Talk:Transmission line/Archive 1

Archive 1

Archive 2

Overhead Lines

Why do high-voltage transmission lines buzz?

No idea. Try asking at Wikipedia:Reference desk. Optim 19:17, 19 Feb 2004 (UTC)

The current in the line induces a magnetic field, which can exert a force on the wire. The higher the current, the greater the force. In an AC power line, the current reverses itself, which will cause the force to reverse itself, which could cause the wire to vibrate at 60hz, or could cause anything magnetic in the field that is loose to vibrate at 60hz. If it is loose, it is probably attached to something, and probably is smacking it at 60hz as it vibrates. That's what you hear. Transformers are especially bad about this; the windings will buzz if they are not glued in well enough. The more they buzz, the more the glue gets smacked around, and the looser the windings get, the louder the buzzing gets, etc... --ssd 04:38, 24 Jul 2004 (UTC)

This may be a part of the cause, but is not the whole cause - there is also the issue of the electrostatic field around the wires directly affecting the air around the wire, and corona effects. The electric field (E field) will repell the air as it builds up, pushing the air. This effect is used in very high dollar speakers (http://www.google.com/search?q=electrostatic+speaker&start=0).In addition, very high voltages cause dielectric breakdown of the air, ionizing it and causing it to expand - this is called corona discharge, and it a sign that the line needs to be serviced (as corona leakage costs power, and causes RF interference as well).The magnetic effects would be dependant upon current only - a line carrying 100 amps at 10V would be just as likely to show them as a line carrying 100 amps at 10,000,000 volts. The electric effects, however, will only show up on high voltage (high tension) lines. N0YKG 14:28, 25 July 2005 (UTC)

How does the electrostatic field repel the air if the air is neutral? - Omegatron 14:35, July 25, 2005 (UTC)

It doesn't. The only relevant explanation amongst all the above is corona discharge (described here (PDF)). --Heron 19:34, 25 July 2005 (UTC)

That's what I figured. See also ionic wind, Lifter (ionic propulsion device), Biefeld-Brown effect. - Omegatron 20:55, July 25, 2005 (UTC)

And plasma arc loudspeaker. --Heron 21:02, 25 July 2005 (UTC)

And magnetostriction - Omegatron 21:06, July 25, 2005 (UTC)

Bzzt. Sorry, wrong. It's called electrostatic induction - for a practical example, take a balloon, rub it with cloth to cause a triboelectric charge on the balloon, and hold it near your hair. The charge on the balloon will induce an opposing charge on your hair and attract it. It is perfectly possible for a charged object to affect an uncharged object, as fundamentally all "uncharged" objects are collections of charged objects. N0YKG 13:22, 26 July 2005 (UTC)

I don't think that's right. Air isn't hair. It's not connected to a charge sink that can give it a net charge in the presence of an electric field. You can induce instantaneous dipoles in the air molecules and attract them that way, but I'm skeptical that the force would be strong enough to have any kind of audible effect. - Omegatron 13:45, July 26, 2005 (UTC)

Did you even *bother* to go look at the link I provided? Once again: http://www.google.com/search?q=electrostatic+speaker&start=0. My brother-in-law has a set of these - they use an electrostatic field to directly move the air. The are NOT plasma speakers - they are electrostatic speakers. Now, go read the link. N0YKG 14:46, 26 July 2005 (UTC)

Try typing in the link again; that one's just a google search for "electrostatic speaker".

Electrostatic speakers work by attracting a charged sheet of conductive plastic. They don't move the air directly. You're probably thinking of plasma arc loudspeakers, which push neutral air with a corona discharge in the same way as the other links I gave above: ionic wind and Lifter (ionic propulsion device) - Omegatron 15:43, July 26, 2005 (UTC)

Corona at 50/60 Hz

What does corona sound like at 50/60 Hz? I dont know, but I bet it sounds like a hissy buzz, or a buzzy hiss!!!--Light current 21:56, 24 September 2005 (UTC)

BTW have you noticed OH lines are much noisier in damp or misty/foggy weather?--Light current 22:04, 24 September 2005 (UTC)

Low frequency (Power) Transmission lines

THis subject has already been disambigged in lead in and linked to. So I suggest it really doesnt need to be here. THoughts?--Light current 14:03, 12 September 2005 (UTC)

Cable impedance variation with frequency

Im pretty sure that OH lines do not act as high freq transmission lines at 50/60 Hz. When wL<<R and wC<<G, then the characteristic impedance of a TL is Z0= sqrt(R/G). At high frequencies where wL>>R, and wC>>G, Z0 =sqrt(L/C). So there are two distinct characteristic impedances for every line. Usually G is very small so the lf impedance is high, whereas the hf impedance is low. The break points in the impedance frequency graph are at w1 = G/C and w2 = R/L (where w=2*PI*f). If R/G>> L/C, it is obvious that w2>>w1. Between these two break frequencies the cable impedance decreases smoothly.

Example:

Take the case of a 50R coax with polyethylene dielectric. R is about 100mohm/m and G<20pS/m (based on measuremnets of leakage resistance in a 1m length). Using L=CZ^2, L can be calculated at about 250nH/m. So,

w2 = R/L = 200krad/s (f2=30kHz).

and w1= G/C= 0.2rad/s (equ to f1=30milliHertz).

At 100Hz the 50ohm coax will have an impedance of about 900 ohms only getting down to 50 ohms at about 30 or 40 kHz. The phase angle of the impedance between the two break frequencies has yet to be determined by me, but I strongly suspect that it is leading (ie cable looks capacitive). I will try to derive the phase angle when I get time.--Light current 16:47, 25 September 2005 (UTC)

OH lines seem to be inductive from what I can gather on the net, and shunt caps are used to compensate them. --Light current 00:22, 26 September 2005 (UTC)

Yes. Ive just confirmed this with my electrical eng friend from the local power company. He says OH lines are generally inductive, but it can depend on the exact configuration and interconnects etc. IOW they dont know what PFC to use until the systems up & running!--Light current 00:54, 28 September 2005 (UTC)

Merge with Telegrapher's Equations

Theres not much you can say about transmission lines if you cant mention the equations! But there's a lot you can say if you do!--Light current 22:42, 12 September 2005 (UTC)

I have attempted to merge this with the telegrapher's equations and I hope that this is satisfactory to everyone. I know there will be some 'tweaking' required but I would appreciate everyone's patience whilst I (or others) do this 'tweaking'!. Thank you all for your understanding my good intentions and for your patience with me !!! --Light current 01:43, 24 September 2005 (UTC)

I think the equations look great now!. I cant attest to the accuracy without checking, but they look very nice!--Light current 20:13, 24 September 2005 (UTC)

Having trouble with the wiki. It seems to have devoured the externallinks I tried to move to page bottom. Wiki very slow at the moment so I'll wait before trying to restore these links etc.

I think all the important matl from telegrahers equations page has now been merwged into this page. Have a look see if I've missed anything before we delete telegraphers equations

Mindstretcher- Paradoxes

#1 The two piece TL problem

Image two identical ideal transmission lines (say pieces of coax). Their central conductors are separated by a single pole ideal lossless switch (say an ideal in line reed switch). Their outers are are connected together at the switch location. Both TLs are open circuit at the ends not connected to the switch. Initially the left hand TL is charged to a voltage V and the right hand TL is discharged. At time t=0, the switch is closed and remains closed. If the capacitace of each piece of coax is C, application of conservation of charge before and after switch closure shows that half of the original energy (0.5CV^2) has been lost. If this energy cannot radiate away and there is no skin loss or dielectric loss in the TLs, where has this energy gone?--Light current 15:00, 8 October 2005 (UTC)

If there are no losses, how does the system get to steady state? If you think of a hydraulic analogy, with pipes and tanks, one empty, one full, and a suddenly-opened valve between them, if there are no losses the water keeps sloshing back and forth, no? --agr 13:54, 9 October 2005 (UTC)

Thats a very interesting observation, Arnold!--Light current 14:27, 9 October 2005 (UTC)

So in your analogy, there is no energy loss and no charge loss either, but this is at odds with experimental results that show a definite energy loss in such systems. How come?--Light current 03:23, 12 October 2005 (UTC)

I'm confused. Your original statement of the problem stated that the system was lossless. That's why he made an analogy to a lossless system. Pfalstad 18:51, 12 October 2005 (UTC)

Yes. Confusing , isn't it? Can you enlighten us?--Light current 02:21, 13 October 2005 (UTC)

Capacitors and transmission lines - Similarities

(moved from Talk:capacitor no 17/10/05)

Original Question

Consider this: Take a piece of 50 ohm coaxial cable with its far end o/c. Using your 50 ohm o/p impedance pulse generator, apply a pulse of voltage/current to the near end for just long enough so that the leading edge of the pulse reaches the far end of the cable. Disconnect the pulse gen leaving both ends of cable o/c. You now have a capacitor charged to half the o/c voltage of the generator. Right? So the capacitor has charged in the time it takes the pulse to travel to the end of the cable. NB PLEASE SEE MODIFIED CONDITIONS IN FOLLOWING POSTS (by me) BEFORE ATTEMPTING ANSWERS--Light current 13:42, 12 September 2005 (UTC) Questions:

1)If you now looked at either end of the cable with your oscilloscope, you would see only a dc voltage, so what has happened to the pulse of energy travelling at 2/3 c. Why does the pulse wave suddenly stop without reflecting off each open end of the wire continuously? (Does it suddenly stop - WHY?)

2) what is the ESR of this capacitor?

3) what is the inductance of this capacitor?

AND the BIG ONE

4) If this tranmission line is a capacitor, are all capacitors transmission lines?

