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

[Alexandre Torres Porres]

""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. And I still do. The reason being that I checked
the code of other objects like [biquad~], and "last" and "prev" where names
used both for input and output operations, the difference being that the
math for the output operation was something like *out++ + coef1 * last +
coef2 * prev  instead of *in++ + coef1 * last + coef2 * prev (like bp~) .

So I feel pretty strong about getting this [fexpr~] right. Is there
anything I did not take into consideration?

One way or another, input or output samples, seems pretty clear to me you
could achieve [bp~] with [biquad~] coefficients. I think the tricky part
now is getting to the coefficients and gain values.

Cheers



2014-04-11 3:23 GMT-03:00 volker böhm <vboehm at gmx.ch>:

>
> On 11.04.2014, at 03:07, Alexandre Torres Porres wrote:
>
> > hey, the code I sent only calculates the coeficients, but I left out an
> important part which is
> >
> > t_sample output = *in++ + coef1 * last + coef2 * prev;
> >
> >         *out++ = gain * output;
> >
> >         prev = last;
> >
> >         last = output;
> >
> >
> >
> > This shows how the filter is done with those coefficients
> >
> >
> >
> > It's easy to implement this with [fexpr~], it goes something like:
> >
> >
> >
> > [fexpr~ $x + (coef1 * $x1[-1]) + (coef2 * $x1[-2])]
>
> "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.
> vb
>
>
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