# Talk:Transmission line

## Input impedance of transmission line

In

${\displaystyle Z_{in}\left(l\right)={\frac {V(l)}{I(l)}}=Z_{0}{\frac {1+\Gamma _{L}e^{-2\gamma l}}{1-\Gamma _{L}e^{-2\gamma l}}}}$

exponent 2 is due to back and forth reflection?

— Preceding unsigned comment added by 82.56.167.93 (talk) 07:48, 4 August 2014 (UTC)

Yes, ${\displaystyle \scriptstyle {2l}}$ is the distance from measurement point to the load and back again and so ${\displaystyle \scriptstyle {e^{-2\gamma l}}}$ is the phase delay for the round-trip, and ${\displaystyle \scriptstyle {\Gamma _{L}e^{-2\gamma l}}}$ is the voltage amplitude of the reflected wave relative to that of the incident wave. --catslash (talk) 17:52, 4 August 2014 (UTC)

## Input impedance of transmission line: the reason(s) why ${\displaystyle Z_{in}\neq Z_{0}}$

In
"Since most transmission lines also have a reflected wave, the characteristic impedance is generally not the impedance that is measured on the line",
for completeness sake, shouldn't it be:
Since most transmission lines also have reflected waves and standing waves...?

I am not an expert but I am just reading here and there and it seems to me that those 2 are different phenomena:

• reflected waves are due to unmatched lines (terminations);
• standing waves may be generated by reflected waves, which thing usually happens;

don't both of them contribute to the reasons why ${\displaystyle Z_{in}\neq Z_{0}}$?

This way, the reasons why ${\displaystyle Z_{in}\neq Z_{L}}$ should be

• losses due to R and G of the line that we often neglect;
• reflected waves;
• standing waves;

— Preceding unsigned comment added by 95.237.220.142 (talk) 09:22, 6 August 2014 (UTC)

The standing wave is due to interference of the incident and reflected wave. It is not a separate phenomenon. The input impedance is a function of distance, ZL, Z0 and γ. The last two are functions of R and G amongst other things, but are not additional parameters needing to be taken into account—they are already included. SpinningSpark 14:24, 6 August 2014 (UTC)
Agree with Spinningspark: Reflected waves are the cause, standing waves are the effect. If you say that the measured impedance is affected by reflected waves, that already implicitly includes anything related to standing waves too. --Steve (talk) 14:53, 6 August 2014 (UTC)

## "Schematic showing how a wave flows down a lossless transmission line" is wrong

Why has no-one noticed that the handy animated diagram at the top of the paper is wrong? Current maxima occur in parallel along the two wires. At the load the current must be at a maximum at both ends of the load simultaneously to satisfy charge conservation. I propose to delete it at a date at least 2 months from today. 77.96.212.249 (talk) 16:27, 24 October 2014 (UTC)

