Incorrect bidirectional Butterworth filtering of analog channels in Mokka/BTK

[Ben van Basten]

Hello,

A very technical issue:

Apparently, Mokka has options to (zero-lag, bidirectionally) filter the analog data in a C3D using a Butterworth filter. As Mokka is based on BTK, I was interested in how this could be done efficiently in BTK and checked the code (see below)

I noticed that Mokka nicely changes the order and cutoff frequency when zero-lag filtering is selected (using the formula's in Winter's book). However, I noticed that Mokka uses the order of the filter instead of the number of passes when calculating the scaling/compensation factor.

Is this correct? When looking at Winter's book, the cutoff frequency correction should only be based on the number of passes... Unless I am mistaking, this would mean that zero-lag filtering in Mokka only works for 2nd order BW filters.

Regards,

Ben van Basten
Motek

PS. I posted this on the BTK forum as well, but from my experience, this forum is very inactive, so I would like to consult the bm-l community as well.

[FONT=Arial]FilterAnalog.cpp:

const double hfs = data->acquisition()->analogFrequency() / 2.0;
int n = dialog.orderSpinBox->value();
bool zeroLagFilter = (dialog.zeroLagComboBox->currentIndex() == 0);


if (dialog.typeComboBox->currentIndex() == 0)
{
     double wn = dialog.lowerCutOffSpinBox->value() / hfs;
     if (zeroLagFilter) btkEigen::adjustZeroLagButterworth(n, wn);
     btkEigen::butter(&b, &a, n, wn, btkEigen::LowPass);
}

IIRFilterDesign.h:

void adjustZeroLagButterworth(int& n, double& wn)
{
    const double c = 1.0 / std::pow(std::pow(2,1.0/static_cast<double>(n))-1.0, 0.25); //BEN: n should be 2 (number of passes)
    wn *= c;
[FONT=Arial]        n /= 2;[/FONT]
};[/FONT]