Doubting Thomases and others of little faith could look here! [1]and here [2] and here [3] :-)

LC: Do these links have anything to do with your question? [4] [5] Alfred Centauri 21:44, 13 September 2005 (UTC)

Might have. I have only seen one of them before in my life- the Catt one in WW 1978.;-)--Light current 00:06, 14 September 2005 (UTC)

Answers

No, they are not. How are they similar? — Omegatron 22:16, September 11, 2005 (UTC)

"You now have a capacitor charged to half the o/c voltage of the generator. Right? So the capacitor has charged in the time it takes the pulse to travel to the end of the cable."

I disagree with this. In the scenario you give above, the transmission line has not settled into steady state before the generator is disconnected. Thus, the voltage across the line is a function of time and position. In fact, for an ideal lossless T-line, the DC voltage at the ends is a square wave with a period of twice the propagation time of the line, a P-P amplitude equal to the open circuit voltage of the generator, and an average value of 1/2 of that voltage. Alfred Centauri 20:31, 11 September 2005 (UTC)

I hadn't considered that one. Do you realise that you've just invented a square wave oscillator!(you may have to keep kicking it - but so what?)--Light current 04:43, 12 September 2005 (UTC)

How is the steady state, as you put it, different from the condition when the pulse has just reached the end of the line. Remember, the pulse is only lon enough to fill the line, no longer.--Light current 20:45, 11 September 2005 (UTC)

In steady state, the voltage and current along the line are constant with time. For the scenario you describe above, steady state will be reached at 2T where T is the propagation time down the line. It takes one T for the initial pulse wave to reach the far end and another 1T for the reflected pulse to reach the source end. Because there is a match at the source end, there is no reflection so the transient is over - the line is in steady state. For t > 2T, the voltage along the line is equal to the open circuit generator voltage and the current in the line is zero. If you disconnect the generator after 2T seconds, nothing happens on the line. There is a charge stored on the line and there is a voltage across the line. The ratio of charge to voltage is then given by the product of the capacitance per unit length with the length of the line. Alfred Centauri 21:44, 11 September 2005 (UTC)

No, I said that the pulse generator was disconnected from the line after the line was filled with energy. So does the line reach a steady state?. What is that steady state?

BTW lets make the pulse twice the electrical length of the line, then the 'capacitor' has charged completely in 2T? to the original o/c amplitude of the pulse generator.Yes?--Light current 21:58, 11 September 2005 (UTC)

Sorry AC, I misunderstood your reply initially. You are in fact Correct if t=2T. THere is charge stored and of course there is voltage across the line. But what has happened to the traveling EM Waves? I thought they could not be stopped!!.--Light current 00:08, 12 September 2005 (UTC)

So you are applying a voltage to one end of a transmission line, and then open-circuiting the source end when the pulse reaches the load? — Omegatron 22:21, September 11, 2005 (UTC)

Well, Originally I was but AC has shown that you need to allow the pulse to return to the sending end before diconnecting the pulse gen. Then you have a beautifully charged capacitor just sitting there. But what has happened to the travelling waves??? Mysterious eh?--Light current 22:38, 11 September 2005 (UTC)

You don't have to disconnect the generator at any given time, although maybe for your capacitor equivalence thing you would have to. Like maybe you could say "under certain conditions during which the generator is removed after x time period and the load is this and the source is that the transmission line will behave similarly to a capacitor".

I think the transmission line article needs some expanding.  :-)

OK but were just having a little fun right now - the serious stuff will come later, believe me!!--Light current 04:20, 12 September 2005 (UTC)

And if you open-source the sending component right as the traveling wave reaches it, the reflected wave will be double the incident wave, since the reflection coefficient is exactly 1, no? In a lossless line the chunk of pulse would just keep reflecting back and forth forever.

Yes! I believe that's what does happen, but you dont see it because the voltage is the same at all points on the line at all times. So the 2 oscillating pulses just add up to a steady dc voltage. Yes?--Light current 04:00, 12 September 2005 (UTC)

In a real, lossy line it tapers off.

Eventually--but since the EM waves are just guided by the conductors maybe you dont get much loss. But hey, its only dc on there isnt it? So what causes the loss mechanism? ;-)Light current 04:00, 12 September 2005 (UTC)

So you're saying that you send a pulse down a line with an open-circuited load and then open-circuit the source right when the pulse reaches it? then your incident pulse from the source would become zero, but the reflected pulse at the source would be 1 in the same direction, leaving the exact same voltage as originally (starts at 0, becomes 1, becomes 1+1, becomes 0+1+1, and will stay at 2 steady state). So the pulses are still there in a sense, just cancelling themselves out to the same value at any point along the line. If you didn't stop the source at exactly the right moment you would still be able to see the timing glitch bouncing back and forth. — Omegatron 03:30, September 12, 2005 (UTC)

Yep! Good isnt it?--Light current 04:00, 12 September 2005 (UTC) So, can you go on to answer questions 2,3 &4?--Light current 04:07, 12 September 2005 (UTC)

Questions what? — Omegatron 06:40, September 12, 2005 (UTC)

In my original post, I asked people four questions. You have answered the first question correctly. Can you, or anyone else, now answer the other three?--Light current 12:14, 12 September 2005 (UTC)

But we haven't agreed on your premise yet, LC, that the T-line is a capacitor. We have a length of line with oscillating electric and magnetic fields trapped inside, somewhat analogous to a resonating, lossless LC circuit. The voltage at either end of the line is a square wave (I think's it's a bit more complex in the middle). The electric field inside an open-circuited capacitor is d.c. The line is storing energy in a different way to the capacitor. --Heron 12:59, 12 September 2005 (UTC)

Apply the pulse(or battery via a switch if you like) for twice the delay time (T) of the cable. The front edge of the energy wave (amplitude V volts say) reaches the end of the cable (after time T). This incident wavefront is reflected back toward the source with preserved polarity and adds in voltage to the forward travelling wave. We therefore have a step of voltage (magnitude 2V) travelling backwards toward the source. (Dont forget that the transmission line equations allow for 2 independent oppositely travelling waves to exist on a line). After total time 2T, this step reaches the source end of the cable. All this time, though, the source has still been pumping energy into the line, so by the time the original wave front gets back to the source (2T) the line is completely filled with voltage at 2V. At this point, if the source is now instantaeously disconnected, voltages on the line are indistinguishable from steady dc on a capacitor. There are no oscillations! (PS please read the previous posts by AC and 'O' for more clarification )--Light current 13:38, 12 September 2005 (UTC)

There are still oscillations; they just sum together to equal a uniform voltage. The comparison with a capacitor is very sketchy. The only similarity I see is that you start by connecting a voltage source for a given period of time to a two-conductor object and end up with a uniform voltage across it when disconnected. You could say a battery is a capacitor by the same logic. The period of time in between these two events is not the same at all. The distribution of voltage across the objects is not the same at all. If you disconnect the source at any different time, the end results will be completely different for a capacitor and t-line. If you plot the voltage over time, the two curves are very different, and so on.

Although yes, all objects are transmission lines sort of, if you want to think of them that way.  :-) — Omegatron 14:24, September 12, 2005 (UTC)

So have you got oscillations or not? They cannot be detected by any instrument you care to think of. So this means that the incoming EM radiation has been perfectly captured by the transmission line. Yes? The comparison with a capacitor is very good actually. In fact it's perfect. There can be no better capacitor (apart from the amount of cable you would need for say 0.1uF). Have you ever seen bits of wire twisted together to form an adjustable capacitor (adjust by snipping bits off). In US its called a 'gimmick' or sucklike. Over here we use the 'posh' term a 'user adjustable cable capacitor'. They can be made with coax as well. The voltage distribution in the transmission line after time 2T is completely even. If you disconnect the source at a time later than 2T, what will happen?. Let's see.... The source has an o/c output voltage of 2V say, and is matched to the line so the amplitude of of the initial pulse is V. When after 2T the pulse has returned, 2V exists all over the line (as you have already seen) and a state of equilibrium exists. At this point you have a source of 2V connected via a resistor to a 'capacitor' (the t-line) charged to 2V. Does it now matter when you disconnect the source?? Also I dont agree with your statement that the results for a capacitor and a t-line will be different. Consider a battery charging a capacitor via a resistor. The voltage across the capacitor will rise logarithmically until it reaches the battery voltage. Do the same thing with a transmission line. The voltage will rise (in steps this time) approximating a logarithmic rise. THe macroscopic effects as shown on an osciloscope are the same. (unless you have avety long cable).--Light current 15:09, 12 September 2005 (UTC)

"So have you got oscillations or not? They cannot be detected by any instrument you care to think of."

Sure. You can think of it either way. But I think it's more helpful to think of it as wavefronts bouncing back and forth and cancelling out to a steady voltage. Think of what would happen if you connected a matched load to the line after it was "charged". Think about standing waves. — Omegatron 16:29, September 12, 2005 (UTC)

Yes!! You can think of it either way!!! Do you realise what you have said here? You have understood that the t-line and capacitor are exactly equivalent!.

No. I meant you can think of it as either a steady voltage across the entire line or as pulses moving down the line and adding up to a steady voltage. They're the same thing. Doesn't make it anything like a capacitor, though. — Omegatron 05:29, September 13, 2005 (UTC)

Your last sentence is also interesting. If you connected a matched load to the end you would then release the genie from the bottle and a pulse of energy would rush out of the cable and dissipate totally in your load.