"Current maxima occur in parallel along the two wires" -- yes, when the electrons are moving most quickly leftward in the bottom wire, at the same moment directly above they are moving most quickly rightward in the top wire. That's what you're saying, right? I believe that this is true in the animation. For example, at the left edge, you can see the electrons entering the image on top at the same time as they exit the image on the bottom, and vice-versa.
The black dots represent electrons. So a violation of charge conservation would be black dots appearing out of nowhere, or disappearing. This does not happen in the animation. Do you agree?
I made the animation, so I'm biased. :-D But that also means I can easily correct it if I made a mistake. (But I don't think I did.) --Steve (talk) 16:38, 24 October 2014 (UTC)
I don't know what to make of the dots; it isn't intuitively obvious what electrons being bunched together has to do with current.
As for the voltage, the caption is unclear. Is a low voltage the most negative voltage, or does it mean low RMS voltage? Jc3s5h (talk) 17:06, 24 October 2014 (UTC)
Stare at just one of the dots, ignoring everything else. Watch as it moves back and forth. Is the dot moving left? That means there are electrons moving left, i.e. there is a current flow. Is the dot stationary? That means that the electrons are stationary, i.e. the current is zero at that location and time. Does that make sense?
I was assuming that most readers would find it intuitively obvious that when electrons move, it's a current. Is there any other way to think about current? That said, if you find it confusing then it's a safe bet that other people do too! So I hope you will help me understand where you're coming from. :-D
The color is instantaneous voltage, not RMS voltage. (RMS voltage does not oscillate :-P ) Do you think it would help if I changed the sentence from "Red color indicates high voltage, and blue indicates low voltage." to "Red color indicates a more positive voltage, and blue indicates a more negative voltage." ? Or is there another wording that would be better? --Steve (talk) 17:59, 24 October 2014 (UTC)
It would be much clearer if the load was electrically small instead of distributed over 5λ/16. If we assume the load is actually electrically small but simply not drawn to scale, then it should carry no current at all (because the upper and lower loops are each precisely λ/4 long, and therefore each present an infinite impedance in series with the load. --catslash (talk) 18:31, 24 October 2014 (UTC)
Diagrams are supposed to make the text easier to understand. If the reader has to ponder the drawing for 15 minutes to figure out what it means, it should probably be deleted. I especially object to complex drawings that move. If I'm having trouble understand something, I want it to stay still while I try to figure it out. Right now, it reminds me of a clay pigeon. The way to deal with those is to avoid thinking; just point and shoot. Jc3s5h (talk) 18:42, 24 October 2014 (UTC)
Catslash - This is a good point. I was imagining not only the load but also those thinner wires are electrically small (but not drawn to scale). Does that make sense? Now that you have pointed it out, I can see how that's very confusing!
Therefore I propose to erase the load altogether, and just have a simpler image with the two thick wires going straight through from one side of the image to the other. Basically I would crop off the right side of the image. Do you think that would help?
Another option is making an electrically small impedance-matched resistive load that is very small (drawn to scale), and with no weird leads. But I'm not sure how to fit it or what it could look like.
Jc3s5h: You are welcome to argue that the animation should be removed entirely. I disagree of course. We can see what other people think. But maybe you can also spend a minute thinking about how to improve the animation and/or caption (or as you might say: make it less bad)? For example, I proposed a small change to the caption text above, can you comment on it? Or suggest other changes that would make it easier to understand? --Steve (talk) 19:26, 24 October 2014 (UTC)
I like animations, but this one confuses me too. For me, it moves too fast to make sense of it. I think what might work better is a single pulse with some dead time before the next pulse.Constant314 (talk) 21:11, 24 October 2014 (UTC)
Yes, it's conventional is some contexts to draw transmission lines in bold and electrically small connections thinner as in your file:Smith_chart_explanation.svg diagram. However in mapping the current distribution, the actual size of the various parts is important, and so the change of scale is confusing. It's particularly unlucky that you chose 160 pixels for λ and 40 pixels for the loops! It's not necessary for the load to resemble a IEC-style resistor symbol, it could just be a small spot, maybe a small spot glowing red hot or even a small spot with a glow pulsing at the power harmonic.
There are a couple of other possible objections to the diagram as it presently stands. Firstly, the use of colour to show the 'voltage'; there is no well-defined electrical potential here as the E-field is far from conservative. In particular there is a potential gradient between the two conductors but not along them (assuming high conductivity), whereas in the diagram there is a prominent longitudinal colour gradient. It's more common to depict the E-field as arrows from one conductor to the other, bunched up where the field is strongest - but I don't know how you would animate that.
The second (lesser), objection is that the moving electrons should be on the surface facing the other conductor and not on the outer side. --catslash (talk) 21:25, 24 October 2014 (UTC)

──────────────────────────────────────────────────────────────────────────────────────────────────── Catslash - a highly conductive (or even infinitely conductive) wire can have an electric potential gradient along it [Update: Sorry, I was wrong here. I should have said "voltage drop down it, sorta".] ... if it's inductive! Which in this case it is! So, I agree that there are situations where there is no well-defined electric potential but I don't think this is one of them, luckily. The longitudinal color gradient is a correct representation of what's going on with the distributed inductance. Maybe the caption could even say this? "Voltage drops between the wires are related to capacitance, while voltage drops along the wires are related to inductance"...

