Exercise 7.6.3: A coating with a prescribed surface reactance

$k_0=209.4rad/m$, $\omega {\epsilon }_0=0.5556S/m$.

so $n_{\mathrm{eff}}=\tau /k_0=1.132$. The field falls to $1/\mathrm{e}$ at $1/\alpha =9.0\mathrm{mm}$ and by $20\mathrm{dB}$ at $\ln 10/\alpha =20.7\mathrm{mm}$ above the coating.

Inside the coating

$H_x\propto \cos hy$ in the coating (metal at $y=0$) and $\propto {\mathrm{e}}^{-\alpha (y-t)}$ above it. Continuity of $H_x$ and of $(1/\epsilon )\mathrm{d}{H}_x/\mathrm{d}y$ at $y=t$ gives

TE${}_1$ starts at

and TM${}_2$ at twice that. At $10\mathrm{GHz}$ only TM${}_0$ propagates.

TM${}_0$ alone needs $V=k_{0}t\mathrm{NA}<\pi /2$, so $t<{\lambda }_0/(4\mathrm{NA})=30/(4\times 1.095)=6.85\mathrm{mm}$. At that thickness, $V=\pi /2$, the equation $w=u\tan u/2.2$ with $u^2+w^2=(\pi /2{)}^2$ gives $u=1.130$, $w=1.091$, so $\alpha =w/t=159.3Np/m$ and

Forward, as in Problem 7.6.2: $V=209.4\times 0.0044\times 1.095=1.010$, $u=0.8833$, $w=\alpha t=0.4889$, and $u^2+w^2=0.780+0.239=1.019=V^2$. Also $w=u\tan u/2.2=0.8833\times 1.218/2.2=0.489$.