[sdiy] tanh() and approximations and warping, why is my maths wrong?

[Tom Wiltshire]

I think what he’s looking for is to know why x/(1+x) gives the right warping to get the correct cutoff frequency.

Since I know very little of how digital filter models are derived from their analog counterparts and the approximations inherent in that process, I’m not the person to ask…

Tom

> On 22 May 2019, at 23:23, Matthias Puech <matthias.puech at gmail.com> wrote:
> 
> Hello,
> Here is a more reasonable approximation between -1 and 1, based on continued fraction expansion of tanh:
> 3x / (3 + x^2)
> More terms will give you less error over a larger interval. Is this what you are looking for?
> Best,
>     -m
> 
> On Wed, May 22, 2019 at 11:52 PM Gordonjcp <gordonjcp at gjcp.net> wrote:
> Hi folks,
> 
> I was giving a bit of thought to how ladder filters tune and of course
> you need to warp Fc a bit for the frequency response to "track" properly
> at higher cutoff settings.
> 
> Looking at various bits and pieces it looks like the trick is to use
> some function of tan() to calculate how much the filter cutoff needs
> "bent down" by.  I can't figure this out, because I suck at maths.
> 
> However if I use the "quick tanh approximation" x/(1+x) to warp my
> normalised cutoff (ie. 6.28 * Fc / Fs) - which is at best a very very
> roughly tanh() shaped function and a very loose approximation - my
> actual filter resonant frequency comes out absolutely bob on.
> 
> Can anyone explain the maths to me?
> 
> -- 
> Gordonjcp
> 
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