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TE field components; 1 GHz in 5 cm waveguide

Short answer

TE10 mode fields: Ey = -jωμ a/π H0 sin(πx/a) e^{-jβz}, Hx and Hz nonzero. Cutoff frequency is fc=c/(2a)=3×10^8/(2×0.05)=3 GHz. Since the 1 GHz signal is below cutoff, the mode is evanescent and cannot propagate.

Define expressions for various field components of a rectangular waveguide in TE mode. Show that a 1 GHz signal cannot propagate in TE10 mode in a rectangular waveguide with a wall separation of 5 cm.

Field components of a rectangular waveguide in TE mode

For a guide with inner dimensions a × b (a > b) and Ez = 0, the solution of ∇t²Hz + kc²Hz = 0 with ∂Hz/∂n = 0 on the walls gives (m, n = 0, 1, 2 …, not both zero):

 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

with

 kc² = (mπ/a)² + (nπ/b)²,   β = √(ω²με - kc²)
 fc = (c/2)√((m/a)² + (n/b)²)
 Z_TE = ωμ/β = η/√(1 - (fc/f)²)
 λg = λ0/√(1 - (fc/f)²)

For the dominant TE10 mode (m = 1, n = 0): only Ey, Hx, Hz exist, and fc = c/(2a).

Can 1 GHz propagate in TE10 with a = 5 cm?

 fc(TE10) = c/(2a) = (3 × 10⁸)/(2 × 0.05)
          = 3 × 10⁹ Hz = 3 GHz

Equivalently, λc = 2a = 10 cm while λ0 = c/f = 30 cm.

Since f = 1 GHz < fc = 3 GHz (λ0 > λc), β = √(ω²με − kc²) is imaginary: the field decays exponentially (evanescent) instead of propagating.

Answer: the TE10 cut-off is 3 GHz, so a 1 GHz signal cannot propagate in this waveguide.

✓ Answer checked by Claude Opus 5.5 (high effort)

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