0

Having read Dave Tweeds excellent description of reflections in electrical transmission lines, I thought I was one step closer to a full understanding of SWR. However, later I was watching w2aew's description of how a directional coupler works and I'm a little confused. In the snippet shown below, you can see where the reflected current is described as having the opposite direction to the incident current.

enter image description here

However, what I understood from reading Dave's answer is that in the case of short-circuit terminations it's a negative voltage that's reflected back along the line. My understanding is that this negative voltage arises from the need to supply "additional" current into the short (2*I behind the reflected voltage wavefront). But then, isn't the current moving towards the short or do I have that wrong? I'm trying to reconcile my understanding of transmission line terminations (intuitively) with w2aew's description and failing owing to my own ignorance!

Buck8pe
  • 1,674
  • 18
  • 44

1 Answers1

1

My understanding is that this negative voltage arises from the need to supply "additional" current into the short (2*I behind the reflected voltage wavefront). But then, isn't the current moving towards the short

Yes, you are correct.

If the termination is purely resistive, with value less than the line's characteristic impedance, then the current at the point of termination will be in the same direction for the forward and reflected waves.

If the termination is purely resistive, with value greater than the line's characteristic impedance, then the current at the point of termination will be in the opposite directions for the forward and reflected waves.

But if you consider points on the line at some distance from the termination you could see the forward wave and reflected wave currents either constructively or destructively interfering in either case.

In the snippet shown below, you can see where the reflected current is described as having the opposite direction to the incident current.

The snippet gives the eqautions

$$I_F = \frac{V_F}{Z_0}$$ and $$I_R = \frac{-V_R}{Z_0}$$

Remember that in the case of a reflection from a lower resistance load (or short), \$V_R\$ will have opposite sign from \$V_F\$ (at the point of reflection), so there is no contradiction.

The Photon
  • 126,425
  • 3
  • 159
  • 304
  • The voltage at the short-circuit has to be zero, so the phase of the reflected wave has to be opposite (-Vr) of the incident wave by property of the boundary condition alone. If there is an open circuit, the phase of the reflected wave has to be the same. – user55924 Oct 18 '19 at 16:22
  • I suppose the issue I have is that in my mind I see that "demand" for additional current rippling backwards along the line. In Dave's example, the reflected voltage wavefront goes negative to produce a lower potential on the shorted end (of any given lumped section). Thus I visualize the current moving towards the shorted end and not back towards the source?? – Buck8pe Oct 18 '19 at 16:23
  • @Buck8pe, just think about the forward travelling wave at some instant in time. On some parts of the line, the current is flowing "forward", at other points on the line the current is flowing "backward". Same thing with the reflected wave. The current won't be flowing "forward" at all points on the line just because it's flowing that direction at the point of reflection. – The Photon Oct 18 '19 at 16:34
  • @The Photon Maybe my confusion stems from the fact that Dave's answer uses a step (or pulse maybe). I'm guessing this produces a different response from the more typical sine wave examples (shown by @user24368). – Buck8pe Oct 18 '19 at 17:05
  • @The Photon Got it (I think)! It helps to visualize a sine wave in this example (as opposed to a pulse as described in Dave's answer). All Dave described is still true, but you get phase reversed signal back at the source as described in the video formula. The incident current and reflected current are in opposition in equal measure, indicating a 100% discontinuity. Same is true for an open, but voltage in-phase and current reversed (same difference). Is that about right? – Buck8pe Oct 18 '19 at 17:53
  • In my opinion it is easier to think about the step. In all cases you first have voltage 0,5*V_in in the line, due to resistor divider formed by the source impedance and the characteristic impedance of the transmission line. When the wave hits short circuit the voltage drops to zero. Thus the voltage from the line is drawn to the short circuit, i.e. current is towards the short circuit. When the wave hits an open circuit, the voltage is doubled (the resistor divider kind of disappears). The current can't go to the open circuit so it has to return to the source. – TemeV Oct 18 '19 at 18:40