Cut-off frequencies of the three lowest modes of a dielectric-filled rectangular waveguide: 22-Elec-A7 Electromagnetics, December 2017, Q4

Question 4: Cut-off frequencies of the three lowest modes of a dielectric-filled rectangular waveguide

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. A rectangular guide whose broad wall is 2.25 cm and narrow wall 1 cm, completely filled with a non-magnetic dielectric.

Given data
Quantity	Symbol	Value
Broad inside dimension	$a$	$2.25\ \text{cm}$
Narrow inside dimension	$b$	$1.00\ \text{cm}$
Relative permittivity of the filling	$\varepsilon_r$	$2.25$
Relative permeability	$\mu_r$	$1$ (ordinary dielectric)
Aspect ratio	$a/b$	$2.25$

Find. The cut-off frequencies of the three lowest-order modes, identified by name.

Approach. Scale the wave velocity by the filling, evaluate the standard cut-off formula for every low index pair, then rank the results numerically rather than trusting a remembered mode order.

Find the wave velocity inside the filling. Because $\mu_r = 1$,$$u = \frac{c}{\sqrt{\varepsilon_r}} = \frac{3 \times 10^{8}}{\sqrt{2.25}} = \frac{3 \times 10^{8}}{1.5} = 2.00 \times 10^{8}\ \text{m/s}$$Every cut-off frequency of the empty guide is divided by $\sqrt{\varepsilon_r} = 1.5$.
Write the cut-off formula. For a rectangular guide the transverse resonance condition gives$$f_{c,mn} = \frac{u}{2}\sqrt{\left(\frac{m}{a}\right)^{2} + \left(\frac{n}{b}\right)^{2}}$$with $m, n$ the number of half-cycles across $a$ and $b$. TE modes exist for any $(m,n) \neq (0,0)$; TM modes require both indices to be at least 1, so every pair with $m,n \ge 1$ supplies a degenerate TE/TM couple.
Evaluate the low-order candidates. Substituting $u = 2 \times 10^{8}\ \text{m/s}$, $a = 0.0225\ \text{m}$ and $b = 0.0100\ \text{m}$: $f_c(\mathrm{TE}_{10}) = u/2a = 4.444\ \text{GHz}$; $f_c(\mathrm{TE}_{20}) = u/a = 8.889\ \text{GHz}$; $f_c(\mathrm{TE}_{01}) = u/2b = 10.000\ \text{GHz}$; $f_c(\mathrm{TE}_{11}) = f_c(\mathrm{TM}_{11}) = 10.943\ \text{GHz}$; $f_c(\mathrm{TE}_{30}) = 3u/2a = 13.333\ \text{GHz}$.
Rank them, and note why the order is what it is. Sorting the five numbers puts $\mathrm{TE}_{10}$, $\mathrm{TE}_{20}$ and $\mathrm{TE}_{01}$ lowest, in that order:$$\boxed{f_c = 4.444\ \text{GHz}\ (\mathrm{TE}_{10}), \quad 8.889\ \text{GHz}\ (\mathrm{TE}_{20}), \quad 10.000\ \text{GHz}\ (\mathrm{TE}_{01})}$$The second mode is $\mathrm{TE}_{20}$, not $\mathrm{TE}_{01}$, because $u/a \lt u/2b$ whenever $a \gt 2b$; here $a/b = 2.25$, so a second half-cycle across the broad wall costs less than the first half-cycle across the narrow wall. Had the aspect ratio been below 2 the order would reverse.
State the single-mode band. The guide carries only the dominant $\mathrm{TE}_{10}$ mode between $4.444\ \text{GHz}$ and $8.889\ \text{GHz}$ — a bandwidth of $4.444\ \text{GHz}$, exactly one octave, which is the usual figure when $\mathrm{TE}_{20}$ closes the band. As a sanity check on the whole calculation, just above cut-off the guide wavelength $\lambda_g = \lambda/\sqrt{1 - (f_c/f)^{2}}$ must exceed the wavelength in the unbounded filling, which it does for every mode listed.

Figure 4.1 — cut-off ladder for the 1 cm × 2.25 cm guide filled with ε(r) = 2.25. The three lowest cut-offs are shown in green.

Final results
Quantity asked	Result
Wave velocity in the filling	$u = 2.00 \times 10^{8}\ \text{m/s}$
Lowest mode	$\mathrm{TE}_{10}$ at $4.444\ \text{GHz}$
Second mode	$\mathrm{TE}_{20}$ at $8.889\ \text{GHz}$
Third mode	$\mathrm{TE}_{01}$ at $10.000\ \text{GHz}$
Next up (degenerate pair)	$\mathrm{TE}_{11} / \mathrm{TM}_{11}$ at $10.943\ \text{GHz}$
Single-mode band	$4.444$ to $8.889\ \text{GHz}$

Topic. Cut-off frequencies and mode ordering in a dielectric-filled rectangular waveguide.

Key relations. $u = c/\sqrt{\mu_r \varepsilon_r}$; $f_{c,mn} = (u/2)\sqrt{(m/a)^{2} + (n/b)^{2}}$; TE needs $(m,n) \neq (0,0)$ while TM needs $m,n \ge 1$; $\mathrm{TE}_{20}$ precedes $\mathrm{TE}_{01}$ whenever $a \gt 2b$; $\lambda_g = \lambda/\sqrt{1 - (f_c/f)^{2}}$.

Why this works. A hollow guide supports a mode only when the transverse dimensions can hold the required number of half-wavelengths, and the cut-off frequency is the frequency at which that transverse pattern exactly fits with no axial propagation left over. Filling the guide slows the wave by $\sqrt{\varepsilon_r}$, so every cut-off scales down by the same factor while the mode pattern itself is unchanged. Which mode comes second is then a pure question of geometry: the index pair that produces the smallest transverse wavenumber wins, and that depends on the aspect ratio rather than on any fixed textbook list.

Common pitfall. Two habits cause trouble: quoting $\mathrm{TE}_{01}$ as the second mode because it appears second in most tables (true only for $a \lt 2b$), and forgetting to divide by $\sqrt{\varepsilon_r}$, which inflates every answer by 50 % here. Including $\mathrm{TM}_{10}$ or $\mathrm{TM}_{01}$ is a third error — those modes do not exist.

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

Question 4 of 8: Cut-off frequencies of the three lowest modes of a dielectric-filled rectangular waveguide

Question 4: Cut-off frequencies of the three lowest modes of a dielectric-filled rectangular waveguide