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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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