> 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@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