Mode count in a 2.5 cm × 1 cm guide at 20 GHz: 22-Elec-A7 Electromagnetics, December 2015, Q4

Question 4: Mode count in a 2.5 cm × 1 cm guide at 20 GHz (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 waveguide of stated internal cross-section, excited at a single frequency well above its dominant cutoff.

Given data
Quantity	Symbol	Value
Broad wall (internal)	$a$	$2.5\ \text{cm}$
Narrow wall (internal)	$b$	$1.0\ \text{cm}$
Signal frequency	$f$	$20\ \text{GHz}$
Filling	—	air, $u = c = 3.00\times10^{8}\ \text{m/s}$

Find. How many modes can propagate at 20 GHz.

Figure Q4. Guide cross-section with the dominant TE10 field pattern, and the cutoff ladder against the 20 GHz operating line. Everything below the red line propagates.

Approach. Compute the cutoff frequency of every candidate index pair from the one universal formula, keep those below 20 GHz, and remember that every index pair with both indices non-zero supports a TM mode as well as a TE mode.

Write the cutoff formula. For a rectangular guide of internal dimensions $a\times b$ filled with a medium of wave speed $u$, $$f_{c,mn} = \frac{u}{2}\sqrt{\left(\frac{m}{a}\right)^{2} + \left(\frac{n}{b}\right)^{2}}$$ The same expression serves TE and TM: the cutoff depends only on the transverse eigenvalue, not on which field component is longitudinal.
Fix the scaling constants. With $a = 0.025\ \text{m}$ and $b = 0.010\ \text{m}$, $$\frac{u}{2a} = 6.00\ \text{GHz}, \qquad \frac{u}{2b} = 15.00\ \text{GHz}$$ so every cutoff is $f_{c,mn} = \sqrt{(6.00m)^2 + (15.00n)^2}\ \text{GHz}$, and the search can stop as soon as $6.00m$ or $15.00n$ alone exceeds 20 GHz.
Enumerate the candidates. Running $m = 0\ldots4$ and $n = 0\ldots1$ gives, in ascending order: $$f_c(\mathrm{TE}_{10}) = 6.00,\quad f_c(\mathrm{TE}_{20}) = 12.00,\quad f_c(\mathrm{TE}_{01}) = 15.00,\quad f_c(\mathrm{TE}_{11}/\mathrm{TM}_{11}) = 16.15$$ $$f_c(\mathrm{TE}_{30}) = 18.00,\quad f_c(\mathrm{TE}_{21}/\mathrm{TM}_{21}) = 19.20\ \text{GHz}$$ all in GHz. The next two candidates fall just outside: $f_c(\mathrm{TE}_{31}) = 23.42$ and $f_c(\mathrm{TE}_{40}) = 23.99\ \text{GHz}$, both above 20 GHz, so the list is complete. ($\mathrm{TE}_{02}$ would need 30 GHz.)
Count TE and TM separately. A mode with $m=0$ or $n=0$ exists only as a TE mode — the corresponding TM field would vanish identically — while every pair with $m\ge 1$ and $n\ge 1$ supports a degenerate TE/TM pair. So the propagating set is $$\underbrace{\mathrm{TE}_{10},\ \mathrm{TE}_{20},\ \mathrm{TE}_{01},\ \mathrm{TE}_{11},\ \mathrm{TE}_{30},\ \mathrm{TE}_{21}}_{6\ \text{TE}} \;+\; \underbrace{\mathrm{TM}_{11},\ \mathrm{TM}_{21}}_{2\ \text{TM}}$$
State the count. Adding the two families, $$\boxed{8\ \text{modes propagate at }20\ \text{GHz}}$$ six TE and two TM. The guide is being operated far into its over-moded region: single-mode operation would require 6.00 GHz $\lt f \lt$ 12.00 GHz, the octave between $\mathrm{TE}_{10}$ and $\mathrm{TE}_{20}$.
Sanity check the dominant mode. At 20 GHz the $\mathrm{TE}_{10}$ mode has $$\beta = \frac{2\pi f}{c}\sqrt{1-\left(\frac{f_c}{f}\right)^2} \;\Longrightarrow\; \lambda_g = \frac{2\pi}{\beta} = 15.7\ \text{mm}$$ against a free-space wavelength of 15.0 mm, the expected modest stretching well above cutoff. A guide wavelength shorter than the free-space one would signal an arithmetic error.

The practical consequence is that this guide cannot be used at 20 GHz for anything requiring a clean single mode. Any bend, iris or misalignment converts power between the eight modes, and because each travels with its own guide wavelength the recombined signal disperses.

Final results
Quantity	Value
$\mathrm{TE}_{10}$	$6.00\ \text{GHz}$ — propagates
$\mathrm{TE}_{20}$	$12.00\ \text{GHz}$ — propagates
$\mathrm{TE}_{01}$	$15.00\ \text{GHz}$ — propagates
$\mathrm{TE}_{11}$, $\mathrm{TM}_{11}$	$16.15\ \text{GHz}$ — both propagate
$\mathrm{TE}_{30}$	$18.00\ \text{GHz}$ — propagates
$\mathrm{TE}_{21}$, $\mathrm{TM}_{21}$	$19.20\ \text{GHz}$ — both propagate
$\mathrm{TE}_{31}$, $\mathrm{TE}_{40}$	$23.42$, $23.99\ \text{GHz}$ — cut off
Total propagating at 20 GHz	8 modes (6 TE + 2 TM)

Topic. Modal spectrum of a rectangular waveguide and the counting of propagating modes.

Key relations. $f_{c,mn} = (u/2)\sqrt{(m/a)^2+(n/b)^2}$ with $u = c/\sqrt{\mu_r\varepsilon_r}$; TE exists for all $(m,n)\ne(0,0)$, TM only for $m,n\ge 1$; $\beta=\sqrt{k^2-k_c^2}$ and $\lambda_g = 2\pi/\beta$; single-mode band $u/2a$ to $\min(u/a,\ u/2b)$.

Why this works. The guide walls impose a transverse standing wave whose half-wavelengths must fit an integer number of times across each dimension. That quantises the transverse wavenumber, and the longitudinal wavenumber is whatever is left of $k^2$ after the transverse part is removed. If the transverse eigenvalue already exceeds $k$, nothing is left, $\beta$ becomes imaginary, and the mode decays instead of travelling. Cutoff is therefore purely geometric, which is why one formula covers both TE and TM.

Common pitfall. Forgetting the TM partners of the $m,n\ge1$ modes and reporting 6 rather than 8; or assuming $\mathrm{TE}_{01}$ is the second mode when $a \gt 2b$ makes $\mathrm{TE}_{20}$ lower. Always rank the numbers rather than trusting the usual textbook ordering, which assumes a 2:1 aspect ratio.

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

Question 4 of 8: Mode count in a 2.5 cm × 1 cm guide at 20 GHz

Question 4: Mode count in a 2.5 cm × 1 cm guide at 20 GHz (20 marks)