Topic: Microwave Engineering
An air filled rectangular waveguide R1 is operating at the frequency 2 GHz and another air filled rectangular waveguide R2 is operating at 4 GHz. The guide wavelengths of these waveguides at their respective frequencies are equal. If the cut-off frequency of waveguide R1 is 1 GHz, what is the cut-off frequency of the waveguide R2 in GHz?
- A. $\sqrt{10}$
- B. $\sqrt{11}$
- C. $\sqrt{12}$
- D. $\sqrt{13}$
Answer: D. $\sqrt{13}$
Explanation: The guide wavelength is given by $\lambda_g = \frac{\lambda}{\sqrt{1 - (f_c/f)^2}}$. For R1: $f_1 = 2\text{ GHz}$, $f_{c1} = 1\text{ GHz}$, so $\lambda_1 = C/2$. The guide wavelength is $\lambda_{g1} = \frac{C/2}{\sqrt{1 - (1/2)^2}} = \frac{C}{2\sqrt{3/4}} = \frac{C}{\sqrt{3}}$. For R2: $f_2 = 4\text{ GHz}$, so $\lambda_2 = C/4$. Equating $\lambda_{g1} = \lambda_{g2}$, we get $\frac{C}{\sqrt{3}} = \frac{C/4}{\sqrt{1 - (f_{c2}/4)^2}}$. Solving this yields $1 - (f_{c2}/4)^2 = 3/16$, leading to $(f_{c2})^2 = 13$, so $f_{c2} = \sqrt{13}\text{ GHz}$.