Question 6.4: DESIGN OF A CIRCULAR CAVITY RESONATOR A circular cavity reso...

From (6.53a) \ f_{nm\ell }=\frac{c}{2\pi \sqrt{\mu \epsilon _{r}} } \sqrt{\left(\frac{p^{\prime }_{nm}}{a} \right)^{2} +\left(\frac{\ell \pi }{d} \right)^{2} } the resonant frequency of the \ TE_{011} mode is

with\ p\prime _{01}=3.832Then, since d = 2a
\ \frac{2\pi f_{011}\sqrt{\epsilon _{r}} }{c}=k=\sqrt{\left(\frac{p\prime _{01}}{a} \right)^{2}+\left(\frac{\pi }{d} \right)^{2} }.
Solving for a gives
\ a=\frac{\sqrt{\left(p\prime _{01}\right)^{2}+\left(\pi /2\right)^{2} } }{k} =\frac{\sqrt{\left(3.832\right)^{2} +\left(\pi /2\right)^{2} } }{151.0} =2.74cm,
so we have d = 5.48 cm.
The surface resistivity of copper at 5 GHz is\ R_{s}=0.018\Omega .

Then from (6.57)\ Q_{c}=\frac{\omega _{0}W}{P_{c}} =\frac{\left(ka\right)^{3}\eta ad }{4\left(p^{\prime }_{nm}\right)^{2}R_{s}} \frac{1-\left(\frac{n}{p^{\prime }_{nm}} \right)^{2} }{\left\{\frac{ad}{2}\left[1+\left(\frac{\beta an}{\left(p^{\prime }_{nm}\right) } \right)^{2} \right] +\left(\frac{\beta a^{2}}{p^{\prime }_{nm}} \right)^{2} \left(1-\frac{n^{2}}{\left(p^{\prime }_{nm}\right)^{2} } \right) \right\} } ,

with n = 0, \ m = \ell = 1, and d = 2a, the unloaded Q due to conductor losses is

where (6.51a)\ \beta _{nm}=\sqrt{k^{2}-\left(\frac{p^{\prime }_{nm}}{a} \right)^{2} }

was used to simplify the expression. From (6.59)\ Q_{d} =\frac{\omega W}{P_{d}} =\frac{\epsilon }{\epsilon ^{\prime \prime }} =\frac{1}{\tan \delta } the unloaded Q due to dielectric loss is

and the total unloaded Q of the cavity is

This result can be compared with the rectangular cavity case of Example 6.3,
which had \ Q_{0}=1927  for the \ TE_{101} mode and \ Q_{0}=2065 for the \ TE_{102} mode.
If this cavity were air filled, the Q would increase to 42,400.