Late response but thanks a bunch you guys :) On Thu, Jun 27, 2019 at 1:55 PM Ethan Duni <ethan.d...@gmail.com> wrote:
> So as Nigel and Robert have already explained, in general you need to > separately handle the spectral shaping and pdf shaping. This dither > algorithm works by limiting to the particular case of triangular pdf with a > single pole at z=+/-1. For that case, the state of the spectral shaping > filter can be combined with the state of the pdf shaper, and so a single > process (with no multiplies!) handles both pdf shaping and spectral shaping. > > For arbitrary order M, you would roll one die at each step and then sum it > with M previous rolls (possibly with some set of signs inverted). So the OP > example is M=1. You have your choice of 2^M spectral shapes, depending on > which (if any) of the previous rolls you invert. For the "highness" output, > you will want to invert every other previous roll. As M increases, the > output gets more Gaussian. > > However for higher orders this multiplier-free algorithm does not produce > attractive spectral shapes. For even orders, the highpass does not have a > zero at z=1. For odd orders, the frequency response has large notches in > the middle of the bandwidth. > > For most applications, a triangular pdf with single zero at z=1 is a > perfectly good dither configuration, and there is no need to go any > further. If you are looking for a higher-order dither algorithm without > multiplies, I think the way to extend this would be to include bit shifts > in the summation. Then you can get some reasonable spectral shapes. The > simple summation approach is too constrained for orders>1. > > Ethan > > On Thu, Jun 27, 2019 at 7:43 AM Alan Wolfe <alan.wo...@gmail.com> wrote: > >> I read a pretty cool article the other day: >> https://www.digido.com/ufaqs/dither-noise-probability-density-explained/ >> >> It says that if you have two dice (A and B) that you can roll both dice >> and then... >> 1) Re-roll die A and sum A and B >> 2) Re-roll die B and sum A and B >> 3) Re-roll die A and sum A and B >> 4) repeat to get a low pass filtered triangular noise distribution. >> >> It says that you can modify it for high pass filtered triangle noise by >> rolling both dice and then... >> 1) Re-roll die A and take A - B >> 2) Re-roll die B and take B - A >> 3) Re-roll die A and take A - B >> 4) repeat to get a high pass filtered triangular noise distribution. >> >> What i'm wondering is, what is the right thing to do if you want to do >> this with more than 2 dice? (going higher order) >> >> For low pass filtered noise with 3+ more dice (which would be more >> gaussian distributed than triangle), would you only re-roll one die each >> time, or would you reroll all BUT one die each time. >> >> I have the same question about the high pass filtered noise with 3+ more >> dice, but in that case I think i know what to do about the subtraction >> order... I think the right thing to do if you have N dice is to sum them >> all up, but after each "roll" you flip the sign of every die. >> >> What do you guys think? >> _______________________________________________ >> dupswapdrop: music-dsp mailing list >> music-dsp@music.columbia.edu >> https://lists.columbia.edu/mailman/listinfo/music-dsp > > _______________________________________________ > dupswapdrop: music-dsp mailing list > music-dsp@music.columbia.edu > https://lists.columbia.edu/mailman/listinfo/music-dsp
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