Question 4 of 7: WR-90 Rectangular Waveguide Modes
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
17-Phys-A3, Electromagnetics — National Exam, December 2019. 3-hour closed-book exam (Casio or Sharp approved calculators only); any FIVE of the seven questions constitute a complete paper and only the first five as they appear in a candidate's answer book are marked, each of equal value, with full justification required for marks. All seven printed questions are solved below as a complete study resource.
Reference texts: Sadiku, Elements of Electromagnetics (7th ed.) — plane waves and boundaries, transmission lines, waveguides, Gauss's and Faraday's laws; Hayt & Buck, Engineering Electromagnetics (9th ed.) — transmission-line transients and bounce diagrams; Pozar, Microwave Engineering (4th ed.) — transmission-line theory, rectangular waveguide cutoff; Balanis, Antenna Theory (4th ed.) — wave polarization and radiation fundamentals.
Given. Rectangular waveguide, inside dimensions $a=22.86$ mm (broad wall) $\times$ $b=10.16$ mm (narrow wall), filled with free space ($\varepsilon_r=\mu_r=1$).
WR-90 cross-section with the dominant-mode TE$_{10}$ field pattern (one half-sine variation across the broad wall $a$).
Find. The dominant mode and its cutoff frequency; the phase velocity, guide wavelength and wave impedance at 10 GHz; the cutoff frequencies (and TE/TM identity) of the next three modes.
Approach. Apply the general TE$_{mn}$/TM$_{mn}$ cutoff formula for a free-space-filled guide, evaluate the low-order $(m,n)$ pairs, then use $f_c$ of the dominant mode in the standard waveguide dispersion relations for $v_p$, $\lambda_g$, $Z_{TE}$.
Part (a) — Cutoff formula and dominant mode. For a free-space-filled guide,
$$f_{c,mn}=\frac{c}{2}\sqrt{\left(\frac{m}{a}\right)^2+\left(\frac{n}{b}\right)^2}.$$
Because $a\ >\ b$, the mode with the fewest variations across the LARGER dimension cuts off lowest: TE$_{10}$ ($m=1,n=0$; TM modes require both $m,n\ge1$, so no TM mode can cut off below TE$_{11}$):
$$f_{c,10}=\frac{c}{2a}=\frac{3\times10^8}{2(0.02286)}=\boxed{6.56\ \text{GHz}}.$$
Part (b) — Phase velocity at 10 GHz.
$$v_p=\frac{c}{\sqrt{1-(f_c/f)^2}}=\frac{3\times10^8}{\sqrt{1-(6.56/10)^2}}=\boxed{3.98\times10^8\ \text{m/s}}.$$
Part (c) — Guide wavelength at 10 GHz.
$$\lambda_g=\frac{v_p}{f}=\frac{\lambda_0}{\sqrt{1-(f_c/f)^2}}=\frac{0.030}{\sqrt{1-(6.56/10)^2}}=\boxed{39.8\ \text{mm}}.$$
Part (d) — Wave impedance (TE) at 10 GHz.
$$Z_{TE}=\frac{\eta_0}{\sqrt{1-(f_c/f)^2}}=\frac{376.7}{\sqrt{1-(6.56/10)^2}}=\boxed{499\ \Omega}.$$
Part (e) — Next three cutoff frequencies. Evaluating the cutoff formula for the remaining low-order pairs:
$$f_{c,20}=\frac{c}{a}=13.12\ \text{GHz (TE}_{20}\text{)},\qquad f_{c,01}=\frac{c}{2b}=14.76\ \text{GHz (TE}_{01}\text{)},$$
$$f_{c,11}=\frac{c}{2}\sqrt{\left(\frac1a\right)^2+\left(\frac1b\right)^2}=16.16\ \text{GHz (TE}_{11}\text{ and TM}_{11}\text{, degenerate}\text{)}.$$
Ranking $6.56<13.12<14.76<16.16$ GHz, the three next-lowest cutoffs above TE$_{10}$ are $\boxed{\text{TE}_{20}\ (13.12\ \text{GHz}),\ \text{TE}_{01}\ (14.76\ \text{GHz}),\ \text{TE}_{11}/\text{TM}_{11}\ (16.16\ \text{GHz})}$; the last is shared by BOTH a TE and a TM mode since $m,n\ge1$ admits both field families at the same cutoff.
