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Rectangular TM equations; TE10 vs TE20

Short answer

TM modes in a rectangular guide have Ez = E0 sin(mπx/a) sin(nπy/b) e^{-jβz} with m,n ≥ 1; the lowest is TM11. TE10 has cutoff fc=c/(2a) and is the dominant mode; TE20 has fc=c/a, twice TE10's cutoff, making it a higher-order mode.

Describe field equations and other related parameters of a rectangular waveguide in TM mode. Compare TE10 and TE20 in terms of cut-off frequency and dominant mode.

Field equations of a rectangular waveguide in TM mode

For a guide of inner dimensions a × b (a > b), TM modes have Hz = 0. Solving ∇t²Ez + kc²Ez = 0 with Ez = 0 on all four walls gives kx = mπ/a, ky = nπ/b (m, n ≥ 1):

 Ez = E0 sin(mπx/a) sin(nπy/b) e^(-jβz)
 Ex = -(jβ/kc²)(mπ/a) E0 cos(mπx/a) sin(nπy/b) e^(-jβz)
 Ey = -(jβ/kc²)(nπ/b) E0 sin(mπx/a) cos(nπy/b) e^(-jβz)
 Hx =  (jωε/kc²)(nπ/b) E0 sin(mπx/a) cos(nπy/b) e^(-jβz)
 Hy = -(jωε/kc²)(mπ/a) E0 cos(mπx/a) sin(nπy/b) e^(-jβz)
 Hz = 0

Related parameters:

  • kc² = (mπ/a)² + (nπ/b)²; β = √(ω²με − kc²)
  • Cut-off frequency: fc = (c/2)√((m/a)² + (n/b)²) (air)
  • Cut-off wavelength: λc = 2/√((m/a)² + (n/b)²)
  • Guide wavelength: λg = λ0/√(1 − (fc/f)²)
  • Phase velocity vp = c/√(1 − (fc/f)²); group velocity vg = c√(1 − (fc/f)²)
  • Wave impedance: Z_TM = η√(1 − (fc/f)²), always less than η
  • Lowest TM mode: TM11 (TM10 and TM01 do not exist because Ez would vanish everywhere).

TE10 vs TE20

 TE10: fc = c/(2a),  λc = 2a
 TE20: fc = 2c/(2a) = c/a,  λc = a
PointTE10TE20
Cut-off frequencyc/(2a)c/a = 2 × fc(TE10)
Half-wave variations across aOneTwo
Dominant?Yes (lowest fc for a > b)No, a higher-order mode
X-band guide (a = 2.286 cm)6.56 GHz13.12 GHz

TE10 is the dominant mode; the band between fc(TE10) and fc(TE20) (if a ≥ 2b) is the single-mode operating range of the guide.

✓ Answer checked by Claude Opus 5.5 (high effort)

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