Exercise 7.8.1: A frequency read from a sliding short

The positions are $2.20\mathrm{cm}$ apart, so ${\lambda }_g=4.40\mathrm{cm}$. With ${\lambda }_c=2a=4.572\mathrm{cm}$,

and $f=c/\lambda =9.463\mathrm{GHz}$.

$d={\lambda }_g/2$, ${\lambda }_g$ and $3{\lambda }_g/2$: TE${}_{101}$, TE${}_{102}$ and TE${}_{103}$ of three different cavities, all at $9.463\mathrm{GHz}$. Check with the cavity formula: $15\sqrt{1/2.286^2+(3/6.60{)}^2}=15\times 0.6309=9.463\mathrm{GHz}$.

At $11\mathrm{GHz}$, $\lambda =2.727\mathrm{cm}$, $f_{co,10}=c/2a=6.562\mathrm{GHz}$ and ${\lambda }_g=2.727/\sqrt{1-(6.562/11{)}^2}=3.398\mathrm{cm}$: resonances at $d=1.699$, $3.398$ and $5.097\mathrm{cm}$.

Each reading gives ${\lambda }_g=2d/p$, so a reading error of $\pm 0.01\mathrm{mm}$ becomes $\pm 0.02/p\mathrm{mm}$ in ${\lambda }_g$: the farthest resonance ($p=3$) is best. From $f^2=f_{co,10}^2+(c/{\lambda }_g{)}^2$,

so $\pm 0.0067\mathrm{mm}$ in ${\lambda }_g$ is $\pm 0.74\mathrm{MHz}$. The first reading alone would give $\pm 2.2\mathrm{MHz}$.

No. $6\mathrm{GHz}$ is below the TE${}_{10}$ cutoff of $6.56\mathrm{GHz}$, so ${\lambda }_g$ is imaginary and no plunger position satisfies $d=p{\lambda }_g/2$. Every resonance of the TE${}_{10p}$ family lies above $f_{co,10}$.

${\lambda }_g>\lambda$ ($4.40>3.17\mathrm{cm}$), as for every guided mode above cutoff. At $11\mathrm{GHz}$ the TE${}_{20}$ mode of WR-90 ($13.1\mathrm{GHz}$) is still cut off, so only TE${}_{10p}$ resonances appear.