[PD] Moving Sum object?

[Matt Barber]

Yep, you're right.

To be clear to anyone else reading, the equation is y[n] = (y[n-1] + x[n] -
x[n-k])/k . I have a mental block when it comes to the filter element
objects and tend to do things the hard way at first. :)

On Tue, Dec 8, 2015 at 6:19 AM, Christof Ressi <christof.ressi at gmx.at>
wrote:

> Hey Matt,
>
> there's no need for the feedback path (and therefore no [block~ 1] ;-))
>
> Just use the following formula:
>
> y[n] = (y[n-1] - x[n-k])/k
>
> where k is the number of samples to be averaged (must be at least 1). see
> the patch I sent to Alex in my last mail.
> it uses [rpole~ 1] for the y[n-1] part and [z~ k] for the x[n-k] part (you
> can replace the latter one with a [delwrite~] [delread~] pair to make it
> purely vanilla).
>
> The funny thing about linear moving average filters is, that although it
> can be implemented as a recursive filter (like in both our patches), it is
> still a FIR filter (and therefore it defeats the notion that recursive
> filters are always IIR filters). The impulse response is just a rectangular
> pulse and therefore finite.
>
>
>
> Gesendet: Dienstag, 08. Dezember 2015 um 07:13 Uhr
> Von: "Matt Barber" <brbrofsvl at gmail.com>
> An: "Alexandre Torres Porres" <porres at gmail.com>
> Cc: "pd-list at lists.iem.at" <pd-list at lists.iem.at>
> Betreff: Re: [PD] Moving Sum object?
>
> Something like this? Almost completely untestsed. :D
>
> On Tue, Dec 8, 2015 at 12:20 AM, Alexandre Torres Porres <porres at gmail.com>
> wrote:
>
> Talking about averages I wonder if we have an object that sums (in a
> moving average fashion) a series of samples
>
> cheers
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