22-Elec-A7 Electromagnetics · May 2018

Question 5 of 8: Behaviour Beyond a Dielectric-to-Air Step in a Rectangular Waveguide

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

Notes on this paper

Paper format. National Examinations, May 2018 — 16-Elec-A7, Electromagnetics. Closed book, three hours, approved Casio or Sharp calculator only. Eight questions, all of equal value; the paper states that any five constitute a complete paper, so each question carries 20 of the 100 marks that are actually graded. All eight are solved here, because this set is a study resource rather than an examination script.

Aids given on the paper. $\varepsilon_0 = 8.85\times10^{-12}\ \text{F/m}$, $\mu_0 = 4\pi\times10^{-7}\ \text{H/m}$, $e = 1.6\times10^{-19}\ \text{C}$. Those two constants fix $c = 1/\sqrt{\mu_0\varepsilon_0} = 2.9986\times10^{8}\ \text{m/s}$ and $\eta_0 = \sqrt{\mu_0/\varepsilon_0} = 376.82\ \Omega$ — which is why Question 1 quotes a 377 Ω line: it is free space itself, rendered as a cable.

Reference texts. D. M. Pozar, Microwave Engineering, 4th ed. (transmission-line transients, stubs, waveguide modes); M. N. O. Sadiku, Elements of Electromagnetics, 7th ed. (Biot–Savart, motional EMF, plane waves); W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed.; F. T. Ulaby and U. Ravaioli, Fundamentals of Applied Electromagnetics, 8th ed.; C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed. (short-element fields and the reactive near-field boundary).

Constants used throughout: the paper's own aid list is taken as authoritative, so $c = 2.9986\times10^{8}$ m/s and $\eta_0 = 376.82\ \Omega$. Question 1 supplies its own propagation velocity ($3\times10^{8}$ m/s) and that stated value is used there. Question 5 is the one place where the difference between $2.9986\times10^{8}$ and $3.00\times10^{8}$ changes the wording of the answer, and both readings are worked through explicitly.

Question 5: Behaviour Beyond a Dielectric-to-Air Step in a 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. One guide, two fillings, and a signal frequency that lands remarkably close to the empty guide’s cut-off.

Given data
Quantity	Symbol	Value
Signal frequency	$f$	6 GHz
Broad wall	$a$	2.5 cm
Narrow wall	$b$	1 cm
Filling before the step	$\varepsilon_r$	2.25 (so $n = 1.5$)
Filling after the step	—	air, $\varepsilon_r = 1$
Amplitude ratio sought	—	10 % of the value at the step

Find. The distance into the empty section at which the amplitude has fallen to one tenth of its value at the transition.

The step, and the amplitude that follows it. Emptying the guide multiplies every cut-off by 1.5, which lands TE10 exactly on the signal frequency instead of above it.

Approach. Rank the cut-offs on each side of the step. If the signal is below cut-off in the empty section it decays as $e^{-\alpha z}$ with $\alpha = \sqrt{k_c^{2}-k^{2}}$ and the answer is $\ln 10/\alpha$; if it is not, there is no decay at all and the honest answer says so. Everything therefore turns on comparing $f$ with $f_c$ — carefully, because they very nearly coincide.

Cut-offs in the filled section. With $u = c/\sqrt{\varepsilon_r} = 2.9986\times10^{8}/1.5 = 1.999\times10^{8}$ m/s, $$f_c(\text{TE}_{10}) = \frac{u}{2a} = 4.00\ \text{GHz}, \quad f_c(\text{TE}_{20}) = \frac{u}{a} = 8.00\ \text{GHz}, \quad f_c(\text{TE}_{01}) = \frac{u}{2b} = 10.0\ \text{GHz}.$$ At 6 GHz only TE$_{10}$ propagates, so the wave arriving at the step is a clean single mode.
Cut-off in the empty section. Removing the dielectric multiplies every cut-off by $\sqrt{\varepsilon_r} = 1.5$: $$f_c(\text{TE}_{10}) = \frac{c}{2a} = \frac{2.9986\times10^{8}}{0.050} = 5.997\ \text{GHz}.$$ That is 0.05 % below the 6 GHz signal — and if the rounded $c = 3.00\times10^{8}$ m/s is used instead, it is 6.000 GHz exactly.
Test the propagation constant rather than trusting the ranking. $$k = \frac{2\pi f}{c} = 125.72\ \text{rad/m}, \qquad k_c = \frac{\pi}{a} = 125.66\ \text{rad/m}.$$ Since $k \gt k_c$, the square root $\sqrt{k_c^{2}-k^{2}}$ is imaginary: the mode is propagating, not evanescent, with $$\beta = \sqrt{k^{2}-k_c^{2}} = 3.79\ \text{rad/m}, \qquad \lambda_g = \frac{2\pi}{\beta} = 1.66\ \text{m}.$$
State the attenuation. A lossless guide above cut-off has no attenuation constant at all: $$\boxed{\alpha = 0\ \text{Np/m}} .$$ With the rounded $c = 3.00\times10^{8}$ m/s the guide sits precisely at cut-off, where $\gamma^{2} = 0$ and both $\alpha$ and $\beta$ vanish; the amplitude is then constant along $z$ as well. Either reading gives the same conclusion.
Answer the question as asked. The amplitude beyond the step behaves as $e^{-\alpha z}$ with $\alpha = 0$, so it never falls: $$d_{10\%} = \frac{\ln 10}{\alpha} \to \infty \;\Rightarrow\; \boxed{\text{no finite distance gives 10 \%}} .$$ The 2.5 cm guide is simply not small enough to cut the 6 GHz signal off once the dielectric is gone — it is right on the boundary.
Say what does happen at the step. The TE wave impedance jumps from $\eta_d/\sqrt{1-(f_c/f)^{2}} = 337\ \Omega$ on the filled side to $12.5\ \text{k}\Omega$ on the empty side, so the step behaves almost like an open circuit: $\Gamma = 0.947$ and about 90 % of the incident power is thrown back down the filled section. The 10 % that does cross creeps forward at a group velocity of only $9\times10^{6}$ m/s — but it creeps forward without dying away, which is what the question asked about.
Show what the answer would have been below cut-off. Had the guide been genuinely evanescent, $\alpha = \sqrt{k_c^{2}-k^{2}}$ and $d_{10\%} = \ln 10/\alpha$. The table below evaluates that for the same guide at lower signal frequencies, and shows how brutally short the distances are — a few centimetres — which is the physical point the question is driving at.

