Rectangular Waveguide Cutoff Relation (Dominant TE10) A rectangular waveguide has broad-wall width a. At cutoff for the dominant TE10 mode, which relation between a and the cutoff wavelength λc is correct?
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Aa = λc / 2
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Ba = λc
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Ca = 2 λc
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Da = λc / 4
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Ea is independent of λc
Answer
Correct Answer: a = λc / 2
Explanation
Introduction / Context:Cutoff conditions in rectangular waveguides determine the lowest frequency that can propagate for a given mode. For the dominant TE10 mode, the broad dimension a controls the cutoff wavelength and hence the usable frequency band for single-mode operation.
Given Data / Assumptions:
- Uniform, air-filled rectangular waveguide with width a (broad wall) and height b (narrow wall).
- Dominant mode is TE10, with one half-sine variation across a and none across b.
- Perfect electric conductor approximation for walls.
Concept / Approach:
The cutoff wavelength for mode TEmn in a rectangular guide is λc = 2 / sqrt((m/a)^2 + (n/b)^2). For TE10, m = 1 and n = 0, giving λc = 2a. Rearranging yields a = λc / 2. This simple proportionality is a foundational design rule for waveguides and is frequently used for selecting a standard WR size given a desired band.
Step-by-Step Solution:
1) Start from general formula: λc = 2 / sqrt((1/a)^2 + 0) = 2a.2) Rearrange: a = λc / 2.3) Therefore, at TE10 cutoff, the broad-wall width equals half of the cutoff wavelength.Verification / Alternative check:
The well-known dominant-mode cutoff frequency is fc10 = c / (2a). Converting to wavelength (λc = c / fc10) again gives λc = 2a, consistent with the above.
Why Other Options Are Wrong:
a = λc or a = 2λc contradict the derived relation; a = λc / 4 is also incorrect; “independent of λc” ignores the defining cutoff equation.
Common Pitfalls:
Confusing λ (operating wavelength) with λc (cutoff wavelength); λ must be shorter than λc for propagation in a given mode.
Final Answer:
a = λc / 2