[PD] 8 point average filter with biquad and raw filters?

[Alexandre Torres Porres]

so, here's the biquad verision of the 8 point average filter

I was actually more interested in the raw filters and I could get there
with the biquad version by converting the coefficients to poles/zeros. All
I did now was the opposite conversion to make the biquads.

but yeah, I fried my brains a bit to be able to make the biquad for the 4
average filters, it was all about trying to find a clever solution, but
then I got stuck by trying to follow the same line of thought to make other
poin averages like 8...

So let me see if I got it straight. All you need to do is divide the circle
in the number of points and you know where the zeros are. For instance, a 2
point average just divide 360 / 2 and you have 180º

8 point average 45º - 90º - 135º - 180º - 225º - 270º - 315º

So, a 3 point average filter is 120º and 240º

right?

cheers



2015-09-09 2:06 GMT-03:00 Alexandre Torres Porres <porres at gmail.com>:

> as long as we're on this, I don't know much about average filters, but is
> it common to have even sample average like 3-point / 5-point?
>
> i was guessing there had to be at least even numbers, but now i have a
> different intuition
>
> cheers
>
> 2015-09-09 2:00 GMT-03:00 Alexandre Torres Porres <porres at gmail.com>:
>
>> I wasn't looking for this, but it's great to know it :) thanks!
>>
>> 2015-09-08 20:28 GMT-03:00 Christof Ressi <christof.ressi at gmx.at>:
>>
>>> Hi, there's actually a nice and easy way of implementing a moving
>>> average filter of ANY length using only an integrator and a samplewise
>>> delay [z~]. The formular for a moving average filter of N points is simply:
>>> y[n] = (x[n] - x[n-N])/N + y[n-1]. I attached an abstraction. Not totally
>>> sure this is what you're looking for.
>>>
>>> Cheers
>>>
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>>
>
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