[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

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

  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.​


Pouneh Baniasad
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