Home exercise 6.3.4: A quarter-wave matched ceramic window

$\Gamma =\frac{1/3-1}{1/3+1}=-\frac{1}{2}$, so $R=25\%$.

$n_2=\sqrt{n_1{n}_3}=\sqrt{3}$, that is ${\epsilon }_{r2}=3$, and with ${\lambda }_0=3$ cm

With ${\eta }_1/{\eta }_3=3$ and ${\eta }_2/{\eta }_3={\eta }_1/{\eta }_2=\sqrt{3}$ the imaginary part of the numerator of ${\Gamma }_d$ vanishes:

$|{\Gamma }_d{|}^2<0.01$ requires $3{\tan }^2{\beta }_{2}d>96$, $|\tan {\beta }_{2}d|>4\sqrt{2}=5.66$, that is $|{\beta }_{2}d-90^{\circ }|<10.02^{\circ }$. Since ${\beta }_{2}d=90^{\circ }f/f_0$:

At $2f_0$, ${\beta }_{2}d=\pi$, $\tan {\beta }_{2}d=0$ and $|{\Gamma }_d{|}^2=\frac{1}{4}$: the layer is a half wave thick, ${\eta }_{\mathrm{in}}={\eta }_3$, and the block reflects exactly like the bare ceramic.

At $f_0$, $\tan {\beta }_{2}d\to \infty$ and $|{\Gamma }_d{|}^2\to 0$ $✓$. At $f\to 0$ the layer vanishes and $|{\Gamma }_d{|}^2\to \frac{1}{4}$ $✓$. The band is symmetric about $f_0$ because $|{\Gamma }_d{|}^2$ depends only on ${\tan }^2{\beta }_{2}d$ (Figure 6.3.16).