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Re: filter, "Q" and resonance

Date1998-01-14 23:24
FromRichard Dobson
SubjectRe: filter, "Q" and resonance
Hans Mikelson wrote:

>
>
> >       Can anyone, please, explain to a begginer what's a nTh order
> >filter. i.e. like a 2nd order lowpass and what is a 12db/24db filter?
>
> The order of the filter corresponds to the order of the differential
> equation which describes the filter.  A first order filter has one
> pole and
> has a roll of of 6 db/octave.  Each order adds another 6 db.  An
> Oberheim
> Matrix has a 2-pole filter so the frequency response rolls off at 12
> db/octave.  A MiniMoog has a 4-pole filter that rolls off at 24
> db/octave.
> In other words frequencies one octave above the cut-off point would be
>
> attenuated (reduced in volume) 24 db with a 4 pole filter.  The
> frequency
> response is the ratio of two polynomials
> (a0+a1*x+a2*x^2...)/(b0+b1*x+b2*x^2...)
>
> The biggest exponent on the polynomials is the same as the number of
> poles.
>  (I think the a's are called zeros and the b's are called poles
> actually
> but lets not get too complicated here...)
>
> Digital filters are a bit different.  Instead of the differential
> equation
> they are based on the difference equation.  The frequency response is
>
> (a0+a1*exp(-i*theta)+a2*exp(-i*2*theta)...)/(b0+b1*exp(-i*theta)+b2*e
> p(-i*2
> *theta)...)
>
> Which looks to me like the zero term is an offset, and each of the
> other
> terms traces out a circle whose radius corresponds to the
> coefficient.  The
> first term would trace out a semi-circle as the frequency goes from 0
> to
> sr/2 (the Nyquist frequency) (theta goes from 0 to pi)  The second
> term
> would complete a full circle, the third term a circle and a half etc.
>
>

To which one could add that in a digital filter, the 'order' identifies
the most delayed sample (referring to the above, where the numerator
gives the order of delayed input samples (giving zeroes) , and the
denominator gives the order of the delayed (fed back) outputs (giving
the poles).

Thus, in practical terms, the higher the order of the filter, the more
samples are involved in the calculation, and the greater the overall
processing time.

Richard Dobson