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@gmail.com>
An: "Alexandre Torres Porres" <porres@gmail.com>
Cc: "pd-list@lists.iem.at" <pd-list@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@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
_______________________________________________
Pd-list@lists.iem.at[Pd-list@lists.iem.at] mailing list
UNSUBSCRIBE and account-management -> http://lists.puredata.info/listinfo/pd-list[http://lists.puredata.info/listinfo/pd-list]
 _______________________________________________ Pd-list@lists.iem.at mailing list UNSUBSCRIBE and account-management -> http://lists.puredata.info/listinfo/pd-list[http://lists.puredata.info/listinfo/pd-list]