- 2071 Magh · 10 marks
TE vs TM mode in circular waveguide
Short answer
TE modes have Hz≠0, Ez=0 with boundary condition Jn'(kc a)=0 giving kc = p'nm/a; TM modes have Ez≠0, Hz=0 with Jn(kc a)=0 giving kc = pnm/a. TE11 is the dominant mode (lowest cut-off, λc=3.41a), TM01 is next (λc=2.61a). TE wave impedance exceeds η, TM wave impedance is less than η.
Describe how TE mode is different from TM mode in a circular waveguide.
In a circular waveguide of radius a, both TE and TM modes are solutions of Bessel's equation, but they differ in which longitudinal field exists, which boundary condition applies and therefore which Bessel roots set the cut-off.
TE modes (Ez = 0)
Hz = (A sin nφ + B cos nφ) Jn(kcρ) e^(-jβz)
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)
Boundary condition Eφ = 0 at ρ = a gives J'n(kc·a) = 0, kc = p'nm/a.
TM modes (Hz = 0)
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 = (jωεn/(kc²ρ)) (A cos nφ - B sin nφ) Jn(kcρ) e^(-jβz)
Hφ = -(jωε/kc) (A sin nφ + B cos nφ) J'n(kcρ) e^(-jβz)
Boundary condition Ez = 0 at ρ = a gives Jn(kc·a) = 0, kc = pnm/a.
Differences
| Point | TE mode | TM mode |
|---|---|---|
| Longitudinal field | Hz ≠ 0, Ez = 0 | Ez ≠ 0, Hz = 0 |
| Boundary condition | J'n(kc a) = 0 | Jn(kc a) = 0 |
| Cut-off wave number | kc = p'nm/a | kc = pnm/a |
| Cut-off frequency | fc = p'nm c/(2πa) | fc = pnm c/(2πa) |
| Lowest mode | TE11 (p'11 = 1.841), λc = 3.41a | TM01 (p01 = 2.405), λc = 2.61a |
| Dominant? | TE11 is dominant mode of the guide | TM01 is the second mode |
| Wave impedance | Z_TE = η/√(1 − (fc/f)²) > η | Z_TM = η√(1 − (fc/f)²) < η |
| Low-loss mode | TE01: wall loss falls with frequency | No such mode |
| Typical use | TE11 general transmission; TE01 long low-loss links | TM01 rotary joints (circular symmetry) |
Mode roots and degeneracy
| Mode | Root | Mode | Root |
|---|---|---|---|
| TE11 | 1.841 | TM01 | 2.405 |
| TE21 | 3.054 | TM11 | 3.832 |
| TE01 | 3.832 | TM21 | 5.135 |
TE01 and TM11 share the root 3.832, so they are degenerate (same cut-off). Also, each mode with n ≥ 1 has two polarizations (sin nφ and cos nφ), another form of degeneracy; this makes TE11 polarization sensitive to bends and ellipticity.
Example
For a = 1 cm (air): fc(TE11) = 1.841 × 3 × 10⁸/(2π × 0.01) = 8.79 GHz; fc(TM01) = 11.48 GHz. Between 8.79 and 11.48 GHz only TE11 propagates.
✓ Answer checked by Claude Opus 5.5 (high effort)
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