Basic physics says that there is never a E-field component directed along the surface of a pure conductor. The E-field is the gradient of the potential only when a potential exists. The condition for a potential to exist is curl(E) = 0, which is not the case in the space round a transmission line where at every point curl E = dB/dt (one of Maxwell's equations). I'm not a fan of descriptive physics but one can think of the 'voltage' induced by a changing magnetic field linking a closed loop even though it's made from a perfect conductor, and this is a similar case where there is a potential across the ends of the loop even though there is no E-field (potential gradient) along the loop. 77.96.212.249 (talk) 12:33, 26 October 2014 (UTC)
The so called retarded potentials are always well defined except in impossible situations like inside a perfect conductor. Constant314 (talk) 20:54, 27 October 2014 (UTC)

Well, I'm going to try re-making the diagram with various changes: Make the waves move slower, make the wavelength longer, and throw out the terminal resistor altogether (notwithstanding Catslash's good suggestions to improve it). I think all these changes will make it easier to see what's going on. I will also put the electrons inside the wires, at the appropriate sides, assuming that it looks OK. Thanks again for the suggestions and feedback :-D --Steve (talk) 23:26, 24 October 2014 (UTC)

Look at the diagram: electrons bunched up means electrons changing direction - zero current. Electrons spaced out means electrons travelling quickly - maximum current. At the feed point, the diagram shows maximum current on one side coinciding with zero current on the other side, a breach of charge-conservation. 77.96.212.249 (talk) 16:07, 25 October 2014 (UTC)

