Effacer les filtres
Effacer les filtres

Fanbeam command projection directions and attributions clarification

3 vues (au cours des 30 derniers jours)
In the fan beam projection in matlab, what is the starting point of the projection acquisitions? Is the source considered placed on the right side on the positive x axis and the detectors on the other side or the opposite?
Also does the source with detectors rotate clockwise or anticlockwise for 360 degrees?
Regarding the detector positions in the sinogram which range from for example -40 to 40 degrees, when the source is at 0 degrees for the fan beam rotation assuming the source is placed horizontally at the beginning (along positive x axis), is -40 attributed to the uppermost detector or the lowest?

Réponse acceptée

Bjorn Gustavsson
Bjorn Gustavsson le 14 Avr 2023
The way I interpret the help and documentation the vertex with the source are located at negative y for rotation-angle zero. For the x-prime coordinate when converting the arc-geometry for the sensors to linear coordinates it is stated that:
This linear spacing is measured on the x-prime axis. The x-prime axis for each column, COL, of F is oriented at FAN_ROTATION_ANGLES(COL) degrees counterclockwise from the x-axis. The origin of both axes is the center pixel of the image.
As to which sensor is to the right on the sensor-array I think this part of one of the examples in the help:
% Convert angular positions to linear distance along x-prime axis
FposArc = D*tan(FposArcDeg*pi/180);
shows that the positive FposArcDeg is to the right in the detector array.
You could always try with a simpler test-image:
TestImg = zeros(128);
TestImg(32:48,32:48) = 1;
TestImg(78:89,64:85) = 5;
TestImg(20:25,89:100) = 3;
That should make it more straightforward to untangle what is where looking in this or that direction.
HTH
  2 commentaires
oumi k
oumi k le 14 Avr 2023
Thank you for your clarifications. Much appreciated!
Bjorn Gustavsson
Bjorn Gustavsson le 15 Avr 2023
You're welcome, I'm happy it helped.

Connectez-vous pour commenter.

Plus de réponses (0)

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by