Home exercise 7.7.2: Dispersion of an FR-4 line

With ${\epsilon }_r-{\epsilon }_{\mathrm{eff}}(0)=1.070$,

At $10\mathrm{GHz}$, $k_0=209.4rad/m$ and $k_0\ell =20.94\mathrm{rad}$ for $\ell =10\mathrm{cm}$:

$f_{T{E}_1}=\frac{c}{4h\sqrt{{\epsilon }_r-1}}=\frac{3\times 10^8}{4\times 1.6\times 10^{-3}\times 1.844}=25.4\mathrm{GHz}$. The transverse-resonance limit is even lower, $19.3\mathrm{GHz}$. At $10\mathrm{GHz}$ the line still works, but with $31.5dB/m$ of loss (Problem 7.7.1 formulas: ${\alpha }_c=1.5dB/m$, ${\alpha }_d=30.1dB/m$) and a phase that is $137^{\circ }$ off over $10\mathrm{cm}$ unless dispersion is modelled, it is a poor choice. A thin low-loss laminate is the better board.

At $f=f_p$ the model gives $4.4-1.070/2.052=3.879$, a little more than half way from $3.33$ to $4.4$. For $f\to \infty$, ${\epsilon }_{\mathrm{eff}}\to {\epsilon }_r$: all the field is in the substrate.