Question 4 of 8: Single-Mode Band of a Dielectric-Filled Waveguide
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, December 2013 — 07-Elec-A7 Electromagnetics. Three hours, closed book, one approved calculator. Eight questions of equal value (20 marks each); the paper states that any five constitute a complete paper, but all eight are solved here so the set works as a complete study resource. Aids printed on the paper: $\epsilon_0 = 8.85\times10^{-12}$ F/m and $\mu_0 = 4\pi\times10^{-7}$ H/m.
Reference texts. Sadiku, Elements of Electromagnetics, 7th ed.; Hayt & Buck, Engineering Electromagnetics, 9th ed.; Ulaby & Ravaioli, Fundamentals of Applied Electromagnetics, 8th ed.; Pozar, Microwave Engineering, 4th ed.; Balanis, Antenna Theory: Analysis and Design, 4th ed.
Free-space quantities follow from the printed aids: $c = 1/\sqrt{\mu_0\epsilon_0} \approx 3\times10^{8}$ m/s and $\eta_0 = \sqrt{\mu_0/\epsilon_0} = 376.8\ \Omega$.
Question 4: Single-Mode Band of a Dielectric-Filled Waveguide (20 marks)
Given. A rectangular guide whose broad wall is 10 mm and narrow wall 4 mm, uniformly filled with a lossless dielectric of relative permittivity 9.
Find. The band of frequencies over which exactly one mode propagates.
Figure 4.1 — Guide cross-section with the half-sine transverse electric field of the TE₁₀ mode, and the cut-off frequencies that bracket the single-mode band.
Approach. Compute the wave speed inside the filling, evaluate the cut-off frequency of every low-order mode, rank them, and take the band between the lowest and the second-lowest.
Wave speed inside the dielectric. Filling the guide slows the wave and scales every cut-off frequency down by the same factor:$$u = \frac{c}{\sqrt{\epsilon_r}} = \frac{3\times10^{8}}{\sqrt{9}} = 1.00\times10^{8}\ \text{m/s}.$$
General cut-off formula. For the TE$_{mn}$ (or TM$_{mn}$) mode of a guide of inside dimensions $a \times b$,$$f_{c,mn} = \frac{u}{2}\sqrt{\left(\frac{m}{a}\right)^{2}+\left(\frac{n}{b}\right)^{2}}.$$Only the three lowest members can matter here.
Cut-off of the dominant mode. With $m=1$, $n=0$ and $a = 10$ mm,$$f_{c,10} = \frac{u}{2a} = \frac{1.00\times10^{8}}{2(0.010)},$$$$\boxed{f_{c,\text{TE}_{10}} = 5.00\ \text{GHz}}$$
The two nearest competitors. The next modes up are the second broad-wall harmonic and the first narrow-wall mode:$$f_{c,20} = \frac{u}{a} = 10.0\ \text{GHz},\qquad f_{c,01} = \frac{u}{2b} = \frac{1.00\times10^{8}}{2(0.004)} = 12.5\ \text{GHz}.$$Because $a \gt 2b$ here (10 mm against 8 mm), TE$_{20}$ cuts on before TE$_{01}$ and is the mode that ends the single-mode band. The lowest TM mode, TM$_{11}$, lies higher still at 13.5 GHz and never competes.
Single-mode band. Below 5 GHz nothing propagates (the guide is cut off); above 10 GHz the TE$_{20}$ mode joins TE$_{10}$. Hence exactly one mode propagates for$$\boxed{5.0\ \text{GHz} \lt f \lt 10.0\ \text{GHz}}$$a full octave, which is the widest single-mode band a rectangular guide can offer and is obtained whenever $a \geq 2b$. In practice a guide is operated over the middle of this range, roughly 6–9 GHz, to stay clear of the high dispersion just above cut-off and of the TE$_{20}$ onset.
