[PD] I'm stuck in a corner, please help! RE: [delta~] object was: Re: Cyclone in vanilla?
brbrofsvl at gmail.com
Fri Apr 25 17:47:35 CEST 2008
For [z~ 1] you can use [rzero_rev~ 0] -- its response is:
y[n] = -a[n] * x[n] + x[n-1]
setting the coefficient to 0 gets rid of the current input and leaves
the one-sample delay.
> Date: Fri, 25 Apr 2008 12:21:54 +0200
> From: IOhannes m zm?lnig <zmoelnig at iem.at>
> Subject: Re: [PD] I'm stuck in a corner, please help! RE: [delta~]
> object was: Re: Cyclone in vanilla?
> To: Enrique Erne <enrique at netpd.org>
> Cc: pd-list at iem.at
> Message-ID: <4811B0C2.20504 at iem.at>
> Content-Type: text/plain; charset=ISO-8859-1
> Enrique Erne wrote:
> > IOhannes m zm?lnig wrote:
> >> Enrique Erne wrote:
> >>> or [biquad~ 0 0 0 1]
> >>> Miller Puckette wrote:
> >>>> I believe z~ is just rzero~ 0.
> >> no.
> >> both of them are equivalent to [z~ 1]
> >> you could also argue that [f] is just the same as [0(
> >> :-)
> > oups, yes ofcorse z~ 1.
> > the output of 1 sample with rzero~ 0, z~ 1 and biquad~ 0 0 0 1 seems to
> > be slightly different. if one wants to be fuzzy about that :) maybe ome
> > rounding problem?
> no, i don't see any rounding errors...
> > and now i even couldn't do the delwrite/read with the subpatch :( :(
> it's generally a good idea to tell [delwrite~] how much space it should
> allocate for the delayline. e.g. [delwrite~ abcd 1000] helped a lot...
> and [rzero~ 0] is not the same as [z~ 1].
> the output of [z~ 1] is y[n]=x[n-1]
> according to [rzero~]s help-patch it does the following:
> since you set a[n] to "0", you just get y[n]=x[n] :-(
> to get [z~ 1], do something like
> | |
> | [rzero~ 1]
> | |
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