[FieldTrip] Source Timecourse

[Robert Oostenveld]

Dear Tony

The lcmv beamformer (by default) estimates the time series of the source, given that the source has an unknown orientation. The result is a 3xNtime matrix as the source activation, which is in source.avg.mom (or in source.trial(i).mom). The three rows of the moment are the strength of the source in the x-, y- and z-direction.

Using ft_sourcedescriptives with the option cfg.projectmom=yes you can project the source to its strongest orientation, i.e. the direction that explains most of the source variance. That is equivalent to taking the largest eigenvector of the source timeseries (which is a phrasing that is often used in papers, also on combining fMRI bold timeseries over multiple voxels). 

You can also do it manually, like this

mom3d = randn(3,100);  % example source dipole moment
[u, s, v] = svd(mom, 'econ');
mom1d = v(1,:);

Alternative to taking the largest eigenvector (which has a sign-ambiguity), you can do something like

mom3d = randn(3,100);  % example source dipole moment
[u, s, v] = svd(mom, 'econ');
mom1d = abs(v(1,:));

to take the absolute value, or

mom3d = randn(3,100);  % example source dipole moment
mom1d = sqrt(sum(mom3d.^2,1));

to take the strength over all three directions for each timepoint.

best
Robert



On 25 Aug 2011, at 19:43, Rojas, Don wrote:

> Tony,
> 
> I don't use the LCMV beamformer, so I can't tell you what the structure of the output is, but what you need to perform a source space projection is the orientation and location information for the sources. In the DICS beamformer output, which I'm familiar with, this information would be in the source structure as follows (assuming your source output structure is called "source"):
> 
> source.pos 		% the position info as an n x 3 matrix, where n is the number of sources, in the same units as your volume conductor model
> source.avg.ori 	% the orientation as an 1 x n cell array, where n is the number of locations and each cell is a 3 x 1 array of x y z orientation and n refers to the 					% same source in each array
> 
> So, in that case, you might find the necessary information for the peak source, in my case, as follows:
> 
> [~,ind]=max(source.avg.pow(:)); 	% index of the source with the maximum power in the dics output, let's say ind = 5000
> ori=source.avg.ori(5000);      		% orientation info
> loc=source.pos(5000,:);         		% location information
> 
> Then, if you get this info, you can search the archives for my posts, one of which has a detailed explanation for getting the source waveform when you have a known source location, source orientation and a volume conductor. If you want to get all the waveforms, you can of course iterate the process over the entire voxel volume, but since the leadfields are highly correlated in adjacent source locations, you will see that many of the waveforms will look essentially the same for sources within a few centimeters.
> 
> Best,
> 
> Don
> -----------------------
> Don Rojas, Ph.D.
> Associate Professor of Psychiatry
> U. of Colorado Denver Anschutz Medical Campus
> Director, UCD Magnetoencephalography Lab
> 13001 E. 17th Pl F546
> Aurora, CO 80045 USA
> 303-724-4994
> 
> On Aug 23, 2011, at 12:50 PM, Passaro, Antony D wrote:
> 
>> Hi Fieldtrippers,
>>  
>> I was looking for a way to extract a time-course from each source estimated by the lcmv beamformer. I searched through the mailing list archives and came across a few emails which talked about the possibility but I was unable to find any examples explaining exactly how this process is done. I am interested in extracting the time-course over the average response rather than the individual trials, if possible. Any help would be greatly appreciated.
>>  
>> Thanks,
>> -Tony
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> 
> 
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