# [music-dsp] Tweakable filters of higher orders?

Frederick Umminger music-dsp at umminger.com
Thu Sep 30 03:37:28 EDT 2004

```----- Original Message -----
From: "Koen Tanghe" <koen at smartelectronix.com>
> Sure, you can factor any filter out in lower order polynomials, but as
> Nigel
> already pointed out in another mail, what I asked was: "how do you people
> usually implement steep higher order filters *that are at the same time
> tweakable* in both cutoff/center frequency and bandwidth/Q". Note the word
> "tweakable", by which I meant: you can change these two parameters
> (knobs/sliders/whatever...) in real-time to change the filters'
> properties.
> It's not about a single fixed design which you can do with the Matlab
> tools
> for example.
>

Well, if you can manage to create a z-domain version of a higher-order
Butterworth filter then you can create higher order tweakable forms of all
of the usual filters (lowpass, highpass, parametric, low shelf, high shelf)
in the same way you do with a 1st order Butterworth; by adding or
subtracting it from a unity gain signal and using some simple fiter
transforms like lowpass to bandpass. The only difficult step is changing the
cutoff frequency.
You can change the cutoff frequency either with a table lookup for the
coefficients or by applying a linear fractional transformation to a
prototype filter. A linear fractional transformation is just a substitution
of variables of the form z -> (a+bz)/(c+dz). The appropriate choice of a,b,c
and d will give the combined result of performing an inverse bilinear
transform, scaling in frequency to get a new cutoff, and performing a
bilinear transform to get back to the z-domain, all for the same cost as a
single bilinear transformation. With the right choice of filter topology you
might not have to actually compute the transformation at all. Frequency
warped filters are an example of this.
In some cases it might be cheap to just recompute the factored filter
coefficients for the new cutoff from first principles. I think this would be
the case with a Butterworth, since the poles are at easily computable known
locations.

-Frederick Umminger

```