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Question:
Published on: 10 August, 2022

The dominant mode TE10 is propagated in a rectangular waveguide of dimensions a=6 cm, b=4 cm. The distance between two successive minima is 4.47 cm. Determine the signal frequency.

Answer:

Given \(\frac{\lambda_g}{4}=4.47, \lambda_g=4\times4.47=17.88\ cm\)

The cut-off frequency in dominant mode is \(f_c=\frac{c}{2a}=\frac{3\times{10}^8}{2\times6\times{10}^{-2}}=2.5\ GHz\)

\(\lambda_g=\frac{\lambda}{\sqrt{1-\left(\frac{f_c}{f}\right)^2}} \lambda_g=\frac{\frac{C}{f}}{\sqrt{1-\left(f_c/f\right)^2}} 17.88=\frac{\frac{3\times{10}^{10}}{f}}{\sqrt{1-\left(2.5\times{10}^9/f\right)^2}}\)

\(319.69=\left(\frac{3\times{10}^{10}}{f}\right)^2+319.69\times\left(2.5\times{10}^9/f\right)^2\)

\(f^2=\frac{1}{319.69}\left[\left(\left(3\times{10}^{10}\right)^2\right)+319.69\times\left(2.5\times{10}^9\right)^2\right]\)

\(f=\frac{1\times{10}^{10}}{319.69}\times28.98=0.90\ GHz\)

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