Circular waveguide TE equations and degenerate modes

Field equations of a circular waveguide in TE mode

For a circular guide of radius a (cylindrical coordinates ρ, φ, z) with Ez = 0, Hz satisfies Bessel's equation and the solution finite at the axis is

 Hz = (A sin nφ + B cos nφ) Jn(kcρ) e^(-jβz)

The transverse fields from Maxwell's equations are:

 Eρ = -(jωμn/(kc²ρ)) (A cos nφ - B sin nφ) Jn(kcρ) e^(-jβz)
 Eφ =  (jωμ/kc) (A sin nφ + B cos nφ) J'n(kcρ) e^(-jβz)
 Hρ = -(jβ/kc) (A sin nφ + B cos nφ) J'n(kcρ) e^(-jβz)
 Hφ = -(jβn/(kc²ρ)) (A cos nφ - B sin nφ) Jn(kcρ) e^(-jβz)
 Ez = 0

Boundary condition Eφ(ρ = a) = 0 gives J'n(kc a) = 0, so kc = p'nm/a.

Characteristic equations:

• fc = p'nm·c/(2πa), λc = 2πa/p'nm
• β = √(k² − kc²), λg = λ0/√(1 − (fc/f)²)
• Z_TE = η/√(1 − (fc/f)²)
• Dominant mode TE11: p'11 = 1.841, λc = 3.41a.

Dominant and degenerate modes of a rectangular waveguide

Dominant mode: the mode with the lowest cut-off frequency (longest cut-off wavelength). For a > b:

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

TE10 is the dominant mode. Example: a = 2.286 cm, b = 1.016 cm gives fc,TE10 = 6.56 GHz, next mode TE20 at 13.12 GHz. So between 6.56 and 13.12 GHz only TE10 propagates, giving single-mode operation (X-band, 8.2–12.4 GHz).

Degenerate modes: different modes (different field patterns) that have the same cut-off frequency. In a rectangular guide, TEmn and TMmn (m, n ≥ 1) always share the same fc, e.g. TE11 and TM11, TE21 and TM21. In a square guide (a = b), TE10 and TE01 are also degenerate.

Example: in the X-band guide above, TE11 and TM11 both have fc = (3×10⁸/2)√((1/0.02286)² + (1/0.01016)²) ≈ 16.16 GHz.

Degenerate modes are undesirable because energy can transfer between them at discontinuities, causing distortion; guides are therefore operated in the dominant-mode band.