A rectangular hollow metal waveguide is required to be so designed to propagate a 9375 MHz signal in its TE 10-mode that the guide-wavelength equals the cut-off wavelength. Calculate the value of 'a' (breadth or the wider dimension of the waveguide). Take b = a/2. Also, calculate the cut-off freque

Question

A rectangular hollow metal waveguide is required to be so designed to propagate a 9375 MHz signal in its [latex]TE _{10}[/latex]-mode that the guide-wavelength equals the cut-off wavelength. Calculate the value of 'a' (breadth or the wider dimension of the waveguide). Take b = a/2. Also, calculate the cut-off frequency of the next higher order mode.

Accepted Answer

For rectangular waveguide.

[latex]TE _{10} ext { mode. } f=9375 MHz[/latex]

b = 9/2
For, follow waveguide,

[latex]\begin{aligned}& u _{0}=3 imes 10^{8} m / sec \&\lambda_{0}=\frac{V_{0}}{f}=3.2 cm\end{aligned}[/latex]

guide wavelength, [latex]\bar{\lambda}=\lambda_{C}=[/latex] cutoff wavelength

[latex]\lambda_{0}=\frac{\bar{\lambda} \lambda_{c}}{\sqrt{\bar{\lambda}^{2}+\lambda_{C}^{2}}}=\frac{\lambda_{C}^{2}}{\sqrt{2} \lambda_{C}}=\frac{\lambda_{C}}{\sqrt{2}}=3.2 \sqrt{2}[/latex]

[latex]\begin{aligned}&\lambda_{C}=2 a=3.2 \sqrt{2}, \Rightarrow a=2.26, \& ext { Since } b a=2 b \&f_{c}=\frac{V_{0}}{2 a} \sqrt{m^{2}+(4 n)^{2}} \& ext { For TE }_{10} ext { and TE }_{20}, f_{c}=\frac{V_{0}}{9}=13.257 GHz\end{aligned}[/latex]

Question 4.5.7: A rectangular hollow metal waveguide is required to be so de...

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