# time domain reflectometry on long cables

Is it possible to use time domain reflectometry on cables with a length of more than 100km? What kind of problems do time domain reflectometers on that long cables face?

• Don't think is much different from a shorter cable. Jan 2, 2021 at 12:50
• @MarkoBuršič I heard that dispersion is a big factor for longer cables Jan 2, 2021 at 12:53

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.

• Why is dispersion at lower frequencies a problem? Is there no dispersion for high frequency signals? Jan 2, 2021 at 17:20
• No - read the answer - at higher frequencies dispersion WOULD be a bigger problem except that the attenuation prevents you seeing it (or indeed any return at all).
– user16324
Jan 2, 2021 at 17:32
• There is dispersion at all frequencies, unless R/L = G/C. However, it is much more pronounced at lower frequencies. Dispersion depends upon the change of phase velocity for a given change of frequency. There is a plateau, usually above 50kHz for typical cables, where impedance and phase velocity stay relatively constant up to somewhere in the vicinity of 50MHz. This is just approximate. In that region, dispersion is relatively small. Jan 2, 2021 at 17:34
• @MathKeepsMeBusy ok thank you thats very interesting. Does dispersion also depend on the cable length? Jan 2, 2021 at 17:40
• Yes, but not exponentially. I'm going to give a broken analogy, in which individual signals have a speed. (Actually if a signal has only one frequency, then it is a sine wave, and extends infinitely back and forward in time). Using the broken analogy, if it takes a 10kHz signal 10 mSec to travel some distance D, and 9 mSec for a 20kHz signal to travel the same distance, then the signals become "dispersed" by 1 ms. (Again, broken analogy). If the distance is increased to 2xD, the signals will be dispersed by 2 ms. It is a linear degradation with length, not exponential. Jan 2, 2021 at 17:58

Dispersion (spreading if pulse) and damping (resistance) are the problems. Because of this, in practical setups, I don’t think you will be able to do TDR with lines >100km.

With extremely long lines faults might be possibly located with some sort of impedance spectroscopy. Noise would be the biggest issue. It would need to be done with high SNR and extremely low frequency.(ELF)

• What if I use a needle like pulse? Jan 2, 2021 at 17:12
• Not much ULF spectrum.... Jan 2, 2021 at 17:52
• You would use a unit step signal. Mar 7, 2021 at 19:28