Hello,
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.
Thanks,
Matt
Date: Fri, 25 Apr 2008 12:21:54 +0200 From: IOhannes m zm?lnig zmoelnig@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@netpd.org Cc: pd-list@iem.at Message-ID: 4811B0C2.20504@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:
y[n]=x[n]-a[n]*x[n-1] since you set a[n] to "0", you just get y[n]=x[n] :-(
to get [z~ 1], do something like
| +--+ | | | [rzero~ 1] | | [-~] |
fmgads.r IOhannes