Question 6.5: RESONANT FREQUENCY AND Q OF A DIELECTRIC RESONATOR Find the ...
RESONANT FREQUENCY AND Q OF A DIELECTRIC RESONATOR Find the resonant frequency and approximate unloaded Q for the\ TE_{01\delta } mode of a dielectric resonator made from titania, with\ \epsilon _{r}=95 and\ \tan\delta = 0.001. The resonator dimensions are a = 0.413 cm and L = 0.8255 cm.
The transcendental equation of (6.70) must be solved for \ k_{0} , with\beta and \alpha given by
(6.65a)\ \beta =\sqrt{\epsilon _{r}k_{0}^{2}-k_{c}^{2}} =\sqrt{\epsilon _{r}k_{0}^{2}-\left(\frac{P01}{a} \right)^{2} }
and (6.66a) \ \alpha=\sqrt{k^{2}_{c}-k^{2}_{0}}=\sqrt{\left(\frac{P01}{a} \right)^{2} -k^{2}_{0}}. Thus,
\ \tan\frac{ \beta L}{2}=\frac{\alpha}{\beta},where
\ \alpha=\sqrt{\left(2.405/a\right)^{2}-k^{2}_{0} },
\ \beta =\sqrt{\epsilon _{r}k^{2}_{0}-\left(2.405/a\right)^{2}},
and
\ k_{0}=\frac{2\pi f}{c}.
Because α and β must both be real, the possible frequency range is from \ f_{1} to f_{2},where
\ f_{1}=\frac{ck_{0}}{2\pi }=\frac{c\left(2.405\right) }{2\pi \sqrt{\epsilon _{r}}a } =2.853GHz,
Using the interval-halving method (see the Point of Interest on root-finding
algorithms in Chapter 3) to find the root of the above equation gives a resonant
frequency of about 3.152 GHz. This compares with a measured value of about
3.4 GHz from reference [2], indicating a 10% error. The approximate unloaded
Q, due to dielectric loss, is