Difficulty: Medium
Correct Answer: 1.706 D
Explanation:
Introduction:
Designing circular waveguides requires accurate cutoff relations for allowed TE/TM modes. TE11 is usually the dominant mode in circular guides (lowest cutoff), analogous to TE10 in rectangular guides. Expressing λc in terms of the physical diameter D helps in quick sizing and band allocation.
Given Data / Assumptions:
Concept / Approach:
For circular guides, the cutoff frequency is f_c = (c / 2π) * (x′mn / a). The corresponding cutoff wavelength is λ_c = 2πa / x′mn. Substituting TE11 data yields λc relative to diameter.
Step-by-Step Solution:
Verification / Alternative check:
Cutoff charts for circular WG list TE11 with λc/D ≈ 1.706; higher modes have smaller λc multiples and thus higher cutoff frequencies.
Why Other Options Are Wrong:
2.11 D corresponds to other roots/modes; 0.82 D and 0.41 D are too small for dominant TE11; 3.54 D is roughly 2× the correct value.
Common Pitfalls:
Confusing TE versus TM roots (x′mn vs xmn) or using radius a where the formula expects diameter D.
Final Answer:
1.706 D.
Discussion & Comments