22-Elec-A7 Electromagnetics · May 2013

Question 4 of 8: Attenuation of a Below-Cutoff Rectangular Waveguide

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

Notes on this paper

Paper format. National Examinations, May 2013 — 07-Elec-A7, Electromagnetics. Three hours, closed book; one of two approved calculators (Casio or Sharp) permitted. Eight questions, all of equal value; any five constitute a complete paper and only the first five presented are marked. All eight are solved here, since the set is a study resource. Aids printed on the cover page: ε₀ = 8.85 × 10⁻¹² F/m and μ₀ = 4π × 10⁻⁷ H/m; Question 2 additionally supplies the quarter-wave relation.

Reference texts (22-Elec-A7 Electromagnetics).

D. M. Pozar, Microwave Engineering, 4th ed. — transmission lines, matching, waveguides.
M. N. O. Sadiku, Elements of Electromagnetics, 7th ed. — plane waves, guided waves, radiation.
W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed. — field theory and transients on lines.
C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed. — short elements, image theory.
F. T. Ulaby and U. Ravaioli, Fundamentals of Applied Electromagnetics, 8th ed. — oblique incidence, surface currents.
S. J. Chapman, Electric Machinery Fundamentals, 5th ed. — rotating-loop induction and torque (Question 3).

Conventions used throughout. The paper itself quotes propagation velocities as $3\times10^{8}\ \text{m/s}$, so free space is taken as $c=3.00\times10^{8}\ \text{m/s}$ and the intrinsic impedance as $\eta_0=\mu_0 c=120\pi=376.99\ \Omega$. Lines are treated as lossless, and "rms" is stated explicitly wherever the exam asks for it.

Question 4: Attenuation of a Below-Cutoff 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. An air-filled rectangular guide of internal cross-section 25 mm by 15 mm is excited at 5 GHz, and the attenuation of its lowest-order mode is required.

Given data
Quantity	Symbol	Value
Broad inside dimension	$a$	25 mm
Narrow inside dimension	$b$	15 mm
Filling	—	air ($\varepsilon_r=1$, $c=3\times10^{8}$ m/s)
Signal frequency	$f$	5.00 GHz
Mode of interest	—	lowest (dominant) mode

Find. The attenuation constant of the dominant mode at 5 GHz, expressed in dB/cm.

Figure 4.1 — Guide cross-section with the TE₁₀ transverse electric field: a half sine across the 25 mm dimension, uniform across the 15 mm dimension.

Approach. Identify the dominant mode from the cutoff frequencies, confirm that 5 GHz lies below that cutoff so the propagation constant is purely real, then evaluate the evanescent attenuation constant and convert nepers to decibels.

Rank the modes by cutoff. For an air-filled guide $f_{c,mn}=\frac{c}{2}\sqrt{(m/a)^{2}+(n/b)^{2}}$, so $$f_{c,10}=\frac{c}{2a}=\frac{3\times10^{8}}{2\times0.025}=6.00\ \text{GHz},\qquad f_{c,01}=\frac{c}{2b}=10.0\ \text{GHz},\qquad f_{c,20}=\frac{c}{a}=12.0\ \text{GHz}$$ The lowest mode is therefore TE₁₀, with a 6.00 GHz cutoff.
Establish that the signal is cut off. Since $f=5.00\ \text{GHz}\lt f_{c,10}=6.00\ \text{GHz}$, the axial propagation constant $$\gamma=\sqrt{k_c^{2}-k^{2}},\qquad k_c=\frac{\pi}{a},\qquad k=\frac{2\pi f}{c}$$ is purely real: the field does not propagate but decays exponentially as $e^{-\alpha z}$. This is a reactive cutoff, not a dissipative loss.
Evaluate the two wavenumbers. $$k_c=\frac{\pi}{0.025}=125.66\ \text{rad/m},\qquad k=\frac{2\pi\times5.00\times10^{9}}{3\times10^{8}}=104.72\ \text{rad/m}$$
Compute the attenuation constant. $$\alpha=\sqrt{k_c^{2}-k^{2}}=\sqrt{15791-10966}=\sqrt{4825}=\boxed{69.46\ \text{Np/m}}$$ The equivalent normalised form gives the same value as a useful check: $$\alpha=\frac{2\pi f_c}{c}\sqrt{1-\left(\frac{f}{f_c}\right)^{2}}=125.66\times\sqrt{1-\left(\tfrac{5}{6}\right)^{2}}=125.66\times0.5528=69.46\ \text{Np/m}$$
Convert nepers to decibels per centimetre. One neper of field decay is $20\log_{10}e=8.686$ dB, so $$\alpha_{dB}=69.46\times8.686=603.2\ \text{dB/m}=\boxed{6.03\ \text{dB/cm}}$$

Physically this is an enormous rate: the field falls to half its value in about 1.2 mm of guide, and a 5 cm length would provide roughly 30 dB of isolation. Below-cutoff guides are used deliberately for exactly this reason — as calibrated piston attenuators and as waveguide-beyond-cutoff feed-throughs in screened rooms — because the attenuation depends only on the geometry and is therefore extremely stable.

Final results — Question 4
Quantity	Value
Dominant mode	TE₁₀
Cutoff frequency of TE₁₀	6.00 GHz (next modes 10.0 and 12.0 GHz)
Cutoff wavenumber $k_c=\pi/a$	125.66 rad/m
Free-space wavenumber at 5 GHz	104.72 rad/m
Attenuation constant	69.46 Np/m
Attenuation of the lowest mode at 5 GHz	6.03 dB/cm (603 dB/m)

Topic. Waveguide cutoff and evanescent (below-cutoff) attenuation.

Key relations. $f_{c,mn}=\frac{c}{2}\sqrt{(m/a)^{2}+(n/b)^{2}}$; $\gamma=\sqrt{k_c^{2}-k^{2}}$, real below cutoff and imaginary above; $\alpha=(2\pi f_c/c)\sqrt{1-(f/f_c)^{2}}$; 1 Np = 8.686 dB.

Why this works. A rectangular guide can only support a field pattern whose transverse variation fits the boundary conditions, and that pattern carries a fixed transverse wavenumber $k_c$. The axial wavenumber is whatever is left over from the total $k$, through $k^{2}=k_c^{2}+\beta^{2}$. When the frequency is too low, $k$ cannot supply the required transverse wavenumber, the remainder turns negative, and $\beta$ becomes an imaginary quantity — that is, a real attenuation constant. The energy is not absorbed; it is stored and reflected, which is why an ideal below-cutoff guide attenuates without dissipating and why the answer is independent of wall conductivity. Because $\alpha$ approaches $k_c$ as the frequency falls well below cutoff, the maximum attenuation obtainable is set purely by the guide width.

Common pitfall. Reaching for a conductor-loss or dielectric-loss formula and reporting a fraction of a dB per metre. The other frequent slip is choosing the wrong dominant mode: with $a\gt b$ the dominant mode is always TE₁₀ across the broad wall, so the cutoff uses 25 mm, not 15 mm.

Source. Pozar, Microwave Engineering, 4th ed., §3.3 (rectangular waveguide, TE₁₀ and cutoff); Sadiku, Elements of Electromagnetics, 7th ed., Ch. 12.

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