You and I seem to be in violent agreement on many things; but not everything.

I think we all know that Zo=50Ω is not the same thing as Rs=50Ω. These quiz questions were stipulated on the basis of ideal lossless transmission lines, although it might not explicitly state it.

> 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.

Power source has nothing to do with it. It does double, because the incoming 1V edge needs to satisfy the 0V boundary condition. No "power source" is needed at that end to generate another 20mA. The power to launch the reflected wave, comes from the incoming step waveform, reflected by the impedance mismatch at the short. The incident power is 100% turned around into a reflected wave, with zero power generated by or used by the short itself.

You do need to separate what happens at the fine level (observing the step waveforms propagating down the line and back), from the steady-state. I think you shouldn't speak about the infinite steady-state current while you are analyzing what transpires at the picosecond level when waves are still bouncing back and forth. Steady-state doesn't happen the first time the incident wave arrives at the load, not even the Nth time, unless N is infinity. Except when the load is matched to the line.

>> 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?

Oh, but it IS fundamentally flawed! That's why we call it an "approximation". The lumped model approximation is only truly valid in the limit when the size of the lumps is zero and the number of lumps is infinite. Otherwise, it is a fair approximation when you don't look at the fine details. But looking at the fine details is precisely what we are doing here.

If you want to look at the instantaneous current when there is an instantaneous incoming step waveform, it is imperative to either take that lumped model to the infinite limit, or abandon it in favor of a better model (SPICE's ideal T-line model is more accurate for lossless lines), or use an analytical approach that makes no such approximations.

> 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.

You can't say it works at one end but not the other end. If one end allows instantaneous changes in current, then so does the other end. Nothing prohibits it, and the boundary conditions demand it.

Before we had lumped models, and before we had SPICE, we had equations to describe exactly how ideal transmission lines work. Most of us plugged through the math a few times, long ago, then didn't look back because there were other tools that made the work simpler for us. Look at those analytical equations. Transmission lines still behave the way they did with those analytical models. You will see that the current does double at the shorted end.

It is also true that the current eventually becomes infinite, but it gets there after an infinite number of reflections back and forth. The current increases step-wise, dependent on the electrical length of the line, and takes theoretically infinite time to reach infinite current.

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.

> 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.

> 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.

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.

@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)?

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.

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 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 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.

Vetinari Clock: Additional Possibilities Max Maxfield11 comments In my previous blog on the topic of my Vetinari Clock project, we pondered a number of possibilities for the positioning of switches on the front panel, and we ended up with a number of ...

The Other Tesla David Blaza5 comments I find myself going to Kickstarter and Indiegogo on a regular basis these days because they have become real innovation marketplaces. As far as I'm concerned, this is where a lot of cool ...

To save this item to your list of favorite EE Times content so you can find it later in your Profile page, click the "Save It" button next to the item.

If you found this interesting or useful, please use the links to the services below to share it with other readers. You will need a free account with each service to share an item via that service.