[FieldTrip] Dimension of lead-field Matrix

Simon Homolle s.homolle at donders.ru.nl
Mon Oct 3 10:23:46 CEST 2016

Dear pooh,

Could you provide more information how you constructed your BEM-model?

best regards,

Simon Homölle
PhD Candidate
Donders Institute for Brain, Cognition and Behaviour
Centre for Cognitive Neuroimaging
Radboud University Nijmegen
Phone: +31-(0)24-36-65059

> On 02 Oct 2016, at 12:22, pooneh baniasad <pooneh.baniasad at gmail.com> wrote:
> 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
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