[Eeglablist] Newtimef parameters - window size and number of cycles

[Katharina Limbach]

Dear all,

I have two questions regarding the description of a time frequency analysis
using  the newtimef function.

This the is command that I am using:

*[ersp itc powbase times frequencies boot1 boot2 tfdata] = pop_newtimef(
EEG, 1, chani, [-1460   596], [3  14] , 'baseline',[NaN], 'freqs', [5 60],
'plotitc' , 'off', 'plotersp', 'off', 'plotphase', 'off','freqscale',
'log', 'scale', 'abs', 'padratio', 1);*

Which leads to the following output in the Command window:

*Computing Event-Related Spectral Perturbation (ERSP) and*
*  Inter-Trial Phase Coherence (ITC) images based on 1031 trials*
*  of 514 frames sampled at 250 Hz.*
*Each trial contains samples from -1460 ms before to*
*  596 ms after the timelocking event.*
*  Image frequency direction: normal*
*Using 3 cycles at lowest frequency to 14 at highest.*
*Generating 200 time points (-1125.3 to 261.3 ms)*
*Finding closest points for time variable*
*Time values for time/freq decomposition is not perfectly uniformly
distributed*
*The window size used is 167 samples (668 ms) wide.*
*Estimating 28 log-spaced frequencies from 5.0 Hz to 60.0 Hz.*
*Processing time point (of 200): 10 20 30 40 50 60 70 80 90 100 110 120*
* 130 140 150 160 170 180 190 200*
*Computing the mean baseline spectrum*
*Note: Add output variables to command line call in history to*
*      retrieve results and use the tftopo function to replot them*


How exactly is the window size computed? In the archives I read a couple of
times, that it should be close to 1000/(lowest frequency) x (number of
cycles). In my example that should lead to a window size of 600 and not
668. Where do those 'extra' 68 ms come from?

How can I calculate how many cycles were used at specific frequencies? I
have 3 cycles at the lowest and 14 cycles at the highest frequency and I am
getting estimates for 28 frequencies. Does that mean that I get  ~ 0.4
cycles more per frequency step? (3.4 cycles at  5.48 Hz; 3.8 cycles at 6.01
Hz)?

Thanks a lot for your help,
Katharina
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