Dear Seung-Goo,
PPI.xn is expressed in microtime resoluton, see the relevant comment in
spm_peb_ppi.m:
> % Save variables (NOTE: xn is in microtime and does not account for
> % slice timing shifts). To convert to BOLD signal convolve with a hrf.
> % Use a microtime to scan time index to convert to scan time: e.g.,
> % k = 1:NT:N*NT; where NT = number of bins per TR = TR/dt or SPM.xBF.T
> % and N = number of scans in the session. Finally account for slice
> % timing effects by shifting the index accordingly.
load SPM.mat
load VOI_xxx.mat
RT = SPM.xY.RT;
dt = SPM.xBF.dt;
NT = round(RT/dt);
fMRI_T0 = SPM.xBF.T0;
N = length(xY(1).u);
k = 1:NT:N*NT;
PPI.xn((k-1) + fMRI_T0)
Best regards,
Guillaume.
On 06/12/16 15:04, SG KIM wrote:
> Hi SPMers,
>
> I wish to confirm sampling time-point of the deconvoluted timeseires
> using PPI-'simple deconvolution'.
>
> Based on the number of elements (i.e.,
>
> numel( [0 : PPI.dt : (numel(PPI.Y)-PPI.dt)] ) == numel(PPI.xn)
>
> and also the shape of it, PPI.xn seems to be "estimated neural
> activation" (or just simply deconvoluted PPI.Y with the canonical HRF
> kernel) sampled at every PPI.dt (sec).
>
> I just wonder how the samples in PPI.xn corresponds to the samples in
> PPI.Y. It looks like that, if we say t=0 for the first sample, the n-th
> sample is at t=(n-1)*PPI.dt sec from the first timepoint. Is it correct?
>
> And also because the last elements of the PPI.xn are zero, so it looks
> like simply deconvoluted after zero-padding at the end of the PPI.Y, right?
>
> I would be grateful very much if anyone can confirm these points!
>
> Best regards,
> ---
> Seung-Goo KIM
--
Guillaume Flandin, PhD
Wellcome Trust Centre for Neuroimaging
University College London
12 Queen Square
London WC1N 3BG
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