Question 4 of 8: Filling permittivity that places the dominant cut-off at 4.2 GHz

Question 4: Filling permittivity that places the dominant cut-off at 4.2 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. A rectangular guide of fixed inside dimensions is to be filled completely with a non-magnetic dielectric chosen so that its dominant mode turns on at a specified frequency.

Find. The relative permittivity $\varepsilon_r$ of the filling, and (as a check on the phrase “lowest propagating mode”) where the next two modes then sit.

Approach. The lowest mode of a rectangular guide with $a \gt b$ is always TE$_{10}$, whose cut-off depends only on $a$ and the filling; invert its formula for $\varepsilon_r$, then rank the neighbouring modes to confirm the identification.

1. Cut-off of the dominant mode. For TE$_{mn}$ in a guide filled with $\varepsilon_r$, $$f_{c,mn} = \frac{c}{2\sqrt{\varepsilon_r}}\sqrt{\left(\frac{m}{a}\right)^{2} + \left(\frac{n}{b}\right)^{2}} ,$$ and the smallest non-zero value is TE$_{10}$ because $a \gt b$: $$f_{c,10} = \frac{c}{2a\sqrt{\varepsilon_r}} .$$
2. Empty-guide value. Before filling, $$f_{c,10}^{\text{air}} = \frac{2.9986\times 10^{8}}{2(0.024)} = 6.247\ \text{GHz},$$ so the dielectric must slow the wave enough to pull 6.247 GHz down to 4.2 GHz.
3. Solve for the permittivity. Since every cut-off scales as $1/\sqrt{\varepsilon_r}$, $$\sqrt{\varepsilon_r} = \frac{f_{c,10}^{\text{air}}}{f_{c,10}} = \frac{6.247}{4.2} = 1.4874 \quad\Longrightarrow\quad \boxed{\varepsilon_r = 2.21} .$$ Back-substituting, $f_{c,10} = 2.9986\times 10^{8}/[2(0.024)(1.4874)] = 4.20$ GHz, as required. (Using the rounded $c = 3.00\times 10^{8}$ m/s that the exam's own arithmetic implies gives $\varepsilon_r = 2.214$ — the same answer to three figures, so nothing here is sensitive to that choice.)
4. Confirm that TE$_{10}$ really is the lowest. With $\varepsilon_r = 2.21$, $$f_{c,20} = \frac{c}{a\sqrt{\varepsilon_r}} = 8.40\ \text{GHz}, \qquad f_{c,01} = \frac{c}{2b\sqrt{\varepsilon_r}} = 10.08\ \text{GHz} .$$ Because $a/b = 2.4 \gt 2$, it is TE$_{20}$ and not TE$_{01}$ that closes the single-mode band, so the guide propagates TE$_{10}$ alone over $4.2\text{--}8.4$ GHz — an octave.
5. Physical reading. A relative permittivity of about 2.2 is exactly what PTFE-based microwave dielectrics provide, so the specification is realisable; the price is that the guide wavelength and the power-handling both fall, and any loss tangent of the filling now contributes attenuation the empty guide did not have.