I think this is my mistake -- as someone else pointed out, it should have been "rzero_rev~", not "rzero 1" -- sorry for the confusion.
Miller
On Fri, Apr 25, 2008 at 04:08:38PM +0100, Andy Farnell wrote:
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 enrique@netpd.org wrote:
IOhannes m zm?lnig wrote:
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] | | [-~] |
thanks iohannes. it looks good now.
-- Use the source
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