unloaded q

Want to know the unloaded Q by HFss simulation....

I can get the eigenmodes of the resonnator, but don't know how to get the unloaded Q. Cannot find it in the list.

The value of f0/BW(3db) is loaded Q, rt?
The calculation of g0g1/Δf from the prototype is also loaded Q, rt?
Any equation for unloaded Q?

Thanks.

I also want to know it.

Nobody knows this?

I have checked HFSS fullbook, but cannot find any clue, could anyone help?

Really want to make this clear.....

Appreciate any help.

I see from the hfss help file: Qu=abs(f)/2/im(f)

Added after 8 minutes:

I have a problem:
when I simulate my cavity in eigenmode, the inner conductor and the shield are silver-plated, and the cavity is filled with air. Do I need to handly set each surface of the shield as silver boundary? If so,is there any convenient way to do it?

Thanks

I am not sure whether I completely understand your meaning. IF you already defined the material (silver) for the inner and outer conductor of your cavity, why you need to set the surface? The surface is perfect E by default in HFSS, right?

I want to know the unloaded Q, so I need set the actual boundary, not pec.