[gate-users] Image Reconstruction of Gate output

[Long ZHANG]

Dear Kris,

Thank you very much. 

It seems that I have mistake the sinogram width (angle position) as sinogram
length (view number, or tangential bin number, or displacement position). Am I
right now?

Regards,

Long

In your mail:
>From:"Kris Thielemans" <kris.thielemans at csc.mrc.ac.uk>
>Reply-To: GATE feedback and helpline for Users <gate-users at lphe1pet1.epfl.ch>
>To: "'Long ZHANG'" <zhanglong99 at tsinghua.org.cn>,
  "'GATE feedback and helplinefor Users'" <gate-users at lphe1pet1.epfl.ch>
>Subject: RE: [gate-users] Image Reconstruction of Gate output
>
>> 
>> BTW, I was pussled by the sinogram width of DST (420 crystal 
>> per ring). The number 249 is larger than the 420/2 = 210. How 
>> can DST get the extra LORs of 39? By interpolation? Could 
>> this help to improve the spatial resolution? Thank you!
>> 
>Notation: N= num_detectors per ring
>
>Why do you think that the max sinogram width is N /2? This is a hardware
>choice (recent CTI scanners tend to have sinogram_width=N/2). The
>'physical' maximum of LORs in 2D sinograms is (N^2-N)/2 (i.e. all lines
>between pairs of detectors, excluding the degenerate cases, and
>identifying det pairs ij and ji). The max number of views is N/2 (you'd
>expect N, but it's halved because of interleaving). So, the max sinogram
>width is something like (N-1)
>
>
>Kris
>
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