Rectangular waveguide TE mode field equations

In a rectangular waveguide (inner dimensions a × b, a > b, propagation along z), a TE (transverse electric) mode has Ez = 0 and Hz ≠ 0. All other field components are derived from Hz.

   y
 b +-----------------+
   |  air (μ, ε)     |   propagation along +z
   +-----------------+
   0                 a  -> x

Derivation

With fields varying as e^(j(ωt − βz)), Hz satisfies the reduced wave equation

 ∂²Hz/∂x² + ∂²Hz/∂y² + kc² Hz = 0,   kc² = ω²με - β²

Separation of variables, Hz = X(x)·Y(y):

 X = A1 cos kx·x + A2 sin kx·x
 Y = B1 cos ky·y + B2 sin ky·y,     kx² + ky² = kc²

The transverse fields come from Maxwell's curl equations:

 Ex = -(jωμ/kc²) ∂Hz/∂y      Hx = -(jβ/kc²) ∂Hz/∂x
 Ey =  (jωμ/kc²) ∂Hz/∂x      Hy = -(jβ/kc²) ∂Hz/∂y

Boundary conditions (tangential E = 0 on walls): Ex = 0 at y = 0, b → ∂Hz/∂y = 0 there; Ey = 0 at x = 0, a → ∂Hz/∂x = 0 there. This gives A2 = B2 = 0 and

 kx = mπ/a,  ky = nπ/b,  m, n = 0, 1, 2 ... (not both zero)

TEmn field equations

 Hz = H0 cos(mπx/a) cos(nπy/b) e^(-jβz)
 Hx =  (jβ/kc²)(mπ/a) H0 sin(mπx/a) cos(nπy/b) e^(-jβz)
 Hy =  (jβ/kc²)(nπ/b) H0 cos(mπx/a) sin(nπy/b) e^(-jβz)
 Ex =  (jωμ/kc²)(nπ/b) H0 cos(mπx/a) sin(nπy/b) e^(-jβz)
 Ey = -(jωμ/kc²)(mπ/a) H0 sin(mπx/a) cos(nπy/b) e^(-jβz)
 Ez = 0

Because Hz uses cosines, one index may be zero (TE10, TE01, TE20 exist), but TE00 gives no field.

Characteristic parameters

• Cut-off wave number: kc = √((mπ/a)² + (nπ/b)²)
• Cut-off frequency: fc = (c/2)√((m/a)² + (n/b)²) (air-filled)
• Cut-off wavelength: λc = 2/√((m/a)² + (n/b)²); for TE10, λc = 2a
• Phase constant: β = √(k² − kc²) = k√(1 − (fc/f)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); vp·vg = c²
• Wave impedance:

 Z_TE = Ex/Hy = ωμ/β = η/√(1 - (fc/f)²)     (η = 377 Ω)

Z_TE is always greater than η.

Dominant TE10 mode

With m = 1, n = 0 (and a > b), TE10 has the lowest cut-off fc = c/2a:

 Hz = H0 cos(πx/a) e^(-jβz)
 Hx = (jβa/π) H0 sin(πx/a) e^(-jβz)
 Ey = -(jωμa/π) H0 sin(πx/a) e^(-jβz)

Only Ey, Hx and Hz exist; E is maximum at the centre (x = a/2) and zero at the side walls.

Example: for a = 2.286 cm at 10 GHz, fc = 6.56 GHz, λg = 3.98 cm and Z_TE ≈ 500 Ω.