[was: CORRECTION: cookbook formulae for audio EQ biquad

Sylvain Rebaud Sylvain at spatializer.com
Mon Nov 16 16:51:57 EST 1998


You're right about calculating the proper filter instead. We just wanted
to come up with
a way to assure that a filter would always be stable whatever
coefficients are spit out
of a design algorithm. It can be Robert's formulae or anything else.
It's just a way to make it stable.
We actually found out afterwards that it was unstable because we were
above 0.5Fs.
So you're right that it makes more sense to verify first that we are not
above Nyquist but 
that's not the point. The point was that if you have an unstable biquad,
you can make it stable
by inverting the conjugate poles and zeros and keep a fairly (very very)
similar response. 
I agree by the way that it's not exactly the same (it's non linear) but
by doing a simple offset
in f=0, you can have something that matches the original frequency
response pretty well.
Since a frequency response is a ration of distances (from z to
poles/zeros),
it seemed to me that inverting the poles & zeros would keep that ratio.
It's true that I didn't
look into the math to verify it. ;-)

Sylvain


> -----Original Message-----
> From:	Chris Weare [SMTP:cweare at netmagic.NET]
> Sent:	Monday, November 16, 1998 12:50 PM
> To:	Multiple recipients of list
> Subject:	Re: [was: CORRECTION: cookbook formulae for audio EQ
> biquad
> 
> Sylvain Rebaud wrote:
> > 
> > Chris,
> > 
> > We are actually inverting both poles and zeros so that the frequency
> > response is exactly the
> > same as before but the filter is stable.
> 
> The response won't be exactly the same.  You are reflecting across an
> arc (the unit circle to be exact),  not a line.  However,  this is all
> academic.  If you know beforehand that your cutoff is above 0.5Fs why
> not just calculate the proper filter?
> 
> dazed and confused,
> -chris






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