Quantization Grid

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Quantization Grid

Reply #50 – 2012-03-20 18:54:05

Quote from: 2Bdecided on 2012-03-20 18:43:36

Those original 8-bit quantisation steps are still easily visible...

It wasn't my intention to put this in dispute.

Quote from: 2Bdecided on 2012-03-20 18:52:56

You thanked him a bit early

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.

Quantization Grid

Reply #51 – 2012-03-20 18:59:06

Quote from: greynol on 2012-03-20 18:54:05

Quote from: 2Bdecided on 2012-03-20 18:43:36
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.

Quantization Grid

Reply #52 – 2012-03-20 19:02:23

Quote from: 2Bdecided on 2012-03-20 18:52:56

Quote from: greynol on 2012-03-20 18:42:41
Quote from: Arnold B. Krueger on 2012-03-20 18:36:47
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.

Quantization Grid

Reply #53 – 2012-03-20 19:02:31

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.

Quantization Grid

Reply #54 – 2012-03-20 19:03:21

Quote from: greynol on 2012-03-20 18:54:05

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.

Quantization Grid

Reply #55 – 2012-03-20 19:08:28

Quote from: 2Bdecided on 2012-03-20 19:03:21

From one extreme to the other.

With such a professorial tone, no less.

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.

Quantization Grid

Reply #56 – 2012-03-20 19:09:28

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.

Quantization Grid

Reply #57 – 2012-03-20 19:30:20

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.

Quantization Grid

Reply #58 – 2012-03-20 20:13:52

Quote from: 2Bdecided on 2012-03-20 18:59:06

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"?

Quantization Grid

Reply #59 – 2012-03-20 20:30:40

It's always present. Dither simply replaces much of it with a different kind of noise.

Quantization Grid

Reply #60 – 2012-03-20 20:39:01

Understood. The point I'm making is that the "surviving grid" is in fact just the quantization error. Which shouldn't be controversial.

Quantization Grid

Reply #61 – 2012-03-20 20:40:57

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.

Quantization Grid

Reply #62 – 2012-03-20 20:43:37

Quote from: drewfx on 2012-03-20 20:39:01

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.

Quantization Grid

Reply #63 – 2012-03-20 21:07:24

Quote from: greynol on 2012-03-20 20:43:37

Quote from: drewfx on 2012-03-20 20:39:01
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 [a href='index.php?act=findpost&pid=789226']early[/a] in this thread. From there it went down, as if there really was some kind of unavoidable 'grid' residue, different from classical quantization error.

Quantization Grid

Reply #64 – 2012-03-20 23:00:50

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.

Quantization Grid

Reply #65 – 2012-03-20 23:34:02

Quote from: 2Bdecided on 2012-03-20 16:33:21

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".

Quantization Grid

Reply #66 – 2012-03-21 00:03:05

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.

Quantization Grid

Reply #67 – 2012-03-21 10:02:36

Quote from: Speedskater on 2012-03-20 23:34:02

Quote from: 2Bdecided on 2012-03-20 16:33:21
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...
[attachment=6976:distribu...nodither.gif]

This is from the 8-bit with dither data, 8x oversampled (i.e. resampled to 352.8kHz) at 16-bits...
[attachment=6977:distribu...thdither.gif]

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.

Quantization Grid

Reply #68 – 2012-03-21 10:03:29

Quote from: googlebot on 2012-03-20 21:07:24

2Bdecided has already explained it all very [a href='index.php?act=findpost&pid=789226']early[/a] in this thread.

Thank you - that's the bit where I explain that it doesn't matter. Worth repeating!

Cheers,
David.

Quantization Grid

Reply #69 – 2012-03-21 11:37:53

At what stage did you add the dither noise David.

Owen.

Quantization Grid

Reply #70 – 2012-03-21 12:10:55

Quote from: KMD on 2012-03-21 11:37:53

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.

Quantization Grid

Reply #71 – 2012-03-21 12:25:50

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

Quantization Grid

Reply #72 – 2012-03-21 13:50:56

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

Quantization Grid

Reply #73 – 2012-03-21 14:07:30

Quote from: KMD on 2012-03-21 12:25:50

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...
[attachment=6978:distribu...nodither.gif]

[attachment=6979:distribu...thdither.gif]

Cheers,
David.

Quantization Grid

Reply #74 – 2012-03-21 14:08:52

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