What are cavity resonators? A rectangular waveguide cavity has the dimension $22.86$ $\mathrm{mm} \times 10.16 \mathrm{~mm} \times 15 \mathrm{~mm}$. Calculate the resonant frequency for the $\mathrm{TE}_{101}$ mode. Consider air as dialectic.

Solution:

A resonator is a device that exhibits resonance at a particular frequency called resonance frequency. Cavity resonators are constructed by sharing both ends of a guide.

When the canty is excited through probes, with a signal frequency equal to the resonant frequency decided by waveguide dimensions, a resonance is created.

The resonant frequency is expressed as for TE

$$ \begin{aligned} & a=22.86 \mathrm{~mm} \\ & b=10.16 \mathrm{~mm} \\ & d=15 \mathrm{~mm} \end{aligned} $$

$\begin{aligned} f_{\text {or }} & =f_{101}=\frac{v_p}{2} \sqrt{\left(\frac{m}{a}\right)^2+\left(\frac{n}\\{b}\right)^2+\left(\frac{p}{d}\right)^2} \\ f_e & \left.=\frac{1}{2 \pi \sqrt{\mu_0 \varepsilon_0}}-\sqrt{\left(\frac{m \pi}\\{a}\right)^2+\left(\frac{n \pi}{b}\right)^2+\left(\frac{p \pi}{d}\right)^2}\right] \\ & =11.96\mathrm{CH}_2\end{aligned}$