Miscellaneous Filter

Hello Hermione,

The cut-off frequency will change as you change the sampling frequency.However,you can identify the parameter in the params tab as pole,so you can change it there by writing directly into it.

It is set that low so that it would have minimal impact on the audio signal.It is just intended to remove the DC.

It is a single precision 1st order HPF. So It doesn't have any Q-factor.

For an alternative,If you use single order double precision HPF set to 0.1Hz.It takes one instruction less than the DC block.You can also identify the parameter in the params tab as A1.So you will save one instruction over Dc block and have total control of the filter parameters.

You can relate pole parameter in Dc as the A1 parameter in First order double precision HPF.

Regards,

Harish

Hello,

As Harish says above, the DC block is a first order filter, so it does not have Q.

For a DC block, the cutoff frequency is very low, so in this case we can just use impulse invariance from s domain to z domain:

z(k) = e^(s(k)*T)

i.e. analog poles s(k) are converted to digital poles z(k)

so, you can calculate the cutoff Fc from the digital pole value:

Fc = -Fs/2pi * ln(pole)

i.e. Fc = -48000/2pi * ln(0.9999) = 0.764 Hz

or calculate the pole value from the Fc:

pole = e^(-2pi*Fc/Fs)

Dear bit,

Your equation is flawed

Since for a pole value of 0.9999,the AD website mentions cut off freq as 0.1Hz whereas your equations indicates 0.764 Hz.

Secondly what value of Pole should be chosen to have a cut-off freq of 0.9Hz

Regards

Hermione

Hello Hermione,

Please follow what Harish suggested and use a double precision 1st order HPF and set it for the cutoff frequency you are looking for.

Dave T

Hello,

0.764 Hz is the cutoff, defined as the -3dB point, see the attached plot.