<p><span style="padding: 3px 10px; border-radius: 5px; color: #ffffff; font-weight: bold; display: inline-block; background-color: #ff0000;"> External Email - Use Caution </span></p><p></p><div><div dir="auto">Hi Alex,</div></div><div dir="auto"><br></div><div dir="auto">I see, it's because the timeseries are the rows, not the columns, of M. Thank you!</div><div dir="auto"><br></div><div dir="auto">Best,</div><div dir="auto">Gladia</div><div><br><div class="gmail_quote"><div>On Sun, Jul 15, 2018 at 05:15 Alexandre Gramfort <<a href="mailto:alexandre.gramfort@inria.fr">alexandre.gramfort@inria.fr</a>> wrote:<br></div><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><p><span style="padding:3px 10px;border-radius:5px;color:#ffffff;font-weight:bold;display:inline-block;background-color:#ff0000"> External Email - Use Caution </span></p><p></p><div><span style="font-size:small;text-decoration-style:initial;text-decoration-color:initial;float:none;display:inline">Gladia,</span><br><div><span style="font-size:small;text-decoration-style:initial;text-decoration-color:initial;float:none;display:inline"><br></span></div><div><span style="font-size:small;text-decoration-style:initial;text-decoration-color:initial;float:none;display:inline">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),</span></div><div><span style="font-size:small;text-decoration-style:initial;text-decoration-color:initial;float:none;display:inline">a linear inverse method estimates the sources S as S = K M where K is often called the imaging kernel.</span></div><div><span style="font-size:small;text-decoration-style:initial;text-decoration-color:initial;float:none;display:inline">So source imaging is a left multiplication.</span></div><div><span style="font-size:small;text-decoration-style:initial;text-decoration-color:initial;float:none;display:inline"><br></span></div><div><span style="font-size:small;text-decoration-style:initial;text-decoration-color:initial;float:none;display:inline">Note that the time series are the rows of M.</span></div><div><span style="font-size:small;text-decoration-style:initial;text-decoration-color:initial;float:none;display:inline"><br></span></div><div><span style="font-size:small;text-decoration-style:initial;text-decoration-color:initial;float:none;display:inline">If you do a temporal transform such as Fourier transform you do M F, where F is matrix that goes</span></div><div><span style="font-size:small;text-decoration-style:initial;text-decoration-color:initial;float:none;display:inline">from time to spectral domain.</span></div><div><span style="font-size:small;text-decoration-style:initial;text-decoration-color:initial;float:none;display:inline"><br></span></div><div><span style="font-size:small;text-decoration-style:initial;text-decoration-color:initial;float:none;display:inline">So to get sources in the spectral domain you can either do:</span></div><div>K (M F) : first do spectral transform than source estimation.</div><div>or </div><div><span style="font-size:small;background-color:rgb(255,255,255);text-decoration-style:initial;text-decoration-color:initial;float:none;display:inline">(K M) F<span style="text-decoration-style:initial;text-decoration-color:initial;float:none;display:inline"><span> </span>: first do source estimation than <span style="font-size:small;background-color:rgb(255,255,255);text-decoration-style:initial;text-decoration-color:initial;float:none;display:inline">spectral transform</span></span></span><br></div><div><span style="font-size:small;background-color:rgb(255,255,255);text-decoration-style:initial;text-decoration-color:initial;float:none;display:inline"><span style="text-decoration-style:initial;text-decoration-color:initial;float:none;display:inline"><span style="font-size:small;background-color:rgb(255,255,255);text-decoration-style:initial;text-decoration-color:initial;float:none;display:inline"><br></span></span></span></div><div><span style="font-size:small;background-color:rgb(255,255,255);text-decoration-style:initial;text-decoration-color:initial;float:none;display:inline"><span style="text-decoration-style:initial;text-decoration-color:initial;float:none;display:inline"><span style="font-size:small;background-color:rgb(255,255,255);text-decoration-style:initial;text-decoration-color:initial;float:none;display:inline">HTH</span></span></span></div><div><span style="font-size:small;background-color:rgb(255,255,255);text-decoration-style:initial;text-decoration-color:initial;float:none;display:inline"><span style="text-decoration-style:initial;text-decoration-color:initial;float:none;display:inline"><span style="font-size:small;background-color:rgb(255,255,255);text-decoration-style:initial;text-decoration-color:initial;float:none;display:inline">Alex</span></span></span></div><div><span style="font-size:small;background-color:rgb(255,255,255);text-decoration-style:initial;text-decoration-color:initial;float:none;display:inline"><span style="text-decoration-style:initial;text-decoration-color:initial;float:none;display:inline"><span style="font-size:small;background-color:rgb(255,255,255);text-decoration-style:initial;text-decoration-color:initial;float:none;display:inline"><br></span></span></span></div></div><br><div class="gmail_quote"><div>On Fri, Jul 13, 2018 at 6:30 PM Gladia Hotan <<a href="mailto:gladiach@gmail.com" target="_blank">gladiach@gmail.com</a>> wrote:<br></div><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><p><span style="padding:3px 10px;border-radius:5px;color:#ffffff;font-weight:bold;display:inline-block;background-color:#ff0000"> External Email - Use Caution </span></p></blockquote></div><div class="gmail_quote"><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><p></p><div><div class="gmail_default" style="font-size:small"><div style="background-color:rgb(255,255,255);text-decoration-style:initial;text-decoration-color:initial">Hi,</div><div style="background-color:rgb(255,255,255);text-decoration-style:initial;text-decoration-color:initial"><br></div><div style="background-color:rgb(255,255,255);text-decoration-style:initial;text-decoration-color:initial">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: <a href="https://mail.nmr.mgh.harvard.edu/pipermail//mne_analysis/2009-December/000336.html" style="color:rgb(17,85,204)" target="_blank">https://mail.nmr.mgh.harvard.edu/pipermail//mne_analysis/2009-December/000336.html</a>. This suggests that it doesn't matter whether you localize first or Fourier transform first.</div><div style="background-color:rgb(255,255,255);text-decoration-style:initial;text-decoration-color:initial"><br></div><div style="background-color:rgb(255,255,255);text-decoration-style:initial;text-decoration-color:initial">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: <a href="https://ccrma.stanford.edu/~jos/st/Matrix_Formulation_DFT.html" style="color:rgb(17,85,204)" target="_blank">https://ccrma.stanford.edu/~jos/st/Matrix_Formulation_DFT.html</a>. This source multiplies filter matrices on the left: <a href="https://www.dsprelated.com/freebooks/filters/Matrix_Filter_Representations.html" style="color:rgb(17,85,204)" target="_blank">https://www.dsprelated.com/freebooks/filters/Matrix_Filter_Representations.html</a>.</div><div style="background-color:rgb(255,255,255);text-decoration-style:initial;text-decoration-color:initial"><br></div><div style="background-color:rgb(255,255,255);text-decoration-style:initial;text-decoration-color:initial">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. </div><div style="background-color:rgb(255,255,255);text-decoration-style:initial;text-decoration-color:initial"><br></div><div style="background-color:rgb(255,255,255);text-decoration-style:initial;text-decoration-color:initial">Could somebody shed some light on this?</div><div style="background-color:rgb(255,255,255);text-decoration-style:initial;text-decoration-color:initial"><br></div><div style="background-color:rgb(255,255,255);text-decoration-style:initial;text-decoration-color:initial">Thanks and Best Regards,</div><div style="background-color:rgb(255,255,255);text-decoration-style:initial;text-decoration-color:initial">Gladia</div><br></div></div></blockquote></div><div class="gmail_quote"><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
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