If you connected a load to the end of your "charged up transmission line", the voltage at the load would suddenly halve, as the wave incident on the load was suddenly absorbed in the load instead of reflected. The wave that was constantly being reflected back to the source would stop, in other words, and you'd see a wave of V volts traveling from load to source, where the entire line used to be 2V. Then when that wave reaches the source, it will reflect again, and a wave of 0 V will move from source to load. I was trying to illustrate how it's more helpful to think of waves constantly in motion adding up to 2V than to just think of the line as having a steady 2V on it. — Omegatron 05:29, September 13, 2005 (UTC)

So the 2 situations are opposites: The pulse charging (in 2T) is just the reverse of the pulse discharge of the cable (also takes 2T to discharge completely). AAhh - -its all becoming very clear to me now (2001/2010);-))--Light current 21:36, 12 September 2005 (UTC)

Yes, that's essentially correct. — Omegatron 05:29, September 13, 2005 (UTC)

It wouldn't be logarithmic at all. It jumps up in two discrete steps. It's not very close to a capacitor at all. — Omegatron 15:54, September 12, 2005 (UTC)

If the charging resistor is much larger than the cable impedance, the rise would be logarithmic.;-) --Light current 16:14, 12 September 2005 (UTC)

1. You said the source is matched to the line

2. You said the source was disconnected after the reflected wave hits the source.

This would mean the voltage at the load jumps up to 2V, and the voltage at the source jumps up to V and then up to 2V. Not logarithmic at all. — Omegatron 16:20, September 12, 2005 (UTC)

I said Ifthe source (charging) resistor is much larger than Z0 (ie unmatched), the rise would approximate the normal charging characteristic of a capacitor in an RC Circuit. If the source is matched to the line then you are correct, this would mean the voltage at the load jumps up to 2V, and the voltage at the source jumps up to V (at t=0) and then up to 2V (after time 2T) after which the capacitor(line) is fully charged. No current then flows into or out of the generator and it can be disconnected at leisure (just like you can do with the battery in an RC cct). The cap is still charged after you un hook the battery. Yes?,. Actually, since no current can flow back into the generator from the line, the near end now looks like an o/c- so no matching problems either. Now you know how a capacitor really charges up (in discrete steps and not smoothly at all!!!!!) Do you get it??--Light current 21:25, 12 September 2005 (UTC)

Yes, all real objects require time for the waves to travel across them and behave like transmission lines, and a capacitor would act like a very very short transmission line as you charge it, but every capacitor is a different shape, and the reflections and transmission times from one point to the next would be very different, so it wouldn't happen in discrete steps. They would all blend together. An object like a capacitor is not as ordered as a real transmission line. Plus there is no dependence on disconnecting the source at a given time, like you were saying.

I think I see what you're saying now. It has nothing to do with disconnecting the source at a certain time. You're not saying all transmission lines are capacitors. You're saying capacitors are transmission lines, which is sort of true, depending on how you look at it.

And it's only open-circuited transmission lines that behave like this, and only for pulses. Short-circuited transmission lines behave more like inductors for pulses, and for steady sinusoids it also depends on the length of the line, which is where all the RF stub filtering comes from. — Omegatron 05:29, September 13, 2005 (UTC)

"There are still oscillations; they just sum together to equal a uniform voltage" I have to disagree 'O'. This business about disconnecting the generator 'just as the reflected wave arrives' is nonsense. Look at it this way, at the 'instant' the reflected wave arrives at the source end, there are only two possibilities - either the generator is there or it isn't. If the generator is there, the source end is matched and there is no reflection period - there are no waves on the line! Or, if you like, there are two waves on the line but the wave fronts have moved off the line so there is nothing to be reflected - the line is in steady state. If the generator isn't there, there is complete reflection at the source and far end so we have wave fronts reflecting off the ends forever if the line is lossless. The voltage along the line oscillates. Alfred Centauri 14:44, 12 September 2005 (UTC)

1. Why is it nonsense?

What does "disconnecting the generator at the instant the reflected wave arrives" mean? As I said there are only two possibilities - either the generator is there or it isn't at the instant the reflected wave arrives. Which one is it? Alfred Centauri 15:02, 12 September 2005 (UTC)

It isn't. The wave traveling from load to source sees an open-circuit. — Omegatron 15:54, September 12, 2005 (UTC)

2. If the generator is there, the wave may be reflected varying amounts, depending on the source impedance. It's not necessarily matched.

"Take a piece of 50 ohm coaxial cable with its far end o/c. Using your 50 ohm o/p impedance pulse generator". The generator is matched in this setup you proposed. Did you change that along the way? Alfred Centauri 15:02, 12 September 2005 (UTC)

Apparently I haven't read the original question very well.  :-) — Omegatron 15:54, September 12, 2005 (UTC)

3. Yes, the generator won't be there at the instant the wave reaches it, which means complete reflection, as you said, and the wave reflects off the ends forever, as you said. But the total voltage when this is happening is uniform. The incident wave from the generator is 0, the reflected wave from the load at the source is V, and the reflected wave from the source at the source is V, leaving 2V everywhere. — Omegatron 14:52, September 12, 2005 (UTC)

LC, that's wrong and you should know that. If the generator isn't there when the reflected wave arrives, the generator was disconnected before the wave arrives - to say anything else is illogical. Even if it is just an instant before, there is nonetheless a second incident wave generated if the generator is disconnected before the reflected wave arrives. This wave travels down to the end and reflects so the voltage changes with time along the line. Alfred Centauri 15:02, 12 September 2005 (UTC)

My apologies! I mistakenly addressed 'O' as 'LC'.Alfred Centauri

I don't see how this situation is impossible. The generator is open-circuited at the exact instant the wave incident from the load arrives. The wave coming from the load sees an open circuit, and the wave being generated by the source becomes 0 at the same instant. — Omegatron 15:54, September 12, 2005 (UTC)

The generator can't be both connected and disconnected the same instant. When the reflected wave arrives, the voltage at the source end is either 1V (generator connected) or 0V (generator disconnected). Alfred Centauri 16:55, 12 September 2005 (UTC)

Yes. When the wave arrives, the voltage generated by the source jumps from 1 V to 0 V. — Omegatron 18:18, September 12, 2005 (UTC)

Hmmm... The generator is disconnected from the line so the voltage generated by the source doesn't change at all. But that's not really the issue here. The issue is this - the disconnection of the generator has to be defined one way or the other. For example, let t = 2T be the instant in time that the reflected wave reaches the source end of the line. Now, you claim that the generator is disconnected at that instant in time. What does this mean? Mathematically, it could mean one of the following:

(a) the generator is connected for 0 <= t < 2T

(b) the generator is connected for 0 <= t <= 2T

If you mean (a), the reflected wave 'see's an open circuit at the source end and is thus reflected. However, there is also a new wave created by the disconnection of the generator at 2T- so the voltage at the source end is the sum of four voltages: +1 (generator connected), -1 (generator disconnected) +1 (arrival of reflected wave) +1 (reflection of reflected wave) = 2V. To see this, let the generator be disconnected at time t = 2T - $\epsilon$ and then let $\epsilon$ tend to zero.

If you mean (b), the reflected wave 'see's' a 50 ohm termination so the voltage at the source end is the sum of two voltages: +1 (generator connected), +1 (arrive of reflected wave) = 2V. Alfred Centauri 20:11, 12 September 2005 (UTC)

Just a bit of clarification here guys. There is no load at the far end of the cable. It is open circuit. Please refer to that location as the far end of the cable and the other end as the source end (or near end).Also, there is only one resistance in the cct. That is the source resistance of the generator (50 ohm)--Light current 16:06, 12 September 2005 (UTC)

The "new wave created by the disconnection of the generator" is a wave of 0 V, which, added to the wave reflecting from the source (V) and the wave incident on the source (V) = 2V.

Sorry, but that's not correct. When the generator is disconnected from the line, a new incident wave is created with a voltage of -1V and and current of -0.02A. This wave added to the first incident wave gives zero volts and zero current. Alfred Centauri 17:07, 13 September 2005 (UTC)

Are you sure? I must be overlooking something about the disconnection, then. So if you had an infinite transmission line and you turned on a generator that created a 1 V pulse down the line, and then disconnected that generator while it was on, you would see a −1 V pulse down the line?

 1 -             *--------*
                 |        |
 0 -             |        *--------
                 |
-1 - ------------*

This doesn't make intuitive sense to me, but I might be missing something. — Omegatron 18:44, 13 September 2005 (UTC)

I believe that Omegatron is correct here. Disconnection of the generator cannot create a negative voltage. The voltage at the sending end just returns to zero and the pulse carries on down the line.--Light current 22:03, 13 September 2005 (UTC)

Then you don't understand TLs. The disconnection of the generator creates a negative voltage pulse that is superimposed with the positive going pulse ahead. Thus, there is no NET negative voltage on the line but then, I never said there was. The measured voltage on the line is always the superposition of all the waves on the line. Please see my diagrams below if you still don't understand what I am saying. Alfred Centauri 22:41, 13 September 2005 (UTC)

The way your diagram is drawn with the -1 level led me to believe you were intending actual negative voltage ( as per convention on this sort of diagram). However, if what you mean is that disconnection of the gen. causes a negative going step to appeear on the forward travelling wave, than I agree with you. I think this is just confusion of terminology actually and that now we all three basically agree on what is happening (although I could be wrong - often am!)--Light current 22:58, 13 September 2005 (UTC)

Now I see where our 'disconnect' is. The voltage measured at any point on the line is always the superposition of whatever waves are on the line. Before the generator is disconnected, you would measure 1V across the line near the source end. When the generator is disconnected, a -1V pulse is generated that moves down the line. At any point on the line where both waves are present, the measured voltage is 1 + (-1) = 0V.

