New subject: I'm stuck in a corner, please help! RE: [delta~] object was: Re: Cyclone in vanilla?

25 Apr 2008


      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

Re: [PD] I'm stuck in a corner, please help! RE: [delta~] object was: Re: Cyclone in vanilla?