Question:

Published on: 12 October, 2024

**When a dominant mode is propagating in an air filled rectangular waveguide, the guide wavelength for a frequency of 9 GHz is 4 cm. Calculate broad wall dimension of the guide.**

Answer:

Given that λ_{g}=4 cm and f=9 GHz

We know that \(\lambda_g=\frac{\lambda}{\sqrt{1-\left(f_c/f\right)^2}}\)

\(\lambda_g=\frac{\frac{C}{f}}{\sqrt{1-\left(f_c/f\right)^2}}\)

\(4=\frac{\frac{3\times{10}^{10}}{9\times{10}^9}}{\sqrt{1-\left(f_c/9\times{10}^9\right)^2}}\)

\(1-\left(f_c/9\times{10}^9\right)^2=\frac{\left(10/3\right)^2}{16}\)

\(\left(f_c/9\times{10}^9\right)^2=1-0.69=0.31\)

\(f_c=0.556\times9\times{10}^9=5.01\ GHz\)

In dominant mode cut – off frequency \(f_C=C/2a\), where C is the velocity of light in free space.

\(a=\frac{c}{2f_c}=\frac{3\times{10}^{10}}{2\times5.01\times{10}^9}=3\ cm\)

Therefore length of the waveguide is a=3 cm

The width of the waveguide is \(b=\frac{a}{2}=\frac{3}{2}=1.5\ cm\)

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