In a typical resonant bandpass filter, resonance is set as none at 0 and full at 1, such that notes ring infinitely at 1 and not at all at 0.

If you have a two-pole resonant bandpass filter with a basic single sample impulse excitation to a maximum amplitude of "1" and a given resonance setting between 0 and 1:

What is the mathematical equation that would allow you to predict the decay time of the resonant filter based on its resonance setting? For example, to an arbitary level of "1/e"? ie. To 36.7879% its original amplitude?

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**1**of**1**- KVRian
- 946 posts since 11 Apr, 2002

The 2-pole output decays as exp( t * Re p ) in analog and |p|^n in digital case, where p is one of the filter's poles.

If the poles are real you have two answers, take the one which decays slower. If you need an exact target level, you'll need to take both then, but you'll need to know their relative amplitudes, off the top of my head I wouldn't be able to write the respective equations, needs some paperwork

If the poles are real you have two answers, take the one which decays slower. If you need an exact target level, you'll need to take both then, but you'll need to know their relative amplitudes, off the top of my head I wouldn't be able to write the respective equations, needs some paperwork

- KVRAF
- 3762 posts since 8 Mar, 2004, from Berlin, Germany

this is the function, i'm using for this:

filter difference equation is:

y[n] = something-with-old-inputs - a1*y[n-1] - a2*y[n-2]

rsLogB is the base-B logarithm i.e. log(x)/log(B). but this formula does not take into account the positions of the zeros and i'm not sure how i would take them into account. maybe by evaluating the magnitude response at the pole-frequencies/angles? or would i need to do the partial fraction expansion to obtain the pole "residues"? dunno - so far, the pole-only based formula was good enough for my purposes.

btw. if i remember correctly, using "+" in the pole-computation instead of "-" (you may recognize the formula as the "p-q-formula") makes sure to select the larger pole (with slower decay) in case of two real poles...edit: wait - i think, i need to actually verify this...the sqrt function will return a positive number in this case, so if -a1 is positive, too, "+" will indeed result in the larger number - i think, using the negative sign convention for a-coeffs (as i do), a1 will indeed be negative, so -a1 should be positive. i hope so...

- Code: Select all
`template<class T>`

T biquadRingingTime(T a1, T a2, T threshold)

{

std::complex<T> p = -T(a1/2.0) + sqrt(std::complex<T>(a1*T(a1/4.0)-a2)); // 1st pole

return (T) rsLogB(threshold, abs(p));

}

filter difference equation is:

y[n] = something-with-old-inputs - a1*y[n-1] - a2*y[n-2]

rsLogB is the base-B logarithm i.e. log(x)/log(B). but this formula does not take into account the positions of the zeros and i'm not sure how i would take them into account. maybe by evaluating the magnitude response at the pole-frequencies/angles? or would i need to do the partial fraction expansion to obtain the pole "residues"? dunno - so far, the pole-only based formula was good enough for my purposes.

btw. if i remember correctly, using "+" in the pole-computation instead of "-" (you may recognize the formula as the "p-q-formula") makes sure to select the larger pole (with slower decay) in case of two real poles...edit: wait - i think, i need to actually verify this...the sqrt function will return a positive number in this case, so if -a1 is positive, too, "+" will indeed result in the larger number - i think, using the negative sign convention for a-coeffs (as i do), a1 will indeed be negative, so -a1 should be positive. i hope so...