[PD] [PD-dev] Filter design for iPhone

[Peter Brinkmann]

On Sat, Mar 24, 2012 at 5:20 AM, Ed Kelly <morph_2016 at yahoo.co.uk> wrote:
> My mistake
>
>> Ah.
>> As often happens, as soon as I have pressed "send" and posted the
>> question, the answer pops out.
>> xb4 = xb4 - xb4 * xb4 * xb4 * 0.166667f;
>>
>> is replaced by...
>>
>> xb4 = xb4 - xb4 * xb4 * xb4 * 0.01f;
>>
>> ...and it works.
>
> and then breaks again

Inf - Inf is NaN.  If the value of xb4 spirals out of control, that
may happen here.  That would also explain why it takes longer to fail
when you reduce the last factor.

>
> Gemnotes-0.1alpha: Live music notation for Pure Data
> http://sharktracks.co.uk/
> ________________________________
> From: Ed Kelly <morph_2016 at yahoo.co.uk>
> To: Mathieu Bouchard <matju at artengine.ca>
> Cc: PD List <pd-list at iem.at>; Joe White <joe at rjdj.me>; pddev <pd-dev at iem.at>
> Sent: Saturday, 24 March 2012, 8:47
> Subject: Re: [PD-dev] [PD] Filter design for iPhone
>
>>> I'm anxious to know what limit is reached in the coefficients of the
>>> filter that causes the undefined result (NaN).
>>
>>I haven't seen the code, but I just want to make you notice that adding
>> together -Infinity and +Infinity results in a NaN ; so does subtracting
>>two infinities of the same sign.
>>
>>So, the NaN might happen when two expressions that are supposed to
>> partially cancel each other, happen to both overflow, in different
>> directions.
>>
>>There are various possible causes for NaN, but with formulas that only
>> involve +, - and *, the possibilities are a lot more limited.
>
> Hmmm.
> I think that's why it puzzles me so. Can you see a / anywhere here?
>
>   while (n--) {
>     i1=(*in++);
>     fc1 = (*fc++);
>     res1 = (*res++);
>     q = 1.0f - fc1;
>     p = fc1 + 0.8f * fc1 * q;
>     fcoeff = p + p - 1.0f;
>     q = res1 * (1.0f + 0.5f * q * (1.0f - q + 5.6f * q * q));
>     i1 -= q * xb4;
>     t1 = xb1;
>     xb1 = (i1 + xb0) * p - xb1 * fcoeff;
>     t2 = xb2;
>     xb2 = (xb1 + t1) * p - xb2 * fcoeff;
>     t1 = xb3;
>     xb3 = (xb2 + t2) * p - xb3 * fcoeff;
>     xb4 = (xb3 + t1) * p - xb4 * fcoeff;
>     xb4 = xb4 - xb4 * xb4 * xb4 * 0.166667f;
>     xb0 = i1;
>     *out++ = xb4;
>  }
>
> No reciprocals, no divisions, just +,-,*.
> I cannot help thinking I must be reaching a "limit" in the calculus sense...
>
> Ed
>
>
>
>
> ______________________________________________________________________
> | Mathieu BOUCHARD ----- téléphone : +1.514.383.3801 ----- Montréal, QC
>
>
> _______________________________________________
> Pd-dev mailing list
> Pd-dev at iem.at
> http://lists.puredata.info/listinfo/pd-dev
>
>
>
>
>
> _______________________________________________
> Pd-list at iem.at mailing list
> UNSUBSCRIBE and account-management ->
> http://lists.puredata.info/listinfo/pd-list
>