I am a licensed radio amateur, and find bewildering the many different explanations, which range from folksy urban myth to Maxwell-Heaviside Equations, of what happens at the termination of a transmission line or feeder. I realise that they all come to the same thing in the end (or should do, pun perfect), but none of them give me a gut feeling for what is going on.

I like diagrams, so an answer in terms of (graphical) phasors for the currents and voltages at the load would suit me best. How, for instance, does a step pulse down the line cause twice the voltage at an open circuit termination? Similarly for current at a short circuit. And how is the reflected step generated by the inductance and capacitance of the line?

Can anyone help, without getting all mathematical, and not telling any "lies to children"?

  • \$\begingroup\$ I like to think of EM waves as water through a hose. If the hose is empty and you open the water valve, the water will run all the way down the hose and see a high impedance and will bounce back \$\endgroup\$
    – hassan789
    Feb 26, 2013 at 15:51
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    \$\begingroup\$ You can model a somewhat similar effect by joining sections of rope/string of different weight and sending waves down them. \$\endgroup\$ Feb 26, 2013 at 15:51
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    \$\begingroup\$ This is the best explanation of transmission lines I've read so far.. \$\endgroup\$
    – m.Alin
    Feb 26, 2013 at 16:13
  • \$\begingroup\$ Thank you @m.Alin, that reference is exactly what I need, and couldn't find for myself. \$\endgroup\$ Feb 27, 2013 at 15:07
  • \$\begingroup\$ @HarryWeston I'm glad it helped. Dave's answer is very good as well. You should consider accepting his answer. \$\endgroup\$
    – m.Alin
    Feb 27, 2013 at 15:11

4 Answers 4


OK, for what it's worth, here's how I visualize it.

As you say, a transmission line has both distributed capacitance and distributed inductance, which combine to form its characteristic impedance Z0. Let's assume we have a step voltage source whose output impedance ZS matches Z0. Prior to t=0, all voltages and currents are zero.

At the moment the step occurs, the voltage from the source divides itself equally across ZS and Z0, so the voltage at that end of the line is VS/2. The first thing that needs to happen is that the first bit of capacitance needs to be charged to that value, which requires a current to flow through the first bit of inductance. But that immediately causes the next bit of capacitance to be charged through the next bit of inductance, and so on. A voltage wave propogates down the line, with current flowing behind it, but not ahead of it.

If the far end of the line is terminated with a load of the same value as Z0, when the voltage wave gets there, the load immediately starts drawing a current that exactly matches the current that's already flowing in the line. There's no reason for anything to change, so there's no reflection in the line.

However, suppose the far end of the line is open. When the voltage wave gets there, there's no place for the current that's flowing just behind it to go, so the charge "piles up" in the last bit of capacitance until the voltage gets to the point where it can halt the current in the last bit of inductance. The voltage required to do this happens to be exactly twice the arriving voltage, which creates an inverse voltage across the last bit of inductance that matches the voltage that started the current in it in the first place. However, we now have VS at that end of the line, while most of the line is only charged to VS/2. This causes a voltage wave that propogates in the reverse direction, and as it propogates, the current that's still flowing ahead of the wave is reduced to zero behind the wave, leaving the line behind it charged to VS. (Another way of thinking about this is that the reflection creates a reverse current that exactly cancels the original forward current.) When this reflected voltage wave reaches the source, the voltage across ZS suddenly drops to zero, and therefore the current drops to zero, too. Again, everything is now in a stable state.

Now, if the far end of the line is shorted (instead of open) when the incident wave gets there, we have a different constraint: The voltage can't actually rise, and the current just flows into the short. But now we have another unstable situation: That end of the line is at 0V, but the rest of the line is still charged to Vs/2. Therefore, additional current flows into the short, and this current is equal to VS/2 divided by Z0 (which happens to be equal to the original current flowing into the line). A voltage wave (stepping from VS/2 down to 0V) propogates in the reverse direction, and the current behind this wave is double the original current ahead of it. (Again, you can think of this as a negative voltage wave that cancels the original positive wave.) When this wave reaches the source, the source terminal is driven to 0V, the full source voltage is dropped across ZS and the current through ZS equals the current now flowing in the line. All is stable again.

Does any of this help? One advantage of visualizing this in terms of the actual electronics (as opposed to analogies involving ropes, weights or hydraulics, etc., etc.), is that it allows you to more easily reason about other situations, such as lumped capacitances, inductances or mismatched resistive loads attached to the transmission line.

