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

Sun Jul 15 10:01:36 EDT 2018
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Hi Alex,

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

Best,

On Sun, Jul 15, 2018 at 05:15 Alexandre Gramfort <
alexandre.gramfort at inria.fr> wrote:

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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:
>
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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,
>>
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