[FieldTrip] questions itc/plf calculation via hilbert transformations

Eelke Spaak eelke.spaak at donders.ru.nl
Fri Dec 6 07:18:47 CET 2013

Hi Russ,

The imaginary part of a complex number (i.e. the 'b' in 'a+b*i') is very
different from the phase (i.e. the 'p' in 'A*exp(i*p)'). When computing
ITC, the expression z = z / abs(z) sets the amplitude (A) to 1, while
preserving the phase. This is not the same as taking the imaginary par.

To be able to later compute the ITC, I would actually just use cfg.hilbert
= 'complex', and then do the exact same computations as with your wavelet
approach. Using cfg.hilbert = 'angle' gives you the angle (in radians)
directly, so a real number. This is more difficult (and slow) to work with
than using the normalized complex representation.

Hope this helps,
On Dec 5, 2013 11:39 PM, "Russell G Port" <russgport at gmail.com> wrote:

> Hi fellow fieldtippers,
> I have a question that brought around by an answer I previously received.
>   For data analyzed with ft_freqanalysis and using the 'wavelet' option I
> can computer ITC/PLF via,
> tmpdat = freq.fourierspctrm;
> tmpdat = tmpdat./abs(tmpdat); ; % this will normalize each trial for its
> amplitude;
> itc = abs(mean(tmpdat)); % this will give the itc
> thank you again Jan-Mathijs for this insight
> I am now trying to calculate ITC via hilbert transformation, putting my VE
> data through
> ft_preproc_hilbert. Can I just use the cfg.option = 'imag', to get the
> imaginary component of
> the analytic signal, since this is what i use with wavelets (i.e. what the
> tmpda =tmpdat./abs(tmpdat) achieves). Then I would just repeat the last
> line itc =
> abs(mean(tmpdat)). I only hesitate because on several of the mailing list
> emails people have
> suggested doing cfg.option = 'angle' (which computes the imaginary
> component (tmpdat./abs(tmpdat)  then uses the angle function (angle()).
> Have I been calculating ITC wrong for my wavelets? Or should i just the the
> cfg.option = 'imag' and mimik my wavelet based ITC.
> Thanks
> Russ
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