I'm following an electromagnetics course and got this question as homework: enter image description here

From earlier examples I tried to piece to together a solution but ran into some problems. I learned that if the a circuit is switched off (after steady state has been reached) the voltage persists and a compensating wave emerges which counteracts the steady state voltage and after some time equals the old steady state voltage with a different sign so they add up to zero. I also learned that after switching a circuit on, it will produce a wave with voltage amplitude equal to the 45V (since there is no internal resistance).

which would simply be 45V since there is no internal resistance.

Does this mean the voltage at z=0 t = 0+ (so right after switching) is the addition of the old steady state voltage plus the 45V of the new source minus some compensating wave voltage?

How do I find this value? I tried to calculate it considering only the switched off circuit and ended up with -13.5V but then the total voltage would be 58.5V. I also tried to simulate the circuit, according to which the voltage never exceeds 45V. But I'm not sure if Multisim accounts for the effects on a nanosecond scale. Could you maybe explain to me the workings of a compensating wave after switching?

Edit to avoid confusion: I appreciate the help I'm already getting, but the reason I want to bring this question back to the compensating wave is that the next question is about its steady state. I'm not sure how I should decompose the voltage at z=0 into the voltage generated by the 45V source, the old steady state voltage and the compensating wave. enter image description here

  • \$\begingroup\$ The voltage at distance Z=0 after the switch closes onto the 45 volt source has no other option than to remain perfectly at 45 volts. \$\endgroup\$
    – Andy aka
    Commented May 15, 2017 at 9:10

2 Answers 2


I also tried to simulate the circuit, according to which the voltage never exceeds 45V.enter image description here

Of course it won't be any different to 45 volts - you have a perfect voltage source (45 volts) and you are measuring it at distance Z = 0. It will always be 45 volts at zero distance because, it's a perfect voltage source and this will override everything except a perfect short circuit (indeterminate solution).

  • \$\begingroup\$ I get that from a circuit theory perspective the new steady state will be 45V, but what happens on smaller level when we consider the EM waves and what happens just after flipping the switch? According to Multisim, the voltage linearly increases from 27V to 45V. And how do I calculate the compensating voltage wave? \$\endgroup\$
    – Jay_ESE
    Commented May 15, 2017 at 9:40
  • \$\begingroup\$ @Jay_ESE Multisim won't ever show an instantenous rise, it numerically always needs takes one timestep, but you can force that timestep to be as small as you want (edit or properties or simulator is the sort of place you'll find the control for that). \$\endgroup\$
    – Neil_UK
    Commented May 15, 2017 at 9:45
  • \$\begingroup\$ @Jay_ESE At t=0 the voltage at that node will become 45 volts and it will stay at the voltage until the switch deselects it. End of story. \$\endgroup\$
    – Andy aka
    Commented May 15, 2017 at 9:46
  • \$\begingroup\$ Thanks for your replies guys! I'm just not sure if we are on the same level of abstraction. I need to answer this question in terms of EM waves and I also want to understand the reasoning not only the final answer. I have difficulties with understanding the concept of the compensating wave that emerges after a switch occurs. Does this mean that after the switch is flicked a compensating wave with amplitude -27V will be sent in the positive z direction? In this case we would get 27V-27V+45V = 45V for Vtotal, which would be in line with your answer. \$\endgroup\$
    – Jay_ESE
    Commented May 15, 2017 at 9:55
  • \$\begingroup\$ @Jay_ESE Listen, all I can see is the one main question and its answer is simply 45 volts. No ifs or buts, it's 45 volts. It's got nothing to do with abstraction, it's quite simply 45 volts and the fine detail is also 45 volts. \$\endgroup\$
    – Andy aka
    Commented May 15, 2017 at 10:05

Today was the deadline of this homework and the professor explained the problems. If anyone else gets stuck on these kind of questions here is the solution:

As answered above in the comment the voltage at t=0 after flicking the switch will be 45V. After the switch the voltage source will send a TEM wave across the transmission line with voltage amplitude 45V, while the other voltage source stops producing waves. Since the reflection coefficient at z=0 changes to -1 as mentioned in the comments, all the waves that made up the old steady state that are reflected at z = 0 change sign and since V = V+ + V- they cancel the old steady state voltage at z = 0.

Because it is very tedious to keep track of all waves the concept of a compensating wave was introduced. We can also see the old steady state voltage as persisting even after the switch was flicked if a compensating wave with voltage 18V (so not 45V as suggested in the comments, this would amount to a new voltage of 72V).

What was interesting to me is that the voltage will oscillate in the first few nanoseconds. I am not entirely sure what happens at exactly z = 0 (because it's the reflection point, I think it will stay constantly 45V) but at other points of the transmission line the voltage will go as high as 51V (eg. at z = 2 after 15ns) and as low as 43V (eg. at z = 1 after 25ns) and converge to 45V across the whole transmission line.

What really helped me understanding the concept of a compensating wave and how the voltages can be decomposed in moving waves was drawing a lot and a lot of bounce diagrams and realizing that compensating waves are a kind of shortcut to not keep track of all the infinite waves that make up the old steady state voltage.

Hopefully this can help someone in the future,

Have a nice day

edit: as for the second question in the OP The compensating voltage converges to 18V and 60 mA (which add up with the old steady state values to the new steady state of the system).

To find V+ and V- we can simply use: V = V+ + V- and I = (V+ - V-)/Z, which gives answer (g)

edit 2: in the comments an explanation for the compensating waves was asked, since comments cannot be long enough here is my attempt:

What really happens is that the infinite amount of waves that make up the steady state voltage before the switch is flicked do not vanish suddenly but still bounce between z=0 and z=L, however since the reflection coefficient is changed to -1 (since the new voltage source has zero impedance) the some values will change, if you add all of those waves just after the switch was flicked, they add up to 0V at z=0 (but still to 27 at z=L). In addition the new voltage source sends out a TEM with amplitude 45V.

If you want to calculate the voltage at a different z and t, you would have to keep track of all those waves from the old steady state and the new waves originating from the new voltage source.

A mathematical equivalent way is to assume that the old steady state voltage stays there forever but a wave with negative amplitude starts to propagate in +z direction. The compensating wave now is this negative wave + the wave that emerges from the new voltage source (-> -27 + 45 = 18). You get the correct results for all V(z,t) if you assume there is a kind of background voltage of 27V and new TEM waves with start off with 18V and slowly decrease as they bounce against the load or the generator. If you draw the bounce diagrams for both ways, you will see that they are identical. So I understand compensating waves as a kind of summary of all the waves that bounce between the load and generator. I'm not sure where the name comes from but I guess the compensate for the changes that happen after the switch is flicked.

  • \$\begingroup\$ I think you should explain what the compensating wave is and what it is compensating. \$\endgroup\$
    – Andy aka
    Commented May 16, 2017 at 12:01
  • \$\begingroup\$ explanation provided in edit2, since comments are too short \$\endgroup\$
    – Jay_ESE
    Commented May 16, 2017 at 12:28

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