Le 12/04/2014 08:45, Alexandre Torres Porres a écrit :
> change the [fexpr~] to something like
> [fexpr~ $x[0] + ($f2 * $y[-1]) + ($f3 * $y[-2])]
f*ck, I'll be damned, now my patch that implements [bp~] with [fexpr~]
seems to work, it's attached. Thanks!
thanks for the share, when I tried to implement filters with expr and
biquad I haven't been able to get it working fully (some clics appeared
when changing frequency)... Your implementation works very well in both
cases, "chapeau bas"
> it's pretty easy to see that from the code you quoted
I can't really see it from the code itself. And, well, remember I
mentioned about the biquad code?
{
t_sample output = *in++ + fb1 * last + fb2 * prev;
if (PD_BIGORSMALL(output))
output = 0;
*out++ = ff1 * output + ff2 * last + ff3 * prev;
prev = last;
last = output;
}
Well, I made a silly confusion mistake and thought the first line was
feedforward (and then equivalent to the bp~). But still, it could be
it for all I can tell. How can you actually see wether is feedback or not?
Anyway, the patch works and I can also make it on biquad, it's all
attached.
> after all it's a resonating filter and therefore needs a feedback path.
I wouldn't know about that, but that's how you convinced me you knew
what you were talking about :)
Thanks again
2014-04-11 16:46 GMT-03:00 volker böhm <vbo...@gmx.ch
<mailto:vbo...@gmx.ch>>:
On 11.04.2014, at 16:48, Alexandre Torres Porres wrote:
> ""last" and "prev" are the last two _output_ samples.
> i don't know fexpr~ very well, but it looks like you try to
access the last _input_ samples."
>
> In [fexpr~] you can access input samples with $x variables and
output samples with $y. So you're correct. I'm going for the input
samples.
>
> But I did it because I believe "last" and "prev" in this formula
are in fact about input samples.
no, and it's pretty easy to see that from the code you quoted:
> > t_sample output = *in++ + coef1 * last + coef2 * prev;
> >
> > *out++ = gain * output;
> >
> > prev = last;
> >
> > last = output;
after all it's a resonating filter and therefore needs a feedback
path.
so it somehow has to take outgoing samples back in.
> So I feel pretty strong about getting this [fexpr~] right. Is
there anything I did not take into consideration?
yes, calculate coef1, coef2 and gain by using the formulas from
the code,
change the [fexpr~] to something like [ fexpr~ $x[0] + ($f2 *
$y[-1]) + ($f3 * $y[-2]) ],
(where $f2 and $f3 would be coef1 and coef2 resp.)
apply the gain factor afterwards,
and you are done.
vb
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