Here's what finally helped me understand termination and reflections: Suppose you have a really, really long coaxial cable, with the far end shorted together. If you put current through it, what will the voltage be?

Because the cable is shorted at the far end, you'd expect the voltage to remain near 0. But, the far end is a long way away - if the voltage was immediately 0 volts, we'd be communicating faster than light! Instead, the signal has to propagate down the cable to the short, then back to the near end again, before we see the short on our end. This is what a reflection is.

What does the signal look like in the time before the reflection arrives? Well, the cable has nonzero resistance, and nonzero capacitance - electrically, it's like a long sequence of series resistors and shunt capacitors - and that will cause it to charge from our current source as the signal propagates. Electrically, this looks like a resistance - this is called characteristic impedance. An infinitely long piece of 50 ohm coaxial cable would look exactly like a 50 ohm resistor, electrically. A shorter one looks like a 50 ohm resistor during the period the signal is propagating down the cable.

In our imaginary scenario, then, applying current to a long cable with a short at the end, the voltage waveform will look like a short peak (with voltage equal to current * characteristic_impedance) followed by a return to (near) 0 volts. If the other end of the cable were an open circuit, it would instead look like a short peak followed by a higher voltage (determined by our current source's maximum voltage).

Suppose we didn't want any reflections. If we terminate the coax with a resistor that has the same value as the cable's characteristic impedance, we're sorted! The coax looks like a 50 ohm resistor while the signal is propagating, and still looks like a 50 ohm resistor once the propagation has finished - because we connected one across it at the far end. This is termination.

Here's what finally helped me understand termination and reflections: Suppose you have a really, really long coaxial cable, with the far end shorted together. If you put current through it, what will the voltage be?

Because the cable is shorted at the far end, you'd expect the voltage to remain near 0. But, the far end is a long way away - if the voltage was immediately 0 volts, we'd be communicating faster than light! Instead, the signal has to propagate down the cable to the short, then back to the near end again, before we see the short on our end. This is what a reflection is.

What does the signal look like in the time before the reflection arrives? Well, the cable has nonzero resistance, and nonzero capacitance - electrically, it's like a long sequence of series inductors and shunt capacitors - and that will cause it to charge from our current source as the signal propagates. Electrically, this looks like a resistance - this is called characteristic impedance. An infinitely long piece of 50 ohm coaxial cable would look exactly like a 50 ohm resistor, electrically. A shorter one looks like a 50 ohm resistor during the period the signal is propagating down the cable.

In our imaginary scenario, then, applying current to a long cable with a short at the end, the voltage waveform will look like a short peak (with voltage equal to current * characteristic_impedance) followed by a return to (near) 0 volts. If the other end of the cable were an open circuit, it would instead look like a short peak followed by a higher voltage (determined by our current source's maximum voltage).

Suppose we didn't want any reflections. If we terminate the coax with a resistor that has the same value as the cable's characteristic impedance, we're sorted! The coax looks like a 50 ohm resistor while the signal is propagating, and still looks like a 50 ohm resistor once the propagation has finished - because we connected one across it at the far end. This is termination.