The top waveform is the incident wave due to the connection of the generator. The middle waveform is the incident wave due to the disconnection of the generator. The bottom waveform is the superposition of the two waves and is the measured waveform. Alfred Centauri 19:18, 13 September 2005 (UTC)

 1 -----------------------------------------*
 *                                          |
 0                                          *-------------------

 0       *------------------------------------------------------
 *       |
-1 ------*

 1       *----------------------------------*
 *       |                                  |
 0 ------*                                  *-------------------

I believe that AC is not correct strictly here because the generator is still in its high state of 2V. How can the disconnection of the generator produce a net negative pulse at the sending end. A step reduction of 1 v is possible and that is in fact what happens--Light current 21:59, 13 September 2005 (UTC)

Then you don't understand TLs. Maybe this will help. If the far end of the TL is shorted, the voltage must be zero there. The only way for this to happen when the 1V pulse arrives is for there to be a reflected voltage wave of -1V. Now, consider that when the generator is disconnected, the current at the source end must be zero. That means that a -0.02A current pulse must propagate down the line to cancel the 0.02A pulse ahead of it. Since the voltage and current of each INDIVIDUAL wave (NOT the superposition) must be related by the characteristic impedance, the voltage associate with the current pulse is -1V. Alfred Centauri 22:33, 13 September 2005 (UTC)

Right..(deep breath). If the far end is s/c you would indeed get a -1V pulse returning to the source.

Now, in the o/c far end case that we are discussing, when the gen is disconnected you will get an effective -1 v pulse travelling down the line. But the effect of this is to cancel the existing 1V on the line. So the voltage on the line progressively becomes zero as your diagram shows. THis is just another way of saying that when the gen is removed, the trailing ecge of the pulse is formed, again as your diagram shows. The only slight quibble I would have here is that when the generator is disconnected, the original wave has completed the round trip, and so the voltage at the source end is 2V (NOT V as in your diagram). That is why I said you were not strictly correct with regard to the question in hand. Otherwise I think you are a fine fellow and no offence meant!!--Light current 03:44, 14 September 2005 (UTC)

BUT the far end of the line is open circuit. Did you not read the question?--Light current 22:45, 13 September 2005 (UTC)

Not if I say I have short there! Apparently you didn't see that I'm giving a separate example of a shorted end to elucidate the point that a negative wave can be created where there is no negative voltage source present - thus the statement 'Maybe this will help'. Perhaps TLs are taught different these days but when I studied them initially, the most important concept taught was that the actual voltage and current on a TL can be understood to be the superposition of individual waves. It is the individual waves that are created by events such as the connection or disconnection of a generator or by reflection at a mismatch. Don't forget, it is the individual waves that are multiplied by the reflection and transmission coefficients, not the wave that you see with an oscilliscope (unless it is the only wave on the line at the time). Thus, it is the -1V wave created by the generator disconnect that is reflected at the open circuited far end, not the 'zero step' you keep referring to. IF there are reflections at both ends of the line, a new wave is added to the line at each reflection. Yet, if you watch the voltage on the line, you might just see a pulse going back a forth. But, this pulse is the superposition of an every increasing number of waves as time moves forward. Alfred Centauri 00:27, 14 September 2005 (UTC)

Ah. Yes, that's exactly what I meant by a "0 V pulse" when the generator is disconnected. That is the "wave" coming from the source when the gen is disconnected; a sudden switch to 0 V.

 1       *----------------------------------
 *       |    -->                             
 0 ------*                                  

I believe that 'O' is again correct here with the above diagram ( assuming leading edge has not been already reflected and come back into the line section represented in the diagram.--Light current 22:03, 13 September 2005 (UTC)

There is also the wave returning from the load, which is 1 everywhere:

 1 -----------------------------------------
 *            <--                             
 0    

I believe that 'O' is unfortunately not totally correct with the above diagram because the reflected wave adds in magnitude to the incident wave. The diagram should show a backward traveling step of magnitude 1V sitting on a pedestal of 1V--Light current 21:59, 13 September 2005 (UTC) And that same wave wave reflecting from the source, which is suddenly open-circuit:

 1 ------*
 *       |    -->
 0       *----------------------------------

I believe that 'O' is not totally correct with this diagram. When the reflected wave reaches the source end it again reflects from the apparent or real o/c present here but adding to the 1V already existing on the line. At this instant, the transient phase is over and no further pertubations take place anywhere on the line.--Light current 22:03, 13 September 2005 (UTC)

The superposition of the three waves is 2 V everywhere. — Omegatron 19:27, 13 September 2005 (UTC) I believe however that 'O' is totally correct in the above statement.--Light current 21:59, 13 September 2005 (UTC)

Here is a simpler method for obtaining the "charged up line":

• Vs = 2 V
• Zs = 50 ohm
• Z0 = 50 ohm
• Zl = ∞

Now just turn on the generator at 2 V. You'll see 1 V at the source, because of the voltage divider between source and line, which will propagate down the line to the open-circuit load. Then it will reflect perfectly, and a wave of 2 V will travel back (1 V incident + 1 V reflected). When the 1 V reflected wave gets to the source, it will be absorbed in the source resistance, leaving the entire line at a steady state of 2 V, which is similar to LC's conditions. You could then disconnect the generator and you would be left with 2 V on the entire line. No need to stop it at an exact time. I believe that this description of events is almost 100% accurate. --Light current 21:59, 13 September 2005 (UTC)

You can simulate all these things in pspice, you know.  :-) — Omegatron 15:13, 13 September 2005 (UTC)

Except for the part about a 2V wave travelling back to the source, I agree. As you say later, there is a 1V reflected wave that travels back to the source end. Alfred Centauri 17:07, 13 September 2005 (UTC)

That's what I meant. There is the 1 V wave incident on the load, and the 1 V wave reflected from the load, and they add up to a 2 V wave which you would see moving from load to source if you were looking at the voltage as a function of position on the line over time. You would see a wavefront moving from load to source at which point the voltage jumps from 1 to 2. — Omegatron 18:44, 13 September 2005 (UTC)

I believe that 'O' is essentialy totally correct on this point also

Not only that, you could actually do the experiment with a reel of coax a sq wave gen and a scope! :-))--Light current 13:55, 13 September 2005 (UTC)

But that requires getting out of the chair! The horror!

(Also I don't have a reel of coax handy.) — Omegatron 15:13, 13 September 2005 (UTC)

Never mind then!. Any way I do believe you have now grasped the strong similarities between t-lines and capacitors aqnd between incident EM radiation and its other disguise: dc! All I have to do now is wait for 'AC' to come round. In the mean time, what about questions 2,3 or my original post?;-)--Light current 15:40, 13 September 2005 (UTC)

No, I mostly disagree with you. Yes, a transmission line can look like a capacitor under some conditions, but that doesn't mean they're the same thing. It can also look like an inductor, resistor, etc. I've said I disagree several times. — Omegatron 18:44, 13 September 2005 (UTC)

A transmission line always looks like a capacitor of simliar physical dimensions Just ponder those words simliar physical dimensions then tell me why you disagree.--Light current 21:01, 13 September 2005 (UTC)

It's typical to refer to the "far end" as the load, and say "the load is an open circuit", etc.

Think of "load" as being a shorthand for "termination impedance". — Omegatron 16:20, September 12, 2005 (UTC)

Yes OK Thanks 'O'--Light current 15:16, 13 September 2005 (UTC)

AC theres been a bit of an edit conflict, so Im not sure what youre accusing me of being wrong about. Please see my last reply to 'O'--Light current 15:31, 12 September 2005 (UTC)

My apologies! I mistakenly addressed 'O' as 'LC'.

Two Terminal Transmission Line

A t-line with the far end open-circuited is not equivalent to a capacitor at all. The t-line has distributed series inductance and distributed parallel capacitance. Replace your pulse generator with a sine wave generator and sweep the frequency up from DC while measuring the voltage across and the current into the t-line. You will find that as the frequency is increased, the impedance looking into the t-line alternates between pure capacitance and pure inductance. No capacitor that I know of behaves this way. Finally, consider this: let the length of your t-line go to infinity. In the limit, your open-circuited t-line looks like a 50 ohm resistor. Alfred Centauri 00:36, 13 September 2005 (UTC)

Yes, but in the context of the original question, ie pulse charging, the cable acts only like a 50 ohm resistor until the cable is fully charged. Also, when charging via a resistor the cable acts as a capacitor. For ac also, if the cable is short enough, it will look like a (low loss) capacitor. Infinite length of cable just means an infinitely large capacitor. What does a real capacitor look like when hit with a very fast edge?? When you gave your very first answer to my (modified) question about the EM waves, you assessed the situation correctly, so I can't see why you cant see that a t-line is a capacitor- it acts like an ordinary capacitor of equivalent dimensions(repeat for emphasis equivalent dimensions)--Light current 01:25, 13 September 2005 (UTC)

What do you mean 'in the context of the original question'? You're original question was 'are all capacitors transmission lines?' I've been waiting for you to come forth with whatever point your are trying to make here. So, rather than wait any longer, here's what I think of all this...

We all know that capacitor can be formed from parallel conducting plates or coaxial conducters, or just two conductors uniformly separated. Further, we know that these structures can also be used as transmission lines for guiding EM waves from one end to the other. Also, we know that if one end of a t-line is open-circuited or short-circuited, the t-line can appear, depending on the frequency, as either a pure capacitance or a pure inductance 'when the line is in steady state'. Finally, we know that there are transients on a t-line and that in some situations these transients do not decay - the line never comes to steady state. This is all well known.

I contend that an open-circuited (short-circuited) t-line is not equivalent to a capacitor (inductor). To be more specific, an ideal capacitor is not a t-line and an ideal inductor is not a t-line. I believe that the point you must be trying to make here is that the structure of a physical capacitor can exhibit t-line effects. If this is your point, then I agree with this. But, hopefully, this isn't your point. After all, physical capacitors are non-ideal in many ways.

