Talk:Reflections of signals on conducting lines

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Voltage sign?[edit]

Shouldn't the voltage in figure 1 be of opposite sign since it's a reflection? — Preceding unsigned comment added by 71.61.20.140 (talk) 12:03, 2 March 2013 (UTC)[reply]

No, the reflection is from an open-circuit and is the same sign as the incident wave, if the termination were a short-circuit then the reflection would be negative. Consider a voltage source that when terminated with an impedance matching its internal impedance has an emf of E at its terminals. Such a source will have an open-circuit voltage of 2E and a short circuit voltage of 0. Thus, when connected to a transmission line, we know the open-circuit voltage at the end of the line must end up as 2E when all the transients have died away. Assuming the line is matched to the source, the incident wave will be E. The reflected wave must therefore be +E in order for the final voltage to be 2E. Likewise, for a short-circuited line the reflected wave must be −E in order for the final voltage to be zero. SpinningSpark 14:50, 2 March 2013 (UTC)[reply]


What is the change in phase of the reflected signal for a purely reactive load relative to? I can't see how the change in phase relative to the incident wave is -2arctan(Xl/Ro). By inspecting the Smith chart, would it not appear that the phase of the complex reflection coefficient at the load is
θ = π - 2arctan(Xl/Ro) for purely inductive loads
θ= -π - 2arctan (Xl/Ro) for purely capacitive loads (where Xl is negative)
So if the reactance of the load is zero (and of course the resistance of the load is zero as this is the case for purely reactive loads) the phase is ±180° and if the reactance is infinitely high the phase is 0°. Also if the normalised impedance (Xl/Ro)= 1 (inductive load) then the phase is 90° and -90° if (Xl/Ro) = -1 (capacitive). Shropshire70 (talk) 22:17, 18 June 2013 (UTC)[reply]
There is a voltage wave and a current wave and their product determines the direction that the power flows. When the wave reflects, power in the reflected wave is going in the opposite direction to the power in the incident wave. To reverse power the product of voltage and current must reverse. In the case of an open termination, the current changes sign and the voltage does not. If the incident wave is traveling left to right and the direction of the current is left to right then in the reflected wave the direction of the current is right to left.Constant314 (talk) 15:22, 24 August 2013 (UTC)[reply]

GA Review[edit]

This review is transcluded from Talk:Reflections of signals on conducting lines/GA1. The edit link for this section can be used to add comments to the review.

Reviewer: North8000 (talk · contribs) 13:54, 22 August 2013 (UTC)[reply]

I am starting a review of this article North8000 (talk) 13:54, 22 August 2013 (UTC)[reply]

Thanks for taking a look, it's much appreciated. SpinningSpark 16:19, 22 August 2013 (UTC)[reply]
My pleasure. I noticed that it had been waiting a long time. North8000 (talk) 17:58, 22 August 2013 (UTC)[reply]

Review discussion[edit]

Preface: One area where I'm a bit tougher than a typical reviewer is in empathy for a typical reader who is trying to learn the subject from the article. But I also try to be helpful and offer to help in this area. North8000 (talk) 21:01, 23 August 2013 (UTC)[reply]

A core item is explaining what situation causes a reflection. And the noun for this in the main statement (at the beginning) is a discontinuity in the transmission parameters, with no explanation of that term just a link. But if one follows that link they end up in an article with a totally different name with no answer insight. What do you think about "discontinuity in the characteristic impedance"?

That suggestion is certainly an improvement, but depicting discontinuity as a change in characteristic impedance does not really cover the case where the discontinuity is caused by a lumped element at the discontinuity. The characteristic impedance of the line is the same on either side of the discontinuity and down its entire length. But I guess the qualification on correct termination which follows can be deemed to cover the case even though the "termination" is not terminating the end of the line. SpinningSpark 00:07, 24 August 2013 (UTC)[reply]
In the short list of things that I thought I remembered from transmission line theory, I thought that characteristic impedance included lumped elements. Or maybe that is just "impedance".  :-) North8000 (talk) 13:09, 24 August 2013 (UTC)[reply]
Ok, I have made the change, but to answer your point...it is true that some authors (particularly in dated books) refer to the characteristic impedance of lumped networks but that's a can of worms. The first worm in the can is that two different things can be meant by characteristic impedance if the network is not symmetrical: iterative impedance or image impedance. More importantly, neither of these impedances, in general, can actually be constructed with a finite network of discrete elements. This is a mathematical restriction, not due to any kind of engineering shortcoming. The "correct termination" of such a network is usually taken to be the network's nominal impedance, a fixed resistance. Take, for instance, the example of a filter connected to the end of a transmission line; part of a radio receiver for instance. The filter is correctly terminated with its nominal impedance and this nominal impedance is equal to the characteristic impedance of the transmission line. Despite this, the filter will still reflect some frequencies back up the transmission line. Lumped element filters built of inductances and capacitances are obliged to reflect the rejected signals since they contain no resistances to absorb them. SpinningSpark 14:31, 24 August 2013 (UTC)[reply]


