Simulation of TDR using LTSpice

I'm trying to simulate a TDR based on the figure below, and one of the scenarios is a short circuit. The simulation should last for 140ns. The parameters are

$Z_A = 50 \Omega$

$T_A = 5ns$

$Z_B = 75 \Omega$

$T_B = 10ns$

Rise time of source signal $\tau_r = 0.1ns$

I took this website as reference to come up with the schematic below: As nothing about the voltage source was given other than the rise time, I didn't really know what to put as its parameters, so I took most of the parameters given in the reference website.

I then probed the wire between R2 and TL-A to get the graph.

First of all, is my simulation even correct? This is the first time I'm using LTSpice, and I'm completely new to this electronics topic.

Second, as I've only seen TDR with one transmission line, I'm not sure how to interpret the graph.

a) I thought this circuit is similar to 3 resistors in series - R2 ($50\Omega$), ZA ($50\Omega$) and ZB ($75\Omega$). So shouldn't the incident voltage at the point between R2 and TL-A be $\frac{50}{50+50+75} * 2V = 0.571V$?

b) I understand that the time taken for the signal to come back to the probe point is twice the transmission delay. Isn't this the case be it 1 or 2 transmission lines?

At the probe point, I thought it takes 2(5ns + 10ns) = 30ns for the reflected signal to first reach the probe point, meaning the voltage level at that point is constant until 30ns? Why is the voltage level changing at 10ns already, and for it to actually increase?

c) Why are the next voltage level changes at time intervals of 20ns? Why not at multiples of 30ns?

We can sort out the times here.

At the probe point, the incident signal will be at 0.5(on) (as it is) until the signal has traversed TLA (it is a 1:1 voltage divider during transit), then had a discontinuity at TLB and returned from that point.

Note that you must use the effective impedance depending on where the signal is; at < 10nsec (for your probe point), TLB does not exist.

As the traversal time (round trip) in TLA is 10ns, that is the time you will see the discontinuity reflected from TLB at the probe point. When the signal is returning from an impedance higher than the source, it returns in-phase. The reflection co-efficient at the TLB entrance is (75-50)/(75+50) = 0.2, and we do indeed see the reflection increase by this amount.

At this time, the signal has already moved through TLB by 5 ns, and therefore it will take another 5nsec (outbound through TLB) + 10 ns (return through TLB) + 5 nsec (return through TLA) = 20 nsec for the next discontinuity to show up at the probe point for a total time of 30 nsec.

Just why you see another discontinuity into something larger than the specified load impedance (reflection co-efficient of -0.8, approxiamtely 5 ohms) is interesting and may have something to do with the specifics of each model.

• When does the reflection happen? The moment the signal hits the transmission line, or does it have to reach the end of the transmission line (i.e. after transmission time)? It appears from the way you describe the reflection from TLB, that it is the former? – Rayne Jan 31 '16 at 14:28
• Using the model of just one transmission line (reflection coefficient = (ZL - Z0)/(ZL + Z0)), at the time when the signal hits TLB, the characteristic impedence of TLB becomes ZL, while the characteristic impedence of TLA is Z0? We ignore the short circuit at this point? This is the way you derived the reflection coefficient of 0.2, right? – Rayne Jan 31 '16 at 14:35
• I will update the answer to answer your comments in the morning. – Peter Smith Feb 1 '16 at 18:07