[PD] [bp~] really equal to [vcf~]?

[Claude Heiland-Allen]

On 25/07/14 16:06, IOhannes m zmölnig wrote:
> On 07/25/2014 03:36 PM, Alexandre Torres Porres wrote:
>>> it's accurate to say that [biquad~] is a real-valued 2-pole/2-zero filter.
>>
>> but why, if the poles and zeros can have complex values?
>
> can they?

Yes, but they occur in conjugate pairs, as the transfer function has 
real coefficients:

        a s^2 + b s + c     (s - z1) (s - z2)
H(s) = --------------- = k -----------------
        e s^2 + f s + g     (s - p1) (s - p2)

          -b +/- sqrt(b^2 - 4 a c)
z1,z2 =  ------------------------
                    2 a

          -f +/- sqrt(f^2 - 4 e g)
p1,p2 =  ------------------------
                    2 e

inside the sqrt is positive -> two real roots
inside the sqrt is negative -> complex conjugate pair

The imaginary parts of the complex conjugate "cancel out" to the 
quadratic polynomial with all-real coefficients.

> afaict, [biquad~] only ever has real-valued coefficients and states.

sure, just different representations of the same thing


Claude
-- 
http://mathr.co.uk