Source localization then Fourier transform vs the other way round

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Hi,

I'm thinking about frequency-domain source space results, specifically
whether doing source localization in the time domain then transforming to
the frequency domain is the same as transforming to the frequency domain
and then doing source localization. I found this archive post which states
that the FFT and other filtering procedures are right matrix
multiplications: https://mail.nmr.mgh.harvard.edu/pipermail/
/mne_analysis/2009-December/000336.html. This suggests that it doesn't
matter whether you localize first or Fourier transform first.

However, other sources say that Fourier transforms and filters are
typically implemented as matrices which are multiplied on the left of the
original signal. For example, this source multiplies the DFT matrix on the
left: https://ccrma.stanford.edu/~jos/st/Matrix_Formulation_DFT.html. This
source multiplies filter matrices on the left: https://www.dsprelated.
com/freebooks/filters/Matrix_Filter_Representations.html.

I'd have thought that if the inverse operator W is multiplied on the left
of the source activity x, then the filter/FFT matrix should also be
multiplied on the left; if the filter/FFT is applied on the right, I'd
expect that we'd have to modify W so that it is also multiplied on the
right.

Could somebody shed some light on this?

Thanks and Best Regards,
Gladia
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Gladia,

if you denote M \in R^{C x T} the M/EEG data (C is number of channels and T
the number of time points),
a linear inverse method estimates the sources S as S = K M where K is often
called the imaging kernel.
So source imaging is a left multiplication.

Note that the time series are the rows of M.

If you do a temporal transform such as Fourier transform you do M F, where
F is matrix that goes
from time to spectral domain.

So to get sources in the spectral domain you can either do:
K (M F) : first do spectral transform than source estimation.
or
(K M) F : first do source estimation than spectral transform

HTH
Alex

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Hi Alex,

I see, it's because the timeseries are the rows, not the columns, of M.
Thank you!

Best,
Gladia