Induced activity

Stanley Klein sklein at BERKELEY.EDU
Wed Apr 7 10:55:41 CEST 2010


This is a most interesting and important thread.  I would think that one
would want to separate the time-locked from the non-time-locked components
independent of whether they were generated by a true additive response,
or by phase resetting or by asymmetric modulation of noise. The reason is
that the ERP/ERF is so simple to show in a standard one-dimensional plot
that one would want to separate it out. Then one would want to display the
rest of the response in some sorts of power and coherence plots. The obvious
thing to do is to subtract off the ERP from each event on a trial by trail
basis as Thomas suggested 7 days ago, and then calculate power and
coherence. I don't see what's wrong with that as a first approximation.

The thing I'd do as a 2nd approximation is to take into account the changing
gain from trial to trial whereby the amplitude (but not phase, for
simplicity) of the evoked response can change from trial to trial.  Suppose:

V(t, k) is the raw data on the kth trial
Ve(t), the evoked response, is the mean of V(t,k) , averaged over all k
trials.
Pe = Ve(t) Ve(t)  is the power of the evoked response. We are using the
     Einstein summation convention of summing over repeated indices (t in
this case) .
f (k)=V(t,k) Ve(t) / Pe is the amplitude of the evoked response on the kth
trial.
The induced response can now be obtained:
Vi(t, k) = V(t, k) - f(k) Ve(t)

By the definition of f(k) the dot product of Vi and Ve is zero for each k.
If one doesn't do this doesn't one get all sort of things that look like
coherence but that are simply the dumb, standard ERP/ERF. I hope I'm not
saying something stupid by forgetting something simple.
Stan
On Wed, Apr 7, 2010 at 12:49 AM, Michael Wibral <michael.wibral at web.de>wrote:

> Hi Thomas,
>
>
> "...
> Maybe I wasn't clear. The trick is to maintain the
> complex components (real and imag) after the wavelet transform, then to
> separate induced and evoked and then to calculate power in the end.
> ..."
>
> From what I understand you suggest to:
>
> (a) take the FFT of a trial :
>
> FFT(trial i)
>
> (b) then to take the average of those FFTs and stay in the complex domain:
>
> 1/n [sum(FFT(trial i))]
>
> (c) to subtract this complex quantity from each trial:
>
> FFT(trial i) - 1/n [sum(FFT(trial i))]
>
> (d) and to take the power and then the average , finally:
>
> 1/n sum {(FFT(trial i) - 1/n [sum(FFT(trial i))])^2}
>
>
> If you transform this, taking the linearity of the FFT into account where
> appropriate you get:
>
> 1/n sum {(FFT(trial i) - 1/n [sum(FFT(trial i))])^2} =
>
> 1/n sum {(FFT(trial i - 1/n [sum(trial i)])^2}=
>
> 1/n sum {(FFT(trial i - ERF))^2 }
>
> In the end you seem to subtract the ERF from each trial, then take the FFT
> compute power and then compute the average. I am  a bit confused here: To me
> this seems to be the same approach as simply subtracting the ERF in the time
> domain before computing power, i.e. a simple version of the old regression
> approach. In my opinion this must be the case. This is because keeping the
> numbers complex, means keeping phase information and computing the average
> over trials in the Fourier domain should then be the same as computing the
> (trivially phase-sensitive) average in the time domain, then taking the
> Fourier transform.
>
> On the other hand, if you really take power as the very last operation:
>
> {1/n sum (FFT(trial i) - 1/n [sum(FFT(trial i))])}^2 =
>
> {1/n sum (FFT(trial i - ERF)) }^2 =
>
> {1/n sum (FFT(trial i)) - FFT(ERF)) }^2 =
>
> {1/n sum (FFT(trial i)) - 1/n n FFT(ERF) }^2 =
>
> {FFT(1/n sum (trial i)) - FFT(ERF)}^2 =
>
> {FFT(ERF) - FFT(ERF)}^2 = 0
>
>
> Could you let me know where I misunderstand that approach?
>
> With regards to something like the ERF being present in every single trial,
> I was thinking of other mechanisms like phase-reset or asymetric modulations
> of oscillation amplitude that may or may not be detected by looking at power
> increases.
>
> Michael
>
>
>
> -----Urspr√ľngliche Nachricht-----
> Von: Thomas Witzel <twitzel at NMR.MGH.HARVARD.EDU>
> Gesendet: Apr 6, 2010 5:56:53 PM
>  An: FIELDTRIP at NIC.SURFNET.NL
> Betreff: Re: [FIELDTRIP] Induced activity
>
> >
> >Maybe I wasn't clear. The trick is to maintain the
> >complex components (real and imag) after the wavelet transform, then to
> >separate induced and evoked and then to calculate power in the end.
> >This can be done with entire TFRs that way. I'm not sure whether this is
> >possible in the regular fieldtrip workflow which might cause confusion
> with
> >terminology here.
> >As for the ERF not reflecting activity that might not be present in this
> >form in the trials, I guess we have a bit of a philosophical question
> >here. The entire premise of an ERF is that the brain response is identical
> >in every trial + some noise. Since EEG/MEG is extremely noisy you can't
> >tell from single trials whats really going on, so averaging all trials
> >could be the best estimation of what the signal in every trial looks like.
> >Now, of course we know that this is not entirely true, because in many
> >experiments we know of systematic trial to trial variation, in which
> >case the whole ERF or for that matter most common analysis methods are
> >inappropriate.
> >Also, even if there is random trial to trial variation, some of it might
> >not be noise, as already described by Schimmel back in 1967 in a nice
> >Science article. This is where the induced signal comes in. For me its
> >signal that can be detected by its respective increase or decrease in
> >power, but its not coherent across trials so it cancels mostly in ERFs.
> >Now subtracting the ERF from every trial brings the assumption back in
> >that the evoked signal is the same in every trial which it might be or
> >might not be. In most of the experiments I have analyzed subaverages
> >(separate even and odd trials, or early and late ones) were very similar,
> >so the assumption that the evoked response is the same in every trial was
> >fair.
> >Practically I found that subtracting the ERF or not, has very little
> >impact on the final outcome, but I didn't test every case, so I'm
> >subtracting where its deemed appropriate....
> >
> >Thomas
> >
>

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