Conny and all -<br><br>For the 'timef()' function -- the default 'cycles' entry [3 0.5] tells the function to compute a DFT spectrogram in which the lowest frequency has length of 3 cycles (at that frequency), and instead of remaining at 3 cycles for all higher frequencies (as in a 'pure' wavelet approach) the number of cycles in the data window used increases half (
0.5) as fast as number of cycles used in the corresponding FFT (for which the data window length is the same for all frequencies, meaning many more cycles are used at higher frequencies).<br> The [3 0.5] default was chosen to produce reasonably uniform time frequency cells (blobs) when plotted as a linear- (not log-) frequency ERSP/ITC/ERCOH image. The pure wavelet approach (
e.g., [3 0]) tends to produce gamma-band 'vertical streaks,' by sacrificing frequency resolution in favor of increasing time resolution at higher frequencies.<br> To obtain low frequency estimates, either the window length has to be long enough to fit 3 cycles, or zero-padding has to be used to extend the length of the window (on both ends without data more actual data). Thus, to estimate power at frequencies down to 3 Hz using the 'cycles', [3
0.5], either the window length must be 1 s, or the 'padratio' must be high enough (> 1) to extend the actual tapered data window out to 1 s.<br> I hope this is clear enough! To use a standard FFT, set 'cycles',0. In this case, considerations of window length and zero padding still apply to determine the lowest frequency estimated and the spacing between frequencies returned.
<br> Fortunately (or otherwise), these arguments were changed somewhat in 'newtimef()'. There, it is also possible to specify a set of DFT frequencies to use (see >> help newtimef), and -- currently -- the 'padratio' argument has changed.
<br><br>Hopefully, we can revisit these functions to provide a maximally clear set of arguments. I also plan to implement a 'test' mode in which the size and shape of the time/frequency cells at different frequencies produced by a given set of input arguments is shown for some simulated data.
<br><br>Scott Makeig<br><br><div><span class="gmail_quote">On 11/15/07, <b class="gmail_sendername">conny kranczioch</b> <<a href="mailto:email@example.com">firstname.lastname@example.org</a>> wrote:</span><blockquote class="gmail_quote" style="border-left: 1px solid rgb(204, 204, 204); margin: 0pt 0pt 0pt 0.8ex; padding-left: 1ex;">
Dear all,<br><br>I was wondering about how to get the frequency resolution for the output frequencies when using timef with wavelets. What I've done so far is to calculate the frequency resolution based on cycles = (output frequency)/ (sigma(frequency)), so for my (lowest) output frequency of
3.75 Hz and a 3 cycle wavelet sigma would be 1.25. I use the function with [wavecycles factor] = [3 0.5], and for each addititonal output frequency (starting from 3.75 Hz) I add 0.5 cycles to my original 3 cycles to estimate the frequency resolution. Does this make sense or am I completely of the track?
<br><br>Best, Conny<br><br><br><br><br> __________________________________ Ihr erstes Baby? Holen Sie sich Tipps von anderen Eltern. <a href="http://www.yahoo.de/clever">www.yahoo.de/clever</a><br><br>_______________________________________________
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To unsubscribe, send an empty email to <a href="mailto:email@example.com">firstname.lastname@example.org</a><br></blockquote></div><br><br clear="all"><br>-- <br>Scott Makeig, Research Scientist and Director, Swartz Center for Computational Neuroscience, Institute for Neural Computation, University of California San Diego, La Jolla CA 92093-0961,