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    \$\begingroup\$ Thank you @Dave Tweed, this is exactly what I was after, a very clear and convincing explanation. Thank you too for putting in the time and trouble for such a lengthy post. \$\endgroup\$ Feb 27, 2013 at 19:47
  • \$\begingroup\$ @Dave, May I know for the open-ended line how the voltage at the load end becomes twice the arriving voltage. \$\endgroup\$
    – Confused
    Jun 16, 2021 at 17:33

Here is a series of experiments or thought experiments if you like.

1)Take a long rope held at both ends by two friends and held taut. Stand in the middle and ask the person on one end to give a quick flick of the rope vertically, sending a pulse down the rope to the other person. As the wave passes you (in the middle) you will notice that the wave just propagates past you. There are no reflections (at that moment). You will notice that the ropes characteristics are identical before and after your location. This is the case of a matched impedance there is no transition, so there is no reflection.

2) take the same rope, tie it to a fixed location at a rigid wall. Ask your friend to send an pulse down the rope and observe the wave at it approaches, it hits the fixed location and then reflects back off. You will notice that when it reflects it inverts. This is equivalent to a short. The rope flicks up, but cannot move because it is anchored, energy is stored in elastic energy which snaps the rope back (inverting the pulse)

3) Take the same rope and tie a very, very light string to it. Again have your two friends stand at each end and hold the rope/string taut and have a pulse launched down it the rope. At the transition between rope/string the pulse will reflect, but not be inverted. This is an example of an open circuit. The rope flicks up, but the energy cannot go into the string (or rather far less energy) because the mass of the string is far less. So the rope end rises, energy is stored in potential energy and then simply dissipates by falling back down sending the wave back down the line.

In a waveguide, the energy is transforming from magnetic (currents) to electrical (voltage) as the wave propagates. At a open termination, current cannot flow so the energy goes in voltage form. At a short, voltage cannot be expressed (it is a short - or equipotential) so the energy goes into local current loops.

  • \$\begingroup\$ Thank you @rawbrawb for a very useful analogy, and for taking the time and trouble to answer so fully. \$\endgroup\$ Feb 27, 2013 at 15:10

I like to think of a transmission line as being a collection of equal weights connected with matching springs. When a compression pulse is injected at one end, each weight ends up pushing on the next weight in such fashion that the push or pull from the "upstream" weight gets precisely balanced out by a pull or push from the "downstream" weight, leaving each weight motionless after the wave passes through.

If the end of the transmission line can't move, the effect is that the spring that can't move "pushes back" twice as hard as it would if it could have moved. Half of that shove counteracts the shove from the previous wave, and the other half serves to push the previous weight in the direction opposite its earlier motion. The net effect is that a compression wave gets transmitted back.

If the end of the transmission line was simply "open", the effect would be that the last weight wouldn't end up moving just to its starting point after transferring its energy to the next weight, but when it reached its starting point it would still have all the energy it had received from the previous weight. At that point, inertia and momentum would cause it to continue past that point, and effectively "tug" at the previous weight with all the energy the previous weight had fed into it. This would effectively generate a tension wave back up the spring.

  • \$\begingroup\$ Thank you @supercat for that explanation, and for your trouble in putting a lengthy and apposite answer together. \$\endgroup\$ Feb 27, 2013 at 15:09
  • \$\begingroup\$ I am trying to understand reflection as well. I am stuck at - why are we adding incident and reflected voltages but subtracting currents. please can someone help me understand. \$\endgroup\$ Mar 7, 2016 at 16:12
  • \$\begingroup\$ @user3551094: To use the spring analogy, think of voltage as the tension on the springs and current as the amount by which the springs move. If one sends a pulse down the spring such that the tension and motion are in the positive direction, then the reflection will either have the same direction of motion but negative tension, or positive tension but the reverse direction of motion. In either case, if the signs of the tension and motion matched in the original wave, then in the reflected wave they will be opposite. \$\endgroup\$
    – supercat
    Mar 7, 2016 at 17:24

This fascinating Bell Labs video beautifully demonstrates wave motion, SWR, and impedance matching sections on a completely mechanical tabletop device without needing mathematics. It's presented in a way that even a layperson can understand these concepts.

  • Reflection of waves from free and clamped ends
  • Superposition
  • Standing waves and resonance
  • Energy loss by impedance mismatching
  • Reduction of energy loss by quarter-wave and tapered-section transformers
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    \$\begingroup\$ Amazing video indeed! A small clarification: here in 2018, we no longer use Adobe Flash so here is the Youtube link to the same video: youtube.com/watch?v=DovunOxlY1k \$\endgroup\$
    – akhmed
    Apr 2, 2018 at 0:52

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