Exercise 7.1.5: Dielectric loss of any guided mode

With ${\tau }^2=k^2-h^2$,

With $\tau =ks$: ${\alpha }_d=\frac{k\tan \delta }{2s}$, the TEM value $k\tan \delta /2$ divided by $s$. Since $\sigma =\omega {\epsilon }^{\prime }\tan \delta$, this is also $\frac{\sigma \eta /2}{s}$, the dielectric attenuation, Eq. (7.2.36) in Section 7.2.

$v=c/\sqrt{2.1}=2.07\times 10^{8}m/s$, $k=2\pi \times 9\times 10^9/v=273.2rad/m$, $s=\sqrt{1-(6/9{)}^2}=0.745$:

against $0.356dB/m$ for a TEM wave at $9\mathrm{GHz}$ in PTFE.

At $f=f_c$, $\tau =0$ and ${\gamma }^2=\mathrm{j}{k}^2\tan \delta$, so $\gamma =k\sqrt{\tan \delta }{\mathrm{e}}^{\mathrm{j}\pi /4}$ and

($k=182.1rad/m$ at $6\mathrm{GHz}$), finite, with an equal phase constant $\beta =2.23rad/m$.

${\alpha }_d\propto f/\sqrt{1-f_c^2/f^2}=f^2/\sqrt{f^2-f_c^2}$. Setting the derivative of its square, $f^4/(f^2-f_c^2)$, to zero gives $f^2=2f_c^2$: the minimum is at $f=\sqrt{2}{f}_c=8.49\mathrm{GHz}$, where ${\alpha }_d=0.475dB/m$.

The exact value at $9\mathrm{GHz}$, $\mathrm{Re}\{\gamma \}$ computed without approximation, is $0.4775dB/m$, the same to four figures. The approximation is still within $0.1\%$ at $6.06\mathrm{GHz}$ ($1.707dB/m$) and within $0.3\%$ at $6.006\mathrm{GHz}$, and fails only in the last few per mille above cutoff, as the graph shows.