An aluminum waveguide with $a=4.2 \mathrm{~cm}, b=1.5 \mathrm{~cm}, \sigma_c=3.5 \times 10^7 \mathrm{mhos} / \mathrm{m}$, filled with telfon $\left(\mu_r=1, \epsilon_r=2.6, \sigma=10^{-15} \mathrm{mhos} / \mathrm{m}\right)$ operates at $4 \mathrm{G} \mathrm{Hz}$. Determine,

(a) $\alpha_c$ and $\alpha_d$ for $T E_{10}$ mode

(b) The waveguide loss in dB over a distance of $1.5 \mathrm{~m}$.

Solution:

Cut off frequency for $\mathrm{TE}_{10}$ mode is,

$ \begin{aligned} & f_{\mathrm{c} 10}=\frac{1}{2 \mathrm{a} \sqrt{\mu \epsilon}}=\frac{1}{\mathbf{2 x} \mathbf{0 . 0 4 2}\\ \sqrt{\boldsymbol{\mu o 2 . 6 \epsilon}}}=2.213 \mathrm{G} \mathrm{Hz} \\\\ & \boldsymbol{\delta}_{\mathbf{s}}=\frac{\mathbf{1}}{\sqrt{\pi f \mu} \sigma_c}=1.35 \times 10^{-6} \mathrm{~m} \\\\ & \mathbf{R}_{\mathbf{s}}=\frac{1}{\sigma \delta s}=\frac{1}{3.5 \times 107 \times 1.35 \times 10-6}=0.02110 …