Exercise 6.3.4: A two-layer matching coating

$n_s=4$: $R=(\frac{4-1}{4+1}{)}^2=0.36$.

$n=\sqrt{n_s}=2$, that is ${\epsilon }_r=4$, and $d={\lambda }_0/(4n)=30/(4\times 2)=3.75$ mm.

$n_{\mathrm{in}}=1$ requires $n_A^2{n}_s=n_B^2$, $n_B/n_A=\sqrt{n_s}=2$. With $n_A=1.5$: $n_B=3$ (${\epsilon }_r=2.25$ and $9$), and

At $f_0$ the impedance seen through $B$ is ${\eta }_B^2/{\eta }_s=\frac{(1/3{)}^2}{1/4}=0.444{\eta }_0$ and through $A$ it is $\frac{(1/1.5{)}^2}{0.444}=1.000{\eta }_0$ $✓$.

At $8$ GHz every quarter-wave layer has $\beta d=0.4\pi$, $\tan \beta d=3.078$. In units of ${\eta }_0$: single layer ${\eta }_{\mathrm{in}}=0.777+\mathrm{j}0.343$, $R=5.1\%$. Two layers: through $B$ $0.414+\mathrm{j}0.071$, through $A$ ${\eta }_{\mathrm{in}}=1.057+\mathrm{j}0.155$, $R=0.64\%$. Solving $R(f)=0.01$ numerically:

At $20$ GHz each layer is a half wave and disappears: both curves return to the bare $36\%$ $✓$ (Figure 6.3.17).