Sometimes the simplest approach is the best approach. Sounds like a good reverb paper to me. Some user evaluation and references to standard papers and 😁
On Sep 29, 2017 8:51 AM, "gm" <g...@voxangelica.net> wrote: > It's a totally naive laymans approach > I hope the formatting stays in place. > > The feedback delay in the loop folds the signal back > so we have periods of a comb filter. > | | | | > |__________|__________|__________|___ > > Now we want to fill the period densly with impulses: > > First bad idea is to place a first impulse exactly in the middle > > that would be a ratio for the allpass delay of 0.5 in respect to the comb > filter. > It means that the second next impulse falls on the period. > > | | > |____|____|___ > > > The next idea is to place the impulse so that after the second cycle > it exactly fills the free space between the first pulse and the period > like this, > exactly in the middle between the first impulse and the period: > > | | | > | | | | | > |_____|_____|__|__|_____|___ > > this means we need a ratio "a" for the allpass delay in respect to the > combfilter loop that fulfills: > > 2a - 1 = a/2 > > Where 1 is the period of the combfilter. > Alternativly, to place it on the other side, we need > > 2a - 1 = 1 - a/2; > > > | | | > | | | | | > |___|___|___|_|_|___ > > This gives ratios of 0.5. 0.66667 and 0.8 > > These are bad ratios since they have very small common multiples with the > loop period. > So we detune them slightly so they are never in synch with the loop period > or each other. > That was my very naive approach, and surprisingly it worked. > > > The next idea is to place the second impulse not in the middle of the free > space > but in a golden ratio in respect to the first impulse > > | | | > | | | | | > |_______|____|____|__|____| > > 2a - 1 = a*0.618... > > or > > N*a mod 1 = a*0.618.. > > or if you prefer the exact solution: > > a = (1 + SQRT(5)) / ( SQRT(5)*N + N - 2) > > wich is ~ 0.723607 and the same as 1/ (1+ 0.382...) or 1/ (N + 0.382) > > where N is the number of impulses, that means instead of placing the 2nd > impulse on a*0.618 > we can also place the 3rd, 4th etc for shorter AP diffusors. > > (And again we can also fill the other side of the first impulse with > 0.839643 > And the solution for N = 1 is 2.618.. and we can use the reciprocal 0.381 > to place a first impusle) > > The pattern this gives for 0.72.. is both regular but evenly distributed > so that each pulse > falls an a free space, just like on a Fibonaccy flower pattern each petal > falls an a free space, > forever. > (I have only estimated the first few periods manually, and it appeared > like that > Its hard to identify in the impulse response since I test a loop with 3 > APs ) > > The regularity is a bad thing, but the even distribution seems like a good > thing (?). > I assume it doesn't even make a huge difference to using 0.618.. for a > ratio though it seemed to sound better. > (And if you use 0.618, what do you use for the other APs?) > > So it's not the solution I am looking for but interesting never the less. > > I believe that instant and well distributed echo density is a desired > property > and I assume that the more noise like the response is as a time series > the better it works also in the frequency/phase domain. > > For instance you can make noise loops with randomizing all phases by FFT > in circular convolution > that sound very reverberated. > > > > > _______________________________________________ > dupswapdrop: music-dsp mailing list > music-dsp@music.columbia.edu > https://lists.columbia.edu/mailman/listinfo/music-dsp
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