Sort of the same situation as at the slotted line article. I'll try a few tweaks with reader empathy particularly in mind. Please feel free to modify or revert. Sincerely, North8000 (talk) 12:01, 9 September 2013 (UTC)[reply]

Could you say what the variables mean when the equations are presented? I think a few (e.g. what are clearly lumped impedances on the diagram) might be obvious enough, but I think 80%-90% need this in order for the reader to be able to learn from the equation. Sincerely, North8000 (talk) 12:18, 12 September 2013 (UTC)[reply]

After a quick pass, most of the variables seem to be identified in the text. I only spotted Z0 and ZL as being undefined (except, as you say, via the diagrams). The meaning of the function u(t) might also be better clarified although Vu(t) is defined as a step and the unit step article is already linked. I will go through more carefully later and add defs as required. SpinningSpark 11:44, 13 September 2013 (UTC)[reply]
Maybe taking an example would help. So if someone is reading along the article and comes to the first equation, (:) what are variables and where do they go to see what they mean? North8000 (talk) 15:27, 13 September 2013 (UTC)[reply]
Sorry for taking so long to get back to this, I needed to work up the enthusiasm. The text already defines κ and x. Vu(t) was also defined in the text but I have now also made it clear that the u(t) part of this is the unit step function and made the link explicit. I would have thought that "the incident voltage...is given by" would be enough to imply that vi is the incident voltage, but have made that explicit also. Similar treatment has been given to other expressions in the article. Not convinced that cluttering the article with definitions that are straightforwardly implied is improving readability, but I think I have now done what you want. SpinningSpark 16:05, 21 September 2013 (UTC)[reply]

Arbitrary break #1[edit]

BTW, what I try to use as a standard on a highly technical article is that it should be understandable by a technical person who does not know the topic of the article. IMHO, if the "language" of the communication to the reader is an equation, and to communicate the variables should be defined if they are not obvious to the non-specialist. BTW, if at any time you prefer, just say so and I would be happy to step back from this, ask for a second opinion, and let that second opinion decide everything. Sincerely, North8000 (talk) 21:13, 24 September 2013 (UTC)[reply]

I think the article currently meets at least your criterion of understandability. Do you think there are still problems in this area? SpinningSpark 21:57, 24 September 2013 (UTC)[reply]
IMHO some areas are very confusing. I noted them below. Ironically, they are in the "simpler" sections. That's mostly it. Sincerely, North8000 (talk) 22:06, 24 September 2013 (UTC)[reply]

I had some areas of question. I guess that (since I could often be wrong in questioning them) in Wikie terms it could come down to specific sourcing for the specific item.

