Exercise 7.8.5: Over- or undercoupled? A one-port cavity

At $9.40\mathrm{GHz}$ $\lambda =3.191\mathrm{cm}$ and $d=1/\sqrt{(2/\lambda {)}^2-(1/a{)}^2}=2.229\mathrm{cm}$. With $R_s=25.3\mathrm{m\Omega }$, Eq. (7.8.24) gives $Q_{\mathrm{th}}=7840$.

$Q_L=9400/2.0=4700$. $\frac{|\beta -1|}{\beta +1}=\frac{1}{3}$ gives $\beta =2$ (overcoupled) or $\beta =0.5$ (undercoupled).

$\beta =2$ would mean $Q_0=Q_L(1+\beta )=14100$, $1.8$ times the theoretical copper value, which is impossible. Hence $\beta =0.5$, and

$1-|S_{11}{|}^2=1-1/9=88.9\%$ is absorbed in the walls.

Critical coupling: $\beta =1$, $Q_{\mathrm{ext}}=Q_0=7050$, $Q_L=Q_0/2=3525$, $\Delta f=9400/3525=2.67\mathrm{MHz}$, and all the incident power is absorbed at $f_0$.

$1/Q_0+1/Q_{\mathrm{ext}}=1/7050+1/14100=3/14100=1/4700$. Detuned, $S_{11}=\frac{(\beta -1)/(\beta +1)-\mathrm{j}x}{1+\mathrm{j}x}$ with $x=2Q_L{\delta }_f$ (plotted), whose magnitude is the same for $\beta$ and $1/\beta$, and the half-power point of the absorbed power $1-|S_{11}{|}^2$ lies at $x=1$.