Problem 7.8.4: Measuring Q with a network analyser

$\Delta f=2\times 0.6=1.2\mathrm{MHz}$, so $Q_L=9000/1.2=7500$.

$\frac{2\beta }{1+2\beta }=0.5$ gives $\beta =0.5$, and then

The line at port 1 sees $g=1/\beta =2$ in parallel with the matched port 2 (normalised conductance $1$), so $y=1+g=3$ and

Of the incident power $|S_{11}{|}^2=25\%$ is reflected, $|S_{21}{|}^2=25\%$ transmitted, and $1-0.25-0.25=50\%$ absorbed in the walls.

With $\beta \to 0$, $Q_L\to Q_0$ and $\Delta f\to 9\mathrm{GHz}/15000=0.6\mathrm{MHz}$. The measured width always includes the loading by the ports. Only weak coupling shows $Q_0$ directly.

$1/15000+2/30000=2/15000=1/7500$. The same model gives $S_{21}=1+S_{11}=0.5$, the measured peak (the shunt element leaves the voltage continuous).

Away from $f_0$ the cavity reflects everything, $|S_{11}|\to 0\mathrm{dB}$ (Figure 7.8.25). At $f_0$ a quarter of the power passes, a quarter returns and half heats the walls: the $6\mathrm{dB}$ loss of this filter is the price of $\beta =0.5$.