  • In the open circuit section, it is discussing the end point of the line where the current is by definition zero. The text describes that there are two currents at the end of the line traveling in opposite directions which cancel each other out. Isn't this confusing at best, and, when talking about either of the individual currents, in conflict with the definition of an open circuit? North8000 (talk) 21:39, 24 September 2013 (UTC)[reply]
    You have pretty much identified the difficulty novice students have with the whole concept of transmission lines, or at least one aspect of it. It is undeniable that there are two electromagnetic waves travelling in opposite directions on the line and both have a current associated with them. Neither of them disappears out of existence at the end of the line. This may be hard to swallow for the student, but it's how it is and they have to live with it. If the line is open circuit at the end the two currents must be equal and opposite at that point. Feel free to improve the explanation of this if you can. SpinningSpark 22:18, 24 September 2013 (UTC)[reply]
Well, all of these things are our way of describing fields and electrons moving as a current. At the end of the line on an open circuit there are no electrons moving as a current. And so, as you point out, one way to describe this zero current is the sum of two equal and opposite currents (each of which individually could not possibly occur). My point was that that might be a confusing way to describe it. But maybe a few "may be visualized as..." type words might be helpful I might give that a try. Sincerely, North8000 (talk) 11:50, 25 September 2013 (UTC)[reply]
I think you are mistaken on several points. Firstly, each of which individually could not possibly occur. They certainly can if the wave manages to launch itself out of the end of the line as it does in an antenna. Admittedly, the equivalent impedance of the line termination would no longer appear to be an ideal short-circuit in such a case, even if the resistance of the shorting wire was close to the ideal zero ohms.
I am not entirely against phrases like "may be visualized as..." but we need to avoid the impression, which this phrase is giving, that the two currents are some kind of analogy. They are not, the two currents really exist. To accept that there are two currents all the way along the line and that they suddenly just disappear at this one point is quite incredible. In fact one can directly and individually measure the two currents by placing a directional coupler right at the end of the line. One will measure the same magnitude of current at all points all the way along the line, even at nodes where the total current is zero.
All descriptions of physical systems are models of reality and not the reality itself, but it is generally agreed in modern electrical science that the more fundamental concept is the electromagnetic wave and that circuit theory is a special case of wave theory when certain restrictions apply. An absolute fundamental of transmission line theory is the idea of travelling waves on the line and the possibility that waves can travel in both directions at the same time. This has been the basis of full duplex telephony on two-wire circuits for more than a century. Waves travelling along a line will induce a current in the line. Two waves travelling on a line will induce two currents on the line. One could conceive of alternative models, for instance it would be possible to represent the system as the sum of a standing wave and a travelllng wave with the short-circuit line being a special case where the travelling wave has gone to zero. Such a model, while giving correct answers, suffers from not being able to directly assign causal phenomena to the two elements as one can do with two travelling wave representation. More to the point from a Wikipedia standpoint, reliable sources do not usually use such a model. SpinningSpark 17:18, 25 September 2013 (UTC)[reply]
Hello Spinningspark. In our discussion we might be blending fields with actual movement of current/electrons in conductors. In the case of an antenna, what actually leaves is a special combination of electric (electrostatic) and magnetic fields, not actually current flow (as in conductors). I'm with you on "waves" (including volts and amps) traveling in opposite directions on a transmission line....where we differ is what happens at a "boundary" condition where voltage or current is "locked" at zero. An analogy is a wave traveling in the ocean. It has both horizontal and vertical (cyclical) components of actual water flow except if it hits against a vertical seawall, there is no horizontal flow at that boundary. I am enjoying this discussion, but don't think you and me deciding this is in the way of a GA review. In the end if a source says "can be visualized as" or "actually is" then that is good enough. Sincerely, North8000 (talk) 18:01, 25 September 2013 (UTC)[reply]
