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 / \ | [z~] | | [-~] | 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 <[EMAIL PROTECTED]> 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
pour1.pd
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