# [PD] convert reson filter bandwidth to decay time

Charles Henry czhenry at gmail.com
Tue Jan 22 21:12:32 CET 2008

```It depends upon the order of the filter... by resonant filter, I
assume you mean a two-pole bandpass filter.  Here's the math for
converting between bandwidth and exponential decay

Take a function, g(t)=1000^-(t/r)
where r is the -60 dB ring time.  I used a base of 1000, because a
factor of 1000 = 60 dB

g(t)=e^-(t*ln(1000)/r)

Now we take a one-sided fourier transform of this function:

G(f)=integral(0, inf;  e^(-2*pi*i*f*t)*e^-|t*ln(1000)/r|)
G(f)=integral(0, inf;  e^((-2*pi*i*f-ln(1000)/r)*t)
G(f)=1/(-2*pi*i*f-ln(1000)/r) * e^((-2*pi*i*f-ln(1000)/r)*t), eval at t=0, t=inf

G(f)= - 1/(-2*pi*i*f-ln(1000)/r)
G(f)=1/(2*pi*i*f+ln(1000)/r)

G(f)=r/ln(1000)  / [2*pi*i*f*r/ln(1000) + 1]

now, it's just a simple matter of finding the -3 dB points to find
bandwidth of this function and interpreting the result of this
function...

|G(f)|^2=(r/ln(1000))^2 / [1 + (2*pi*f*r/ln(1000))^2]

0.5 = 1 / [1 + (2*pi*f*r/ln(1000))^2]
(2*pi*f*r/ln(1000))^2 = 1

f = +/-  ln(1000)/(2*pi*r)
bandwidth
bw= 2 * ln(1000)/(2*pi*r)

conversions:
bw = ln(1000) / (pi*r)
r = ln(1000) / (pi*bw)
r is in seconds and bw is in Hz

Try out these relations and see if they work.

Chuck

On Jan 22, 2008 12:11 PM, Libero Mureddu <libero.mureddu at gmail.com> wrote:
> Hi,
> I have  a simple patch with a [click~] object connected to a [reson~]
> filter. I'd like to know how it is possible to convert the bandwith
> parameter of the filter to decay lenght (in milliseconds).
> In SuperCollider there is a particular version of the "resonz" filter
> called "ringz", here is the description:
> Ringz.ar(in, freq, decaytime, mul, add)
>
> This is the same as Resonz, except that instead of a resonance
> parameter, the bandwidth is
> specified in a 60dB ring decay time.
>
> My knowledge of filters is too little to be able to convert the
> between the two parameters by myself, any help appreciated!
>
> thanks,
>
> libero
>
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```