Dominant mode in rectangular waveguides: Which mode is the first to propagate (lowest cutoff) in a standard air-filled rectangular waveguide?

Difficulty: Easy

Correct Answer: TE10

Explanation:


Introduction:
Every rectangular waveguide has a dominant mode: the mode with the lowest cutoff frequency, which therefore propagates first as frequency increases. Identifying it is essential for single-mode operation and component design.


Given Data / Assumptions:

  • Air-filled rectangular waveguide of width a and height b (a > b).
  • Standard boundary conditions on perfectly conducting walls.
  • TE/TM cutoff relationships apply: fc ∝ sqrt((m/a)^2 + (n/b)^2).


Concept / Approach:

For TEmn and TMmn modes, the lowest combination is m = 1, n = 0 in TE family for a typical guide with a > b. This gives TE10 as the dominant mode. Its cutoff frequency is fc = c / (2a), which is lower than other combinations for the same geometry.


Step-by-Step Solution:

Write cutoff: fc(m, n) = (c/2) * sqrt((m/a)^2 + (n/b)^2).For TE10: fc = c / (2a) (no dependence on b).Compare with TE01: fc = c / (2b) which is higher because b < a.All other modes (e.g., TE20, TE11, TM11) include larger indices → higher fc.


Verification / Alternative check:

Waveguide charts and standards (WR series) are based on TE10 dominance, ensuring single-mode bandwidth between TE10 cutoff and the next higher-order mode cutoff.


Why Other Options Are Wrong:

  • TE01, TE20, TE11, TM11 have higher cutoff frequencies for a standard aspect ratio.


Common Pitfalls:

Confusing dominant with circular-guide dominance (TE11 is dominant in circular guides). In rectangular guides, TE10 is dominant.


Final Answer:

TE10

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