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<p class=MsoNormal><font size=2 color=navy face=Arial><span style='font-size:
10.0pt;font-family:Arial;color:navy'>Thanks, this was helpful. We’re
still not quite clear on the time bins though – is it the case that the timesout
matrix gives the midpoints of the time bins, and if so, why is it that the time
bins are not evenly spaced? (They seem to be shifted by 8 or 12 ms.)<o:p></o:p></span></font></p>
<p class=MsoNormal><font size=2 color=navy face=Arial><span style='font-size:
10.0pt;font-family:Arial;color:navy'><o:p> </o:p></span></font></p>
<p class=MsoNormal><font size=2 color=navy face=Arial><span style='font-size:
10.0pt;font-family:Arial;color:navy'>Thanks again!<o:p></o:p></span></font></p>
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<p class=MsoNormal><b><font size=2 face=Tahoma><span style='font-size:10.0pt;
font-family:Tahoma;font-weight:bold'>From:</span></font></b><font size=2
face=Tahoma><span style='font-size:10.0pt;font-family:Tahoma'> Arnaud Delorme
[mailto:arno@ucsd.edu] <br>
<b><span style='font-weight:bold'>Sent:</span></b> Thursday, August 06, 2009
2:20 PM<br>
<b><span style='font-weight:bold'>To:</span></b> Sacks, Annette E<br>
<b><span style='font-weight:bold'>Cc:</span></b> '<st1:PersonName w:st="on">eeglablist@sccn.ucsd.edu</st1:PersonName>'<br>
<b><span style='font-weight:bold'>Subject:</span></b> Re: [Eeglablist]
newcrossf coherence matrix</span></font><o:p></o:p></p>
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<p class=MsoNormal><font size=3 face="Times New Roman"><span style='font-size:
12.0pt'><o:p> </o:p></span></font></p>
<p class=MsoNormal><font size=3 face="Times New Roman"><span style='font-size:
12.0pt'><o:p> </o:p></span></font></p>
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<p class=MsoNormal><font size=3 face="Times New Roman"><span style='font-size:
12.0pt'>On Aug 5, 2009, at 9:21 AM, Sacks, Annette E wrote:<o:p></o:p></span></font></p>
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<br>
<o:p></o:p></span></font></p>
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<p class=MsoNormal><font size=2 color=black face="Courier New"><span
style='font-size:10.0pt;font-family:"Courier New";color:black'>I've been doing
a channel cross-coherence analysis in EEGlab v6.01b using newcrossf() with a
sampling freq of 250Hz, epoch start/end [-400 1896], and frequency range [0
100], which outputs a 102 x 200 coherence matrix. This is the call I used:<u1:p></u1:p></span></font><font
color=black><span style='color:black'><o:p></o:p></span></font></p>
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<p class=MsoNormal><font size=2 color=black face="Courier New"><span
style='font-size:10.0pt;font-family:"Courier New";color:black'><u1:p> </u1:p></span></font><font
color=black><span style='color:black'><o:p></o:p></span></font></p>
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<p class=MsoNormal><font size=2 color=black face="Courier New"><span
style='font-size:10.0pt;font-family:"Courier New";color:black'>>> [coh,mcoh,timesout,freqsout,cohboot,cohangles]
= pop_newcrossf(EEG, 1, 1, 2, [-400 1896], 0, 'freqs', [0 100], 'alpha', .05);<u1:p></u1:p></span></font><font
color=black><span style='color:black'><o:p></o:p></span></font></p>
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<p class=MsoNormal><font size=2 color=black face="Courier New"><span
style='font-size:10.0pt;font-family:"Courier New";color:black'><u1:p> </u1:p></span></font><font
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<p class=MsoNormal><font size=2 color=black face="Courier New"><span
style='font-size:10.0pt;font-family:"Courier New";color:black'>Which gives this
output:<u1:p></u1:p></span></font><font color=black><span style='color:black'><o:p></o:p></span></font></p>
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<p class=MsoNormal><font size=2 color=black face="Courier New"><span
style='font-size:10.0pt;font-family:"Courier New";color:black'><u1:p> </u1:p></span></font><font
color=black><span style='color:black'><o:p></o:p></span></font></p>
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<p class=MsoNormal><font size=2 color=black face="Courier New"><span
style='font-size:10.0pt;font-family:"Courier New";color:black'>Adjust min freq.
