Band Artefacts in Detector Block Scanners

Hello Everyone,

I have reconstructed an image of a cylindrical source using a block scanner geometry (see the images below). There is a negligible gap (approximately 1mm) between detector blocks. In my simulation data, two-photon events are saved based on their crystal hit positions and IDs (i.e., as soon as two photons hit a pair of crystals, their IDs/positions are saved) without considering depth of interaction (the mean DOI is set to -1 in the header file).

However, I am observing band/strap artefacts in the reconstructed image. These artefacts appear symmetrical and run from the middle of one detector block to the middle of the opposite detector block (see the attached image).
I have not done any smoothing or convolution.

Are these types of artefacts typical in this context? Additionally, given that my crystals have 100% efficiency and the sensitivity has been computed, is normalization still necessary?

Thank you for your insights.

Best regards,
Seyyed

The reconstructed image:

My Scanner:

Hi,

Yes it is totally expected.

Yes you still need normalization as the normalization aims at normalizing efficiencies of LORs and not crystals. The less the number of sides in your polygonal geometry, the more the amplitude of differences between LORs efficiencies mostly due to solid angles and crystal penetration in your case.

Best
Simon

1 Like

Thank you, Simon!

May I ask why setting the “mean depth of interaction” to 0 (as I have 0mm depth of interation) in the scanner header file results in a deteriorated reconstructed image?

Kind regards,
Seyyed

It is hard to say.

How did you know that you have 0mm depth interaction? Are you stopping the photons in the simulation right at the surface of your detectors?

Simon

Yes, exactly! Then I write the position of the center of the surface (the inner surface) of each crystal into the look-up-table. Any small but non-zero mean DOI causes the image to deteriorate (even values as small as 0.0001 or smaller).
Therefore, I set the mean depth of interaction to -1, which works perfectly.

Seyyed

If you are using a LUT to build your scanner, you are supposed to provide the position of the centroid of crystals as well as a direction vector defining the “depth” direction. In your case, I understand that you provide the entrance of the crystal.

A value of -1 for mean depth of interaction means that the position of interaction will be set at the center of the crystal, which means at the position you provided in the LUT. If you change this value, the depth will be affected based on the direction vector provided in the LUT and the length of the crystals provided in the header of the LUT file.

So, if you change the value of mean depth of interaction, the position will be: centroid_position_from_lut - crystal_depth/2 + direction_vector_from_lut * mean_depth_of_interaction.

For a zero or almost zero value, the position will thus be:
centroid_position_from_lut - crystal_depth/2

This could cause the worsen image quality.

All of that to say that if you provide the entrance position in the LUT and that is what you want to be used in the reconstruction, then you must set the mean depth of interaction to -1.

Cheers
Simon

| s_mousavi
October 4 |

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Yes, exactly! Then I write the position of the center of the surface (the inner surface) of each crystal into the look-up-table. Any small but non-zero mean DOI causes the image to deteriorate (even values as small as 0.0001 or smaller).
Therefore, I set the mean depth of interaction to -1, which works perfectly.

Seyyed

1 Like

Thank you Simon for your explanation.

Best wishes,
Seyyed

I was looking into this and I have a question. In my simulation, I stop each photon right at the inner side/edge of each crystal and record the crystal number, module number, etc. Hence my crystal are 100% efficient. In addition, most of the empty space in my scanner is vacuum. Since two crystals define a line of response, then the lines of response efficiencies should be almost 100%. I was assuming that the sensitivity image without the normalization and the sensitivity image with the normalization are the same. Can you please let me know who the sensitivity is computed and what will be left the the normalization will take care in my simulation?

Best,
Seyyed