damping in biquads for butterworth filter?

On Apr 8, 1:45&#4294967295;pm, Jerry Avins <j...@ieee.org> wrote:
> esfield wrote:
> > Yeah, I just realized that.... so what are the other constraints on these
> > values that force them to be the ones Vlad gave?
>
> Math. All low-pass Butterworth filters have all derivatives zero at f=0.
> That gives a 4th-order filter the response 1 + x^4 where x = w/w0. Solve
> for the roots.
>
> Jerry
> --
> Engineering is the art of making what you want from things you can get.
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Actually a small correction is needed.

A Nth order Butterworth filter has a response of 1/sqrt(1+omega^2N)

So we actually start with 2*N roots of -1 and just use the roots on
the left hand plane to form our polynomial.

For example, a 2nd order butterworth

We start with the 4 roots of -1 on the unit circle. These are at
angles

90,135,225, and 315 degrees.

The ones on the LHP are simply at 135 and 225 degrees or in radians

these are 3pi/4 and 5pi/4

So we combine the poles in the left hand plane as conjugate pairs so
as to result in a polynomial with real valued coefs. For our 2nd order
case we only have one pair to combine. But our poly is simply the
product of (s-root) terms. For our specific case:

(s-exp(j3pi/4))(s-exp(j5pi/4)


which simplfies to

s^2+sqrt(2)s+1


For odd order cases you see that one of the roots is -1 so we have a
(s-(-1)) term. Of course this simplifies to s+1 which is a factor of
all odd order normalized Butterworth filters.

IHTH,

Clay

esfield wrote:
> Thanks again Jerry.  Please bear with me a bit longer!
> 
> That gives pole locations at sqrt(j), sqrt(-j), -1/sqrt(j), -1/sqrt(-j).
> 
> But, aren't these the pole locations for a second order butterworth? 
> There are four poles but we only use the left half plane, so this only
> gives two poles so this is a second order filter, right?
> 
> Also, it's not obvious to me how the Q values come from the pole
> locations.  Can you please enlighten me?

I omitted a sqrt. Go with Clay's formulation.

> Clay, thanks for the link.  I wish I'd had that reference when I started! 
> Just a quick question: to transform the lpf into hpf, all I do is change
> the '1' in each numerator to 's^2', right?

Jerry
-- 
Engineering is the art of making what you want from things you can get.
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