Problem 7.3.1: Cutoff frequencies of a 6\times3 cm guide

$a=0.06\mathrm{m}$, $b=0.03\mathrm{m}$, $(1/a{)}^2=277.8{\mathrm{m}}^{-2}$, $(1/b{)}^2=1111.1{\mathrm{m}}^{-2}$.

The first three TM modes are TM${}_{11}$, TM${}_{21}$ and TM${}_{31}$.

The first three TE modes are TE${}_{10}$, then TE${}_{01}$ and TE${}_{20}$ together at $5\mathrm{GHz}$.

Every mode with $f_{co}<6\mathrm{GHz}$ propagates: TE${}_{10}$, TE${}_{01}$, TE${}_{20}$, TE${}_{11}$ and TM${}_{11}$, five modes. TE${}_{10}$ is alone between $2.5$ and $5\mathrm{GHz}$, the single-mode band. The first TM mode starts only at $5.59\mathrm{GHz}$.

TE${}_{mn}$ and TM${}_{mn}$ with $m,n\ge 1$ share $h_{mn}$, so every TM cutoff of (a) reappears in the TE list of (b). In units of $f_{co,10}$ the cutoffs are $\sqrt{m^2+4n^2}$: $1$, $2$, $2$, $\sqrt{5}$, $\sqrt{8}$, $3,\dots$, the table of Section 7.3.4. TE${}_{01}$ is TE${}_{10}$ turned through $90^{\circ }$, with $E$ along $x$: it needs $b=\lambda /2$, hence $5\mathrm{GHz}$.