Responding to _.249, The driver, if it is balanced, pulls electrons off one wire and puts them on the other wire. As a result a maximum of the electron density (electrons bunched up) is matched by a minimum of electron density (electrons spaced out) on the other wire as the signal travels down the transmission line. The minimum and the maximum meet at the load and cancel each other.Constant314 (talk) 23:14, 26 October 2014 (UTC)
77.96.212.249 - What Constant314 said, and also: The electrons at the center of a bunch are not stationary. On the contrary, an electron at the center of a bunch has the maximum possible rightward velocity. I hope this is easier to see in the new version -- take another look. :-D --Steve (talk) 14:37, 27 October 2014 (UTC)
Most people tend to think of a voltage signal traveling down a transmission line, but it is just as valid to think of it as current signal and they are related by the characteristic impedance. In the usual situation where the characteristic impedance is real, the peaks of voltage and the peaks or current are coincident. Constant314 (talk) 20:54, 27 October 2014 (UTC)
Yes, inductance must exist. A lossless, inductanceless line implies infinite wave velocities, which is impossible of course. Any geometry that gives rise to capacitance, must give rise to inductance through duality. SpinningSpark 00:12, 25 October 2014 (UTC)
Before re-making the diagrams you might want to review some field equations and reconsider your assertion that there is a longitudinal potential gradient (that is a longitudinal E-field and so not a TEM wave). There is of course inductance, but almost all of this is in the dielectric and hardly any in the metal (because the volume of the skin is very small compared with the volume of the dielectric). --catslash (talk) 00:24, 25 October 2014 (UTC)
It is easy to suppose that if there is a voltage drop straight across between the terminals of a coil of wire, then the same voltage drop must exist between the terminals following the path of the wire, but this is Kirchhoff-think and isn't so. The voltage drop for the round trip across the gap between the terminals and back along the path of the wire is equal to the rate of change of magnetic flux linking the path and is in general non-zero. If the wire is a (hypothetical) perfect conductor, then the voltage drop along the portion of the path following the wire is zero, while the voltage across the gap is non-zero. The same is true for the transmission line; the fact that the voltage between the two conductors is different at different points along the length does not require a counterbalancing voltage drop along the conductors. It can be very irritating when people presume to tell you what you think, so please forgive me if this wasn't your line of reasoning. --catslash (talk) 02:08, 25 October 2014 (UTC)
I think what happens is that there is an electric potential and a magnetic potential and that the E-fields produced by these potentials cancel each other. So there is an electric potential distribution along the wire but the E field tangent to the wire is zero. I have not run the numbers, but the directions of the E-fields produced by the two potentials are in opposite directions. Or it's too late at night for me to think straight. :) Constant314 (talk) 06:34, 25 October 2014 (UTC)
Yes, it is of course possible to define a vector potential ${\displaystyle \scriptstyle {\mathbf {A} }}$ and scalar potential ${\displaystyle \scriptstyle {V}}$ such that
${\displaystyle \mathbf {E} +{\frac {\partial \mathbf {A} }{\partial t}}=-\nabla V}$
and
${\displaystyle \mathbf {B} =\nabla \times \mathbf {A} }$
but many readers will not be sufficiently familiar with classical electromagnetism to understand the voltage in the animation in this way. --catslash (talk) 11:31, 25 October 2014 (UTC)
Yes, very few (I would guess less than 1%) would understand the role of the vector potential and I don’t suggest try to add it to the animation. I only mention it because it tends to validate the Steve’s animation which shows an electric potential that varies along the wire. The animation is, I believe, correct, even if the typical reader will not understand all its implications. That is the case with most of E&M.; it is complicated and it is difficult for the animator to get all the details into the animation. I think that, in this case, it would be sufficient to say that although there is a non-zero component of voltage gradient along the wire, there is no E-field component along the wire due to magnetic effects.
But, looking at it a different way, a voltage drop across the inductance is assumed in the derivation of the telegrapher’s equations. Constant314 (talk) 15:29, 25 October 2014 (UTC)
(1) If you naively look at the animation and compare it to the equation V=L*dI/dt, it works perfectly. This cannot just be a coincidence. It suggests that this plot of V reflects some kind of deep truth about how the transmission line works, even if it's not immediately obvious what the details are. :-D
(2) I definitely agree that there is no electric field along the wires (TEM wave).
(3) There is a question "What is the exact voltage distribution inside an inductor (the coil and core) and how is voltage defined there?" This question does not come up in normal circuit theory education because you only care about the effect of the inductor on the outside world (the rest of the circuit); and in the outside world there is always a well-defined V with ∇V=E. Unfortunately, this is not good enough for distributed inductance. :-P
(4) My #1 vote right now is not throwing out the colors but rather adding clarification / cautionary text to the caption. Maybe as simple as "Voltage drops between the wires are related to capacitance (Q=CV), while voltage drops along the wires are related to inductance (V=L*dI/dt)." At least that might get people in the mindset of thinking about lumped-component models (cf Telegrapher's equations) where "voltage drop" would be a literally correct description. --Steve (talk) 17:13, 25 October 2014 (UTC)
If you have a voltage drop along the wires, by duality you would have a current drop between the wires. Curent drop between the wires also shows up as part of the derivation of the telegrapher's equations. Constant314 (talk) 18:20, 25 October 2014 (UTC)
Here is a snap shot of a differential Gaussian pulse as I understand it:
A differential Gaussian pulse on a two-wire transmission line showing electron density, currents, electric potential, magnetic vector potential
Constant314 (talk) 18:20, 25 October 2014 (UTC)
HERE is a new version draft with just those minor changes I mentioned above.
It looks correct to me. Constant314 (talk) 22:36, 25 October 2014 (UTC)
As I said, I think it is correct, but it is visually difficult for me. I think that it is because the red and blue areas are propagating, but the electrons are oscillating about their rest position. All of which is correct. I have two ideas about. First, just make the electrons smaller so its easier to see the propagation. Visually, for me, the electron motion is dominant. The other possibility is to add another animation in the same picture, above the present transmission line that is the same except it has no electrons. That way you can see the propagation of the voltage clearly and then your eye can look down and see what the electrons are doing. But, good work. Constant314 (talk) 20:20, 27 October 2014 (UTC)
Constant314 - I agree with that. --Steve (talk) 22:18, 25 October 2014 (UTC)
Smaller electrons HERE (it's also smoother in general because I switched to subpixel-accurate rendering). But I'm not sure it helps much with the issue you mentioned (that it's not obvious that there is something moving right). So I also tried THIS, with a bit shorter wavelength, I think it's better but you're entitled to disagree. :-D --Steve (talk) 12:05, 28 October 2014 (UTC)
I liked them both better but prefer the second one. It's easy for me to think of things for you to do, but I think an animation of a pulse would be more elucidating. Constant314 (talk) 21:16, 28 October 2014 (UTC)
I like having an animation of the electron motion. It is such a common misconception that electrons travel at enormous speed down the wires. Animations like this help to dispel that misconception. But I do agree with Constant314 that a graphical representation of the voltage would be more intelligible than coloured voltages (if that is what he meant). I don't think it really matters whether it is sinusoid or pulsoid. There are two problems with the coloured voltages as I see it. One is that there is an implication that there is a definite voltage between each conductor and some external reference. This is not necessarily the case, a transformer coupled line is isolated from earth or any other external reference. The other problem is that information that relies on colour has accessibility issues. I also agree that it is easy to think of things for other people to do and I would be happy with the diagram staying as it now is. SpinningSpark 01:23, 29 October 2014 (UTC)

