# About current/voltage reflection in a long coaxial wire.

1. Jun 2, 2010

### Cactor

If I have a long coaxial wire with open end and apply a short voltage pulse, that pulse begins propagating through it until it reaches the end. Then, it reflects back to where it came from. I understand that it behaves like a normal wave, maintaining its energy and bouncing when it reaches a wall. What I don't understand is that the measured voltage at the end of the wire is DOUBLE the applied voltage.

http://www.aps.anl.gov/Science/Publications/lsnotes/ls156/Images/ls156_f1.gif

In books, I read that the voltage at the end of the wire is always the sum of the incident and the reflected one, like:

V=Vi+Vr

But they never explain where does that formula come from. Does anyone know WHY is it the sum of the other two?

Edit: Should I ask this in general forum?

2. Jun 2, 2010

### Staff: Mentor

Welcome to the PF. The EE forum here is fine for this question of yours.

You get 2x the incident voltage because the reflected pulse is in-phase with the incident pulse. The Transmission Line Equations (and info on the Reflection Coefficient) are a good place to start:

http://en.wikipedia.org/wiki/Transmission_line

.

3. Jun 2, 2010

### vk6kro

If you could remove the signal source before the pulse returned, (so that both ends of the cable were open ended) it would bounce from end to end of the cable until it was completely attenuated by losses in the cable.

Because there is only one pulse, there is no incident pulse following the first one, so it doesn't get to combine with a new pulse.

Even in a continuously driven line, the interaction between the incoming wave and the reflected one is not always additive. It depends on the time it takes for the pulse to return compared with the period of the incoming waveform.