There is no difference in the behaviour of fields or electron flow in this discussion. An em field will induce a current flow in the conductor, or alternatively you can take the current as the principle phenomenon and believe the current induces a (magnetic) field. Either way when one is zero so is the other. The point of the antenna example has nothing to do with what is radiated from the antenna. The point was that current is injected into the antenna from the line but does not get reflected back (or at least only a small part of it). That is, it is perfectly possible for a current wave to be flowing in only one direction at the end of the line. Why do you believe that the current node at the end of the line is any different from any other node anywhere along the line? Presumably because there is a physical short there, just as there is a physical barrier in the sea-wave example. But in both cases there is no objective difference for anything one can measure. Wave energy will be measured going in both directions at all points. By the way, you are incorrect that ocean waves have a horizontal current component. They do not, it is all vertical (but waves roll up the beach because of drag on the bottom of the wave as they hit shallow water) parcels of water just move up and down, it is the energy of the wave that moves forward. If one were to confine the sea-waves to a narrow channel the reflected wave would cause standing waves just like on a transmission line. There would be a series of nodes where the water was not moving up and down but wave energy would still be passing through that point (in both directions). As for sources Connor: ...a discontinuity exists at the end of the line and reflected waves of voltage and current will be set up on the line and they travel back towards the sending end; Carr: Because there are both reflected and incident waves, we find that V and I are actually the sum of incident and reflected voltages and currents; Pai & Zhang: That is in a uniform transmission line, both the voltage and current in general are composed of two travelling waves going respectively in opposite directions. Not all the sources are going entirely my way - Bowick et al: voltage, current, or power emanating from a source impedance and delivered to a load can be considered to be in the form of incident and reflected waves travelling in opposite directions along a transmission line but I would say that they are definitely in the minority, the others don't feel the need to use such circumspect language. SpinningSpark 23:55, 25 September 2013 (UTC)[reply]
I disagree with some things in the first 60% of your post. Answering your one question, because a node anywhere else on the line is a midpoint of a an electrical circuit where there are conductors and (distributed) capacitances attached to those conductors on either side of it and the node at the end of the line isn't. On water waves, I agree with your statement that there is no net forward current, but that is not what I was talking about and so not the "false dichotomy" conclusion that such means there is no horizontal motion to a particle of water away from the boundary. Various sources say that there is a circular motion (e.g. [1]). But on wiki the point at hand, if you have even ONE source that either talks about current flow at the END point of an open circuit, or describes the zero current condition at the END point of an open circuit as being the sum of opposing currents, then I'm cool with it. IMHO from what was given neither of the two that you gave seem to say that.....both seem to be talking about the transmission line and not about the point of an open circuit termination of it. (Connor about what happens in the transmission line as a result of the reflections form the open circuit, not about what is happening at the point of the open circuit) Sincerely, North8000 (talk) 12:30, 26 September 2013 (UTC)[reply]
<sigh>The sources I quote simply do not make a distinction between the end of the line and other points, that in itself should be telling you something. They are plainly describing the situation I have given. I restricted the quotes above to sources in the article, but if we look for other sources we can easily find many more. Here is a small sample;
This is my final word on this matter. If you still find it unpalatable, go ahead and fail the article and I'll see you at GAR. You may be confused by the concept, but the article states what the concept is very clearly so I don't see how this can be an issue of clarity in terms of the requirements of GA criterion 1a. Regards, SpinningSpark 13:56, 26 September 2013 (UTC)[reply]
One more point. The thing that really would need citing is the entirely unphysical idea that these currents exist everywhere on the line, even at an infinitesimal distance from the termination, but not at the termination itself, even though that is the very thing that has created the reflected wave in the first place. SpinningSpark 14:07, 26 September 2013 (UTC)[reply]
Hello Spinningspark. Two of the sources that you just linked me to use that method to explain it. That was the "wiki" question at hand and so that resolves it. I didn't understand the "see me at GAR, if it were non-passed it could be simply resubmitted and someone else would review it. But I already offered to ask for a second opinion, and step aside and let them decide the result. If you think that might be better I'd be happy to do so. But we might have just gotten past the biggest open item. North8000 (talk) 14:56, 26 September 2013 (UTC)[reply]
All the sources support this view. All the original sources in the article and all the ones I just gave you. You are simply refusing to accept that is what the sources mean if there is any possible way to misinterpret them and wriggle out of it because you are having difficulty accepting this model is correct. The last source (which I assume is one you have discounted) is particularly telling. I have no idea how you can fail to interpret "waves fill the entire transmission element" as not including the ends. SpinningSpark 16:44, 26 September 2013 (UTC)[reply]

