22-Elec-A7 Electromagnetics · December 2018

Question 3 of 8: The two lowest propagating modes at 15 GHz and their velocities

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

Notes on this paper

Paper format. National Examinations, December 2018, 16-Elec-A7 Electromagnetics. Three hours, closed book, one approved Casio or Sharp calculator. Eight questions, all of equal value; the rubric says any five constitute a complete paper, so a candidate answers five — but all eight are worked here, because the set is a study resource. Aids printed on the paper: $\varepsilon_0 = 8.85\times10^{-12}$ F/m, $\mu_0 = 4\pi\times10^{-7}$ H/m, $e = 1.6\times10^{-19}$ C.

Reference texts. M. N. O. Sadiku, Elements of Electromagnetics, 7th ed.; W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed.; F. T. Ulaby and U. Ravaioli, Fundamentals of Applied Electromagnetics, 8th ed.; D. M. Pozar, Microwave Engineering, 4th ed.; C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed.

Constants used throughout. Question 6 hands us $\omega = 2\pi\times10^{10}$ rad/s together with $k = (2\pi/3)\ \text{cm}^{-1}$, so the setter’s phase velocity is $\omega/k = 3.000\times10^{8}$ m/s exactly; Question 3’s cut-off arithmetic only closes on round numbers under the same choice. Accordingly $c = 3\times10^{8}$ m/s and $\eta_0 = 120\pi = 376.99\ \Omega$ are used here. The paper’s own aid list would give $c = 2.99863\times10^{8}$ m/s and $\eta_0 = 376.83\ \Omega$ — a 0.05 % shift that changes no answer except at one deliberate boundary case, flagged in Question 3.

Question 3: The two lowest propagating modes at 15 GHz and their velocities (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 the standard two-to-one-ish aspect ratio, excited well above its dominant cut-off.

Given data
Quantity	Symbol	Value
Broad inside dimension	$a$	2.5 cm
Narrow inside dimension	$b$	1.0 cm
Filling	—	air, $\varepsilon_r = 1$
Signal frequency	$f$	15 GHz
Free-space velocity	$c$	$3\times10^{8}$ m/s

Find. Which two modes actually propagate at 15 GHz, and the velocity at which each carries the signal.

Every cut-off below 20 GHz, ranked against the 15 GHz signal. TE01 lands on the operating frequency itself, so only two modes are left inside the propagating band.

Approach. Rank the cut-off frequencies of every candidate mode, keep only those strictly below 15 GHz, then evaluate the group and phase velocities from the usual dispersion factor.

Cut-off frequencies. For a rectangular guide, $$f_{c,mn} = \frac{c}{2}\sqrt{\left(\frac{m}{a}\right)^{2} + \left(\frac{n}{b}\right)^{2}} .$$ With $a = 0.025$ m and $b = 0.010$ m this gives, in order: $f_c(\text{TE}_{10}) = c/2a = 6.000$ GHz, $f_c(\text{TE}_{20}) = c/a = 12.000$ GHz, $f_c(\text{TE}_{01}) = c/2b = 15.000$ GHz, $f_c(\text{TE}_{11}) = f_c(\text{TM}_{11}) = 16.155$ GHz, and $f_c(\text{TE}_{30}) = 3c/2a = 18.000$ GHz.
TE01 is excluded, and that is the point of the question. Its cut-off falls exactly on the signal, so $$\beta = \sqrt{k^{2} - k_c^{2}} = 0 \quad\Rightarrow\quad v_g = 0, \qquad \lambda_g \to \infty .$$ A mode at its own cut-off stores energy but transports none along the guide, so it is not a propagating mode. Ranking the numbers to three decimals rather than reciting the textbook order is what catches this.
The surviving set. Only $\text{TE}_{10}$ and $\text{TE}_{20}$ have cut-offs strictly below 15 GHz, so the answer to the first half is $$\boxed{\text{TE}_{10}\ (f_c = 6\ \text{GHz}) \ \text{and}\ \text{TE}_{20}\ (f_c = 12\ \text{GHz})} .$$ Note the aspect ratio at work: because $a \gt 2b$, it is the second broad-wall harmonic $\text{TE}_{20}$ rather than $\text{TE}_{01}$ that closes the single-mode band.
Velocities for TE10. The dispersion factor is $\sqrt{1 - (f_c/f)^{2}} = \sqrt{1 - (6/15)^{2}} = 0.9165$, hence $$v_g = c\sqrt{1 - (f_c/f)^{2}} = 2.750\times10^{8}\ \text{m/s}, \qquad v_p = \frac{c}{\sqrt{1 - (f_c/f)^{2}}} = 3.273\times10^{8}\ \text{m/s},$$ with guide wavelength $\lambda_g = v_p/f = 2.182$ cm.
Velocities for TE20. Here $\sqrt{1 - (12/15)^{2}} = 0.600$, so $$v_g = 1.800\times10^{8}\ \text{m/s}, \qquad v_p = 5.000\times10^{8}\ \text{m/s}, \qquad \lambda_g = 3.333\ \text{cm}.$$ Both satisfy the identity $v_g v_p = c^{2}$, which is the quickest check that neither has been inverted; and the superluminal phase velocity is no paradox, since it is $v_g$ that carries the signal.
Which velocity does the question want? ‘Propagation velocity’ for a signal is the group velocity, so the working answer is $$\boxed{v_g(\text{TE}_{10}) = 2.750\times10^{8}\ \text{m/s}, \qquad v_g(\text{TE}_{20}) = 1.800\times10^{8}\ \text{m/s}} ,$$ with the phase velocities quoted alongside because the higher mode is markedly more dispersive — a 15 GHz pulse split between the two arrives smeared by the difference in $v_g$.