I tried making some pulses, for example SAMPLE PULSE. So far I don't find it any easier to "read" than the oscillation animation. But other people are welcome to disagree...

Thanks for the effort. After I study it, it makes more sense to me. But, why not have both?Constant314 (talk) 22:29, 5 November 2014 (UTC)

### Arbitrary editing break

On voltages: If we want to say nothing at all about voltage, it is very easy to do: Just rewrite the caption to say "colors represent charge densities". Anyone can do this right now, without changing the animation.

So then the question is: Should we eliminate all mention of voltage in the caption? Personally, I would say no. I think that mentioning voltage (with very careful wording) creates more enlightenment than confusion. But that's mainly a judgment call related to pedagogy, and I could be wrong.

I have always understood the colors as representing relative voltages between the conductors.Constant314 (talk) 22:33, 5 November 2014 (UTC)

Spinningspark, I don't understand your position on this. You wrote "there is a [false] implication that there is a definite voltage between each conductor and some external reference", which suggests that you think we shouldn't mention voltage at all. But you also wrote, "a graphical representation of voltage would be more intelligible than coloured voltages", which suggests that you don't mind discussing voltage. Which is it? --Steve (talk) 21:38, 5 November 2014 (UTC)

I very much think we should be discussing voltage and I further think that it should be represented in any diagram aiming to give an understandable representation of what happens on a transmission line. My issue is that the colours do not, Constant314's comment notwithstanding, represent the differential voltage between conductors. It is only the differential voltage that is of any relevance here and at any one point along the line there can only be one differential voltage. However, each point along the line in the diagram is showing two voltages, a red and a blue (positive and negative?). This implies they are red and blue with respect to some external reference. One could argue, of course, that they are red and blue with respect to each other, but I would humbly suggest that that is totally confusing to the reader. That would mean you would be showing the same voltage twice, referenced to two different points in two different colours.
A possible way round this is to show one voltage, in one colour, between the two conductors. However, as I said in my original comment, I don't think this is the right way to display the voltage. It would be much clearer, and more understandable, to show it as a plot with distance colinear with the line. SpinningSpark 23:53, 5 November 2014 (UTC)
Can we fix it with a few words, such as saying the red and blue indicates voltage relative to the common mode voltage and if the common mode voltage happens to be zero, then the colors represent voltage with respect to ground. Of course if it were coax we would probably say the shield is ground and we might see both red and blue on the center conductor. Constant314 (talk) 01:27, 6 November 2014 (UTC)
It's always possible to talk yourself around it, but as I said, it is an unnecessary complication and detracts from the clarity that having a diagram is supposed to bring in the first place. It is also less than general as it does not properly represent a transformer isolated line. SpinningSpark 01:38, 6 November 2014 (UTC)
I'm coming around to your point of view. The colored area should be the dielectric. It should be blue when there is an excess of electrons at the top and red when the excess is at the bottom. Obviously the assignment of red or blue is arbitrary.Constant314 (talk) 04:49, 6 November 2014 (UTC)
Thanks everyone, this is very helpful, I think we are making progress.
How about if I get rid of all the colors and put little arrows between the wires, which point up or down based on the electric field? (A bit like the arrows here) --Steve (talk) 13:35, 6 November 2014 (UTC)
I like the arrows. You could have both and make them coloured. The diagram is surprisingly good at depicting a forward travelling wave, I wasn't expecting that. SpinningSpark 13:56, 6 November 2014 (UTC)
Arrows are what I see in most textbooks. That is, they focus on the fields instead for the voltages and currents. Regarding the pulse, I think it might be better if he pulse width were smaller and as a refinement perhaps pulses of alternating polarity so that is compatible with transformer coupling.Constant314 (talk) 14:08, 6 November 2014 (UTC)

