Dear Claude
> Sorry to insist on that issue, but there seem to be a controversy
> around the method to calculate the cut-off period in fMRI studies. Lets
> take a paradigm in which A, B are active tasks (7 scans), N being a
> neutral task (7 scans) and r being the rest period (3 scans). One scans
> last 5.8 seconds. Subjects will be exposed to rArBrNrArBrN (notice that
> ABN order, called the experimental period, is repeated twice). There
> are 3 methods that have been suggested to me.
>
> 1) Take the longest on-off cycle (rest + task): 2 * (3+7) * 5.8 seconds
> = 116
>
> 2) Take the longest on-off cycle used in the substraction (task +
> neutral): 2 * ( 7+7) * 5.8 = 162.4
>
> 3) Take the repeated experimental period ( in this case it is r+
> A+B+N): 2 * (3+7+7+7) * 5.8 seconds = 278.
>
> In our case, we ARE NOT using the rest period. We are doing A-N and
> B-N. Is the method 2 more appropriate?
>
> Even if method 1 is still recommended, what's the origin of those
> alternative methods? What's the rational behind them?
I would use method 1. It does not matter whether the rest period
enters into the contrast or not, it still defines the fundamental
frequency of the experiment. Consider the epoch sequence R A1 R A2 R
A3.... Any difference among the A1...An represent an interaction
between R vs. A and time. As the interaction is a multiplication in
time, it is a convolution in frequency space. This means that the
interaction (i.e. differences among the As) has the same frequency as R
A R A... cycle smoothed by the time factor (in frequency space).
Although some of the power will be 'convolved' into lower frequencies
most will not. As a simple example consider the comparison of the
first and last activation conditions A1 - An. Although a small
component of this contrast will be collinear with a low frequencies
most of the variance will not. To see this graphically cut and paste
the following into a MatLab window.
With best wishes - Karl
% create a contrast
%--------------------------------------------------------------------------
X = kron([0 1 0 0 0 0 -1],ones(1,8))';
n = length(X);
% create a high pass filter matrix
%--------------------------------------------------------------------------
K = spm_dctmtx(n,3);
K = eye(n) - K*K';
% filter contrast
%--------------------------------------------------------------------------
Xc = K*X;
% display results
%--------------------------------------------------------------------------
subplot(2,1,1)
x = [1:n];
plot(x,X,x,Xc,'-.')
title('contrast (before and after filtering)')
xlabel time
subplot(2,1,2)
plot(x,abs(fft(X)).^2,x,abs(fft(Xc)).^2,'-.')
title('spectral density (before and after filtering)')
xlabel frequency
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