Further, since all physical electrical devices are not infinitely small, all physical electrical devices must exhibit some t-line effects for the simple reason that EM waves do not propagate infinitely fast. The lumped element approximation for circuit analysis is the assumption that the wavelengths of the waves in the circuit are much longer than the physical dimensions of the circuit and circuit elements. If this isn't true for your circuit, you can no longer use the lumped element approximation. Thus, PC motherboard designers now use SPICE or some equivalent program to model the copper tracks on the PCB as t-lines as do high-speed IC designers. Alfred Centauri 17:59, 13 September 2005 (UTC)

Agreed on every point. — Omegatron 18:44, 13 September 2005 (UTC)

Trapped EM waves (moved from Talk:capacitor)

Chapter 1

I also disagree that EM wave is fundamental and current and electric/magnetic field are result of it. Other way around if anything. The capacitor is a perfect example of that. There is no radiation when the cap is charging steadily, but the field increases and the energy density between the plates increases, in absence of any EM wave. There is no standing wave in the capacitor. Pfalstad 02:07, 6 October 2005 (UTC)

I didnt say there was radiation from the cap. In fact there is never any radiation from the cap simply because the EM energy cannot escape. It is trapped in the transmission line formed by the capacitor plates. See earlier posts on this where AC and I appear to have agreed on this one. There are two counter propagating waves in the capacitor just like an O/C transmission line (same thing). These two waves cancel to give dc (a form of standing wave). So simple!--Light current 02:18, 6 October 2005 (UTC)

Why would the EM energy be trapped? Even if there were a standing wave in there, which there isn't, it could easily get out through the sides. Why would there be waves anyway, with a constant current? dc is not a form of standing wave. A standing wave has a frequency. it oscillates. Pfalstad 03:04, 6 October 2005 (UTC)

The counter propagating waves cannot escape thro' the open sides. They are reflected in phase due to the complete mismatch of impedances between the Z0 of the capacitor (very low) and the impedance of free space (377 ohms). As I have explained above, the arrival of each electron on the plates must be accompanied by a very small amount of EM radiation between the capacitor leads. Otherwise, no energy could be transferred to the capacitor. Energy flows in the space between the wires and is given by the Poynting vector ExH. Two counter propagating waves add up to a steady voltage. This steady voltage is what I'm calling the standing wave.--Light current 03:16, 6 October 2005 (UTC)

Why can't they escape through the open sides? What's keeping them from doing so? And what is the frequency of these waves? There is certainly E and M field between the plates, but this isn't radiation since it doesn't oscillate or have any other wave properties. Pfalstad 03:31, 6 October 2005 (UTC)

I believe I have explained above why the EM waves cannot escape (well not much can any way). As to the frequency, this is a little difficult to answer in the case of the constant current source. I must ponder that one. In the case of pulse charging of a TL from another TL of course, one can easily work out the wavelength, as twice the length of the cable and V=f*lambda. Remember, there are two counter propagating waves that add up to a steady dc voltage so you cant see them with a scope or anything. If you were to connect a resistive load suddenly to the end of the charged TL what would you see across the resistor on your oscilloscope?--Light current 05:29, 6 October 2005 (UTC)

Oh, well in the capacitor is filled with dielectric, I could see that much of the wave might stay in. Not so for a vacuum capacitor. (But what happens in a vacuum capacitor, in your model? I don't see how the wave can stay in, in that case.) So you're saying there are two waves that add up to a "steady dc voltage"? The capacitor is filled with a nearly uniform electric field, is that what you mean? Standing waves do not create constant electric fields; the fields oscillate. If you add two waves together you do not get a constant field. Pfalstad 06:30, 6 October 2005 (UTC)

Consider two parallel rectangular plates in outer space one on top of the other with very small spacing. There is nothing between the plates. This is a vacuum capacitor. Now for any decent amount of capacitance between these plates, they are going to need to be very close together (depending on the area). This necessarily means that the characteristic impedance of the transmission line formed by these plates will be very low (sqrt(L/C). EM waves travelling between the plates when reaching the open end will see a much higher impedance then they have been used to (377 ohms to be exact). Now under these conditions only a miniscule anount of power can radiate from the open end due to the impedance mismatch. This is why in the limit of an infinite csa of the plates (giving infinite capacitance), no radiation whatsoever can escape from the open end.

Look, there are no standing waves inside a steadily charging capacitor! Please refer me to any text that says otherwise. If you want to know what the fields look like in a capacitor, read the Feynman Lectures book II, 23-2. There are standing waves if the frequency is 60 GHz or so, where the wavelength is on the order of the capacitor dimensions. At high frequencies, the capacitor can act as a transmission line as well. Or a waveguide, to be more precise. But not at low frequencies. Waveguides have cutoff frequencies. So a steady DC current will not cause waves to propagate inside one. Pfalstad 20:40, 8 October 2005 (UTC)

Not that you can see, but that's because they are counter propagating and adding to make dc. Can you not see this? When you charge a transmission line from a battery via a resistor equal to its Z0, a travelling wave of energy flows into the line at 'c' for the line, reflects off the open end and returns to the start. Then the line is fully charged. Have the travelling waves been stopped? If so how? Or are they continuing to travel up and down in opposition? BTW the text books dont cover this but its obvious to anyone who has been involved in the design of HV pulsers. [6] for radar or such pulsed energy sources. --Light current 20:58, 8 October 2005 (UTC)

PS You dont need to refer to any text, this problem can be solved with just a little thinking!--Light current 21:00, 8 October 2005 (UTC)

Chapter 2

My dear Light Current, now we are getting somewhere! I get what you are saying now. I thought you were nuts. I bet if you suddenly put a DC voltage across a capacitor, you will get some standing waves like you describe. Just like transmission lines. But they will die out quickly. And I doubt you will get much of that with a steady ramp. But you might get some small amount. Which might be all you were saying. Now, when you charge a transmission line like you describe, the wave goes out and reflects back and then stops, because you hit the original resistor and that removes the travelling wave. There is no more wave. I tried this in a circuit simulator. You can try it too, because I might have done it wrong. With a steady ramp, you might get lots of little waves.

OK now I must proceed very carefully in order not to destroy the insight you have just gained. So I will comment on your paras individually to ensure youre still with me. If you put dc straight across a cap/transmission line with no source resistor, you'll get a surge current given by V/Z0 wher z0 is the 'surge' impedance of the line. (Im going to use the word 'line' or 'TL' from now on, but it can be understood generally to include a parallel plate capacitor of similar dimensions.) THe condition I was recently describing was where you include a source resistor equal in value to the Z0 of the line. In other words the system is matched. When the wave (having reflected off the far end) reaches the near end, the voltage there is the same as the source voltage, so no energy can travel back into the source. Yes? Now you say that at this time, the 'resistor' removes the travelling wave. I dont really think I would put it that way, but the resistor appears as an open circuit to the returning wave because it is of the same voltage as the source (I=V/R and v=0, so no current back into source, so looks o/c. yes?)

Not sure I would say that the resistor appears as an open circuit. An open circuit would reflect the wave. Take out the resistor and it would reflect the wave. In the Feynman Lectures, he says that a properly terminated TL looks like one that is infinitely long. The resistor looks like the next segment of the line. Anything else would cause a discontinuity and thus a reflection. Pfalstad 18:21, 9 October 2005 (UTC)

The only reason I say the resistor appears as an o/c is that no current can flow thro' it because the pd across it is zero when the reflected wave arrives. Do you agree? Well, OK, take out the resistor if you like. It makes no difference now because the TL has charged up to the applied voltage. You say there is no more wave, but do you agree that the TL is now charged?--Light current 21:08, 9 October 2005 (UTC)

yes the TL is charged when the wave comes back. Pfalstad 23:20, 9 October 2005 (UTC)

Good! I agree on that one. So theres only dc now. But there were traveling waves. How come?--Light current 01:03, 10 October 2005 (UTC)

Well there were waves when you connected the battery at one end, because the TL wasn't charged, and the information/energy/current/voltage needed to be propagated to the other side. Then the information that the other end wasn't terminated needed to be propagated back to stop the current through the battery. The wave stopped there because the impedance was matched to the line. Pfalstad 01:45, 10 October 2005 (UTC)

Now you also say that the travelling wave has been stopped dead in its tracks! Do you know of any mechanism in EM theory which allows this?? Have a think about the above before we move on.--Light current 15:49, 9 October 2005 (UTC)

Yes, it's called transmission line theory. What do you think happens if you put dc across a transmission line that is terminated by a resistor matched to the line? There is no reflection. I don't know if there are any cases where that can happen with EM waves in space. But it is predicted by transmission line theory, which is derived from EM theory. Pfalstad 18:21, 9 October 2005 (UTC)

If you were to remove the resistor as you suggested above, then you say the wave would reflect from the open circuit. I agree, and this is effectively what happens. If you are worried about the resistor, try this: remove the resistor at exactly the time when the reflected wave reaches the near end. What happens now? Does the wave get reflected or absorbed or what?--Light current 21:08, 9 October 2005 (UTC)

Remove the resistor and replace it with an open circuit? Not sure what would happen, because there is current flowing across the resistor just before the wave hits it. Replace it with a wire? Not sure what would happen there either. I'd have to try it in a simulator. Pfalstad 23:20, 9 October 2005 (UTC)

There is NO current flowing in the resistor when the reflected wave hits it cos there is no voltage across the resistor. Just another little hint here. The voltage source is still putting out V volts say and the reflected wave also has amplitude V volts. What's the current thro the resistor? (I=Vdiff/R). Also Dr Feynman was correct when he said in that a terminated line does not produce reflections. But what we have here is a line that is effectively open circuit. (I say effectively cos no current can flow in the resistor) A terminated line would require that the resistor was connected to ground at the sending end -- it isnt. It would require there to be no voltage applied to its end -- there is. So the cable is NOT terminated at the sending end-- It's Open circuit! You dont need to simulate it, just think about it!--Light current 00:05, 10 October 2005 (UTC)

