Circular waveguide TM field strength and dominant mode

TM field expressions for an air-filled circular waveguide

For a circular guide of radius a (coordinates ρ, φ, z), a TM mode has Hz = 0 and Ez ≠ 0, with fields varying as e^(j(ωt − βz)).

        .-------.
      /     ρ     \
     |   o----->   |   radius a, air (μ0, ε0)
      \           /    propagation along z
        '-------'

Step 1 – wave equation for Ez:

 ∂²Ez/∂ρ² + (1/ρ)∂Ez/∂ρ + (1/ρ²)∂²Ez/∂φ² + kc² Ez = 0
 kc² = k² - β²,  k = ω√(μ0ε0)

Step 2 – separation of variables: Ez = R(ρ)·Φ(φ)·e^(−jβz) gives

 Φ'' + n²Φ = 0          ->  Φ = A sin nφ + B cos nφ
 ρ²R'' + ρR' + (kc²ρ² - n²)R = 0   (Bessel's equation)
 R = C·Jn(kcρ) + D·Yn(kcρ)

Yn is infinite at ρ = 0, so D = 0:

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

Step 3 – boundary condition: tangential Ez = 0 at the wall ρ = a:

 Jn(kc a) = 0   ->   kc = pnm / a

where pnm is the m-th zero of Jn(x).

Step 4 – transverse fields from Maxwell's equations (with Hz = 0):

 Eρ = -(jβ/kc²) ∂Ez/∂ρ        Hρ =  (jωε/(kc²ρ)) ∂Ez/∂φ
 Eφ = -(jβ/(kc²ρ)) ∂Ez/∂φ     Hφ = -(jωε/kc²) ∂Ez/∂ρ

giving the TMnm field strengths:

 Ez = (A sin nφ + B cos nφ) Jn(kcρ) e^(-jβz)
 Eρ = -(jβ/kc) (A sin nφ + B cos nφ) J'n(kcρ) e^(-jβz)
 Eφ = -(jβn/(kc²ρ)) (A cos nφ - B sin nφ) Jn(kcρ) e^(-jβz)
 Hρ =  (jωε0 n/(kc²ρ)) (A cos nφ - B sin nφ) Jn(kcρ) e^(-jβz)
 Hφ = -(jωε0/kc) (A sin nφ + B cos nφ) J'n(kcρ) e^(-jβz)
 Hz = 0

Characteristic relations (air):

 fc = pnm·c/(2πa),   λc = 2πa/pnm
 β = √(k² - (pnm/a)²)
 Z_TM = Eρ/Hφ = β/(ωε0) = η0 √(1 - (fc/f)²)

Dominant modes in TE and TM

• TM: the smallest zero of Jn is p01 = 2.405, so TM01 is the lowest TM mode (fc = 2.405c/(2πa), λc = 2.61a).
• TE: the smallest zero of J'n is p'11 = 1.841, so TE11 is the lowest TE mode (λc = 3.41a).
• Since 1.841 < 2.405, TE11 is the dominant mode of the circular waveguide; TM01 is the second mode. Example: a = 1 cm gives fc(TE11) = 8.79 GHz and fc(TM01) = 11.48 GHz.