Question 4 of 8: Rectangular Waveguide Cutoff Frequencies

Question 4: Rectangular Waveguide Cutoff Frequencies (equal value)

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. Rectangular waveguide, inside dimensions $a=2.25$ cm (broad wall) $\times$ $b=1$ cm (narrow wall), filled with a lossless dielectric $\varepsilon_r=2.25$ ($\mu_r=1$).

Find. The cutoff frequencies of the three lowest-order propagating modes.

Approach. Use the dielectric-filled rectangular-waveguide cutoff formula for TE$_{mn}$/TM$_{mn}$ modes and evaluate it for the low-order $(m,n)$ combinations, keeping in mind $a>b$ makes the $x$-direction (along $a$) the "easy" one.

1. Cutoff formula. For a guide filled with dielectric $\varepsilon_r$ (non-magnetic), $$f_{c,mn}=\frac{1}{2\pi\sqrt{\mu_0\varepsilon_0\varepsilon_r}}\sqrt{\left(\frac{m\pi}{a}\right)^2+\left(\frac{n\pi}{b}\right)^2}=\frac{c}{2\sqrt{\varepsilon_r}}\sqrt{\left(\frac{m}{a}\right)^2+\left(\frac{n}{b}\right)^2}.$$ With $\sqrt{\varepsilon_r}=\sqrt{2.25}=1.5$, the prefactor is $c/(2\times1.5)=3\times10^8/3=1\times10^8$.
2. Dominant mode, TE$_{10}$ ($m=1,n=0$, along the wide wall $a=2.25$ cm $=0.0225$ m): $$f_{c,10}=10^8\times\frac{1}{0.0225}=\boxed{4.44\ \text{GHz}}.$$
3. Next modes. TE$_{20}$ ($m=2,n=0$): $$f_{c,20}=10^8\times\frac{2}{0.0225}=8.89\ \text{GHz}.$$ TE$_{01}$ ($m=0,n=1$, along the narrow wall $b=1$ cm $=0.01$ m): $$f_{c,01}=10^8\times\frac{1}{0.01}=10.00\ \text{GHz}.$$ TE$_{11}$/TM$_{11}$ ($m=1,n=1$), for comparison: $$f_{c,11}=10^8\sqrt{\left(\frac{1}{0.0225}\right)^2+\left(\frac{1}{0.01}\right)^2}=10.94\ \text{GHz}.$$
4. Rank the modes. $4.44<8.89<10.00<10.94$ GHz, so the three lowest propagating modes are TE$_{10}$, TE$_{20}$, and TE$_{01}$ (TE$_{11}$/TM$_{11}$ is the fourth).