I've made it work with this design in fexpr, but it's only good up to half
the block size


2015-12-08 15:21 GMT-02:00 Alexandre Torres Porres <[email protected]>:

> As I say in the attached patch; problem is that whenever you change the
> number of samples without clearing or clearing the filter, it goes crazy...
>
> not sure how to fix this.
>
> cheers
>
>
> 2015-12-08 10:40 GMT-02:00 Christof Ressi <[email protected]>:
>
>> the filter is working perfectly well for me. send me your testing patch.
>>
>>
>>
>> Gesendet: Dienstag, 08. Dezember 2015 um 13:36 Uhr
>> Von: "Alexandre Torres Porres" <[email protected]>
>> An: "Christof Ressi" <[email protected]>
>> Cc: "[email protected]" <[email protected]>
>> Betreff: Re: Re: Re: [PD] Moving Sum object?
>>
>> not sure this is working either, I'm getting negative values when testing
>> it with count~ (only positive values)
>>
>> 2015-12-08 10:30 GMT-02:00 Christof Ressi <[email protected]>:It
>> certainly is and I have no idea how that happened :-D. but it has no effect
>> whatsoever. just delete the connection.
>>
>>
>> Gesendet: Dienstag, 08. Dezember 2015 um 13:20 Uhr
>> Von: "Alexandre Torres Porres" <[email protected][[email protected]]>
>> An: "Christof Ressi" <[email protected][[email protected]]>
>> Cc: "[email protected][[email protected]]" <[email protected][
>> [email protected]]>
>> Betreff: Re: Re: [PD] Moving Sum object?
>>
>> what was the purpose of the output from the max object into rpole~ in
>> your patch? that's a bug, right?
>>
>> cheers
>>
>> 2015-12-08 9:50 GMT-02:00 Christof Ressi <[email protected][
>> [email protected]]>:
>>
>> Or are you talking about a object that outputs the sum as a message? In
>> that case just use [cmavg~] or [maverage~] with [snapshot~]. If you just
>> want the sum of a signal vector, there are some objects in zexy that will
>> do the job, like [avg~] or [pack~]+[sum]
>>
>>
>> Gesendet: Dienstag, 08. Dezember 2015 um 12:41 Uhr
>> Von: "Christof Ressi" <[email protected][[email protected]][
>> [email protected][[email protected]]]>
>>
>> An: "Alexandre Torres Porres" <[email protected][[email protected]][
>> [email protected][[email protected]]]>
>> Cc: "[email protected][[email protected]][[email protected][
>> [email protected]]]" <[email protected][[email protected]][
>> [email protected][[email protected]]]>
>> Betreff: Re: [PD] Moving Sum object?
>>
>> Well, a linear moving average filter is just something that  sums a
>> series of samples. If you don't want the average but rather the true sum,
>> either multiply the output by the number of samples or take my abstraction
>> and get rid of the [/~] object.
>>
>>
>> Gesendet: Dienstag, 08. Dezember 2015 um 12:21 Uhr
>> Von: "Alexandre Torres Porres" <[email protected][[email protected]][
>> [email protected][[email protected]]]>
>> An: "Christof Ressi" <[email protected][[email protected]][
>> [email protected][[email protected]]]>
>> Cc: "Matt Barber" <[email protected][[email protected]][
>> [email protected][[email protected]]]>, "[email protected][
>> [email protected]][[email protected][[email protected]]]" <
>> [email protected][[email protected]][[email protected][
>> [email protected]]]>
>> Betreff: Re: Re: [PD] Moving Sum object?
>>
>> cool guys, but i was asking for an average "sum" object :)
>>
>> 2015-12-08 9:19 GMT-02:00 Christof Ressi <[email protected][
>> [email protected]]>: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" <[email protected][[email protected]]>
>> An: "Alexandre Torres Porres" <[email protected][[email protected]]>
>> Cc: "[email protected][[email protected]]" <[email protected][
>> [email protected]]>
>> 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 <
>> [email protected][[email protected]]> 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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>

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