Exercise 7.3.8: A square-guide polariser

$f_{co,10}=f_{co,01}=c/2a=7.5\mathrm{GHz}$ and the next modes, TE${}_{11}$ and TM${}_{11}$, start at $\sqrt{2}\times 7.5=10.61\mathrm{GHz}$: the band is $7.5$–$10.61\mathrm{GHz}$.

$k=2\pi \times 9\times 10^9/c=188.5rad/m$, $\pi /a=157.1rad/m$, $\pi /b=174.5rad/m$:

$\Delta \tau =33.0rad/m$ and $L=\frac{\pi /2}{33.0}=4.76\mathrm{cm}$. In the section $f_{co,01}=c/2b=8.33\mathrm{GHz}$ and TE${}_{11}$ starts at $11.2\mathrm{GHz}$, so it still carries the two modes only.

Each component lags by $\tau L$, and $E_x$ (TE${}_{01}$, smaller $\tau$) lags less: $E_x$ leads $E_y$ by $90^{\circ }$. At a fixed point $E_x\propto \cos \omega t$ and $E_y\propto \sin \omega t$, so $E$ turns from $x$ to $y$. With the thumb along $+z$ this is right-handed circular polarisation (RHCP).

With the same $L=4.76\mathrm{cm}$: at $8.6\mathrm{GHz}$ $\Delta \tau L=119.0^{\circ }$, at $9.4\mathrm{GHz}$ $75.2^{\circ }$, errors of $+29.0^{\circ }$ and $-14.8^{\circ }$.

At $9\mathrm{GHz}$: $33.0\times 0.0476=1.571\mathrm{rad}=90.0^{\circ }$. The error grows towards the TE${}_{01}$ cutoff ($8.33\mathrm{GHz}$), where ${\tau }_{01}$ changes fastest.