[PD] I'm stuck in a corner, please help! RE: [delta~] object was: Re: Cyclone in vanilla?
padawan12 at obiwannabe.co.uk
Fri Apr 25 17:08:38 CEST 2008
I've attached again an example of a patch that demonstrates the
practical difference between different one sample differentiators.
Try replacing [fexpr~ $x1 - $x1[-1]] in the water flow generator with
which is fine.
Then try implementing the same with [rzero~ 1]
It sounds very different and I have not found a way to correct
the accumulating DC error. Try the obvious [rzero~ 0.99999999]
etc to hear that the behaviour is still not right.
Fundamentally, [z~] is a *very* useful primitive to have
On Fri, 25 Apr 2008 07:04:10 -0400
Enrique Erne <enrique at netpd.org> wrote:
> IOhannes m zmölnig wrote:
> > Enrique Erne wrote:
> >> IOhannes m zmölnig 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]
> > | |
> > [-~]
> > |
> thanks iohannes. it looks good now.
Use the source
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