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Short note: Dominant mode in waveguide

Short answer

The dominant mode has the lowest cutoff frequency, enabling single-mode operation without mode conversion. In rectangular waveguide (a>b) TE10 with fc=c/(2a) is dominant; in circular waveguide TE11 (fc=1.841c/(2πa)) is dominant. It offers lowest attenuation and simple field coupling.

Write a short note on dominant mode in waveguide.

The dominant mode of a waveguide is the propagating mode with the lowest cut-off frequency (longest cut-off wavelength). Below its cut-off no energy propagates; between its cut-off and that of the next higher mode, only the dominant mode can travel, which gives clean single-mode operation.

Rectangular waveguide (a > b):

 fc,mn = (c/2) √((m/a)² + (n/b)²)
 TE10: fc = c/(2a),  λc = 2a   (dominant)

TE10 has only Ey, Hx and Hz components; E is maximum at the centre of the broad wall and zero at the side walls. TM modes start at TM11, which is always higher.

Circular waveguide (radius r): dominant mode TE11 (p'11 = 1.841), fc = 1.841c/(2πr), λc = 3.41r. The lowest TM mode is TM01 (λc = 2.61r).

Why the dominant mode is used:

  • Single-mode operation avoids mode conversion and distortion.
  • Lowest attenuation for a given size (in rectangular guides) and simple field pattern for coupling with probes.
  • Easy design of components (tees, couplers) for one known field pattern.

Example: WR-90 guide (a = 2.286 cm, b = 1.016 cm): fc(TE10) = 6.56 GHz, next mode TE20 at 13.12 GHz. It is used in the X-band (8.2–12.4 GHz), where only TE10 propagates.

✓ Answer checked by Claude Opus 5.5 (high effort)

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