Question 11.2: A distortionless line has Zo = 60 Ω, α = 20 mNp/m, u = 0.6c,...

For a distortionless line

RC=GL   or   G=\frac{RC}{L}

and hence

Z_{o}=\sqrt{\frac{L}{C}}                                         (11.2.1)

\alpha=\sqrt{RG}=R\sqrt{\frac{C}{L}}=\frac{R}{Z_{o}}                  (11.2.2a)

or

R=\alpha Z_{o}                                             (11.2.2b)

But

u=\frac{\omega}{\beta}=\frac{1}{\sqrt{LC}}                                  (11.2.3)

From eq. (11.2.2b)

R=\alpha Z_{o}=(20\times 10^{-3})(60)=1.2 \Omega/m

Dividing eq. (11.2.1) by eq. (11.2.3) results in

L=\frac{Z_{o}}{u}=\frac{60}{0.6(3\times 10^{8})}=333 nH/m

From eq. (11.2.2a)

G=\frac{\alpha^{2}}{R}=\frac{400\times 10^{-6}}{1.2}=333 \mu S/m

Multiplying eqs. (11.2.1) and (11.2.3) together gives

uZ_{o}=\frac{1}{C}

0r

C=\frac{1}{uZ_{o}}=\frac{1}{0.6(3\times 10^{8})60}=92.59 pF/m

\lambda=\frac{u}{f}=\frac{0.6(3\times10^{8})}{10^{8}}=1.8 m