[PD] can [bp~] be obtained with biquad coefficients?

[Alexandre Torres Porres]

there are incremnents to do still, but thanks :)


2014-04-12 9:25 GMT-03:00 patrice colet <colet.patrice at free.fr>:

>
> 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>:
>
>>
>> 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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