A rectangular waveguide is filled with air with dimension $2.28 \mathrm{~cm} \times 1.01 \mathrm{~cm}$ is operated...

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written 2.3 years ago by

prajapatijaimin • 3.3k

Solution:

Air-filled rectangular waveguide,

$\begin{aligned} \therefore \mu_r & =1, \varepsilon_r=1 \text {. } \quad \mu=\mu_0, \quad \varepsilon=\varepsilon_0 \\ a \times b & =2.28 \mathrm{~cm} \times 1.01 \mathrm{~cm} \\ f & =9.2 \mathrm{c}_{\mathrm{Hz}} \end{aligned}$

mode of wave $=$ dominant le. $T E_{10}$ $\begin{aligned} & T E_{m n}=T E_{10} \ & m=1, n=0 \end{aligned}$ a) Calculation of cut-off frequency (fe),

$f_c=\frac{1}{2 \pi \sqrt{\mu \varepsilon}}\left[\sqrt{\left(\frac{m \pi}{a}\right)^2+\left(\frac{n \pi}{b}\right)^2}\right]$

$=\frac{1}{2 \pi \sqrt{4 \pi \times 10^{-7} \times 8.854 \times 10^{-12}}}\left[\sqrt{\left(\frac{\pi}{2.28 \times 10^{-2}}\right)^2+(0)^2}\right]$

$=6.58 \mathrm{GHz}^2$

b) Calculation of guide wavelength $\left(\lambda_9\right)$

$\begin{aligned} \lambda_c & =c / f_c=0.0455 \mathrm{~m} \\\\ \lambda & =c / f=0.0326 \mathrm{~m} \\\\ \lambda_g & =\frac{\lambda}{\sqrt{1-\left(\lambda / \lambda_c\right)^2}}=0.046 \mathrm{~m}\\ \end{aligned}\\$

c) $\quad \eta=\frac{\eta}{\sqrt{1-(/ \pi)^2}}=540,42 \Omega$

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