Possibly my last post wasn't clear. It is resolved and fine. Sincerely, North8000 (talk) 17:45, 26 September 2013 (UTC)[reply]

Arbitrary break #2[edit]

  • This may be better left unexplored, but I believe that the first picture might mislabel power distribution lines as telephone lines. And on power distribution lines the effect of that crud on the line would be negligible. North8000 (talk) 21:39, 24 September 2013 (UTC)[reply]
  • I think you are right, those lines look very much like the power distribution from a star transformer of 3 phases plus star-point neutral. I had not looked too carefully at it before and took the image description at face value. Pity, it is a nice picture. (Although when I wrote wildlife can cause discontinuities I had in mind giant African rats which have a real taste for electric cables for some reason). SpinningSpark 22:28, 24 September 2013 (UTC)[reply]
  • The discussions and diagrams later in the open circuit section seem perplexing. There is talk of power lines and generators (presumably power generators) but the symbols appear to be those of a(n ideal) voltage source. The example isn't really discussing the regular voltage (50Hz or 60Hz sinusoidal) it is discussing phenomena where there is a switching interruption and the generator is not even in the circuit. Also it show the circuit with a series impedance (= not the type which is the subject of this article) and a voltage division (halving) which is not the case for power lines. It is also describing & showing a case where the generator is prodcing twice the voltage what is being delivered to the termination, a situation that I don't think occurs in either type of transmission line. Finally, if it is a(n ideal) voltage source as the symbols seem to show, the voltage would be by definition the voltage of the voltage source and described as such. North8000 (talk) 21:53, 24 September 2013 (UTC)[reply]
    As an undergraduate student I had a professor who undertook some research work for London Underground. At a certain point on the District Line a large number of light bulbs were regularly blowing. My professor suspected some kind of transmission line effect but this was not believed by the railway engineers. The railway was powered by DC so it was surely not affected by this sort of stuff they thought; that only applies at RF and in any case it is all semi-mythical nonsense in their eyes. But in actual fact that was exactly what was happening; a discontinuity existed as the train transited between two sections of track, a doubling of voltage occured and all the lights blew. Well not all of them, but on average any one bulb would survive only a limited number of transits. For the most part, transmission line effects can be ignored on power distribution, but that does not mean that they do not exist. Two things need pointing out here. Firstly, a perfect DC voltage step does not contain only DC power. It actually contains some power at all frequencies up to infinity as Fourier analysis will readily show. Secondly, when the train hits the discontinuity the reflected wave generated does not "know" that it is connected to a low-impedance source, it only knows about the characteristic impedance of the line at that immediate locality. True enough a couple of hundred microseconds later the wave will return with that "information" that it is actually connected to Lots Road Power Station and the voltage will suddenly realise "crikey, I shouldn't be doing that, I'll stop now" but in the interval the equipment has been under severe overload.
The diagrams with a series resistor are meant to represent the equivalent circuit of a source, or in the kind of case described above, the equivalent circuit of a section of line, not the actual arrangement of components in a source. Nevertheless, in telecommunications faced with matching an ideal voltage source to a line that is exactly what we would do, place it in series with a resistor. During the sort of disturbance described for power lines it is a legitimate equivalent circuit of the line. Generator here means the ideal circuit element of a voltage (or current) generator, not an actual power station. Honestly, your comments are beginning to make me think that the article would be clearer without mentioning power lines at all. They are not essential to the article although I think it is interesting to mention them.
The voltage generators have been assigned 2V in the diagrams in order to make the normal voltage on the line simply V. This is just an arbitrary choice. We could just as well make the generator V, then the normal voltage on the line would be 1/2 V. It would still be the case that the voltage would double if the line went open circuit. SpinningSpark 00:08, 25 September 2013 (UTC)[reply]
(dealing only partially with your thorough comments on this item) It certainly is a pleasure and a privilege to work with you on these items. I think that your "the article would be clearer without mentioning power lines at all" is true with respect to talking about 50Hz/60Hz sinusoidal power. I think that just confuses matters in a realm where transmission line theory is not very relevant. But I think that behavior on power distribution during disturbances is a good/practical/important sidebar, especially once one introduces it with an explanation of transients getting produced ("sourced") by interruption of an inductive load. Sincerely, North8000 (talk) 00:58, 25 September 2013 (UTC)[reply]
I guess one example is where it is identified as a power distribution circuit. Above you indicated that the source symbol represented and ideal voltage source and the in line impedance symbol represented the internal impedance of the generator. But the description shows the load equal to the internal impedance of the generator and indicated that it is "matched". While matching is done on many circuits for maximum power transfer and minimum reflection, such is not the case on power distribution where such would be a 20x or 50x overcurrent. Sincerely, North8000 (talk) 18:18, 26 September 2013 (UTC)[reply]
What "it" are you referring to? None of the diagrams identify themselves as being power distribution circuits. Nor are they meant to be and nor is the text description. You might want to quote the exact text you have an issue with. The comment about power circuits ("This doubling of voltage...a voltage spike results"), while still true, is meant to be an aside and the following text is not part of that discussion. I think I already agreed somewhere else that this comment about power circuits could be removed as it is tangential to the article. SpinningSpark 23:23, 26 September 2013 (UTC)[reply]
I'm going to take one more look that the referencing isn't too skinny and then pass this article. So this item is just conversation. That said I'll answer in a minute. North8000 (talk) 00:13, 27 September 2013 (UTC)[reply]
The text I was referring to is (and the referenced figure 2):
This doubling of voltage is a problem frequently arising when switching operations are carried out on power grids. A brief open circuit while switching is in progress will cause such a reflection and a voltage "spike" results. This counter-intuitive result may become clearer if the circuit voltages are considered when the line is so short that it can be ignored for the purposes of analysis. The equivalent circuit of a generator matched to a load \scriptstyle Z_0 to which it is delivering a voltage \scriptstyle V can be represented as in figure 2. That is, the generator can be represented as an ideal voltage generator of twice the voltage it is to deliver and an internal impedance of \scriptstyle Z_0
Sincerely, North8000 (talk) 00:18, 27 September 2013 (UTC)[reply]
Yes, I see the problem, the diversion into power spikes makes it look like the whole paragraph is about power spikes which it is not. I have solved the problem by removing the diversion altogether. SpinningSpark 00:26, 27 September 2013 (UTC)[reply]
  • In the short circuit paragraph, it is discussing the end point of the line where the voltage is by definition zero. The text describes that there are two voltages at the end of the line traveling in opposite directions which cancel each other out. Isn't this confusing at best, and, when talking about either of the individual voltages, in conflict with the definition of an short circuit? North8000 (talk) 21:59, 24 September 2013 (UTC)[reply]
    As with the open circuit case, there really are two waves travelling in opposite directions. These two waves can be separately detected and measured using such devices as directional couplers or circulators. Why should this be in conflict with the definition of a short circuit? What definition are you using? One definition is a circuit element that has zero volts across it no matter how much current is passed through it. It certainly meets that definition, the two voltage waves cancel at the short-circuit. In fact the short-circuit actually causes the reflected wave to exist in order to carry on being a short-circuit. One cannot simply say that the short-circuit causes the voltage to be zero everywhere on the line. That leaves completely inexplicable the fact that a short-circuited line energised by an AC source has a non-zero voltage everywhere on the line except for at the short-circuit (and a finite number of nodes). SpinningSpark 00:29, 25 September 2013 (UTC)[reply]
We can consolidate this at the open circuit post/discussion ......same as that except here is voltage instead of current. Sincerely, North8000 (talk) 11:53, 25 September 2013 (UTC)[reply]

