Single-mode bandwidth of an empty rectangular waveguide: 22-Elec-A7 Electromagnetics, May 2015, Q5

Question 5: Single-mode bandwidth of an empty 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 2.5 cm by 1 cm, so $a = 2.5\ \text{cm}$ (broad wall) and $b = 1\ \text{cm}$.

Given data
Quantity	Symbol	Value
Broad internal dimension	$a$	$2.5\ \text{cm} = 0.025\ \text{m}$
Narrow internal dimension	$b$	$1\ \text{cm} = 0.010\ \text{m}$
Filling	—	air (vacuum), $\varepsilon_r = \mu_r = 1$
Wave speed in the filling	$u$	$3\times10^{8}\ \text{m/s}$
Aspect ratio	$a/b$	$2.5$

Find. The band of frequencies over which exactly one mode propagates — that is, from the dominant-mode cutoff up to the cutoff of the next mode.

Figure 5.1 — cutoff ladder for the 2.5 cm × 1 cm empty guide. TE₂₀, not TE₀₁, closes the band.

Approach. Compute the cutoff frequency of every low-order mode from the general formula, rank them, and take the interval between the lowest and the second-lowest.

General cutoff formula. For a hollow rectangular guide, the TE$_{mn}$ and TM$_{mn}$ modes cut off at $$f_{c,mn} = \frac{u}{2}\sqrt{\left(\frac{m}{a}\right)^{2} + \left(\frac{n}{b}\right)^{2}}, \qquad u = \frac{c}{\sqrt{\varepsilon_r\mu_r}} = 3\times10^{8}\ \text{m/s}$$ TM modes need both indices non-zero, so the lowest few candidates are TE$_{10}$, TE$_{20}$, TE$_{01}$ and TE$_{11}$.
Dominant mode TE$_{10}$. With $m=1,\,n=0$ the formula collapses to $$f_{c,10} = \frac{u}{2a} = \frac{3\times10^{8}}{2(0.025)} = 6.00\times10^{9}\ \text{Hz} = 6.00\ \text{GHz}$$ This is the lowest cutoff of any mode, which is what makes TE$_{10}$ dominant, and it sets the bottom of the band.
The competing modes. Evaluating the same formula, $$f_{c,20} = \frac{u}{a} = 12.00\ \text{GHz}, \qquad f_{c,01} = \frac{u}{2b} = 15.00\ \text{GHz}$$$$f_{c,11} = \frac{u}{2}\sqrt{\left(\tfrac{1}{0.025}\right)^{2} + \left(\tfrac{1}{0.010}\right)^{2}} = 16.16\ \text{GHz}$$
Rank them and identify what closes the band. In ascending order the cutoffs are 6.00, 12.00, 15.00 and 16.16 GHz. Because the aspect ratio is $a/b = 2.5 \gt 2$, TE$_{20}$ falls below TE$_{01}$ and is therefore the mode that ends single-mode operation. (Had $a/b$ been less than 2, TE$_{01}$ would have closed the band instead.)
State the band. Exactly one mode — TE$_{10}$ — propagates for $$\boxed{6.00\ \text{GHz} \lt f \lt 12.00\ \text{GHz}}$$ a bandwidth of 6.00 GHz, which is one full octave (the ratio $f_{c,20}/f_{c,10} = 2$ whenever $a \geq 2b$).

Below 6 GHz nothing propagates at all: every mode is evanescent and the guide behaves as a reflective, reactive stub rather than a transmission medium. Above 12 GHz the guide still works, but the signal now divides between two modes with different guide wavelengths and group velocities, which disperses pulses and makes the two-port behaviour dependent on how the guide is excited. Practical guides are operated over roughly the middle 60–70 % of the single-mode band — here about 7.5 to 11 GHz — to keep away from both the high attenuation just above cutoff and the moding risk at the top.

Final results
Quantity	Symbol	Value
Cutoff of TE$_{10}$ (dominant)	$f_{c,10}$	$6.00\ \text{GHz}$
Cutoff of TE$_{20}$	$f_{c,20}$	$12.00\ \text{GHz}$
Cutoff of TE$_{01}$	$f_{c,01}$	$15.00\ \text{GHz}$
Cutoff of TE$_{11}$/TM$_{11}$	$f_{c,11}$	$16.16\ \text{GHz}$
Single-mode range	$\Delta f$	$6.00\!-\!12.00\ \text{GHz}$
Bandwidth	—	$6.00\ \text{GHz}$ (one octave)

Concept & Theory

Topic. Cutoff frequencies and the single-mode (dominant-mode) band of a hollow rectangular waveguide.

Key relations. $f_{c,mn} = \dfrac{u}{2}\sqrt{(m/a)^2+(n/b)^2}$ with $u = c/\sqrt{\varepsilon_r\mu_r}$; TE$_{10}$ dominant when $a \gt b$; $\beta = \sqrt{k^2-k_c^2}$, so $f \lt f_c$ gives evanescence, not loss.

Why this works. A rectangular guide supports only field patterns that fit an integer number of half-wavelengths across each wall, which quantises the transverse wavenumber $k_c$. Propagation requires $k \gt k_c$, so each pattern has a threshold frequency and the guide behaves as a high-pass filter per mode. Single-mode operation is therefore the gap between the first and second thresholds, and choosing $a = 2b$ (the standard aspect ratio for commercial guide) makes that gap exactly an octave while keeping attenuation and power handling reasonable.

Common pitfall. Assuming TE$_{01}$ is always the second mode. It is second only when $a \lt 2b$; for the common $a \geq 2b$ geometry TE$_{20}$ arrives first. Also, quoting the band as “above 6 GHz” without an upper limit answers a different question.

Source. Pozar, Microwave Engineering, 4th ed., §3.3 (rectangular waveguide, Table 3.2 of standard guide bands); Sadiku, Elements of Electromagnetics, §12.2–12.4.

Question 5 of 8: Single-mode bandwidth of an empty rectangular waveguide

Question 5: Single-mode bandwidth of an empty rectangular waveguide (20 marks)