Exercise 7.8.3: A copper filter on a hot day

$f_{101}=\frac{c}{2a}\sqrt{2}$ gives $a=\frac{c}{\sqrt{2}f}=2.121\mathrm{cm}$. Then $\delta =1/\sqrt{\pi f{\mu }_0{\sigma }_c}=0.661\mu \mathrm{m}$ and $Q_0=a/3\delta =10700$. The band is $\Delta f=f_0/Q_L=5\mathrm{MHz}$.

$\Delta f_0=-17\times 10^{-6}\times 40\times 10\mathrm{GHz}=-6.8\mathrm{MHz}$, more than the whole bandwidth. At $10.000\mathrm{GHz}$ the filter is now detuned by ${\delta }_f=6.8\times 10^{-4}$, $2Q_L{\delta }_f=2.72$:

Invar: $\Delta f_0=-1.2\times 10^{-6}\times 40\times 10^{10}=-0.48\mathrm{MHz}$, $2Q_L{\delta }_f=0.192$, extra loss $10{\log }_{10}(1+0.192^2)=0.16\mathrm{dB}$. For $0.5\mathrm{dB}$, $(2Q_L\alpha \Delta T{)}^2=10^{0.05}-1$, that is $2Q_L\alpha \Delta T=0.349$:

At three times the size $f$ falls to $3.33\mathrm{GHz}$ and $\delta$ grows by $\sqrt{3}$, so $Q_0=a/3\delta$ grows by $3/\sqrt{3}=\sqrt{3}$: $Q_0=10700\times 1.732=18500$.

Directly, a cube of $6.36\mathrm{cm}$ at $3.33\mathrm{GHz}$ has $\delta =1.145\mu \mathrm{m}$ and $a/3\delta =18500$. The loaded $Q$ is set by the couplings, which scale with the cavity, so $Q_L=2000$ is kept in (b) and (c).