Rectangular waveguide cutoff condition — TE20 mode Let a be the broad-wall width and b be the narrow-wall height of a rectangular waveguide. What is the cutoff wavelength λ_c for the TE20 mode?
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Aa
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Ba + b
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Ca - b
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Da2 + b2
Answer
Correct Answer: a
Explanation
Introduction / Context:Each TEmn mode in a rectangular waveguide has a specific cutoff wavelength and frequency determined by the cross-sectional dimensions a (broad wall) and b (narrow wall). Recognizing these standard results is essential for band selection and single-mode operation.
Given Data / Assumptions:
- Metal rectangular waveguide with dimensions a and b.
- Mode: TE20 (m = 2, n = 0).
- Lossless ideal walls for cutoff calculation.
Concept / Approach:The general formula is λ_c = 2 / sqrt((m/a)^2 + (n/b)^2). For TE m0, this simplifies to λ_c = 2a/m. Substituting m = 2 gives λ_c = a.
Step-by-Step Solution:
Start: λ_c = 2 / sqrt((m/a)^2 + (n/b)^2).For TE20 → m=2, n=0 → λ_c = 2 / (2/a) = a.Therefore, the cutoff wavelength for TE20 equals the broad dimension a.Verification / Alternative check:Special cases: λ_c(TE10) = 2a and λ_c(TE01) = 2b; by the same pattern, TE20 → a. Handbooks and mode charts confirm this relationship.
Why Other Options Are Wrong:
- a + b or a − b: no linear combination arises from the correct formula.
- a2 + b2: does not match the inverse-root dependence on dimensions.
Common Pitfalls:Confusing TE10 dominance with TE20 cutoff; forgetting dependence on the m index across the broad wall.
Final Answer:a