Topic: Microwave Engineering
A square waveguide carries $TE_{11}$ mode whose axial magnetic field is given by $H_z = H_0 \cos(\frac{\pi x}{\sqrt{8}}) \cos(\frac{\pi y}{\sqrt{8}})\text{ A/m}$ where waveguide dimensions are in cm. What is the cut-off frequency of the mode?
- A. 5.5 GHz
- B. 6.5 GHz
- C. 7.5 GHz
- D. 8.5 GHz
Answer: C. 7.5 GHz
Explanation: Comparing the given $H_z$ expression with the standard form $H_z = H_0 \cos(\frac{m\pi x}{a}) \cos(\frac{n\pi y}{b})$, we identify $m=1$, $n=1$, and the dimensions $a = b = \sqrt{8}\text{ cm} = \sqrt{8} \times 10^{-2}\text{ m}$. The cut-off frequency is $f_c = \frac{c}{2} \sqrt{(\frac{m}{a})^2 + (\frac{n}{b})^2}$. Substituting the values, $f_c = \frac{3 \times 10^8}{2} \sqrt{\frac{1}{8 \times 10^{-4}} + \frac{1}{8 \times 10^{-4}}} = \frac{3 \times 10^8}{2} \sqrt{\frac{2}{8 \times 10^{-4}}} = 7.5\text{ GHz}$.