Question 6.19: The attenuation on a 50 W distortionless transmission line i...
The attenuation on a 50W distortionless transmission line is 0.01 (dB/m). The line has a capacitance of 0.1 x 10– 9 (F/m).
a) Find the values of the other transmission line elements.
b) Find the velocity of wave propagation.
a) Since this is a distortionless transmission line, we write
\begin{aligned}Z_{C} &=\sqrt{\frac{\hat{L}}{\hat{C}}}=50 \Omega \Rightarrow \hat{L}=(50)^{2}\left(0.1 \times 10^{-9}\right)=2.5 \times 10^{-7} H / m \\\\ \alpha &=\hat{ R } \sqrt{\frac{\hat{C}}{\hat{L}}}=0.01 dB / m =\frac{0.01}{8.69} Np / m =0.0012 Np / m \end{aligned}
From this, we write
\hat{ R } \sqrt{\frac{\hat{ C }}{\hat{ L }}}=\frac{\hat{ R }}{50}=0.0012 \Rightarrow \hat{ R }=0.0575 \Omega / m
We also have the distortionless line criteria
\frac{\hat{R}}{\hat{ L }}=\frac{\hat{ G }}{\hat{ C }} \Rightarrow \hat{ G}=\frac{0.0575}{50^{2}}=2.3 \times 10^{-5} S / m
b) The phase velocity is
v =\frac{1}{\sqrt{\hat{L} \hat{C}}}=\frac{1}{\sqrt{\left(2.5 \times 10^{-7}\right)\left(0.1 \times 10^{-9}\right)}}=2 \times 10^{8} m / s