# [Mne_analysis] Source localization then Fourier transform vs the other way round

Alexandre Gramfort alexandre.gramfort at inria.fr
Sun Jul 15 05:14:19 EDT 2018
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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

On Fri, Jul 13, 2018 at 6:30 PM Gladia Hotan <gladiach at gmail.com> wrote:

>         External Email - Use Caution
>
> 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,
>
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