Question 4 of 8: Shortest and Longest Guide Wavelengths in a Rectangular Waveguide
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
98-Phys-A3, Electromagnetics — National Exam, May 2016. 3-hour closed-book exam (Casio or Sharp approved calculators only); any FIVE of the eight questions constitute a complete paper and only the first five as they appear in a candidate's answer book are marked, each of equal value. Aids given on the paper: ε0 = 8.85×10-12 F/m, μ0 = 4π×10-7 H/m. All eight printed questions are solved below as a complete study resource.
Reference texts: Sadiku, Elements of Electromagnetics (7th ed.) — transmission lines, waveguides, plane waves, antennas; Hayt & Buck, Engineering Electromagnetics (9th ed.) — transmission-line transients; Pozar, Microwave Engineering (4th ed.) — Smith-chart-free impedance transformation and waveguide cutoff; Balanis, Antenna Theory (4th ed.) — short-dipole far field.
Question 4: Shortest and Longest Guide Wavelengths in a Rectangular Waveguide (equal value)
Check: the source prints the frequency as "20 GHz (2×108 Hz)" — the parenthetical is an internally-inconsistent typographical slip in the printed paper (20 GHz = $2\times10^{10}$ Hz, not $2\times10^{8}$ Hz = 200 MHz). The headline "20 GHz" is used throughout.
Given. Waveguide inner dimensions $a=2.25$ cm (broad wall), $b=1$ cm; operating frequency $f=20$ GHz; free space inside the guide ($c=3\times10^8$ m/s).
Given data
Quantity
Symbol
Value
Broad-wall dimension
$a$
2.25 cm
Narrow-wall dimension
$b$
1 cm
Operating frequency
$f$
20 GHz
Free-space wavelength
$\lambda_0=c/f$
1.5 cm
Find. Among all TE/TM$_{mn}$ modes that propagate at 20 GHz, the shortest and longest guide wavelengths $\lambda_g$.
Approach. Compute the cutoff frequency $f_{c,mn}=\dfrac{c}{2}\sqrt{(m/a)^2+(n/b)^2}$ for the low-order modes, keep only those with $f_{c,mn} < f$, then rank their guide wavelengths $\lambda_g=\lambda_0/\sqrt{1-(f_{c,mn}/f)^2}$ — the lowest cutoff gives the shortest $\lambda_g$, the highest sub-cutoff mode gives the longest.
Cutoff frequencies of the candidate low-order modes.
$$f_{c,10}=\frac{c}{2a}=6.667\ \text{GHz},\quad f_{c,20}=\frac{c}{a}=13.33\ \text{GHz},\quad f_{c,01}=\frac{c}{2b}=15.0\ \text{GHz},$$
$$f_{c,11}=f_{c,TM11}=\frac{c}{2}\sqrt{\left(\frac{1}{a}\right)^2+\left(\frac{1}{b}\right)^2}=16.41\ \text{GHz},\quad f_{c,30}=\frac{3c}{2a}=20.0\ \text{GHz}.$$
Checking the next candidate, TE$_{21}$: $f_{c,21}=\dfrac{c}{2}\sqrt{(2/a)^2+(1/b)^2}=20.07$ GHz.
Select the propagating modes at $f=20$ GHz. A mode propagates only if $f_{c,mn} < f$ strictly. TE$_{30}$ sits exactly at $f_{c}=20.0$ GHz (cutoff, evanescent, not propagating) and TE$_{21}$ at $20.07$ GHz is just above cutoff, so neither propagates. The propagating set is
$$\{\text{TE}_{10},\ \text{TE}_{20},\ \text{TE}_{01},\ \text{TE}_{11}/\text{TM}_{11}\},$$
with TE$_{10}$ the lowest cutoff (dominant mode) and TE$_{11}$/TM$_{11}$ the highest cutoff that still propagates.
Guide wavelength of the dominant mode (shortest $\lambda_g$). A lower cutoff frequency, being further from $f$, gives a guide wavelength closer to $\lambda_0$ — the smallest $\lambda_g$ of the set:
$$\lambda_{g,10}=\frac{\lambda_0}{\sqrt{1-(f_{c,10}/f)^2}}=\frac{1.5\ \text{cm}}{\sqrt{1-(6.667/20)^2}}=\boxed{1.591\ \text{cm}}.$$
Guide wavelength of the highest sub-cutoff mode (longest $\lambda_g$). TE$_{11}$/TM$_{11}$, being closest to cutoff among the propagating modes, gives the largest $\lambda_g$:
$$\lambda_{g,11}=\frac{\lambda_0}{\sqrt{1-(f_{c,11}/f)^2}}=\frac{1.5\ \text{cm}}{\sqrt{1-(16.41/20)^2}}=\boxed{2.626\ \text{cm}}.$$
Waveguide cross-section, 2.25 cm × 1 cm; TE10 gives the shortest guide wavelength (1.591 cm), TE11/TM11 the longest among modes propagating at 20 GHz (2.626 cm).
