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

[Alexandre Torres Porres]

> Having only real coeffecients for biquad forces the poles (zeros) to take
this form.

yeah, that makes sense. The tricky part seems to be how a complex conjugate
pair does that too. Is it because you'd add them up and then the imaginary
part becomes zero? That's what I can think of...


2014-07-25 10:49 GMT-03:00 Claude Heiland-Allen <claude at mathr.co.uk>:

> On 25/07/14 14:36, 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?
>>
>
> If the poles (zeros) occur in as a complex conjugate pair (with identical
> real parts and opposite imaginary parts), or both poles (zeros) are real,
> then the filter will have real output for real input.  Having only real
> coeffecients for biquad forces the poles (zeros) to take this form.
>
>
> Claude
> --
> http://mathr.co.uk
>
>
>
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