Help with dipole fitting
Paul Czienskowski
paul_c at GMX.DE
Fri Jul 23 08:06:46 CEST 2010
Hi Ludwig,
On 23.07.2010 02:56, Ludwing Torres wrote:
> Hi, I'm very sorry for using capital letters, please excuse me.
Never mind, I was not too offended at this, but wanted you to take note
of the fact, that somebody could be :)
> I'm new at fieldtrip, I don't know much about this and I'm searching
> for help desperately.
Then I hope I can help you.
> Well, I am trying to compute a fordward model by now, like this:
>
> V=A*J
>
> where:
>
> V= a matrix of potentials measured in the scalp surface by each
> electrode, for example 32 electrodes or channels and 5 seconds at 256
> samples per second, (which is equal to 5*256=1280, which comes to a
> 32x1280 matrix, one epoch I guess, and one trial, just measure, no
> events, *please correct me if i'm wrong*)
This sounds quite correct, I think, but AFAIK you won't compute this
directly but use some function like the ft_dipolesimulation function we
talked about last time. This does exactly what you want to do, compute
the electrode scalp potentials due to a given dipole distribution.
Spares you all the stuff you are now to consider.
> J= a matrix of the sources in the brain that produce the potentials
> above, with its time course, which are more than the electrodes placed
> there.
> Let me understand this: ¿these sources and its time course, are
> the same dipole moments?
> If so, ¿The sources (dipoles) have a time component in the three
> dimensions in the dipoles?
> Thus, if we consider 128 sources, ¿ are we considering 128
> dipoles, and J matrix would be 128x1280 with the time course?
> or ¿Are we considering 128 sources of 3 dimensions and J matrix
> would be of 128x(1280x3)?
I think it's rather the latter, but see below.
> A= the lead field matrix. I've seen the function compute_leadfield of
> FieldTrip, This uses the electrodes positions, the dipoles (or
> sources) positions, and the standard volume of sphere and
> conductivities to perform the computing of a leadfield matrix that has
> dimensions numch x (3xnums) where numch=nuber of channels or
> electrodes (32 in this case) and nums=number of sources (in this
> case 128) ¿why is the number of sources multiplied by 3? ¿ it is
> because they compute the leadfield matrix to each of the cartesian
> coordinates? and if so, ¿How should I take the J matrix to compute the
> fordward model? ¿ should I take this each row having the time course
> of the sources, in the order: row 1:3 dipole n1 (x;y;z) , row 4:6
> dipole n2 (x;y;z) , row 7:9 dipole n3 (x;y;z) , row 10:12 dipole
> n4 (x;y;z) and so on? ¿ so, I'd have a (128x3)x1280 matrix, (128x3)
> rows and 1280 columns, and not like above 128x(1280x3)?
Please see the documentation (reference for compute_leadfield
<http://fieldtrip.fcdonders.nl/reference/compute_leadfield>)
/The forward solution is expressed as the leadfield
matrix (Nchan*3), where each column corresponds with the potential or field
distributions on all sensors for one of the x,y,z-orientations of the
dipole./
This appears to be the leadfield for a dipole at a certain position.
You'd multiply the moment of the dipole you are trying to simulate with
the very matrix. At least mathematically I think you'd build the
lf-matrix for an number of dipoles as an (NchanX(Ndip*3)) matrix where
every three columns represent the contribution of one dipole to the
surface potential. The first of the three columns is the contribution of
the x-moment of the dipole and same for 2nd and 3rd column representing
the y- respectively z-moment. Note that this is - by now - only a
consideration of time independent dipoles. What the lf-matrix does is
mapping dipoles to electrodes. Nchan is the number of electrodes. For
your 128 dipole-example (which could make a problem when fitting) the
lf-matrix would be a 32x384-matrix with fixed positions for the dipoles.
Where - as I said before - every row stands for an electrode and every
triple of three cols for the contribution of one dipole to the
electrodes. If you'd want to simulate the dipoles at a single point in
time you'd build a vector
dip = [d1x d1y d1z d2x d2y d2z ... d128x d128y d128z]'
where dnx is the x-moment of the n-th dipole and same for y and z. This
would be a column vector or matrix with a single column. If you'd make a
(384x1280)-matrix with the rows being the dipoles and the columns the
temporal change of the dipoles and multiply the lf matrix with this
you'd get a matrix 32x1280 with the rows being the electrodes and the
columns the time course. So the source matrix is not a (128x(1280*3))-
but rather a ((128*3)x1280)-matrix.
> And, once I have performed the forward model and I compute the V
> matrix, *¿Which function could I use to perform the inverse problem,
> i.e. , could I use a function that allows me recovering or regain the
> J matrix from the A and V matrices, and compare the new J with the old
> J to see if the inverse problem could be performed well?
> *
Still I'd recommend the use of the dipolesimulation and dipolefitting
functions rather than using the lf directly. I think this is quite more
flexible. I don't really know how to perform the inverse solution in
this case.
> *Please, If you could suggest me some article or book or paragraph
> that I can read to clear these doubts, or if you could clear these to
> me yourselves, I would be very much thankful to you, please I'm
> begging you help please. Sorry for all the possible fouls and lacks of
> politeness, sincerely.Thanks for your attention.
> *
See for example Michel et al.: EEG source imaging
<http://www.lemanic-neuroscience.ch/PENSTrainingCenter/articles/EEG-imaging.pdf>.
Hope this helps.
Cheers,
Paul
--
Paul Czienskowski
Björnsonstr. 25
12163 Berlin
Tel.: (+49)(0)30/221609359
Handy: (+49)(0)1788378772
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