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Dominant mode and 2.254 cm waveguide at 6 GHz

Short answer

The dominant mode has the lowest cutoff frequency. Rectangular guide: TE10 with fc=c/(2a); circular guide: TE11 with fc=1.841c/(2πr). For a=2.254 cm, TE10 cutoff is 6.655 GHz, so a 6 GHz signal is below cutoff and does not propagate.

What is dominant mode in a waveguide? What are the dominant modes for rectangular and circular waveguides? Does a rectangular waveguide of width 2.254 cm support the propagation of a signal having frequency 6 GHz?

Dominant mode

The dominant mode of a waveguide is the mode with the lowest cut-off frequency (longest cut-off wavelength). It is the first mode to propagate as frequency rises, so a guide is normally operated where only this mode can travel.

Dominant modes of rectangular and circular waveguides

  • Rectangular waveguide (a > b): TE10, with fc = c/(2a), λc = 2a.
  • Circular waveguide (radius r): TE11, with fc = 1.841c/(2πr), λc = 3.41r.

Does a 2.254 cm wide guide pass 6 GHz?

For the dominant TE10 mode, with a = 2.254 cm = 0.02254 m:

 fc = c/(2a) = (3 × 10⁸)/(2 × 0.02254)
    = 6.655 × 10⁹ Hz = 6.655 GHz

The signal frequency f = 6 GHz is below the cut-off frequency (6 < 6.655 GHz). Equivalently, λ0 = 5 cm > λc = 2a = 4.508 cm.

Answer: No. The lowest cut-off of the guide is 6.655 GHz, so a 6 GHz signal is evanescent (attenuated) and does not propagate.

✓ Answer checked by Claude Opus 5.5 (high effort)

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