I'm not sure that I follow. If Q'(s+d)-d == Q'(s), then that almost implies that d==0
Juggling the terms, Q'(s+d) == Q'(s)+d ... and I'm not sure how that could be true, given that quantizing throws away precision. Along similar lines, I was thinking that since subtractive dither can improve S/N despite quantization, then would it be possible to just handle the subtractive dither *before* putting data into the bit-depth-limited stream? My hunch is that the bit depth of the stream would have to increase in order for the benefits of subtractive dither to be realized. But maybe that's not what your Q'() is trying to solve. Brian Willoughby On Jan 8, 2022, at 14:55, robert bristow-johnson <r...@audioimagination.com> wrote: > This is true: > > Q(s + d) - d != Q(s) > > i think this is also true: > > Q'(s + d) - d = Q'(s) > > for a hypothetical "quantizer" Q'() that also has error variance of > \Delta^2/12 which is 4.77 dB better than what Q'(s+d) is. > > -- r b-j > > ------ Original message------ > From: B.J. Buchalter > Date: Sat, Jan 8, 2022 2:59 AM > To: MUSIC-DSP@LISTS.COLUMBIA.EDU; > Cc: > Subject:Re: about subtractive dither, for audio and other use (also > scientific) > > > On Jan 8, 2022, at 2:42 AM, Brian Willoughby > wrote: > > > > Where there is quantization, subtractive dither does not help overall. Yes, > > it improves S/N on paper, but the resulting correlated noise is *more* > > audible than the uncorrelated noise, so it become a perceptual loss of > > quality despite being a numerical improvement to S/N. > > I think what you are missing in the subtractive dither proposal/analysis is > that > > Q(s + d) - d != Q(s) > > Q = Quantizer > s = signal > d = dither > > You can see this by thinking about s being a constant value between > quantization levels; Q(s) will be a constant value; Q(s+d) - d will not be a > constant signal, and will average out over time to s. > > Applying appropriate dither before quantization removes the correlated > distortion. Subtracting the dither signal from the quantized dithered signal > does not add the distortion back. > > Q(s + d) and Q(s+d)-d will both be noisier than s (well, at least assuming > the noise in s is below the quantization point), but the argument is that > Q(s+d)-d will be less noisy than Q(s+d). > > B.J. Buchalter > Metric Halo > > http://www.mhlabs.com >