GA criteria final checklist[edit]

Well-written

  • Meets this criteria. Pretty good job on an immensely-difficult-to-explain topic. North8000 (talk) 18:20, 26 September 2013 (UTC)[reply]

Factually accurate and verifiable

  • Meets this criteria. North8000 (talk) 01:30, 27 September 2013 (UTC)[reply]

Broad in its coverage

  • Meets this criteria. North8000 (talk) 14:30, 28 August 2013 (UTC)[reply]

Neutral: it represents viewpoints fairly and without bias, giving due weight to each

  • Meets this criteria. North8000 (talk) 20:11, 25 August 2013 (UTC)[reply]

Stable: it does not change significantly from day to day because of an ongoing edit war or content dispute

  • Meets this criteria. North8000 (talk) 13:03, 23 August 2013 (UTC)[reply]

Illustrated, if possible, by images

  • Meets this criteria. Has 7 images. None are non-free so no article-specific use rationales are required. North8000 (talk) 12:59, 23 August 2013 (UTC)[reply]
  • Rechecked substituted image. North8000 (talk) 01:33, 27 September 2013 (UTC)[reply]

Result[edit]

Congratulations. This has passed as a Wikipedia good article. Sincerely, North8000 (talk) 01:37, 27 September 2013 (UTC) Reviewer[reply]

Thank you so much for reviewing this article. It can be really difficult to find reviewers willing to take on highly technical articles, although I hope this one is now more easily understandable through your efforts. Thanks again. SpinningSpark 10:03, 27 September 2013 (UTC)[reply]
It is a privilege and a pleasure working with you. Even when we see/saw things a little differently. Wikipedia and its readers are fortunate to have folks like you writing these difficult articles. Sincerely, North8000 (talk) 12:03, 27 September 2013 (UTC) Reviewer[reply]

Congratulations. This article has passed as a Wikipedia Good Article[edit]

This is "duplicated" here for when the review is no longer transcluded.

Congratulations! This article has passed as a Wikipedia Good Article. Sincerely, North8000 (talk) 01:42, 27 September 2013 (UTC) Reviewer[reply]

Is "wave" correct terminology[edit]

Is "wave" correct terminology for pulses being reflected in a conducting line? I see that term used extensively in the literature, but I feel it is inaccurate and in use simply because the pulse's movement would resemble a wave. Please advice. — Preceding unsigned comment added by 2001:700:300:1724:0:0:0:63 (talk) 17:59, 7 October 2015 (UTC)[reply]