What the 10 % distance would be if the signal were below the empty guide’s 5.997 GHz cut-off
Signal frequency	$\alpha$ (Np/m)	Attenuation (dB/cm)	Distance to 10 %
5.5 GHz	50.1	4.35	4.60 cm
5.0 GHz	69.4	6.03	3.32 cm
4.0 GHz	93.6	8.13	2.46 cm

Check: this answer sits on a knife edge and the wording depends on the value taken for $c$. Using the paper’s own aids ($\varepsilon_0 = 8.85\times10^{-12}$, $\mu_0 = 4\pi\times10^{-7}$) gives $c = 2.9986\times10^{8}$ m/s and an empty-guide cut-off of 5.997 GHz, so the 6 GHz signal propagates with $\lambda_g = 1.66$ m. Using the rounded $3.00\times10^{8}$ m/s puts the cut-off at exactly 6.000 GHz, where $\beta = \alpha = 0$. Both readings give $\alpha = 0$, so the conclusion — no finite 10 % distance — is robust; only the description of the mode (barely propagating versus exactly at cut-off) changes. A guide just 0.05 % narrower, $a = 2.4988$ cm, would tip it into evanescence.

Final results
Quantity asked	Result
$f_c$(TE$_{10}$), filled section	4.00 GHz
Modes propagating in the filled section at 6 GHz	TE$_{10}$ only
$f_c$(TE$_{10}$), empty section	5.997 GHz (6.000 GHz with $c = 3\times10^{8}$)
Attenuation constant in the empty section	$\alpha = 0$
Distance for the amplitude to reach 10 %	none exists — the wave does not decay
Guide wavelength beyond the step	1.66 m ($\beta = 3.79$ rad/m)
Reflection coefficient at the step	0.947 (about 90 % of the power reflected)
Formula that would apply below cut-off	$d_{10\%} = \ln 10/\sqrt{k_c^{2}-k^{2}}$

Topic. Cut-off, evanescence, and what changing the filling does to a guide.

Key relations. $f_c(\text{TE}_{mn}) = \tfrac{u}{2}\sqrt{(m/a)^{2}+(n/b)^{2}}$ with $u = c/\sqrt{\varepsilon_r}$; $\gamma = \sqrt{k_c^{2}-k^{2}}$, real ($=\alpha$) below cut-off and imaginary ($=j\beta$) above; $d_{10\%} = \ln 10/\alpha$; $Z_{\text{TE}} = \eta/\sqrt{1-(f_c/f)^{2}}$.

Why this works. A guide below cut-off does not absorb the signal — it refuses to carry it. The transverse boundary conditions demand a minimum transverse wavenumber $k_c = \pi/a$, and if the total $k$ available is smaller, the axial constant must come out imaginary, giving a field that decays with distance while carrying no power at all. Removing a dielectric raises $k_c$ relative to $k$ by $\sqrt{\varepsilon_r}$, which is why an innocuous-looking change of filling can switch a guide from transparent to opaque. Here it lands the guide exactly on that boundary, so the correct answer is a statement about the boundary rather than a number.

Common pitfall. Computing $\alpha$ from $\sqrt{k_c^{2}-k^{2}}$ without first checking the sign of the bracket, which silently returns a real number from a negative argument in many calculators and produces a confident, wrong distance. The second pitfall is treating this as a loss problem — conductor or dielectric loss — when evanescent decay is purely reactive and has nothing to do with either conductivity or loss tangent.

Source. Pozar, Microwave Engineering, 4th ed., §3.3 (rectangular waveguide, cut-off and evanescent modes); Hayt and Buck, Engineering Electromagnetics, 9th ed., §13.3.

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