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