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@gmx.ch mailto: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
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