Topic: Microwave Engineering
An air filled rectangular waveguide with dimensions $a=75$ mm, $b=37.5$ mm has same guide wavelength at frequencies $f_1$ and $f_2$ when operated at TE10 and TE20 modes respectively. If the frequency $f_1$ is $\sqrt{13}$ GHz, what is frequency $f_2$ in GHz?
- A. 10
- B. 5
- C. $\sqrt{13}/2$
- D. $2\sqrt{13}$
Answer: B. 5
Explanation: The cutoff frequency for the $TE_{10}$ mode is $f_{c10} = c/(2a) = 3\times10^{10}/(2\times7.5) = 2$ GHz. The cutoff frequency for the $TE_{20}$ mode is $f_{c20} = c/a = 3\times10^{10}/7.5 = 4$ GHz. The guide wavelength is given by $\lambda_g = \frac{\lambda}{\sqrt{1 - (f_c/f)^2}} = \frac{c}{\sqrt{f^2 - f_c^2}}$. Equating the guide wavelengths for both modes yields $f_1^2 - f_{c10}^2 = f_2^2 - f_{c20}^2$. Substituting the values: $(\sqrt{13})^2 - 2^2 = f_2^2 - 4^2 \implies 13 - 4 = f_2^2 - 16 \implies 9 = f_2^2 - 16 \implies f_2^2 = 25$. Thus, $f_2 = 5$ GHz.