E-field arrows, 1st draft, at http://sjbyrnes.com/anim602.gif. Feedback? :-D --Steve (talk) 14:27, 7 November 2014 (UTC)

The arrows are great. They really demonstrate that the voltage or E-field is sinusoidal. I am having trouble tracking the electrons relative to the voltage. It takes almost the full period for me to visually lock on a bunch of electrons. By the time I acquire a bunch, they are just about ready the move off of the end of the transmission line. Maybe if it was twice as long would be easier for me to follow because I would have be able to focus on a bunch of electrons for a longer time. But great work. Constant314 (talk) 15:29, 7 November 2014 (UTC)
I think it's fine. Constant, the electrons never move off the end of the line (except for those that are very close to the end to begin with, and then they only move off temporarily). They only shuffle back and forth around a reference position. Making the line longer won't make that any clearer. What might help you, is that I found it much easier to track individual electrons if I magnified the diagram. Anyway, it's certainly an accurate depiction. SpinningSpark 16:42, 7 November 2014 (UTC)
Spark, I know what the electrons do, but I see that my comment might suggest otherwise. I'm not actually watching the electrons; I'm watching peaks in the electron density wave. I am trying to see it as a person who doesn't know what the electrons are doing and I am trying to validate the animation by seeing if the current peaks where it should and is zero where it should be. I'm glad you can verify that the animation is accurate. I think it is accurate, but I cannot verify it. Usually there is no animation from a reliable secondary source to cite as justification for an animation. You would know more about that than I would. I suggest that an animation should be verified by at least two editors and I am trying to be the second one.Constant314 (talk) 17:13, 7 November 2014 (UTC)
Ah sorry about that. Actually, I was quite surprised that you were apparently having the difficulty I thought you were having. I should have known better. I would still recommend magnifying the diagram to get a better view. Windows magnifier works quite well if you have it. Perhaps Steve can make the native image larger so that it can be magnified within Wikimedia. SpinningSpark 17:47, 7 November 2014 (UTC)
Arrows are much better! --catslash (talk) 19:18, 7 November 2014 (UTC)
Constant314 -- In the version I uploaded, I made the wavelength slightly smaller so it's slightly more visually obvious where the high-electron-density regions are. Maybe that will help? I would increase the whole image size, but the manual of style frowns on images wider than 300px for some reason (I bet it's something to do with how the article looks on a smartphone.) --Steve (talk) 00:49, 10 November 2014 (UTC)
Yes, I can see the features quite well. It's been a pleasure collaborating with you.Constant314 (talk) 04:01, 10 November 2014 (UTC)

There is also this image in the article. SpinningSpark 02:03, 10 November 2014 (UTC)

Yes, I'm planning to redo the open+short image in the same style :-D --Steve (talk) 13:06, 10 November 2014 (UTC)
Finished: [1]. Then I also made one more, [2], at the suggestion of Constant314. I added the latter to the smith chart article, but I don't see a good place for it here. --Steve (talk) 15:18, 15 November 2014 (UTC)

## Language version

I'm not sure that I can agree with marking this article as American English. The relevant passage of WP:RETAIN says;

When no English variety has been established and discussion cannot resolve the issue, the variety used in the first non-stub revision is considered the default. If no English variety was used consistently, the tie is broken by the first post-stub contributor to introduce text written in a particular English variety.