Generally wave means a solution of the wave equation. Although we focus on current and voltage, there is literally an electromagnetic wave propagation in the dielectric between the conductors.Constant314 (talk) 20:22, 7 October 2015 (UTC)[reply]
As far as my understanding of this "wave" goes, it propagates due to fact that it has to charge all the capacitances in the "infinite" amount of exact-PI equivalents in series. It cannot fill C2 before C1 is full, and so on. Further, the relationship between the cable capacitance and inductance yields the propagation speed. An electromagnetic wave should be traveling in air, like visible light or radio signals etc. In this topic, I think that the correct terminology is "propagating (voltage) pulse", "reflection of (voltage) pulse" etc. The term "wave" should be left out altogether.2001:700:300:1726:0:0:0:1D (talk) 00:25, 8 October 2015 (UTC)[reply]
Waves propagate readily in the dielectrics such as glass and Teflon and even weakly into metal. There is a full featured electromagnetic wave in the dielectric that fully satisfies Maxwell’s equations and the wave equation. The insulation is in effect transparent at the frequencies of the pulse. The transmission line analysis happens to exist right on the border between circuit theory and full electromagnetic field theory. The transmission line can be solved entirely by electromagnetic field methods, but thanks to Heaviside’s learned insight the fundamental mode can also be solved as an infinite series of identical infinitesimal pi-sections. But the ultimate arbitrators on Wikipedia are reliable sources. There are literally hundreds of text books that call it a wave.Constant314 (talk) 00:47, 8 October 2015 (UTC)[reply]
Thank you for clarifying. 2001:700:300:1726:0:0:0:28 (talk) 19:38, 8 October 2015 (UTC)[reply]

Signal reflection[edit]

Would someone be able to merge some of this article to signal reflection? It would help to have some of the explanatory material from this manifestation of the phenomenon (especially since the other article doesn't have any sources), but I don't feel familiar enough with it to do it myself. Thank you. — Sasuke Sarutobi (talk) 12:49, 14 December 2015 (UTC)[reply]

I believe that article should be merged here. fgnievinski (talk) 01:53, 28 March 2016 (UTC)[reply]
  • Oppose merge. This article is specifically about reflections on conductors. Signal reflection is a broader topic and includes material on optical fibres which would be inappropriate to merge here. In principle, that article could be expanded to include non-electromagnetic waves in other media. In addition, this page is listed as a good article and we shouldn't be merging unsourced information into it. SpinningSpark 07:25, 28 March 2016 (UTC)[reply]
  • Oppose merge. ditto. Signal reflection has no citations even though it was noted since 2007. Perhaps it is a candidate for deletion.Constant314 (talk) 04:48, 29 March 2016 (UTC)[reply]

"Transmission coefficient" equations seem to be wrong (Section "Discontinuity along line")[edit]

Just tried doing some hand proofs and I thought the initial equation for T is wrong. I changed it from:

To:

As I understand: Reflection Coeff + Transmission Coeff = 1

Can someone confirm this? I'm not sure how to modify the next T equation. I'm trying to find sources.

(static ip) 14.203.193.219 (talk) 06:34, 3 June 2016 (UTC)[reply]

I think that is wrong. Transmission Coeff = 1 + Reflection Coeff
Test case, shorted line → reflection coef = -1, Transmission Coeff = 0 Constant314 (talk) 08:56, 3 June 2016 (UTC)[reply]
Yes it is. " Transmission Coeff = 1 + Reflection Coeff " is indeed the correct equation, which leads to Z02 on top, not Z01.
Thanks for reverting.
(static ip) 14.203.193.219 (talk) 10:28, 3 June 2016 (UTC)[reply]

Real Operator[edit]

@Spinningspark: Have you looked at what it looks like? Did it always look that way?

That looks like the symbol for the set of real numbers instead of the real part operator. My small sample of seven (Jackson, Kuo, Balanis, Kraus, Harington and two by Hayt) books, all use "Re" for the real operator instead of a script R. I had not noticed it before. Has something changed with LayTex? Constant314 (talk) 17:18, 22 July 2017 (UTC)[reply]

The symbol for the set of real numbers is usually the blackboard bold . The code \Re gets a Fraktur font in all three browsers I have tried it in. You may have some odd setting in your preferences or CSS if you are getting something different. Try logging out and see if you still have the same problem. Using \Re is the right thing to do here because it has the right semantic meaning in the code. Any rendering problem needs fixing elsewhere rather than using code kludges. I might agree that Re is a more intuitive rendering, but my biggest objection to your edit was that it was not self-consistent throughout the article. SpinningSpark 18:55, 22 July 2017 (UTC)[reply]
I thought that I got them all, but now I see that I missed a couple. So,as long as I get all instances (I see four now), you would not object?Constant314 (talk) 19:12, 22 July 2017 (UTC)[reply]
I don't really have any strong opinion, but it might be wise to discuss this more generally first. I seem to remember that I started off writing Re() and Im() (but not in this article apparently) and later on other editors changed it. I suspect that the issue will be maintaining the semantic content of the code as I said above. Give me a day or so and I might be able to come up with examples of articles where that change was made. SpinningSpark 19:29, 22 July 2017 (UTC)[reply]
It was a recent edit to plane wave that got my attention. I noticed that the article on complex numbers and phasors uses \operatorname{Re}.Constant314 (talk) 22:16, 22 July 2017 (UTC)[reply]
I'm not really coming up with anything, I can't remember precisely what started me formatting it this way. Go ahead and change it if you want. SpinningSpark 10:03, 23 July 2017 (UTC)[reply]
I think I got them all changed. I found six instances. Constant314 (talk) 18:18, 29 July 2017 (UTC)[reply]

