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Modes in rectangular waveguide

Short answer

A rectangular waveguide supports TE_mn (E_z = 0, m,n not both zero) and TM_mn (H_z = 0, m,n ≥ 1) modes. TEM cannot propagate. Each mode has cut-off frequency f_c = (c/2)√((m/a)²+(n/b)²). The dominant mode TE_10 has the lowest cut-off c/(2a).

Explain the modes supported by a rectangular waveguide.

A hollow rectangular waveguide of cross-section a × b (a > b) supports a set of discrete field patterns called modes. Each is labelled by integers m and n, the number of half-wave variations of the field along x (width a) and y (height b). Waves travel along z.

TE modes (transverse electric)

  • E_z = 0; the electric field has no component along the propagation direction.
  • H_z = H₀ cos(mπx/a) cos(nπy/b) e^{−jβz}
  • m = 0, 1, 2, ...; n = 0, 1, 2, ...; but m and n are not both zero.
  • Lowest: TE₁₀. Wave impedance Z_TE = η/√(1 − (f_c/f)²).

TM modes (transverse magnetic)

  • H_z = 0.
  • E_z = E₀ sin(mπx/a) sin(nπy/b) e^{−jβz}
  • m = 1, 2, ...; n = 1, 2, ... (both must be non-zero, otherwise all fields vanish).
  • Lowest: TM₁₁. Wave impedance Z_TM = η√(1 − (f_c/f)²).

Cut-off frequency

A mode propagates only when f > f_c:

f_c = (c/2) √[(m/a)² + (n/b)²]

Below f_c the wave is evanescent (it attenuates rapidly).

TEM mode

A hollow guide has a single conductor, so it cannot support a TEM wave (E_z = H_z = 0): a TEM wave would need two conductors.

Dominant mode

TE₁₀ has the lowest cut-off frequency f_c = c/2a and is the dominant mode. A guide is normally used in the band f_c(TE₁₀) < f < the next cut-off (TE₂₀ or TE₀₁), where only one mode exists.

Degenerate modes

TE_mn and TM_mn (m, n ≥ 1) have the same cut-off frequency and are called degenerate modes.

✓ Answer checked by Claude Opus 5.5 (high effort)

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