[Neuroimaging] [Nipy-devel] [dipy] PIESNO
Samuel St-Jean
stjeansam at gmail.com
Fri Jul 24 17:31:42 CEST 2015
Well, it actually depends on your own scanner. Actually, the Siemens
scanner in Sherbrooke is using a 12 channel head coils array, but in the
way is it operated, it is the same as a 4 channel head coils array,
hence we use N=4.
This is because Siemens reconstruction algorithm (GRAPPA) yields a
noncentral chi distribution of the noise due to the algorithm working in
fourier space, as opposed to Philips and GE which are using a
reconstruction in the image space (SENSE), which yields Rician
distributed noise (and hence N=1 in this case).
The parameter N is really related to the algorithm used by the
reconstruction (and it depends on the number of coils of the receptor
antenna), less than the grappa/sense factor itself. Problems is, you
kind of need to know your scanner to use it. Some people [1,2] are using
maximum likelihood estimation form the background to get N, but I have
not tried it myself. See [3] if you want to know more about the various
reconstruction algorithm and the noise distribution they yield.
Samuel
-------------------------
[1] Varadarajan, D., Haldar, J., 2015. A Majorize-Minimize Framework for
Rician and Non-Central Chi MR Images. IEEE Trans. Med. Imaging 0062,
1–1. doi:10.1109/TMI.2015.2427157
[2] Aja-Fernández, S., Vegas-Sánchez-Ferrero, G., Tristán-Vega, A.,
2014. Noise estimation in parallel MRI: GRAPPA and SENSE. Magn. Reson.
Imaging 32, 281–90. doi:10.1016/j.mri.2013.12.001
[3] Dietrich, O., Raya, J.G., Reeder, S.B., Ingrisch, M., Reiser, M.F.,
Schoenberg, S.O., 2008. Influence of multichannel combination, parallel
imaging and other reconstruction techniques on MRI noise
characteristics. Magn. Reson. Imaging 26, 754–62.
doi:10.1016/j.mri.2008.02.001
Le 2015-07-24 11:18, Arnaud Boré a écrit :
> Dear dipy experts,
>
> I try to figure out how to use PIESNO to estimate the noise. In the
> example
> http://nipy.org/dipy/examples_built/piesno.html#example-piesno we
> don't know about the grappa factor. Is it 3 so we have N = 12 (number
> of coils) / 3 (grappa factor) ?
>
> Last question if we don't use grappa then N = 12 ?
>
> Thank you
>
> --
> Arnaud BORE
> Research assistant
> Cellulaire : (001) 514-647-8649
>
>
>
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