Just before the wave hits the resistor, there is current flowing in it. There is a voltage difference across it. Removing the resistor would cause another wave to be propagated down the wire. Just after the wave hits the resistor, there is no current in it, you're right. Removing the resistor at that point would make no difference, because the wave has already been absorbed. For termination purposes, it makes no difference if the resistor is grounded or whether there is a voltage source in series with it. It only matters what the impedance is. A ground or a voltage source will affect the steady state, the DC response, but not the reflection of waves. An open circuit is one where no current can flow, with possibly a voltage difference. Just because there's no voltage or current across it does not make it an open circuit. Pfalstad 01:34, 10 October 2005 (UTC)

You say the wave has been absorbed (by the resistor I assume), but has it? What if you remove the resistor just at the precise moment the wave reaches it. Is the wave absorbed now?--Light current 01:45, 10 October 2005 (UTC)

I'm not sure how you define the moment the wave reaches it. The wave is a steep ramp. If you remove the resistor at the top of the ramp, when there is no voltage across the resistor, there is no reflection. If you remove the resistor just before the ramp, when there is voltage/current across it, then there is. You may be right when you say the resistor has not absorbed the wave. I'm not sure about that now. But the presence of the resistor is what caused the wave to be absorbed. Pfalstad 01:53, 10 October 2005 (UTC)

Im not sure where you get the 'steep ramp' from. This is a square leading edge propagating down the line from a voltage source then reflecting back to the near end. Maybe youve been thinking current source? BTW your last sentence is not logical. I said remove the resistor just as the reflected wave gets there.--Light current 01:59, 10 October 2005 (UTC)

I'm describing the square leading edge as a steep ramp. An instantaneous change of voltage from ground to 5 V is not realistic. Every wire has some inductance in it, which prevents an instantenous change of current. There must be a ramp, however short. In that case it's meaningless to say "just as the wave gets there", it's either before, after, or during. As for the last sentence, I said that if you remove the resistor just before the wave gets there (or during) you get a reflection, but if you remove it just after, you don't. Pfalstad 02:13, 10 October 2005 (UTC)

Aha OK. In that case can you just pretend that the wavefront is infinitely steep? I dont really think it makes any difference to the overall argument because your not going to absorb all of the wave unless you were to terminate the line properly and leave the term on for twice the delay the of the line. I think I see where your going with this idea but I dont think youre right. Just absorbing the leading edge aint gonna do much -- youve got all that travelling wave behind to deal with!!--Light current 02:21, 10 October 2005 (UTC)

Well I don't think you're right. So where does that leave us? I think what I call "the wave" or "the ramp" is what you're calling "the leading edge". But I think that's all there is. Take an infinitely long TL. After the leading edge passes the line is charged and nothing changes in the portion that is charged. There no oscillation, no traveling wave, except for the leading edge. Back to your example; the fact that the source impedance is matched to the line constitutes proper termination in my view, and means that the wave will be absorbed when it comes back, leaving the line fully charged with no current, after twice the delay of the line. After that you can remove the source and it doesn't change anything. Pfalstad 02:35, 10 October 2005 (UTC)

When you say there's no traveling wave except at the leading edge, what about the magnetic portion (or the current if you like to think of it that way) of the EM wave behind the leading edge? Why does that stop travelling if you've only dealt with the front edge?--Light current 11:20, 10 October 2005 (UTC)

that's not a wave. a wire with a steady current has a magnetic field but it's not a wave. anyway on the trip back, the portion behind the leading edge has no magnetic field since there's no current. on the way there the leading edge has magnetic field behind it but that doesn't mean it's a wave. if the line were properly terminated then it would just stop at the end leaving the magnetic field there, with the current, without any wave motion at all. Pfalstad 09:32, 11 October 2005 (UTC)

OK So what youre saying is that once you have absorbed the leading edge/ramp whatever, the travelling wave has stopped and all we have is dc on the line. Now ponder this: If you now, at some slightly later time, connect a matched load to the end of the cable and look across this load with the scope you will see a pulse of half the dc voltage and lasting twice as long as the cable delay. Yes? Does this pulse of energy being dissipated in the load represent a traveling wave coming out of the TL?--Light current 02:52, 10 October 2005 (UTC)

This is after we disconnected the source, right? I think it will last only for the cable delay, not twice. You could call it a travelling wave if you want. Pfalstad 03:11, 10 October 2005 (UTC)

OKAY! Now lets not worry too much about the length of the pulse. But the situation we have is as follows:

1.We send a travelling wave into a cable and remove the source (the line remains charged of course)

2.We connect a matched load and get a travelling wave out. Yes?

So we have a travelling wave entering and a traveling wave exiting.Yes?

So why, when the wave is inside the cable has it decided to stop travelling> I'll leave you to think about that till tomorrrow. Good night!! --Light current 03:23, 10 October 2005 (UTC)

The wave is not "stuck" inside the cable. When you charge the cable and the wave has died out, there is no wave, just a charge. When you connect a load, that action causes a wave to go down the line. It doesn't "wake up" a wave that's already there. take a properly terminated TL. connect a battery to it. a traveling wave goes to the end and then stops. why did it decide to stop traveling? Because the line was terminated. No mystery there. Why'd the wave start traveling? Because you connected the battery. Waves are created and destroyed all the time. Pfalstad 09:32, 11 October 2005 (UTC)

In answer to your question about stopping, the wave doesnt actually stop: it is absorbed by the termination and energy is transferred to the load from the battery. As long as the battery is connected, it seems as if the EM wave continues. Otherwise, how does the energy get into the load. (not thro the wires). In fact this is the only way to kill an EM wave: fully absorb it. In my example, there is no termination (or we arrange for no termination to be present) when the reflected wave reaches the near end again and it is has no where to go except to be reflected. Waves are launched by connecting a source of EM energy (such as a battery) to a transmission line.Waves do not stop, cannot stop. They can have their energy absorbed by resistve load, but thers no stopping or slowing (in free space) You are trying to separate the steady state situation from the initial conditions and treat them differently. Is there a good reason for you to do that? If we accept your argument, there must be a distinct time at which the system changes from behaving like an EM transmission line to behaving like a simple circuit representation. What is that time, and how does the system know what it is?--88.110.58.202 14:48, 11 October 2005 (UTC)

The energy gets into the load through the wires. The electrons are moving. They have electrical potential energy, kinetic energy. That's how the energy gets to the load. In the steady state, if you just connect a resistor to a battery (forget the TL), there is no wave, just a current and a magnetic field. Nobody has detected any wave. Do you think there's a wave? What does it look like? What frequency? The fields are completely static. No oscillation, no wave. The moving electrons transfer energy. In the terminated TL example, there is no time when the system stops behaving like a transmission line. It behaves like one all the time. it's just that the wave (the changing part of the field) is absorbed when it hits the terminated end. At that point, there is no wave left. If we use the transmission line equations to predict the behavior, they predict steady-state because there is no wave present; a static-field solution prevails from then on. Pfalstad 05:21, 16 October 2005 (UTC)

• PS Does this make you wonder about the real nature of steady voltage at all? IE what is it really? Also, I though we needed accelerating charges to create EM radiation. Where are the accelerating charges in the battery?--Light current 15:43, 11 October 2005 (UTC)

The battery makes electrons move. To get them to move, it must accelerate them. This causes some EM radiation. I don't know much about how batteries work. I know that with a generator, the electromotive force is caused by induction. Induction causes electromotive force, forcing electrons to move. Pfalstad 05:21, 16 October 2005 (UTC)

Chapter Three

Define "wave". You've said that a battery connected to a terminated transmission line has a steady-state wave going down the TL. My understanding is that waves involve oscillation of the EM fields. Does a wire with current in it have a steady-state "wave"? Pfalstad 22:34, 11 October 2005 (UTC)

Short answer - yes. Im going to think carefully about the best way to answer fully your last question. Please wait.....--Light current 23:34, 11 October 2005 (UTC)

My defn of "wave":

An sinusoidal EM wave is an electromagnetic disturbance propagating thro a medium at the speed of light in that medium. The electric and magnetic field vectors are orthogonal and in phase. The energy transmitted by such a wave is given by the Poynting vector E x H and this energy flows in the dielectric betwen the guides (wires).

A composite EM wave consisting of a fundamental plus any number of harmonics at any phase relationship to the fundamental may also constitute a travelling EM wave. Thus, travelling EM waves may be of any practically realisable shape including 'square' or 'pulse' waves down to (but not inluding) zero rise and fall times.

--Light current 00:09, 12 October 2005 (UTC)

sounds good.. Pfalstad 04:59, 12 October 2005 (UTC)

When we think of the term 'steady state' sometimes we do not consider what process has had to occur in order to obtain the steady state. This can sometimes lead to erroneous conclusions. I think that this is one of those cases. In the case in question, what has happened is that a pulse (containing all frequencies- see NOTE)) of width twice the length of the TL has been sent down the line and then the generator (effectively) disconnected. So at this time, counter-propagating waves are extant on the line. The fact that you may observe their sum at any point on the line to be steady (dc) is neither here nor there. Now bearing in mind these two waves are traveling in opposition, one can easily see that associated with each wave there must be both a travelling magnetic field and a travelling electric field. Because the waves are counter propagating, it so happens that the magnetic fields cancel resulting in zero resultant current in the cable. However, the electric fields do not cancel but add to give (all points on the line) the effect of a steady dc voltage equal to the sum of the electric field components of each wave. So, even a wire with no current in it can support counter-propagating EM waves. Wires with resultant non-zero currents obviously have more current flowing one way than the other due to impedance mismatch at one end of the line. So, finally, the answer to your last question is that, yes, a wire with a current in it can also be considered to have EM waves travelling up and down it. Does this explanation make any sense to you?