wrong formula in the specific cases section[edit]

In the specific cases section, step voltage is applied to ONE END of a lossless line, and is a function of time t. However, the given formula and are using as if were a function of a distance, which is quite a nonsense.

Remember that indicates voltage AT one end of a line in a time-varying domain.

The correct formula for the incident voltage and current at the other end of the line which is distance x away are

. — Preceding unsigned comment added by Widefoot (talkcontribs) 23:03, 20 February 2019 (UTC)[reply]
I can see both points of view. We really don't care about the units of the unit step, but it clearly specified in the preceding lines as a function of time. I agree that
is probably an abuse of notation. Constant314 (talk) 23:43, 20 February 2019 (UTC)[reply]
Constant314, it is not a matter of abuse but correct or incorrect. Widefoot (talk) 00:57, 21 February 2019 (UTC)[reply]
Looks like this article has been around for a while. I am quite surprised that this simple math problem has not been reported by anyone so far. And, I wonder whether Wikipedia has a means to check when someone raises a serious question about correctness of information. For Wikipedia to gain more authoritative status, it will need groups of knowledgeable people in different areas who are willing to participate in checking validity issues when questions are raised. Widefoot (talk) 07:42, 21 February 2019 (UTC)[reply]
Widefoot, there is no body experts because WP has been bitten by so called experts who were not who they said they were or did not have the credentials that they claimed. WP is not an authority; it is an aggregate of what is said in reliable secondary sources.Constant314 (talk) 19:26, 21 February 2019 (UTC)[reply]
Please pay attention to the terminology used here. The article does not define the unit step function as a function of time. It is no more true to say that is a function of time, than to say the quadratic equation is a function of time. can be a function of any variable we choose. Certainly is a function of time, and the applied voltage at the input of the line is defined as a function of time. But we are next interested in the value of incident voltage at some distance from the start of the line. For this a function of distance is more appropriate. For that, is not an incorrect expression. SpinningSpark 12:10, 21 February 2019 (UTC)[reply]
I support because, to me, it more obviously says that is zero until a time such that Constant314 (talk) 19:26, 21 February 2019 (UTC)[reply]
SpinningSpark, The article DOES define the unit step function as a function of time : " is the unit step function with time ." If you are in doubt, I suggest that you go ask some math professor what this statement means.
Ok, I didn't want to go this far, but I will try to explain.
describes the incident voltage at the one end of the line in consideration. Let's say the other end of the line in consideration is distance away from the one end. Also, if the speed of propagation is , then it takes time for a change at the one end to reach the other end. That is, the voltage seen at the other end is the voltage seen time ago at the one end. Therefore, in math form, the voltage at the other end is . — Preceding unsigned comment added by Widefoot (talkcontribs) 20:26, 21 February 2019 (UTC)[reply]
I was never doubting the correctness of that expression, only whether it is the best way to present it. I've decided it's not really worth getting into a stew over this and withdraw my objection. SpinningSpark 20:29, 21 February 2019 (UTC)[reply]
Widefoot, you have a talk page. To access it, look above where you will find your name in red followed by "(talk)". Click on talk to see your talk page. Constant314 (talk) 20:42, 21 February 2019 (UTC)[reply]

Merge with three (or more) articles[edit]

There is one stub article

and two sub-articles

which appear to have a great deal of overlap. I suggest that Reflections of signals on conducting lines and Signal reflection should be merged, perhaps under the shorter title used by the stub. It also seems to me that at least part of the explanatory text in Transmission coefficient and Reflection coefficient should be merged into this article (or its successor) if not already present, and replaced with a "see also" notice referring to this article. Astro-Tom-ical (talk) 02:57, 5 December 2019 (UTC)[reply]

This was proposed before (and failed). See further up the page for the rationale. SpinningSpark 11:24, 11 December 2019 (UTC)[reply]