I'll try to make this as concise as possible. I'm trying to get my head around how signals behave in coaxial cables. If you set up a circuit like this: Then the earthed end will force a node at that end, right? So you'll end up with your reflections having a phase shift of pi and looking like this: And you find that a standing wave arises with a node at the earthed end. I also understand that if that end were not earthed but left open (infinite resistance) then the signal would reflect without any phase shift, and a similar standing wave would arise but with an antinode at the open end of the circuit. But what if one were to construct a circuit like the one below? With the switch open, it should be just the same as an open circuit: a standing wave should arise with an antinode at the open end. With the switch closed, the sine waves should hit one another in phase and cause an identical standing wave to occur. However, I have measured the electric field strength of the parallel wires in the circuit above with both an open and closed switch, but the standing waves produced are not only of different amplitudes, but also out of phase: Could somebody explain to me why this happens? I feel I'm missing something blindingly obvious. Many thanks in advance for any help!