[PD] can [bp~] be obtained with biquad coefficients?
patrice colet
colet.patrice at free.fr
Sat Apr 12 14:25:59 CEST 2014
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 <vboehm at gmx.ch
> <mailto:vboehm at 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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