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,

$\begin{aligned} & f_{\mathrm{c} 10}=\frac{1}{2 \mathrm{a} \sqrt{\mu \epsilon}}=\frac{1}{\mathbf{2 x} \mathbf{0 . 0 4 2}\\ \sqrt{\boldsymbol{\mu o 2 . 6 \epsilon}}}=2.213 \mathrm{G} \mathrm{Hz} \\\\ & \boldsymbol{\delta}_{\mathbf{s}}=\frac{\mathbf{1}}{\sqrt{\pi f \mu} \sigma_c}=1.35 \times 10^{-6} \mathrm{~m} \\\\ & \mathbf{R}_{\mathbf{s}}=\frac{1}{\sigma \delta s}=\frac{1}{3.5 \times 107 \times 1.35 \times 10-6}=0.02110 \mathrm{hms} \\\\ & 1-\left(\frac{f_c}{f}\right)^2=1-\left(\frac{2.213}{4}\right)^2=0.6939 \\\\ & \mathbf{\eta}=\mathbf{\eta}_0 \frac{\sqrt{\boldsymbol{\mu}} r}{\sqrt{\varepsilon} r}=233.8 \mathrm{ohms} \\\\ & \end{aligned}$

$\boldsymbol{\alpha}_c=\frac{2 R_s}{b \eta} \times\left[\left\{\mathbf{1}+\frac{b}{a}\right\}\left(\frac{f_c}{f}\right)^2+\left\{\mathbf{1}-\left(\frac{f_c}{f}\right)^2\right\}\right]=0.0133 \mathrm{~Np} / \mathrm{m} \\ $

$\alpha_d=\frac{\sigma d \eta_o}{2 \sqrt{1-\left(\frac{f_c}{f}\right)^2}}=\frac{10-15 \times 377}{2 \times \mathbf{1 . 6 6}}=1.4 \times 10^{-13} \mathrm{~Np} / \mathrm{m} \quad\\ $

$\alpha_c \gg \alpha_d, \alpha_d$ can be neglected The wave propagating in the $+z$ direction in the rectangular waveguide vary as $\mathrm{e}^{-} \boldsymbol{\alpha}_c \boldsymbol{z}=\mathrm{e}^{-(0.0133)(1.5)}=0.9802 \mathrm{~Np}$ or $=20 \log _{10}(0.9868)=-0.1737 \mathrm{~dB}$