 07:43


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Speedskater
post Mar 21 2012, 00:34
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QUOTE (2Bdecided @ Mar 20 2012, 12:33) *
For a low-ish frequency input signal, ignoring noise etc, most of the instantaneous output voltages will be essentially "on" the quantization steps. You can see this very easily at low amplitudes.
Cheers,
David.

So you look at this instantaneous output voltage and if it is NOT "on" a quantization step - you say "ignore that, it's noise".
And if this instantaneous output voltage is "on" a quantization step - you say "see I told you so".

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greynol
post Mar 21 2012, 01:03
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Yes, noise (thermal, shot, interference, etc.) that causes the signal to deviate from one of the discrete quantization levels as opposed to quantization error which is the reason that the signal is clamping to one of the discrete quantization levels.

Other than that the operative is "most" and how much is dependent on the highest frequency present in the signal.

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2Bdecided
post Mar 21 2012, 11:02
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QUOTE (Speedskater @ Mar 20 2012, 23:34) *
QUOTE (2Bdecided @ Mar 20 2012, 12:33) *
For a low-ish frequency input signal, ignoring noise etc, most of the instantaneous output voltages will be essentially "on" the quantization steps. You can see this very easily at low amplitudes.
Cheers,
David.

So you look at this instantaneous output voltage and if it is NOT "on" a quantization step - you say "ignore that, it's noise".
And if this instantaneous output voltage is "on" a quantization step - you say "see I told you so".
It's hardly cherry picking.

Here are two graphs showing the distribution of sample values.

This is from the 8-bit without dither data, 8x oversampled (i.e. resampled to 352.8kHz) at 16-bits...
Attached Image


This is from the 8-bit with dither data, 8x oversampled (i.e. resampled to 352.8kHz) at 16-bits...
Attached Image


In both cases, the dominance of the original 8-bit quantisation steps is clearly visible: those huge peaks match the original 8-bit quantisation steps. The skirts around them are due to the effect of the resampling. In a real system, noise would broaden the skirts and reduce the peaks - though not noticeably for 8-bits! More importantly, an AC-coupled system will see the values drifting around with any DC in the sampled waveform.

EDIT: These distribution graphs are for full 20s 20Hz-1kHz log sweep -40dB signals - the same ones from which the waveform images in this post were taken.

Cheers,
David.

This post has been edited by 2Bdecided: Mar 21 2012, 11:06
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2Bdecided
post Mar 21 2012, 11:03
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QUOTE (googlebot @ Mar 20 2012, 21:07) *
2Bdecided has already explained it all very early in this thread.
Thank you - that's the bit where I explain that it doesn't matter. Worth repeating!

Cheers,
David.

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KMD
post Mar 21 2012, 12:37
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At what stage did you add the dither noise David.




Owen.
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2Bdecided
post Mar 21 2012, 13:10
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QUOTE (KMD @ Mar 21 2012, 11:37) *
At what stage did you add the dither noise David.
Steps:

1. Create new project: 44.1kHz 16bit.
2. Generate sine wave, 20s, 20Hz-1kHz, log sweep, 0dB
3. amplitude change: -40dB
4. convert to 8-bit with or without dither <--------------- here. dither shape = TPDF dither amplitude = 1-LSB RMS (2-LSBs peak-to-peak)
5. convert to 16-bit
6. convert to 352800Hz 16-bit

Cheers,
David.
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KMD
post Mar 21 2012, 13:25
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The fact that you are sweeping the frequency adds a layer of complexity. We don't quite know what artifacts may allready be in a sweep synthesised by cooledit. For simplicity can you do those steps , and the statistical analysis on a static 50Hz sine wave, as in the original cooledit images.


Cheers


Owen

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pdq
post Mar 21 2012, 14:50
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What would be the point? We have established that low level signals can be severely distorted by quantization, and that a nominal application of dither removes much, but not all, of that distortion. What more do you want to know?

Edit: typo

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2Bdecided
post Mar 21 2012, 15:07
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QUOTE (KMD @ Mar 21 2012, 12:25) *
We don't quite know what artifacts may allready be in a sweep synthesised by cooledit.
I can assure you, there are none with the sine wave sweep. (The square wave is atrocious, but I didn't use that).

I'd turned off the default dithering of all functions, so there is truncation quantisation distortion in the 16-bit representation, even though CEP is able to generate 32-bit float and dither to 16-bit normally by default when performing operations.

Just using 50Hz makes little difference...
Attached Image


Attached Image


Cheers,
David.
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KMD
post Mar 21 2012, 15:08
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pdq -look at the 2nd image in 2bdecided post68 above . On first looking the dither seems to have not removed the distortions at all. I think it needs to be assertained what cool edit is actually doing.

David just seen your last post, will read it now

This post has been edited by KMD: Mar 21 2012, 15:38
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