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Dear All,

I am designing a fast kicker-septum system for the beam switchyard of a CW X-FEL. Electron beams are distributed by a series of fast-kickers in vertical direction and then pass though a DC Lamberson septum. Between the kicker and septum there are several quadrupoles for beam matching. However, the kick angle is so small that the two direction of beam must share the same quadrupoles. That means there must be at least one direction of beam cannot pass though the quads on axis. The angle of the kicked beam is not only from the kickers but also from the dipole field of the quads. Before the DC Lamberson Septum, there is another DC septum to correct the kicked angle. A schematic view of such a system is seen below.
My question is, how to calculate the real value of the vertical offset of the kicked beam relative to the un-kicked beam. I use the &floor-coordinate option to get the survey coordinate of Y. From the ".cen" file I can get the Cy values. Then how do I set the Dy value of the quads? Should it be Y+Cy or just Y is OK? The two cases with Y, Y+Cy can be seen below.
Best,
Chen

Chen,

I assume your beamline is defined for the straight-through case, except for some steering (VKICK, HKICK, etc) or kicking (BUMPER) elements. Once the beam has passed through the quadrupole between the two septum magnets, you can transition to a new coordinate system centered on the extracted beam path. The offset between the two beamlines is just the Cy value of the beam at the exit of the last quadrupole in your diagram.

If you want to continue tracking the extracted beam, you can use a CENTER element to recenter the beam on x=xp=y=yp=0, then proceed with a new beamline that is centered on that trajectory. The Lambertson would be modeled as a bending magnet, not a steering element.

If you want to get survey data for the two beamlines relative to each other, it becomes more complicated. For the straight-ahead case, just turn off all the steering and kicking elements, septa, etc. For the extracted case, you'll need to replace all the steering and kicking elements with bending magnets, and give the middle quadrupole an offset of -Cy, where Cy is the position you computed from the previous simulation.

I'm not sure I've answered your question, but hope this helps.

--Michael

Hi, Michael

Thanks a lot for your prompt reply.

I almost understand your method but still one question. For getting the survey coordinate \Delta_y and setting Dy, what is the role of the column Y from the floor_coordinate option?

I am afraid my previous description is no so clear. In my case, the straight line passes though the quads on axis (Dy=0), the extract line passes the same quads off axis (Dy=-\Delta_y). I need to calculate the correct value of \Delta_y but I don't understand why the floor coordinate Y is absent?

Best,
Chen