dipole time course
Floris de Lange
florisdelange at GMAIL.COM
Tue Jul 7 14:34:12 CEST 2009
Dear Jan-Matthijs and other fieldtrippers,
Thanks for your reply. I tried out your first suggestion. Unfortunately, the
dipole time courses look considerably worse than what I get from picking the
planar sensors that have the strongest signal. I've attached the results for
2 conditions in the left and right auditory cortex, for comparison. On the
top left is (one of the 3 matrices of) the relevant leadfield, and on the
right is the planar ERF. On the bottom are the time courses of the pinv'ed
leadfield * ERF time course, which are then pca'ed. I have pasted the code
used below. Any idea why the dipole time course looks so much worse?
% calculate leadfields for the two dipole locations
cfg = [];
cfg.grad = erf{1}.grad;
cfg.vol = vol;
cfg.channel = {'MEG'};
cfg.grid.pos = source.dip.pos;
[grid] = prepare_leadfield(cfg);
% multiply pseudo-inverse of leadfield with erf, then do a pca
dip{i}{cond} = pinv(grid.leadfield{i}) * erf{cond}.avg;
[tmp1 tmp2] = pca(dip{i}{cond});
Any input appreciated,
Best wishes,
Floris
Dear Floris,
>
> Dipolefitting uses the pseudo-inverse of the leadfield as its inverse
> operator. Using prepare_leadfield with the right input should give
> you the leadfields of the two locations of interest, pinv'ing the
> concatenated leadfields would give you the inverse operator (and
> multiplying the sensor level evoked field with the inverse operator
> gives you the dipoles' time courses (each in 3D)). The individual
> timecourses could be reduced to 1D by a pca. Alternatively, you could
> multiply each of the individual leadfields with the average
> reconstructed dipole orientation, so that you end up with a
> (concatenated across the two dipoles) forward model of 275x2,
> pinv'ing this one, and multiplying it with your sensor data directly
> gives you time courses in 1D.
>
> Cheers,
>
> JM
>
> On Jul 6, 2009, at 5:48 PM, Floris de Lange wrote:
>
> > Dear Fieldtrippers,
> >
> > I have a fairly basic Fieldtrip question; I hope you can point me
> > in the right direction.
> > I'm running a dipole model on the early auditory response (60-90 ms
> > after the click). The dipoloe fit works very well. Just for
> > completeness, these are the lines of code I used:
> >
> > cfg = [];
> > cfg.numdipoles = 2;
> > cfg.symmetry = 'y';
> > cfg.model = 'regional';
> > cfg.latency = [-0.44 -0.41];
> > cfg.vol = vol;
> > cfg.inwardshift = 0;
> > source = dipolefitting(cfg,erf{1});
> >
> > % visualize where the dipoles are
> > figure(1); clf; headmodelplot(cfg,erf{1})
> > hold on;
> > plot3([source.dip.pos(1,1)],[source.dip.pos(1,2)],[source.dip.pos
> > (1,3)],'ro');
> > plot3([source.dip.pos(2,1)],[source.dip.pos(2,2)],[source.dip.pos
> > (2,3)],'bo');
> >
> > I have fitted the dipole over a 30 msec window. But now, I would
> > like to plot the response of these dipoles for the whole time
> > window sampled.
> > In other words, I would like to use the coordinates obtained from
> > the dipole fitting procedure, and use them as 'virtual sensors' to
> > reduce the dimensionality of my data from 275 channel time courses
> > to 2. I don't want to use the dipole moments, because they're only
> > defined within the 30 ms that I used to calculate my dipole.
> > I suppose I have to project my data through the dipoles, but I'm
> > not sure how. Any help is greatly appreciated!
> >
> > Best wishes,
> > Floris
>
--
Floris de Lange
http://www.florisdelange.com
----------------------------------
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