Question 5 of 8: Propagating modes and longest guide wavelength at 15 GHz
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2014 — 07-Elec-A7 Electromagnetics. Three hours, closed book; one of two approved calculators permitted. Eight questions, all of equal value (20 marks each). The rubric states that any five questions constitute a complete paper and that only the first five presented are marked — all eight are worked here, because the set is a study resource rather than a sat examination. Aids printed on the paper: $\varepsilon_0 = 8.85\times10^{-12}\ \text{F/m}$ and $\mu_0 = 4\pi\times10^{-7}\ \text{H/m}$, from which $\eta_0=\sqrt{\mu_0/\varepsilon_0}\approx 377\ \Omega$ and $c = 3\times10^{8}\ \text{m/s}$.
Reference texts. D. M. Pozar, Microwave Engineering, 4th ed. (transmission lines, waveguides); M. N. O. Sadiku, Elements of Electromagnetics, 7th ed.; W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed. (statics, Faraday's law, plane waves); F. T. Ulaby, Fundamentals of Applied Electromagnetics, 8th ed. (polarisation, radiation); C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed. (short elements, image theory).
Question 5: Propagating modes and longest guide wavelength at 15 GHz (20 marks)
Given. A hollow rectangular guide with air filling, operated at a single frequency; the broad wall is the larger dimension.
Given data
Quantity
Symbol
Value
Broad-wall dimension
$a$
$2.25\ \text{cm}$
Narrow-wall dimension
$b$
$1.00\ \text{cm}$
Filling
$\varepsilon_r,\ \mu_r$
$1,\ 1$ (air)
Operating frequency
$f$
$15\ \text{GHz}$
Free-space wavelength at $f$
$\lambda_0$
$2.00\ \text{cm}$
Find. Which modes propagate at 15 GHz, and the longest guide wavelength among the propagating set.
Guide cross-section with the TE$_{10}$ field pattern, and the cutoff ladder against the 15 GHz operating line. TE$_{01}$ lands exactly on the operating frequency.
Approach. Compute every cutoff frequency from $f_{c,mn}=(c/2)\sqrt{(m/a)^{2}+(n/b)^{2}}$, keep the modes whose cutoff is strictly below 15 GHz, then evaluate $\lambda_g$ for each survivor — the guide wavelength grows without limit as $f_c$ approaches $f$, so the highest-cutoff propagating mode wins.
Cutoff formula for an air-filled guide. $$f_{c,mn}=\frac{c}{2}\sqrt{\left(\frac{m}{a}\right)^{2}+\left(\frac{n}{b}\right)^{2}},\qquad a=0.0225\ \text{m},\ b=0.0100\ \text{m}.$$
Evaluate the low-order modes. $f_{c,10}=c/2a=6.667$ GHz; $f_{c,20}=c/a=13.333$ GHz; $f_{c,01}=c/2b=15.000$ GHz; $f_{c,11}=16.415$ GHz (TE and TM alike); $f_{c,30}=20.000$ GHz; $f_{c,21}=20.074$ GHz. Note that TM modes require both indices non-zero, so the lowest TM mode is TM$_{11}$ at 16.415 GHz.
Select the propagating set. A mode propagates only when $f\gt f_c$. Here $f_{c,01}=15.000\ \text{GHz}=f$ exactly, so TE$_{01}$ sits at cutoff: $\beta=\sqrt{k_0^{2}-k_c^{2}}=0$, its group velocity vanishes and it carries no power. It is therefore excluded, leaving $$\boxed{\text{only TE}_{10}\ \text{and TE}_{20}\ \text{propagate at 15 GHz}} .$$
Guide wavelength of each survivor. $$\lambda_g=\frac{\lambda_0}{\sqrt{1-\left(f_c/f\right)^{2}}},\qquad \lambda_0=\frac{c}{f}=2.00\ \text{cm}.$$ For TE$_{10}$: $\sqrt{1-(6.667/15)^{2}}=0.8958$, so $\lambda_g=2.233$ cm. For TE$_{20}$: $\sqrt{1-(13.333/15)^{2}}=0.4581$, so $\lambda_g=4.366$ cm.
Identify the longest. Since $\lambda_g$ increases as $f_c\to f$, the mode nearest cutoff has the longest guide wavelength: $$\boxed{\lambda_{g,max}=\lambda_g(\text{TE}_{20})=4.37\ \text{cm}} .$$
Cross-check through the phase constant. For TE$_{20}$, $k_c=2\pi/a=279.3\ \text{rad/m}$ and $k_0=2\pi/\lambda_0=314.2\ \text{rad/m}$, giving $\beta=\sqrt{k_0^{2}-k_c^{2}}=143.9\ \text{rad/m}$ and $\lambda_g=2\pi/\beta=4.366$ cm, as above.
Two features of this guide deserve comment. First, the aspect ratio $a/b=2.25$ exceeds two, so the second broad-wall mode TE$_{20}$ arrives before TE$_{01}$ and closes the single-mode band; the useful single-mode range is therefore 6.67 GHz to 13.33 GHz, and at 15 GHz the guide is unavoidably overmoded. Second, the guide wavelength always exceeds the free-space wavelength, and diverges at cutoff; a mode operated close to its cutoff has a long guide wavelength but also a low group velocity and a high sensitivity to dimensional tolerance, which is why practical components are not run there.
Check: TE$_{01}$’s cutoff, $c/2b=15.000$ GHz, coincides with the stated operating frequency to the full precision of the given dimensions. It is treated here as non-propagating ($\beta=0$), which is the correct limiting reading; note that a narrow-wall dimension only 0.1 % larger than 1.00 cm would make it marginally propagating with a very long guide wavelength, so the answer is deliberately stated with that boundary case identified rather than hidden. The same sensitivity applies to the constants: the paper's own aids, $c=1/\sqrt{\mu_0\varepsilon_0}=2.9986\times10^{8}$ m/s, put $f_{c,01}$ at 14.993 GHz, 7 MHz below the signal, so TE$_{01}$ would formally propagate with $\lambda_g\approx 66$ cm. The round numbers of the question (1 cm against 15 GHz) show the setter intended $c=3\times10^{8}$ m/s and the at-cutoff reading, which is the answer given; a candidate taking the aids literally should state that assumption and quote the TE$_{01}$ case as well.
