Exercise 6.3.7: Reading a slab from its reflection nulls

The null spacing is $\Delta f=c/(2d\sqrt{{\epsilon }_r})=1.5$ GHz. The peak value gives the impedance step:

$6/1.5=4$: four half wavelengths (${\lambda }_2=2.5$ cm at $6$ GHz). The sheet is also transparent at $1.5$, $3.0$ and $4.5$ GHz.

${\beta }_{2}d=\pi f/\Delta f$. At $5$ GHz, ${\beta }_{2}d=3.333\pi$, ${\sin }^2{\beta }_{2}d=0.75$:

At $6.375$ GHz, ${\beta }_{2}d=4.25\pi$, ${\sin }^2{\beta }_{2}d=0.5$: $|{\Gamma }_d{|}^2=4.5/20.5=22.0\%$.

No. The null frequencies fix only the electrical thickness $d\sqrt{{\epsilon }_r}=10$ cm. A $7.07$ cm sheet with ${\epsilon }_r=2$ or a $3.33$ cm sheet with ${\epsilon }_r=9$ would have the same nulls. The depth of the pattern, set by the impedance ${\eta }_0/\sqrt{{\epsilon }_r}$, separates ${\epsilon }_r$ from $d$.

At $6.375$ GHz the layer formula with $\tan {\beta }_{2}d=1$ and ${\eta }_2={\eta }_0/2$ gives ${\eta }_{\mathrm{in}}=(0.4-\mathrm{j}0.3){\eta }_0$ and ${\Gamma }_d=\frac{-0.6-\mathrm{j}0.3}{1.4-\mathrm{j}0.3}=-0.366-\mathrm{j}0.293$, $|{\Gamma }_d{|}^2=0.220$ $✓$. At $6.75$ GHz the sheet is $4.5$ half waves: ${\eta }_{\mathrm{in}}={\eta }_2^2/{\eta }_0={\eta }_0/4$ and ${\Gamma }_d=-0.6$ $✓$.