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An air-filled rectangular waveguide has inner dimensions of 3 cm × 2 cm. The wave impedance of the TE\(_{20}\) mode of propagation in the waveguide at a frequency of 30 GHz is (free-space impedance \(\eta_0 = 377\ \Omega\))

1. 308 Ω
2. 355 Ω
3. 400 Ω
4. 461 Ω

Correct answer: 400 Ω

Solution

For a rectangular waveguide, the cutoff frequency of TE\(_{mn}\) is \(f_c=\frac{c}{2}\sqrt{(m/a)^2+(n/b)^2}\). For TE\(_{20}\), \(f_c=c/a=10\) GHz, so at 30 GHz the wave impedance is \(Z_{TE}=\eta_0/\sqrt{1-(10/30)^2}\approx 400\ \Omega\).

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