time domain reflectometry on long cables
If I have done my math correctly, in the region where the characteristic impedance is relatively constant with frequency, and assuming the leakage conductance is negligible, the attenuation parameter for a transmission line is given by
\$\alpha \approx \frac{\displaystyle R}{\displaystyle 2Z_0}\$
If we look at, for example 100\$\Omega\$ cat 5e cable, it has a \$Z_0\$ of 100
\$\Omega\$ and an R of 0.188\$\Omega\$/m.
That gives an attenuation constant of
\$\alpha = \frac{\displaystyle 0.188}{\displaystyle 2\cdot 100} = 0.00094\$ nepers/meter
Over 100km, the attenuation factor should be 100000 x 0.00094 = 94 nepers.
That means that a 1V signal applied to the near end of the cable will be
\$e^{-94} \approx 1.5 \cdot 10^{-41}\$ volts when it reaches the far end of the cable.
By the time it returns, it will have been attenuated to \$1.5^2 \cdot 10^{-82} = 2.25 \cdot 10^{-82}\$ volts.
So, it is not feasible to use TDR for 100km of cat 5e cable using a frequency in that region.
But high voltage AC power lines can transmit power over 100km without that kind of attenuation. What if we use a lower frequency?
At lower frequencies, dispersion will be the major problem. For your enjoyment, I leave you with the story of the first transatlantic cable. In order to distinguish dots from dashes which were spread due to dispersion, they had to use a very slow transmission rate. According to this account of the first transatlantic cable it took 16 hours to send the first message of 98 words. Longer cable, (3200km) but you get the idea.