Question 5 of 8: Guide Wavelength, and What Happens When the Dielectric Is Removed
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, May 2014 — 07-Elec-A7 Electromagnetics. Three hours, closed book, one approved Casio or Sharp calculator. Eight questions of equal value (20 marks each); the paper states that any five constitute a complete paper and that only the first five in the answer book are marked, 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, from which $\eta_0 = \sqrt{\mu_0/\epsilon_0} = 376.7\ \Omega$ and $c = 1/\sqrt{\mu_0\epsilon_0} = 3.00\times10^{8}$ m/s.
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.
Question 5: Guide Wavelength, and What Happens When the Dielectric Is Removed (20 marks)
Find. (i) the guide wavelength of the 5 GHz signal in the filled guide, and (ii) the attenuation in dB/cm if the same signal is launched into the guide once the dielectric has been taken out.
Figure 5.1 — Guide cross-section with the TE10 electric field, and the cutoff ladder for the filled guide. At 5 GHz only the dominant mode propagates.
Approach. Compute the dominant-mode cutoff for each filling, use the dispersion relation for the guide wavelength while the signal is above cutoff, and — on discovering that removing the dielectric puts 5 GHz below cutoff — evaluate the evanescent decay constant instead.
Cutoff of the filled guide. For TE$_{mn}$ the cutoff frequency depends on the guide dimensions and the filling velocity $u = c/\sqrt{\epsilon_r}$:$$f_{c,10} = \frac{u}{2a} = \frac{c}{2a\sqrt{\epsilon_r}} = \frac{3\times10^{8}}{2(0.025)(1.5)} = 4.00\ \text{GHz}.$$The next modes sit at $f_{c,20} = 8.00$ GHz and $f_{c,01} = 10.0$ GHz, so at 5 GHz the guide is comfortably single-moded.
Wavelength in the unbounded filling. Before the walls are taken into account,$$\lambda = \frac{u}{f} = \frac{c}{f\sqrt{\epsilon_r}} = \frac{3\times10^{8}}{(5\times10^{9})(1.5)} = 4.00\ \text{cm}.$$
(i) Guide wavelength. The waveguide dispersion relation stretches that wavelength:$$\lambda_g = \frac{\lambda}{\sqrt{1 - (f_c/f)^{2}}} = \frac{4.00}{\sqrt{1 - (4/5)^{2}}} = \frac{4.00}{0.600},$$$$\boxed{\lambda_g = 6.67\ \text{cm}.}$$The guide wavelength always exceeds the unbounded one, and diverges as the signal approaches cutoff.
Take the dielectric out and check the cutoff again. With air inside, the filling velocity rises to $c$ and so does the cutoff:$$f_{c,10}' = \frac{c}{2a} = \frac{3\times10^{8}}{0.05} = 6.00\ \text{GHz}.$$The 5 GHz signal is now below cutoff. Nothing propagates, and the question's word “attenuation” refers to evanescent decay, not to dissipation.
Evanescent decay constant. Below cutoff the propagation constant is purely real:$$\alpha = \sqrt{k_c^{2} - k_0^{2}},\qquad k_c = \frac{\pi}{a} = 125.7\ \text{rad/m},\qquad k_0 = \frac{2\pi f}{c} = 104.7\ \text{rad/m},$$$$\alpha = \sqrt{125.7^{2} - 104.7^{2}} = 69.5\ \text{Np/m}.$$Equivalently $\alpha = k_0\sqrt{(f_c/f)^{2}-1} = 104.7\sqrt{0.44}$, which gives the same figure.
(ii) Convert to decibels per centimetre. One neper is 8.686 dB, so$$\alpha_{\text{dB}} = 69.5 \times 8.686 = 603\ \text{dB/m},$$$$\boxed{\alpha = 6.03\ \text{dB/cm}.}$$A 5 cm length of the emptied guide would therefore knock the signal down by about 30 dB.
Figure 5.2 — With the dielectric removed the field no longer travels; it decays with distance at 6.03 dB/cm, falling to $1/e$ in 1.44 cm.
Check: below cutoff the decay is reactive, not dissipative. The evanescent field stores energy and returns it to the source rather than heating the walls, so the guide behaves as a very good reflector; this is precisely how waveguide-below-cutoff attenuators and microwave oven door screens work. Ohmic wall loss, which the question does not ask about, would add only a few hundredths of a dB/cm.