NOTE: If we think of a repeating square pulse of voltage or current etc, having a defined length and repetition frequency, then according to Fourier analysis, this can be considered to be the sum of number of sinusoidal waves.--Light current 02:36, 12 October 2005 (UTC)

I disagree here. Show me the math. How can a wire with two counter-propagating waves have a steady dc voltage? there should be oscillating voltage. what do the two separate waves look like, mathematically? Pfalstad 04:59, 12 October 2005 (UTC)

I disagree also. I believe LC is very confused. The claim that a physical signal generator connected for some finite time to a physical TL has produced a pulse consisting of all frequencies is utter nonsense. The Fourier analysis of a traveling wave involves the integration of that signal over all time and over all space. The sinusoidal traveling wave components LC refers to in his ending note must have existed for all time to the present and will exist for all time into the future and further, must exist over all space. That's a pretty tall order for your typical signal generator and finite length TL. Alfred Centauri 22:49, 12 October 2005 (UTC)

I was being charitable by ignoring that. I already explained to LC that an ideal square edge is physically impossible due to the parasitic inductance of any real wire. I don't think his theory necessarily depends on a square edge. (Which is good because you can't have gamma rays in a transmission line.) Pfalstad 03:34, 13 October 2005 (UTC) Wait, no, I think it does.. Pfalstad 21:53, 15 October 2005 (UTC) Yes, this is a big flaw, AC. LC's idea seems to require perfectly straight edges. Unfortunately this is impossible in real life, since it requires combining waves of all frequencies, including gamma rays, taking us well out of the realm of classical electromagnetism. The perfectly straight edges are necessary so he can put the boundaries of the TL right on top of the edges, thus giving the necessary vagueness to where the edges are. If the wave edges are just inside the TL, then they would not behave like DC. If they are just outside, then what we have is DC, not a wave. Pfalstad 00:13, 16 October 2005 (UTC)

Please AC, nobody asked you, so give it a rest. I am not confused! I have shown the figures. You are struggling to find any objection to my ideas. Just give me a chance to develop them. If you have an objection to my idea please present it properly.

BTW How is your house looking now? Do you think you'll get a buyer soon?--Light current 01:43, 13 October 2005 (UTC)

Alrighty then - here is a properly presented objection to your idea. If I understand you idea correctly, you are claiming that the voltage across the TL in steady state is a result of the superposition of the electric fields of the traveling waves on the line adding constructively to a constant. Further, the reason that there is no magnetic field (or current) that we usually associate with a traveling wave on a TL is that the superposition of the magnetic fields (or currents) of the traveling waves add destructively to zero. Fine, let's go with that for a moment.

You are using the principle of superposition which is a powerful principle used successfully in all fields of physics so you seem to be on safe ground here. However, there is a subtle point that I believe you have overlooked. Let the TL in your example above be constructed from conductors with non-zero resistance. In steady state, the net current through the conductors is zero. But, according to you, this is "Because the waves are counter propagating, it so happens that the magnetic fields cancel resulting in zero resultant current in the cable.". OK, so there is zero resultant current since equal and opposite currents cancel, right? But, and this is a big but, power cannot be superimposed as it is a non-linear function of current. That is, even if the currents are equal and opposite so that the net current is zero, the power does not cancel. Since this loss is not observed when the steady state resultant current is zero, your idea is refuted thus.

I'm sorry, It is I who have forgetten superposition. The superpostion sum must be taken before calculating non-linear quantities such as power. Alfred Centauri 14:36, 13 October 2005 (UTC)

I think a diagram may be clearer.Both waves are of course rectangular (but can be made up from an infinite series of sine or cosine waves). V is the voltage amplitude of the waves. THe arrows show the direction of the wavefronts.

Mathematically, what is the form of the individual standing waves(sine/cosine) on the TL? Not the square edge; what do the components look like, I mean. Pfalstad 21:53, 15 October 2005 (UTC)

(Well each Fourier component of the sq wave is reflected in phase at the o/c Then these components recombine to give the sq wave again. But its obviously easier to think of edges rather than split it all up in to separate components. The only trouble with this analysis is that some people may think its not rigorous enough. Hence my Heaviside function research!--Light current 22:19, 15 October 2005 (UTC))

No I mean mathematically. I think for standing waves, the currents look like sin(n pi x) for x = 0 .. 1. And the voltages look like cos(n pi x). The only way to get DC out of those standing waves is to take n=0, which gives you zero current and constant voltage. But that's not a wave. Pfalstad 00:13, 16 October 2005 (UTC)

You are talking about the Fourier components I assume. It must be possible to analyse it that way, but since we start with a step function, its going to be easier to analyse using step functions I think. Just got to find out how to do it in travelling wave form.--Light current 02:52, 16 October 2005 (UTC)

Have a look at this animation while I think about Heaviside step functions [7]

Lets take the instant when the bottom wave has just reached the near end having been refelected from the far end

At a time 2T after first application of 2V to the line, the situation is as shown below.

     near end                        line length = T seconds(say)
     of line ---> ---------------------------------------------------------
                  *********************************************************
              --> * 
                  *                          V                            * -->
                ***                                                       ****
  
                  *********************************************************
              <---*                           ^                           * <--
                  *                           V                           *  
                ***                                                       ***

                  ---------------------------------------------------------

The top wave is the incident wave travelling to the right, and the bottom one is the reflected wave traveling to the left. As you know, the total line voltage is the sum of the 2 waves but I have shown them separately. Now because the near end has been made o/c at this instant, the bottom wave must reflect off the open end and in doing so, it follows the tail of the top wave. There is, of course, no gap between them. Similarly, the existing rightward traveling wavefront at the far end of the line sees an open cct and again reflects in phase joining itself to the end of the bottom wave.

At this point we therefore have both waves extant, traveling in opposite directions, adding up in voltage to 2V but having a resultant (net) current of zero (which is what you would expect). As you have probably noticed, the situation is exactly similar to a steady dc voltage on the line with no current. But would you say the waves are still there or not?--Light current 05:50, 12 October 2005 (UTC)

It's not accurate to say that both waves are extant. What exists are the fields. Looking at the fields, there is a constant voltage and no current. Even if this can be expressed as two counter-propagating waves, this does not reflect any physical reality. It is physically indistinguishable from no wave at all, just a constant field. It's certainly not a standing wave, because standing waves can be detected and measured, and they have mathematical forms which this field does not fit. They cause an oscillating field which has physical reality (even if it's too quick or small to detect). This has a field which is, by your own admission, completely static, undetectable even in theory. So there's no wave. Pfalstad 05:34, 16 October 2005 (UTC)

Ahh Ive just seen this post. This will happen if they are inserted out of order. Anyway, yes the 2 counter waves are indeed physically indistinguishable from no wave at all. Do not call it a standing wave if yuo do not wish to. That may be a misnomer. I was mistaken to call them standing waves and I apologise for the confusion.(I was thinking of the square wave version of 'standing waves' which I think they are, NOT the sine wave version) So to sum up, Thers is no detectable wave.-- or there are 2 waves counter propagating. The two representations are identical! yes?--Light current 14:55, 20 October 2005 (UTC)

I'm not sure I understand the situation. How do these two waves evolve? Do they just stay like this, constantly being reflected? Pfalstad 21:30, 15 October 2005 (UTC)

Well that is the $64K question isnt it? What do you think? I dont really know. All I know is they started out as waves and if you connected a *matched) load, waves come out. I suppose they could stay like they are because there are no losses due to skin effect etc because there's no current! Warning: you are now are leaving the world of physics/engineering and entering what is known as 'philosophical no mans land! ;-)--Light current 22:06, 15 October 2005 (UTC)

Well that's the problem with just "thinking" and not doing the math, isn't it? Pfalstad 00:13, 16 October 2005 (UTC)

Maybe, but you have to think first in order to know what sort of math to use. Also in many cases, proper thinking means you dont need any math at all! (esp vector calculus!) BTW Have U heard of the relativistic explanation of EM theory. Thats not got much math in it. I think a page on that could be good!--Light current 00:26, 16 October 2005 (UTC)

Yes, that's true too. A page on relativistic explanation of EM would be good. Pfalstad 00:57, 16 October 2005 (UTC)

I guess you could think of it that way, but why?? Why not just treat it electrostatically? Pfalstad 16:59, 12 October 2005 (UTC)

LC, Omegatron, and I have been over this before, Pfalstad. LCs description of the this scenario is deceptive for the following reason: the statement "the near end as been made o/c at this instant" is ill-defined. There are only two possibilities: Either the generator (with source impedance = Zo) is connected or it isn't when the reflected wave front arrives.

If it is connected, the reflected wave is absorbed and the TL is in steady state - disconnecting the generator at any time later has zero effect.

If the generator is not there then it must have been disconnected before the reflected wave front arrived. Thus, there is a new incident wave created by the disconnection of the generator before the reflected wave front arrives which implies that there is a time varying voltage along the line. This time varying voltage is in the form of voltage pulse moving back and forth along the line. The time average voltage at any point on the line is then less than the open circuit generator voltage.

In the limit, as the time between the disconnect of the generator and the arrival of the reflected wavefront goes to zero, the width of this pulse goes to zero and the time average voltage goes to the open circuit generator voltage. We can imagine disconnecting the generator an instant before the reflected wavefront arrives such that the pulse width is arbitrarily small but disconnecting the generator at the instant the reflected wavefront arrives is, IMHO, undefined.

Let the time when the reflected wavefront arrives be To. Now, let us say that the generator is disconnected at To. What does this mean? It either means that the generator is connected for t < To or it means that the generator is connected for t <= To. If the former, there is a new incident wave created and the reflected wavefront is re-reflected. If the later, there is no new incident wave and no re-reflection.