Check: TE01 sits on a knife edge. Its cut-off is $c/2b$, which is exactly 15.000 GHz when $c$ is taken as $3\times10^{8}$ m/s — the value the rest of this paper's arithmetic implies. Using instead the constants in the paper's own aid list ($\varepsilon_0 = 8.85\times10^{-12}$ F/m, giving $c = 2.99863\times10^{8}$ m/s) drops the cut-off to 14.993 GHz, marginally below the signal, so TE01 would technically propagate — but at $v_g = 9.05\times10^{6}$ m/s, some thirty times slower than TE10, and with $\lambda_g = 0.66$ m. Either reading gives the same practical answer: TE10 and TE20 are the two usable modes. The other four cut-offs shift by under 0.05 % between the two constants, so only this boundary mode is at risk.

Final results
Quantity asked	Result
Two lowest propagating modes	$\text{TE}_{10}$ and $\text{TE}_{20}$
TE10 cut-off and group velocity	6.000 GHz; $v_g = 2.750\times10^{8}$ m/s
TE10 phase velocity and guide wavelength	$v_p = 3.273\times10^{8}$ m/s; $\lambda_g = 2.182$ cm
TE20 cut-off and group velocity	12.000 GHz; $v_g = 1.800\times10^{8}$ m/s
TE20 phase velocity and guide wavelength	$v_p = 5.000\times10^{8}$ m/s; $\lambda_g = 3.333$ cm
TE01	cut-off 15.000 GHz — at the signal, $\beta = 0$, excluded

Topic. Modal cut-off and dispersion in a rectangular waveguide.

Key relations. $f_{c,mn} = \tfrac{c}{2}\sqrt{(m/a)^{2} + (n/b)^{2}}$; $\;\beta = \sqrt{k^{2} - k_c^{2}}$; $\;v_g = c\sqrt{1 - (f_c/f)^{2}}$; $\;v_p = c/\sqrt{1 - (f_c/f)^{2}}$; $\;v_g v_p = c^{2}$; $\;\lambda_g = v_p/f$.

Why this works. A hollow guide supports a mode only when the transverse resonance it represents fits inside the cross-section at the operating frequency. Below cut-off the axial wavenumber is imaginary and the field decays without transporting power; above it the mode propagates, and the further above cut-off it sits the closer its velocities come to $c$. The two velocities are reciprocal about $c$ precisely because the wave can be pictured as a plane wave zig-zagging between the broad walls: the group velocity is its axial progress, the phase velocity the sweep speed of its wavefront intersection with the wall.

Common pitfall. Two habits cause trouble. One is assuming TE01 is always the second mode; that holds only when $a \lt 2b$, and here it is the third. The other is reading ‘propagation velocity’ as the phase velocity and reporting $5\times10^{8}$ m/s for TE20 without comment, which invites the reader to conclude that a signal outruns light.

Source. Pozar, Microwave Engineering, 4th ed., §3.3 (rectangular waveguide); Sadiku, Elements of Electromagnetics, 7th ed., §12.2–12.4.

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