22-Elec-B9 Electromagnetic Field, Transmission Lines, Antennas, and Radiation · May 2016

Question 4 of 8: Maximum Power in an Air-Filled Rectangular Waveguide

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 4: Maximum Power in an Air-Filled Rectangular Waveguide (20 marks)

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. An air-filled rectangular guide of internal broad dimension $a = 2.25\ \text{cm}$ and narrow dimension $b = 1.00\ \text{cm}$, operating at $f = 10\ \text{GHz}$, matched at both ends so that only a forward travelling wave exists, with an electric-field ceiling of $E_0 = 10^{6}\ \text{V/m}$ peak anywhere inside.

Find. The largest time-average power the guide can carry without exceeding that field.

Approach. Identify which mode or modes can propagate at 10 GHz, write the transmitted power of the dominant mode in terms of its peak field, and evaluate.

1. Find the cut-off frequencies and confirm single-mode operation. For a rectangular guide, $f_{c,mn} = \dfrac{c}{2}\sqrt{(m/a)^{2} + (n/b)^{2}}$, so with $a = 22.5\ \text{mm}$ and $b = 10\ \text{mm}$: $$f_{c,10} = \frac{c}{2a} = 6.662\ \text{GHz}, \qquad f_{c,20} = \frac{c}{a} = 13.32\ \text{GHz}, \qquad f_{c,01} = \frac{c}{2b} = 14.99\ \text{GHz}.$$ At 10 GHz only TE₁₀ is above cut-off; every other mode is evanescent and carries no average power. Note that $a/b = 2.25 > 2$, so TE₂₀ and not TE₀₁ is what closes the single-mode band — a ranking worth checking rather than remembering, since it reverses for $a/b < 2$.
2. Write down the TE₁₀ field. The single transverse electric component varies as a half sine across the broad wall and is uniform across the narrow one: $$E_y(x) = E_0 \sin\!\left(\frac{\pi x}{a}\right), \qquad H_x = -\frac{E_y}{Z_{TE}}.$$ The stated ceiling therefore applies at the centre line $x = a/2$, where the sine peaks.
3. Evaluate the wave impedance at the operating frequency. The dispersion factor appears in every guided quantity, so compute it once: $$\sqrt{1 - \left(\frac{f_c}{f}\right)^{2}} = \sqrt{1 - \left(\frac{6.662}{10}\right)^{2}} = 0.7458,$$ $$Z_{TE} = \frac{\eta_0}{\sqrt{1 - (f_c/f)^{2}}} = \frac{376.73}{0.7458} = 505.2\ \Omega.$$ The same factor gives the guide wavelength $\lambda_g = \lambda_0/0.7458 = 4.020\ \text{cm}$, which is useful for checking that the guide is being operated sensibly.
4. Integrate the Poynting vector over the cross-section. The time-average power is $$P = \frac{1}{2}\int_0^{b}\!\!\int_0^{a} \frac{|E_y(x)|^{2}}{Z_{TE}}\,dx\,dy = \frac{E_0^{2}\,b}{2Z_{TE}}\int_0^{a}\sin^{2}\!\left(\frac{\pi x}{a}\right)dx = \frac{E_0^{2}\,a\,b}{4 Z_{TE}},$$ the factor of four collecting one half from the time average and one half from the mean of $\sin^{2}$ across the guide.
5. Substitute and evaluate. $$P_{\max} = \frac{(10^{6})^{2}\,(0.0225)(0.0100)}{4(505.2)} = \frac{2.25\times10^{8}}{2021}$$ $$\boxed{P_{\max} = 1.114\times10^{5}\ \text{W} \approx 111\ \text{kW}}$$
6. Sanity-check the result physically. The stated ceiling of $10^{6}\ \text{V/m}$ is about a third of the dry-air breakdown strength of roughly $3\times10^{6}\ \text{V/m}$ at sea level, so it is a plausible engineering limit with a safety margin rather than an arbitrary number. Real guides derate further for humidity, altitude, surface roughness and any discontinuity that concentrates the field, and a standing wave of ratio $S$ raises the peak field by the factor $S$ — which is exactly why the question specifies "with no reflections".