Thanks as always Martin, I got 3/4 - not bad considering the last time I realy looked at this theory was 30 years ago or so. A little more explanation in the answers would have been nice, or a web reference.

Question 1: You launch the 1V step into the line. That 1V edge moves along the line until it reaches the end.

If the end is open, the impedance mismatch makes the edge double to 2V, as a reflected wave of the same magnitude and polarity starts to move back towards the source.

If the end is shorted, obviously the voltage there is 0. (There is also a reflected wave of magnitude 1V and opposite polarity.)

If the end is terminated in Zo=50Ω, the 1V edge reaches it, the boundary conditions are satisfied, there is no reflection, and it stays at 1V.

Question 2: The initial current = voltage / Zo = 1V / 50Ω = 20mA.

The initial current has no 'knowledge' of what's on the end of the line. It is 20mA regardless of the load there. It doesn't change until a reflection comes back, by which point it isn't the "initial current" anymore.

Question 3: Probably a trick question for many, who always think of voltages, not currents.

The initial edge is a voltage of 1V and a current of 20mA, moving down the line. At the end, the mismatch causes a reflected wave to begin, of -1V & -20mA, moving in the opposite direction. For the current, -20mA, heading in the opposite direction, means the current flow is in the same direction as the initial edge, so the two currents add to 40mA.

Question 4: Current in the signal path and its return path directly underneath it always exactly match; equal and opposite.

This happens because the electro-magnetic fields around the transmission line make the return current happen instantaneously.

At the moment you connect the 1V source to the transmission line, the transmission line is indistinguishable from a 50 ohm resistor, connected to the source. The current into on one end of that resistor equals the current out the other.

@Andy_I: Thanks for the detailed explanations for all the questions!! I am little confused here about the polarity of the reflected voltage wave you mentioned. Would it be (-1)V voltage wave moving backward? Then why would the resultant voltage be 2V and not 0V (1V-1V)?

I have a problem only with Q3. Current flow with a shorted line at the far end.

The initial edge is a voltage of 1V and a current of 20mA, moving down the line. At the end, the mismatch causes a reflected wave to begin, of -1V & -20mA, moving in the opposite direction. For the current, -20mA, heading in the opposite direction, means the current flow is in the same direction as the initial edge, so the two currents add to 40mA.

In the first question, we said that if the far end is shorted, the voltage at that far end is 0 when the pulse arrives. I agree with that.

This same 0 voltage applies in this question too, does it not? And also look at Q2. Isn't the instantaneous current always 20 mA at the far end as soon as the step arrives? I think both of those answers are correct.

So in Q3, I'd say that the current will start at 20 mA, instantaneously, when the step arrives, and then will tend to infinity (if the power source is a perfect voltage source and the transmission line has zero resistance). Remember too that a transmission line is modeled with a series of inductors wired in series, along the forward path, and a series of capacitors wired between the forward path and the reverse path, between each pair of inductors. So it's not possible to instantaneously double the current. If current is 20 mA at the shorted far end the instant the step function arrives, it can't instantaneously increase to 40 mA. But it will gradually tend to infinity, if that 1 V source remains connected, and the voltage throughout will tend to 0.

On the other hand, if the far end is open, the current will instantaneously be 20 mA as well, when that step function arrives, but then drops to 0. What happens here is that since you cannot instantaneously change current flow through an inductor, and at the open end that current has "no place to go," it will keep charging the (modeled) capacitor at the end of the line, then discharge back upstream. At steady state, current is 0 and voltage is 1 V, throughout the transmission line.

> Would it be (-1)V voltage wave moving backward? Then why would the resultant voltage be 2V and not 0V (1V-1V)?

It would be 0V and not 2V.

If you wondered if the direction matters to the voltage, it doesn't. If you add two voltages, their values add. It makes no difference where those "waves" might be moving.

Current is different because current has a direction.

> This same 0 voltage applies in this question too, does it not?

It does.

> And also look at Q2. Isn't the instantaneous current always 20 mA at the far end as soon as the step arrives? I think both of those answers are correct.

If you look a micron in from the far end, then you will first see the 20mA there, an instant before the edge reaches the far end. A moment later, the reflected wave has begun, with another 20mA flowing in the same direction as the first 20mA, for a total of 40mA.

If you look right at the far end short, you see the current instantaneously jumping from 0 through 20mA to 40mA (with no time spent at 20mA). This all assumes an instantaneous edge = infinite rise time, which is physically impossible, but an interesting theoretical idea nonetheless. Real signals obviously have rounded edges.

> So in Q3, I'd say that the current will start at 20 mA, instantaneously, when the step arrives, and then will tend to infinity (if the power source is a perfect voltage source and the transmission line has zero resistance).

