Home exercise 7.7.1: A 50 Ω line on alumina

This is below $2$, so the first branch is the right one: $w=0.975\times 0.635=0.62\mathrm{mm}$. Check: $p=0.04(1-0.975{)}^2\approx 2.5\times 10^{-5}$, ${\epsilon }_{\mathrm{eff}}=5.4+4.4(1+12/0.975{)}^{-1/2}=5.4+4.4\times 0.2742=6.606$ and

${\epsilon }_{\mathrm{eff}}=6.61$. At $10\mathrm{GHz}$, ${\lambda }_g=30\mathrm{mm}/\sqrt{6.606}=11.7\mathrm{mm}$.

The surface wave is the lower limit. Allowing a margin, and remembering that TM${}_0$ surface waves (no cutoff) are excited more and more as $h/\lambda$ grows, the line is usable up to about $30$ to $35\mathrm{GHz}$.

The rule of thumb, $w\approx h$ for $50\Omega$ on alumina, is confirmed. ${\epsilon }_{\mathrm{eff}}$ lies between $({\epsilon }_r+1)/2=5.4$ and ${\epsilon }_r=9.8$.