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Short note: TE and TM modes
Short answer
TE modes (E_z = 0) have H_z = H_0 cos(mπx/a) cos(nπy/b) e^{-jβz}, wave impedance Z_TE = η/√(1-(f_c/f)²). TM modes (H_z = 0) have E_z = E_0 sin(mπx/a) sin(nπy/b) e^{-jβz}, Z_TM = η√(1-(f_c/f)²). Cut-off f_c = (c/2)√((m/a)²+(n/b)²). Dominant TE_10, lowest TM_11.
Explain the transverse electric (TE) and transverse magnetic (TM) modes used in rectangular waveguides.
In a rectangular waveguide (a × b, a > b, wave travelling along +z) the fields cannot be TEM, so they are split into two families defined by which longitudinal field is zero.
TE (transverse electric) modes
The electric field has no component along the direction of travel: E_z = 0, H_z ≠ 0.
H_z = H₀ cos(mπx/a) cos(nπy/b) e^{−jβz}
- m = 0, 1, 2, ...; n = 0, 1, 2, ... (not both zero).
- TE₁₀ is the dominant mode of the guide.
- Wave impedance Z_TE = η/√(1 − (f_c/f)²), which is greater than η.
TM (transverse magnetic) modes
The magnetic field has no component along z: H_z = 0, E_z ≠ 0.
E_z = E₀ sin(mπx/a) sin(nπy/b) e^{−jβz}
- m = 1, 2, ...; n = 1, 2, ... (both at least 1, since E_z must be zero at the walls).
- TM₁₁ is the lowest TM mode.
- Wave impedance Z_TM = η√(1 − (f_c/f)²), which is less than η.
Common relations
f_c = (c/2)√[(m/a)² + (n/b)²]
β = β₀ √(1 − (f_c/f)²)
λ_g = λ₀/√(1 − (f_c/f)²)
v_p = c/√(1 − (f_c/f)²), v_g = c√(1 − (f_c/f)²)
where η = 377 Ω in air and v_p v_g = c². TE_mn and TM_mn with the same m, n have the same cut-off frequency. Below cut-off, β is imaginary and the mode does not propagate.
✓ Answer checked by Claude Opus 5.5 (high effort)
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