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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.

Can anyone illuminate the correct way to do this?

Edit: I'm looking to get high and low pass filtering with more dice, but am fine with the distribution changing from triangle to gaussian or other shapes.

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To achieve high-pass shaping , the “DC term” must be 0. Note that in the 2-dice high-pass case, every roll appears twice, once with a positive sign and once with a negative sign, which enforces the DC-free condition. Any scheme involving more than 2 dice would still need to meet this condition. The triangular PDF condition is satisfied by the fact that each sum or difference combines 2 rectangular PDF’s. I don’t know how you can get a triangular PDF by combining 3 rectangular PDF’s, regardless of how they are weighted.

Bob Adams

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  • $\begingroup$ Ah, so you are thinking it needs to be an even number of dice $\endgroup$ – Alan Wolfe Jun 30 at 14:15
  • $\begingroup$ To get a zero at DC you need an even number. To get triangular PDF I can’t see how any number other than 2 would work, but I might be missing something. $\endgroup$ – Bob Jun 30 at 19:52

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