No can do. The power source to the "left" of that short is, first of all, a length of 50Ω transmission line. Even if the voltage source at the input end of the line is perfect, the short can't convey that it is a short until the reflected wave propagates back to the source. Anyway, there was only 20mA available on the transmission line, so it can't jump to infinite immediately. It will only tend towards infinite current after bouncing back and forth and back and forth ... forever.

> Remember too that a transmission line is modeled with a series of inductors wired in series, along the forward path, and a series of capacitors wired between the forward path and the reverse path, between each pair of inductors. So it's not possible to instantaneously double the current.

Remember that the lumped model is only a crude approximation. Take the limit as the number of lumps goes to infinity and the inductance per lump goes to zero and the capacitance per lump goes to zero, and it models how the line really behaves. You can instantaneously double the current on a 0.0 Henry inductor.

That's one reason why I dislike using the lumped model approximation. It leads to misconceptions such as this one.

How did you answer Q2? How did the initial current launched into the line get to 20mA? The finitely-lumped model doesn't handle that. That's because it's wrong.

> If current is 20 mA at the shorted far end the instant the step function arrives, it can't instantaneously increase to 40 mA.

It can as quickly as the input signal does. For a perfect step input (and a lossless non-dispersive line), the current does instantaneously increase to 40mA, just as it instantaneously went to 20mA at the input end of the line.

> But it will gradually tend to infinity, if that 1 V source remains connected, and the voltage throughout will tend to 0.

For a real (rounded) "step" input edge, with risetime faster than the transmission line's round-trip delay, the current at the shorted end will increase from 0mA to 40mA exactly as quickly as the input signal went from 0V to 1V. Then on the next reflection (one round-trip delay later), and if the input signal is a perfect voltage source, it will rise to 80mA ... then to 120mA ... and so forth. In infinite time these steps add up to approaching infinite current ... in stairstep fashion.

No can do. The power source to the "left" of that short is, first of all, a length of 50Ω transmission line. Even if the voltage source at the input end of the line is perfect, the short can't convey that it is a short until the reflected wave propagates back to the source. Anyway, there was only 20mA available on the transmission line,

True, the short would not be "conveyed" instantly, to the left side of the line. It propagates back (reflection).The important point is that a 50 ohm characteristic impedance transmission line is NOT the same as a line with 50 ohm resistance. Just to be clear where I'm coming from, here's an experiment for you. Take any length of coax, be it 50 ohm or 75 ohm or anything else, stick a DC voltage source on one end, an ammeter at the other end (or just short the far end and put the ammeter in series wth the voltage source), and measure the current after just a few fractions of a second. The current will not be limited to V/(characteristic impedance). At steady state, with DC voltage applied, that transmission line might as well be any wire pair, like lamp cord. (Of course, if the center conductor of the coax is very very small, then sure. It will represent sone possibly significant resistance.)

A perfect transission line has no resistance. It's just distributed inductrance in series, and distributed capacitance in parallel. With a DC source, the inductors eventually simply conduct, and the capacitances charge up to their equilibrium voltage (1 V in this case), and stop conducting. So, much like lamp cord.

Take the limit as the number of lumps goes to infinity and the inductance per lump goes to zero and the capacitance per lump goes to zero, and it models how the line really behaves. You can instantaneously double the current on a 0.0 Henry inductor.

But this is not how the entire length of transmission line behaves! There is no way the approaching 20 mA current can suddenly double when it reaches the short circuit, because there is no power source at that end. Instead, you have 0 V, because of the short. This also drains any voltage from the distributed capacitance at the way far end, and the voltage drains out of all capcitances going back upstream, after having reached the shorted end. The infinite current source is at the opposite end of a long, distributed inductance, and as the distrbuted capcitance of the line loses all of its charge, that voltage source will try desperately to maintain 1 V. By pumping out more and more current.

How did you answer Q2? How did the initial current launched into the line get to 20mA? The finitely-lumped model doesn't handle that. That's because it's wrong.

You must know that the model can't be that fundamentally flawed, right? It would be worthless as a model. The reason it works is that at the left end, very close to the power source, you have almost no amount of series inductance, you have an uncharged, even if very small, capacitance to the return line, AND you have this infinite current 1 V power source. This means that current can indeed begin to flow instantaneously, A different situation from the shorted far end, with no power source and no voltage.

If the far end is open, the 20 mA current reaches that far end and eventually stops flowing. If the far end is shorted, the 20 mA current will reach that end and keep climbing. It's all very symmetrical.

NASA's Orion Flight Software Production Systems Manager Darrel G. Raines joins Planet Analog Editor Steve Taranovich and Embedded.com Editor Max Maxfield to talk about embedded flight software used in Orion Spacecraft, part of NASA's Mars mission. Live radio show and live chat. Get your questions ready.
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