Home exercise 7.3.4: A dielectric phase shifter

$k_0=2\pi f/c=209.4rad/m$, $\pi /a=137.4rad/m$.

$\Delta \tau =112.6rad/m=64.5^{\circ }$ per cm, and $90^{\circ }$ needs $L=\frac{\pi /2}{112.6}=1.40\mathrm{cm}$.

No. With $v=c/\sqrt{2.1}$, $f_{co,20}=v/a=9.06\mathrm{GHz}<10\mathrm{GHz}$, while TE${}_{01}$ ($v/2b=10.19\mathrm{GHz}$) stays below cutoff. TE${}_{10}$ and TE${}_{20}$ can propagate in the filled section.

TE${}_{20}$ has $E_y\propto \sin (2\pi x/a)$, odd about the centre plane $x=a/2$, while the incident TE${}_{10}$ field and a full-width insert are both even about it. A structure symmetric about $x=a/2$ cannot convert an even field into an odd one, so TE${}_{20}$ is not excited. TE${}_{30}$, which is even, is below cutoff ($13.6\mathrm{GHz}$ in PTFE).

${\lambda }_g$ in PTFE from $1/{\lambda }_g^2={\epsilon }_r/{\lambda }_0^2-1/{\lambda }_c^2=2.1/9-1/4.572^2=0.1855\mathrm{c}{\mathrm{m}}^{-2}$: ${\lambda }_g=2.32\mathrm{cm}$.