The version pointed to by user:Jc3s5h is undoubtedly a stub, being merely a semi-automated copy-paste from the glossary of Federal Standard 1037C, along with hundreds of other stubs created in the same way at that time. The first non-stub contribution is this edit by user:Heron. That contribution is unarguably written in British English. SpinningSpark 03:53, 12 May 2015 (UTC)

If you wish to change it, feel free. As you know, it's not that easy to spot which is the first non-stub version to use a national variety of English, what with all the wiki markup and author names that the spell-checker doesn't recognize. Jc3s5h (talk) 17:14, 12 May 2015 (UTC)
Thanks for replying. I'll leave it for a day or two to see if anyone else comments. SpinningSpark 19:19, 12 May 2015 (UTC)
• Use Oliver's English. Glrx (talk) 20:36, 12 May 2015 (UTC)

## Telegrapher's equation: Figure transmission line element error

The figure of the transmission line element in the section of the Telegraphers equation is wrong. The conductance of the line is modelled as a resistance, therefore the value of the conductance should be (1/G)*dz. — Preceding unsigned comment added by Qvandenbrande (talkcontribs) 09:48, 5 August 2016 (UTC)

It is correct is shown. The symbol is a resister, but the text clearly states that G is the conductance per unit length. We can specify a resister by its resistance or its conductance, but we use the same symbol regardless of whether we are specifying resistance or conductance. Constant314 (talk) 14:01, 5 August 2016 (UTC)

## ABCD parameters

I have just removed this addition to the article

I have a number of problems with this:

• Primarily, it is uncited, especially problematic is the numerical values of line length at which the various approximations apply. We definitely need a reference for that.
• The parameters Y, Z Y', Z' are all undefined.
• I might have been inclined to sort out the problems with the table formatting if that was the only problem, but as it stands, I though it best to remove it from the article.
• The addition opens with "we can also model..." The ABCD parameters are not an alternative to two-port modelling, but one specific instance of them among many. The AD-BC=1 requirement is a statement that the network is reciprocal and the A=D requirement is a statement that the network is symmetrical. Analogous requirements exist for other forms of two-port parameters. The requirements for reciprocity and symmetry could be stated more generally, and more clearly, without the need to invoke ABCD parameters. SpinningSpark 11:28, 23 November 2016 (UTC)

## Transmission line approximations

This recent section seems to be specific to power transmission lines and the article is specifically not about power transmission lines. Certainly 10km is a long line for audio and its short approximation is RC. 1km is a long line for video. 100m is long for RF. Perhaps this information should to into an article about electric power transmission. Constant314 (talk) 23:13, 7 December 2016 (UTC)

I've just reverted it again, partly for the reason you give (), widely spaced in air, chunky power transmission lines may well be dominated by series inductance, but typical telecomm cables are not. The lumped approximation we used to use for short audio lines (couple hundred metres) was series resistance and shunt capacitance. That worked pretty well. I also suspect this is a copyvio, probably from this book. The image in the "long line" section certainly is. Commons have already deleted it once, and seem about to delete it again. In any case, the "long line" section has exactly the same analysis as already in the article so is quite redundant.
On the wider question of whether power lines should be included on this page, I don't see any overriding reason why not. The theory and analysis is exactly the same, even if the numbers and approximations are different. We just need to be clear what is being presented. SpinningSpark 00:24, 8 December 2016 (UTC)
Currently the disambig sentence at the beginning essentially says that it's not about power lines. They really are two different meanings of "transmission line", even if they are technically unify-able. If they were in the same article, I think that terminology (2 meanings) would need to covered.North8000 (talk) 18:16, 30 January 2017 (UTC)

## Single Wire Transmission line

I've noticed that in the examples there is a "Single-wire line" subsection. It points to unbalanced line, and also discribes the line as still being a two-line TL, just with the earth (or some other external thing) as the return. I feel that this should also mention single-wire transmission lines. Although one could argue that they are not truely transmission lines as they are more similar to dielectric waveguides (at least from what I understand), they should perhaps be mentioned as many authors cover them as "SWTs" or single wire transmission lines, and a reader could thus look them up and find them here, with an incorrect discription (For that specific type). There are a couple of IEEE references that could easily be pointed to, as well as perhaps a link to the article on the Goubau line? TheUnnamedNewbie (talk) 16:26, 30 January 2017 (UTC)