hi peter,

it's because fsl's temporal filtering is implemented in terms of sigmas in volumes

----

>" -bptf  <hp_sigma> <lp_sigma> : (-t in ip.c) Bandpass temporal filtering;
>nonlinear highpass and Gaussian linear lowpass (with sigmas in volumes, not
>seconds); set either sigma<0 to skip that filter"

----

technically this comes from

sigma in volumes = FWHM(window in seconds)/ (TR*sqrt(8*log(2)))

where,

sigma in secs = FWHM(seconds)/(sqrt(8*log(2)))

sigma in volumes = sigma in secs / TR

the denominator is a matter of conversion of fwhm to sigma and they approximate the real number sqrt(8*log(2)) ~ 2.35... with 2. 

for fslmaths:

sigma = fwhm/(2*TR)

or 

sigma = 1/(2*TR*cutoff_in_hz)

cheers,

satra

On Tue, May 22, 2012 at 3:23 PM, Peter Fried <[log in to unmask]> wrote:

Hi FSL group,

I saw another post on this question, but no answer so far.

In the FEAT main Data tab, if I set the high pass filter cutoff to 100 sec (with high pass temporal filtering selected in Pre-stats),

the Pre-stats report states"
       "high pass temporal filtering (Gaussian-weighted least-squares straight line fitting, with sigma=50.0s)."

and the log states:
       fslmaths prefiltered_func_data_intnorm -bptf 20.0 -1 prefiltered_func_data_tempfilt


If I divide the cutoff (100) by the TR (2.5), I get

(100 sec / 2.5 vol/sec) = 40 vol

Why is the -bptf flag in fslmaths using a sigma of 20 instead of 40? Am I missing something or is it being divided in half?

The reason I'm asking is that I would like to substitute the high pass filter for a bandpass filter of 0.009-0.08 Hz. I have calculated the following values:

hp_sigma = (1/0.009)/2.5 = 44.44
lp_sigma = (1/0.08)/2.5 = 5

Are these the right values for fslmaths, or do they need to be divided in half?

Thanks.

Cheers,
Peter Fried

Boston University School of Medicine