to 0.98 Hz to match FFT output frequencies Adjust max<u1:p></u1:p></span></font><font
color=black><span style='color:black'><o:p></o:p></span></font></p>
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<p class=MsoNormal><font size=2 color=black face="Courier New"><span
style='font-size:10.0pt;font-family:"Courier New";color:black'>freq. to 99.61
Hz to match FFT output frequencies Generating 200 time<u1:p></u1:p></span></font><font
color=black><span style='color:black'><o:p></o:p></span></font></p>
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<p class=MsoNormal><font size=2 color=black face="Courier New"><span
style='font-size:10.0pt;font-family:"Courier New";color:black'>points (-143.6
to 1639.6 ms) The window size used is 128 samples (512<u1:p></u1:p></span></font><font
color=black><span style='color:black'><o:p></o:p></span></font></p>
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<p class=MsoNormal><font size=2 color=black face="Courier New"><span
style='font-size:10.0pt;font-family:"Courier New";color:black'>ms) wide.
Estimating 102 linear-spaced frequencies from 1.0 Hz to 99.6 Hz.<u1:p></u1:p></span></font><font
color=black><span style='color:black'><o:p></o:p></span></font></p>
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<p class=MsoNormal><font size=2 color=black face="Courier New"><span
style='font-size:10.0pt;font-family:"Courier New";color:black'><u1:p> </u1:p></span></font><font
color=black><span style='color:black'><o:p></o:p></span></font></p>
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<p class=MsoNormal><font size=2 color=black face="Courier New"><span
style='font-size:10.0pt;font-family:"Courier New";color:black'>My question is:
how do I calculate exactly how the time and frequency bins are being split up
in the 102x200 (freq x time) coh matrix ?<u1:p></u1:p></span></font><font
color=black><span style='color:black'><o:p></o:p></span></font></p>
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<p class=MsoNormal><font size=2 color=black face="Courier New"><span
style='font-size:10.0pt;font-family:"Courier New";color:black'>ie, we don't
know the exact beginning and ending time/frequency that is being collected in
each entry of the coh matrix.</span></font><font color=black><span
style='color:black'><o:p></o:p></span></font></p>
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<p class=MsoNormal><font size=3 face="Times New Roman"><span style='font-size:
12.0pt'><o:p> </o:p></span></font></p>
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<p class=MsoNormal><font size=3 face="Times New Roman"><span style='font-size:
12.0pt'><o:p> </o:p></span></font></p>
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<p class=MsoNormal><font size=3 face="Times New Roman"><span style='font-size:
12.0pt'>The <i><span style='font-style:italic'>freqsout</span></i> and <i><span
style='font-style:italic'>timesout</span></i> matrices contain the actual time
latencies and frequencies. When you use FFT, even if you specify actual
frequencies 'freqs', [1 2 3 4 5 6 7.... ] (which is possible), the FFT
decomposition is run based on your window size and the function then searches
for frequencies closest to the one you have asked for. When you use wavelets,
you get the decomposition at the frequency you have actually asked for. <o:p></o:p></span></font></p>
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<p class=MsoNormal><font size=3 face="Times New Roman"><span style='font-size:
12.0pt'><o:p> </o:p></span></font></p>
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<p class=MsoNormal><font size=3 face="Times New Roman"><span style='font-size:
12.0pt'>To obtain round frequencies for FFT, just choose a window size multiple
of the sampling rate. For instance 250, 125, etc...<o:p></o:p></span></font></p>
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<p class=MsoNormal><font size=3 face="Times New Roman"><span style='font-size:
12.0pt'>Hope this helps,<o:p></o:p></span></font></p>
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<p class=MsoNormal><font size=3 face="Times New Roman"><span style='font-size:
12.0pt'><o:p> </o:p></span></font></p>
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<p class=MsoNormal><st1:place w:st="on"><font size=3 face="Times New Roman"><span
style='font-size:12.0pt'>Arno</span></font></st1:place> <o:p></o:p></p>
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<br>
<o:p></o:p></span></font></p>
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<p class=MsoNormal><font size=2 color=black face="Courier New"><span
style='font-size:10.0pt;font-family:"Courier New";color:black'>Thanks,<u1:p></u1:p></span></font><font
color=black><span style='color:black'><o:p></o:p></span></font></p>
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<p class=MsoNormal><font size=2 color=black face="Courier New"><span
style='font-size:10.0pt;font-family:"Courier New";color:black'>Annette Sacks</span></font><font
color=black><span style='color:black'><u1:p></u1:p><o:p></o:p></span></font></p>
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<p class=MsoNormal><font size=1 color=black face=Helvetica><span
style='font-size:9.0pt;font-family:Helvetica;color:black'><ATT00001.txt><o:p></o:p></span></font></p>
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