[FieldTrip] Dimension of lead-field Matrix
pooneh baniasad
pooneh.baniasad at gmail.com
Sun Oct 2 12:22:55 CEST 2016
Dear FieldTrip community
I'm using the forward model to simulating EEG signal although it seems the
dimension of the lead-field matrix is not correct. Here is a review of the
procedure.
First I constructed a BEM headmodel for EEG source analysis and then by
loading the template cortex, I put the dipoles with specific current source
on that. I expect the dimension of the lead-field matrix will be m*n which
m=electrode's number and n=3*dipole's number but 'm' is different.
Since I used the template electrode 'standard_1020.elc', m = 97 according
to:
chanpos: [97x3 double]
chantype: {97x1 cell}
chanunit: {97x1 cell}
elecpos: [97x3 double]
label: {97x1 cell}
type: 'eeg1010'
unit: 'mm'
while the dimension of lead-field matrix is: 2000x122880
I use this function for calculating lead-field matrix:
LF = ft_compute_leadfield(DipPos, elec, VolBEM);
I do not understand why the number of raws are different!
On the other hand I guess that there is a similarity between the number of
raws in the volume head model and LF matrix due to the dimension of
headmodel matrix is: 2000x8000 double .
I will be so thankful if anyone can help me.
--
Bests
Pouneh Baniasad
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://mailman.science.ru.nl/pipermail/fieldtrip/attachments/20161002/01546d44/attachment-0001.html>
More information about the fieldtrip
mailing list