[gate-users] Image Reconstruction of Gate output

Mutanga Theodore muathe02 at student.umu.se
Mon Jan 3 17:30:24 CET 2005


Hi Long,
it seems Sadek is back, have you written to him about the code ?

Cheers

Theodore

By the way I was simulating the ECAT exact hr+ and I would like to compare
my results with someone
Here is what I had for the resolution measurements , NEMA 2001 !

transverse @ 1cm = 8.3 mm
axial @1 cm 11.5 mm


transverse radial @ 10cm = 10.59mm
transverse tangential@ 10cm = 10.66mm

axial @10cm 10.31 mm


I wish to find out from the GATE users if these look reasonable, I was
planning to do measurements myself on the ecat but I dont have acceess to
one since initially I was simulating the DST. Also in the NEMA 2001 there
is no mention about matrix size for reconstruction , doesnt this affect
the resolution ? I did not correct for anythin in the data from GATE ,
would it be better to correct for randoms, scatter ect , if so HOW (using
GATE ?)?


Cheers

'Theo

> 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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>>gate-users at lphe1pet1.epfl.ch
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>
>
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