@melter If I understand correctly what your abstraction is doing, you would have to apply its output (the desired rms value) to the analyzed signal after you have divided it by the actual rms value ....

BUT: That's much too complicated! You don't even have to analyze the signal with [env~]. Under the link that you posted there's a table from which you can see, that doubling the distance corresponds to a decrease of sound pressure by 6dB, which is 1/4 power or 1/2 amplitude. So I guess all you have to do is to multiply the signal with d1/d2.

@jameslo said:

Teach a man to fish: you know this because you looked at source code, or some other way?

I just compared the output of both filters, and it turned out to be exactly the same.

Oh look, and they take the cube root for vcf_hp6~ when they chain 3 filters. I wasn't aware that Q worked that way. Do you know what they are doing with the cutoff freq signal, [iem_cot4~]?

Not sure if you can say in general that Q works this way. It might depend on how the filter is normalized. [vcf~], for example, has unit gain at the resonant frequency, but other filters produce higher gains at the resonant frequency. So if you put them in series the gains potentiate, and there has to be some compensation for that. What the [iem_cot4~] does, I don't know. Unfortunately I can't read C code.

@jameslo With just vanilla objects that's gonna be difficult, because you'd have to use the raw filters like [cpole~]. Since the iemlib filter is a series of two 2nd order filters, my first guess would be to do the same with ELSE's [highpass~], which is also 2nd order. Its exact formula might be different though.

Edit: It actually seems to be the same, but you also have to take the square root of the Q factor!

So here's the fixed version: velvet-noise-fixed.pd

It wasn't just the issue with small floating point numbers, but more importantly it now didn't work with integer periods. In this case, of course, the period must never be reduced by one sample!

@ben.wes Thanks again for testing, really helpful!

Yes, it's true, that I didn't care about consecutive non-zero samples. For now I'm not sure what's more important: to comply to this constraint or to behave properly above nyquist.

The problems with frequencies above nyquist, that you found, probably have to do with the representation of small floating point numbers. I should have used [expr~ int($v1)] to get real zero values. Can't fix it right now, but I'll try tomorrow ...

@porres said:

I'm trying to find a better and more sophisticated idea for the algorithm. We need to find the number of samples in a period and then randomly choose where to place it.

I'm not in C, but I think I kind of did exactly that in my latest version, trying to solve the problem with non-integer periods. Here it is (commented in the patch) ...

velvet-noise.pd

Could certainly be optimized in some ways (especially using ELSE objects). But does it even work? I don't have a good method to test it, unfortunately.

@ben.wes Thanks! So no surprise that it doesn't work ....

@seb-harmonik.ar said:

I do think @manuels original idea of having a sample-increment offset still works without missing periods?
Why don't you think it works with non-integer period lengths?

a phasor~ is always guaranteed to have its last sample be within the last sample increment regardless of whether an integer-period or not, and [wrap~] will pretty much perfectly wrap the phasor~'s phase..

With non-integer period lengths you have effectively a changing number of samples per period. That in itself I don't think would be a problem, but adding a random value and wrapping the result can give you sample-increments that are either smaller or bigger than the calculated value.

Maybe an extreme example can help to clarify: Consider a period length of 2.5 samples. The sample increment is in this case 0.4. If you have a period with the sample values 0.3 and 0.7, add 0.8 as a random value and wrap, what you get is 0.1 and 0.5, so the last sample doesn't get above 1 when you add the calculated sample-increment of 0.4. Am I missing something?

@ben.wes Did you do the testing with non-integer fractions of your sampling frequency? You mentioned 3kHz at 48kHz SR as an example, which shouldn't make problems ....

@manuels said:

By multiplying the frequency with 1/SR you get the phase increment of the phasor. Since there is exactly one sample in each cycle where the sum of the current phase and the increment gets bigger (or equal to) 1, you get the impulse. Now all you have to do is to randomize the phase to get the impulse at a random position.

Just realised, that I didn't consider one important thing here: If the period isn't an integer number of samples, then the method of adding a random number and wrap doesn't work anymore!

@porres said:

Never liked using the samplerate to add an increment, and wrap~ and everything, so what we need is just being able to catch the transition above 1 and make it an impulse out of it, so [op~ >= 1] into [status~] solves that!

Well, it may not be nice, but it was my attempt to solve the problem of missing periods (see further up in the thread). If you have a phasor cycle of, say, 0.1 - 0.3 - 0.5 - 0.7 - 0.9 - 0.1 - etc., then adding small random values to it won't give you any transition above 1.

Edit: ... and the critical case of a cycle 0 - 0.2 - 0.4 - 0.6 - 0.8 - 0 - etc. may illustrate why you need >= instead of > in the [op~].