I am trying to model a very lossy coax transmission line. As shown in the circuit, the top TL model includes the shield line; the bottom lumps the shield terms into the center conductor and looks more like the traditional model.

My error is in wave propagation. (Ignoring the ringing from using a segmentation of 8)

Notice that the complex model single ended signal (top plane) is shown to have instantaneous conduction across the network, as indicated by the three red arrows. Not shown is that the shield path also has this compliment, which makes the differential signal look correct (bottom plane).

Can anyone help shed light on this error? Obviously in real life this would be breaking the laws of physics.

Edit for clarity The exact problem I am trying to indicate is the short ~200mV step that appears to propagate instantly from one side to the other (in the single ended plot).

Edit for Clarity, Attempt 2 New figure added at the bottom, demonstrating more clearly the error of instantaneous end to end response time across the transmission line simulation. The section indicated by pink annotations should remain identical in all three plots, regardless of what happens after this window.

Figure 1

Figure 2

Figure 3

  • \$\begingroup\$ What do you consider as error? Can you explain what puzzles you? (I do not see any problems with these waveforms.) \$\endgroup\$ Oct 4, 2018 at 4:05
  • 1
    \$\begingroup\$ You plotting V(i), instead of the differential voltage V(i,i'), such as in yuor 2nd case, which means the voltage is referenced to ground, thus changing (shorting) the symmetry of the model -- you should expect the results to be entirely different, or weird, if you prefer. \$\endgroup\$ Oct 4, 2018 at 7:22
  • \$\begingroup\$ @StefanWyss If you observer the portion of of the graphs pointed at by the red arrows, they should all look identical. That short little step (which happens instantly in the top plot) cannot actually happen in real life. The wave has not propagated down the TL at that point and it is still at a potential of zero. \$\endgroup\$
    – Bob
    Oct 4, 2018 at 17:03
  • \$\begingroup\$ @aconcernedcitizen The only difference between the single ended and the double ended response is that the settling peak should be slightly smaller in the single ended model, as some of the signal starts to come out deferentially across the out ports. Just to reiterate, the top graph does NOT match a real life transmission line. \$\endgroup\$
    – Bob
    Oct 4, 2018 at 17:06
  • \$\begingroup\$ @Bob Why should the waves be identical? The plots are recorded from different points on the TX line, so they must not be identical. Take for example the Vin step at 15ns. A little later (~18ns) this step arrives at point B (red plot). So to me, these plots make sense. \$\endgroup\$ Oct 4, 2018 at 19:27

1 Answer 1


I'll consider only one stage, to make things more simplified and visible. The whole thing is a two-port network, with no direct connections between pins. Therefore, differential voltage probing matter depending on the points. Here is an attempt at ilutrating the equivalent topologies for different probings:


The upper-right schematic shows what happens if you probe the voltage at points V(X,B) (which is what puzzles you). In your case, B is grounded, so this results in V(X). The network is no longer a PI-lowpass (symmetric or not), it's a lowpass followed by a bandstop, therefore a notch-lowpass. Due to the zero in the transfer function, you will have an apparent advance in time (which we all know it's not).

The lower-right shows the equivalent topology for V(X,Y) probing, where the network retains its PI-lowpass structure. The output obeys the lowpass response.

Here's the frequency response of the two probings (node B is grounded):


What I'm trying to say is that the transfer function changes, so you no longer have an all-pole t.f., as it would be with cascaded PI sections, but a pole-zero t.f., and the zeroes are the ones that cause the apparent time advance. The similarities are only in the passband, the zeroes appear in the stopband -- somehow visible in my example, but clearly seen in your comment's example. Also, due to the symmetric model, the zeroes are not purely imaginary, except for the first stage. Here's the slightly expanded version of your probing for 3 stages:


You probe between X and B, look what that does to your topology: you no longer have cascaded PI sections, you have a complex network that has a simple, cap first LC lowpass (L1, C1), with what would be a notch made of C2 and L2, but C2 has in parallel the input to the next lowpass (C3, L3), followed by the second would-be notch (C4, L4), but whose C4 also has in parallel the 3rd lowpass, etc. Now do you understand why I say the way you probe makes all the difference?

Also, that's why the link in the comment doesn't look like a FIR, or inverse Pascal or Chebyshev, or Cauer, at 128 stages. Speaking of which, here's an (unlikely) step response of a 4th order Cauer, 1dB ripple, 20dB attenuation, which means two stages:


The transfer function has zeroes and the output's response is clearly (almost) instantaneous for a small part (the 20dB part = 0.1V linear attenuation), despite the fact that the transfer function has two stages. The effect is there for any attenuation, but at 20 is more visible, for 40dB it would have been 0.01V, this is solely for exemplification. That's due to the zeroes which, in this case, are purely imaginary, hence the exaggerated "instantaneous" effect. Also, the zeroes are in the stopnad, yet they clearly influence the response.

What does this have to do with your case? Correlate this last explanation with the equivalent circuits in the 1st picture. If you don't probe where you're supposed to probe, you change the transfer function, thus the response. The way you probed is not the correct way because, due to the reference voltage, you no longer have cascaded PI sections, but something else.

  • \$\begingroup\$ I understand what you are saying, but I do not understand the relevance to this particular problem. What you say is also only true for a single element model, if you run the AC sweep for a much larger element count, you find that the transfer functions of x and x-y are almost the same, and only begin to diverge around where the model starts to break down. If you see here is a simulation of a 128 element model, which begins to break down at around 12MHz. \$\endgroup\$
    – Bob
    Oct 9, 2018 at 23:34
  • \$\begingroup\$ I understand now, improper probing changes the lumped element model in a way that would not manifest in a real transmission line. I was not thinking about the poles and zeros before, just about wave propagation. Thank you very much for your persistence in this topic. \$\endgroup\$
    – Bob
    Oct 10, 2018 at 18:09
  • \$\begingroup\$ Glad it worked out. The proof could have been just as well a reductio ad absurdum: if V(X,B) was fine, then probing V(X,N), with N being anywhere in the network, would have been just as right as V(X,Y), which results in a fallacious statement, but going the direct way does tend to make things clearer, at least for me. \$\endgroup\$ Oct 11, 2018 at 6:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.