Why does a coaxial cable terminated in a load ZL that differs from the characteristic impedance Z0, have an input impedance which generally depends on the length of the cable?

  • \$\begingroup\$ The Z(f) changes with f when termination is mismatched and \$length > 1/10 \lambda\$ \$\endgroup\$ – Tony Stewart EE75 Jan 11 '17 at 22:49
  • \$\begingroup\$ How good at following math are you? \$\endgroup\$ – Andy aka Jan 12 '17 at 8:59

Non-maths simplification for lossless transmission lines:

Electrical waves can travel through a transmission line in both directions.

The generator produces a wave that travels towards the load. If the load has the same impedance Z_0 as the line, then the load appears as if the line continues, and so no wave returns. The fact that the load is matched to the line means that the voltage-current relationship of the wave arriving at the load is the same as the voltage-current relationship dissipated by the load. This holds for resistances (Ohm's law) or more in general for complex impedances.

Now, if the load is unmatched, then the arriving wave will have more or less current (and maybe at a different phase) than will be consumed by the load.

The difference in current will produce a backwards-travelling wave.

So in the unmatched case, you have a wave travelling forwards towards the load, and another wave travelling backwards.

The impedance seen at the generator will be the voltage-current relationship at that point, which will be a superposition of both waves.

It is easy to see that this superposition will be different for different line lengths; there will be a delay of 2 line lengths between the generated and the reflected wave at the generator. This delay affects the relative phase at which the current waves add up.


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