Question 11.11: At 10 GHz, a microstrip line has the following parameters: h...

The ratio w/h=0.8. Hence from eqs. (11.70):

\varepsilon_{eff}=\frac{(\varepsilon_{r}+1)}{2}+\frac{(\varepsilon_{r}-1)}{2\sqrt{1+12h/w}}

and (11.71):

Z_{o}=\begin{cases}\frac{60}{\sqrt{\varepsilon_{eff}}}\ln\left(\frac{8h}{w}+\frac{w}{4h}\right) & w/h\leq1 \\ \frac{1}{\sqrt{\varepsilon_{eff}}}\frac{120\pi}{[w/h+1.393+0.667\ln(w/h+1.444)]} & w/h\geq1\end{cases}

\varepsilon_{eff}=\frac{7.6}{2}+\frac{5.6}{2}\left(1+\frac{12}{0.8}\right)^{-1/2}=4.5

Z_{o}=\frac{60}{\sqrt{4.5}}+\ln\left(\frac{8}{0.8}+\frac{0.8}{4}\right)=65.69\Omega

The skin resistance of the conductor is

R_{s}=\frac{1}{\sigma_{s}\delta}=\sqrt{\frac{\pi f\mu_{o}}{\sigma_{c}}}=\sqrt{\frac{\pi\times 10\times 10^{9}\times 4\pi\times 10^{-7}}{5.8\times 10^{7}}}=2.609\times 10^{-2}\Omega/m^{2}

Using eq. (11.75):

\alpha_{c}\simeq 8.686\frac{R_{s}}{wZ_{o}}

we obtain the conduction attenuation constant as

\alpha_{c}=8.686\times\frac{2.609\times 10^{-2}}{0.8\times 10^{-3}\times 65.69}=4.31 dB/m

To find the dielectric attenuation constant, we need \lambda:

\lambda=\frac{u}{f}=\frac{c}{f\sqrt{\varepsilon_{eff}}}=\frac{3\times10^{8}}{10\times10^{9}\sqrt{4.5}}=1.414\times10^{-2}m

Applying eq. (11.76):

\alpha_{d}\simeq 27.3\frac{(\varepsilon_{eff}-1)\varepsilon_{r}}{(\varepsilon_{r}-1)\sqrt{\varepsilon_{eff}}} \frac{\tan\theta}{\lambda}

we have

\alpha_{d}=27.3\times \frac{3.5\times 6.6\times 10^{-4}}{5.6\times \sqrt{4.5}\times 1.414\times 10^{-2}}=0.3754 dB/m