Question 4 of 10: A road tunnel as a rectangular waveguide

Question 4: A road tunnel as a rectangular waveguide

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A 7 m wide by 4.5 m high air-filled tunnel with perfectly conducting walls, entered by a 1 MHz AM signal whose vertical field is 0.025 V/m on the centre-line at the portal.

Find. The dominant mode and its cut-off frequency; a sketch of that mode's electric field over the cross section; and a quantitative explanation of the rapid decay of the AM signal down the tunnel.

Approach. Rank every low-order cut-off frequency to identify the dominant mode, then compare the 1 MHz carrier with it: because the carrier is far below cut-off the field is evanescent, and the decay rate follows from $\alpha = \sqrt{k_c^2 - k^2}$.

1. (a) Rank the cut-off frequencies. For an air-filled rectangular guide $$f_{c,mn} = \frac{c}{2}\sqrt{\left(\frac{m}{a}\right)^2 + \left(\frac{n}{b}\right)^2}$$ With $a = 7$ m and $b = 4.5$ m and $c = 3\times10^8$ m/s this gives $f_c(\mathrm{TE}_{10}) = c/2a = 21.43$ MHz, $f_c(\mathrm{TE}_{01}) = c/2b = 33.33$ MHz, $f_c(\mathrm{TE}_{11}) = f_c(\mathrm{TM}_{11}) = 39.63$ MHz and $f_c(\mathrm{TE}_{20}) = c/a = 42.86$ MHz. The smallest belongs to the widest dimension, so $$\boxed{\text{dominant mode } \mathrm{TE}_{10}, \quad f_c = \frac{c}{2a} = \frac{3\times10^8}{14} = 21.43\ \text{MHz}}$$ Because $a/b = 1.56 \lt 2$ here, $\mathrm{TE}_{01}$ — not $\mathrm{TE}_{20}$ — is the runner-up; that ordering flips once $a \ge 2b$, so it is worth computing rather than remembering.
2. (b) Describe the dominant-mode field. For $\mathrm{TE}_{10}$ the only electric-field component is $$E_y(x,y) = E_0\sin\!\left(\frac{\pi x}{a}\right)e^{-j\beta z}, \qquad E_x = E_z = 0$$ so the electric field is vertical everywhere, independent of $y$, zero on the two side walls at $x = 0$ and $x = a$ (as the perfect-conductor boundary condition demands), and maximum on the centre-line $x = a/2$ — exactly where the question specifies the 0.025 V/m measurement. The figure below draws that vector field: equally spaced vertical arrows whose lengths trace a half sine across the tunnel width.
3. (c) Compare the carrier with cut-off. The AM signal sits at 1 MHz against a 21.43 MHz cut-off, a factor of 21.4 below it. The axial propagation constant is $\gamma = \sqrt{k_c^2 - k^2}$ with $k_c = \pi/a$ and $k = 2\pi f/c$, and when $f \lt f_c$ this is real: the field does not propagate at all but decays exponentially, $E \propto e^{-\alpha z}$. So the tunnel behaves as a high-pass filter that the AM band cannot pass.
4. Quantify the evanescent decay. With $k_c = \pi/7 = 0.44880$ rad/m and $k = 2\pi(10^6)/(3\times10^8) = 0.020944$ rad/m, $$\alpha = \sqrt{k_c^2 - k^2} = \sqrt{0.201420 - 0.000439} = 0.44831\ \text{Np/m}$$ $$\boxed{\alpha = 0.4483\ \text{Np/m} = 3.894\ \text{dB/m}}$$ so the field falls by a factor of ten every $\ln 10/\alpha = 5.14$ m. Starting from 0.025 V/m at the portal, the field is only 3.19 µV/m at 20 m in and the loss over 50 m is 195 dB — utterly unreceivable, which is precisely the observation the question describes.
5. Note what the mechanism is not. This attenuation is reactive, not dissipative: below cut-off the guide presents an almost purely imaginary wave impedance, so the incident energy is reflected back out of the portal rather than absorbed in the walls. That is why $\alpha$ contains no conductivity term at all — it depends only on the tunnel width and the frequency — and why lining the tunnel with a better conductor would not help. The engineering fix is a leaky feeder (radiating coaxial cable) or in-tunnel repeaters, which is standard practice in Canadian road and rail tunnels.