I suppose it is LCs intent to show that in the limit, these two scenarios must give the same average voltage so that he can claim that the steady state case is really the superposition of travelling waves. From there, I suppose he will claim that the voltage between the terminals of battery are really the result of trapped EM waves and from there, my guess is that he will claim that the electric charge (or the electron) is simply trapped EM waves. After all, when an electron and a positron combine what comes out - light, of course! (Actually, this isn't entirely true. If an electron and a positron collide with enough energy, you just might get a neutral vector boson...) Alfred Centauri 01:37, 13 October 2005 (UTC)

OK, if you can predict my argument, surely you can offer a rebuttal. I notice you have not done so. Why is that?

Im afraid that you have missed my point once again, Alfred. It does not matter when the generator is disconnected as long as there is enough time for the line to chaqrge up fully. There is no new wave introduced by the disconnexion of the generator. Indeed, as I have stated before many times, the generator can be disconnected at any time folloing the complete charging of the line (2T). This does not alter the situation on the line.

Well Alfred , I can see that you are a very intelligent fellow. BTW do you play chess by any chance? You assume (partly correctly) my plan. So can you trash my argument before I waste a lot of time in developing it?

I shall not reveal my modus operandi to you or any one else at this juncture. Please bear in mind tho' that my aim is not to win hollow arguments at any cost, but merely to seek the truth. I hope you will communnicate with me on that basis and not as an adversary.--Light current 02:10, 13 October 2005 (UTC)

If you argument is that the electron is a trapped EM wave or waves, then I shall offer the following points in rebuttal: The electron has a property called electric charge while an EM wave does not. The electron has a property called weak isospin while an EM wave does not. The electron has a property called mass while an EM wave does not. The electron is a quantum particle while a classical EM wave is not.

It appears that it is you that continue to miss my point regarding the disconnection of the generator. Your claim is that the disconnection of the generator does not launch a second incident wave AND that the reflected wave from the far end is re-reflected at the source end. These are mutually exclusive. IF the reflected wave is re-reflected at the source end, it MUST be because the source end is open. But this means that the generator MUST have been disconnected BEFORE the reflected wave arrived which means a NEW incident wave was launched BEFORE the reflected wave arrived.

Do we agree that if the generator is connected when the reflected wave arrives at the source end that the line is not only in steady state but that there are no waves on the line? Alfred Centauri 16:15, 13 October 2005 (UTC)

No. THere are still waves on the line. How can EM waves be stopped without absorption of their energy?.You should know this!--Light current 23:31, 13 October 2005 (UTC)

'Displacement' current in transmission lines

I have a new question on the transmission line. Is there any current from inner to outer (or vice versa) whilst the line is charging (or subsequently)?--Light current 16:12, 13 September 2005 (UTC)

If you are referring to electric charge current, then the answer depends on the material separating the conductors. This what the 'G' is in the transmission line equations - the distributed conductance in parallel with the conductors.

If you are referring to a displacement current, the answer is yes while charging. Alfred Centauri 18:05, 13 September 2005 (UTC)

Since the Catt is now out of the bag so to speak ;-), you know what my next question is, don't you? Where does this displacement current come from, bearing in mind the TEM structure of the waves travelling up and down the line?

Q: Where does displacement current come from? A: Displacement current is defined as the time rate of change of the electric flux density. So, whatever creates a changing electric flux density creates displacement current.

Now, consider a t-line with a voltage pulse propagating to the right along what I'll call the x-axis. The electric field associated with this voltage pulse is directed along the y-axis. This y directed E field is then given by:

$E_{y}=E_{0}u(ct-x)\,$

The displacement current is then given by:

$j_{D}={\frac {d\epsilon E_{y}}{dt}}=\epsilon cE_{0}\delta (ct-x)\,$

The magnetic flux density must be normal to the x-axis (TEM wave) and must be zero in front of the pulse. Now, imagine a closed contour in the x-z plane within the dielectric of the t-line and that encloses the displacement current. The integral of B along this contour is given by the flux of displacement current through the area enclosed by this contour. Only the leg of the countour inside the pulse and along the z-axis has a non-zero contribution. Thus, the magnetic flux density is along the z-axis and is given by:

$B_{z}={\frac {E_{y}}{c}}\,$

Alfred Centauri 03:29, 14 September 2005 (UTC)

Yeah, but where does the current come from? Do your equations say that it cmes from the EM wave, the conductors or what. Unless I'm missing something here, all you seem to have shown is that we have mag flux along the Z axis. So what -- isnt that what an EM wave does?. You have shown the existence of a time varying B field and are equating it to displacement current!?!? --Light current 04:07, 14 September 2005 (UTC)

I thought I was quite clear. Displacement current is defined as the time rate of change of electric flux density. Thus, if the electric flux is changing with time there is, by definition, a displacement current. Is this so hard to understand? What if I said 'the source of that displacement current was a changing magnetic flux current which was due to a changing displacment current which was due to a...', does that help? I didn't think so. Yet this is precisely how a TEM wave propagates in free space.

Is there any difference between this idea and the idea of an electric current? Do we not say that (due to charge conservation, I might add) an electric current is the time rate of change of electric charge? If the total electric charge in a region of space is changing with time, there is by definition an electric current into (or out of) that region of space. What is the source of that current - ultimately? For example, if I said 'the source of that electric current was due to an electric field which was due to a concentration of charge which was due to a current which was due to...', well, I hope you get the idea by now.

You said "So what -- isnt that what an EM wave does?". Why yes and that is exactly my point. A TEM wave within the dielectric of a t-line is no different from a TEM wave in free space except for the boundary conditions. A TEM wave in free space is 'unguided'. A TEM wave within a t-line is both guided and bounded. This guiding and bounding is due to a charge density wave in the conductors that make up the t-line.

You said "You have shown the existence of a time varying B field and are equating it to displacement current!?!?". Actually, no. I have used the well known Maxwell equation equating the closed contour integral of H to the the flux of displacement current through the surface bounded by the contour. Thus, a time-varying B field is equated to a time-varying displacement current. And frankly, the !?!? at the end of your 'question' is silly. If you don't believe in Maxwell's equations then just say so. If you believe I have applied Maxwell's equations improperly then say so. Alfred Centauri 12:14, 14 September 2005 (UTC)

Maybe we have slightly crossed wires here.

a) I do believe in Maxwells equations as applied to EM radiation.

b) I do believe that you have applied Maxwells equations correctly.

c) So what you are saying is that this entity that you call 'displacement current' actually does come from the propagating EM wave. Yes I would agree with that also.

d)Recall my original question: Is there a current passing from inner to outer during charging or subsequently?. I maintain that the answer to this is NO. There is no source of energy apart from the travelling wave that induces currents in the wires. BTW is this what you are calling the 'charge' current?

A charge current is a flow of charge - an electron current is a flow of electrons - an air current is a flow of air - a displacement current is a flow of displacement. What is displacement? See my reply to (e). Alfred Centauri 16:01, 14 September 2005 (UTC)

d) !?!? at the end of questions merely means asking the question with an expression of surprise (probably with a rising intonation in the voice).

e) When talking about displacement current (I didnt actually use that term in my question) I take it to mean what Maxwell took it to mean. Is that what youre calling 'charge current' (ie current consisting of charge carriers like electrons). --- What Maxwell was trying to explain away when he used the term was how a conduction cuurent could pass through vacuum. That current was what Maxwell termed displacement current. --Light current 15:15, 14 September 2005 (UTC) Correction: above parentheses should read "(I didnt actually mean to use that term....)" Sorry.--Light current 16:54, 14 September 2005 (UTC)

A charge current is a flow of charge - oh yeah, I've already said that. Maxwell thought that the vacuum wasn't a vacuum and that it could be polarized just like a dielectric. In some sense, he was right. Alfred Centauri 16:01, 14 September 2005 (UTC)

In an EM wave you are, of course, correct that there are both time varying electric and magnetic fields. Your belief is that because there is a magnetic field there must be a current to cause it. I personally dont see any source of current in the conventional sense of moving charges --- I just prefer to call EM radiation an unexplained physical phenomenon of the universe. Sure, you can write down all sorts of equations to describe its effects, but no one has been able to explain the physical basis of EM waves as far as Im aware. I dont think, either, that anyone has claimed to have measured the current(not its effects etc, the actual current) that is supposed to cause the mag field in an EM wave. Maxwells 'Displacement' current is what is supposed to flow between the plates of a vacuum capacitor when subjected to an alternating voltage. This sort of displacement current doesnt exist. Your sort--- well, I'm not sure-- nobodys measured it , or have they?--Light current 16:46, 14 September 2005 (UTC)

I suggest that you, if you haven't already, take a look at the quantum field version of electrodynamics (QED) to see what might be flowing between the plates of a charging capacitor. Regarding your concern that 'no one has been able to explain the physical basis of EM waves' - what is the physical basis of electric charge? Alfred Centauri 00:51, 15 September 2005 (UTC)

Sorry Alfred, I just would not know where on earth to look that up. Do you have any references I could use please? BTW personally, I think physical charge is actually an EM wave!!--Light current 01:59, 15 September 2005 (UTC)

If you have little or no background in quantum theory, expect a steep learning curve. Loosely speaking, QED posits that ultimately, all 'things' are 'fields'. For each type of subatomic particle there is an associated field. Particles such as the electron are quantum excitations in the electron field. Electrically charged particles 'couple' to another quantum field that is responsible for EM interactions. The quanta of this field are photons. That is, electrically charged particles interact by exchanging photons. Try this book for an introduction: [8] Alfred Centauri 02:38, 15 September 2005 (UTC)

Transmission line fed from both ends

Consider the following setup. A coaxial, or twin-lead, or parallel plate transmission line. Connect one side of your generator to a conductor on one end of the t-line. Connect the other side of the generator to the other conductor at the other end of the t-line. How does this 'thing' behave?

1.Is it a capacitor?

2.Is it a t-line?

What about a dielectric t-line? When I say dielectric t-line, think fiber optic cable. Alfred Centauri 03:34, 14 September 2005 (UTC)

3.Is it a capacitor?

4.Is there a charge current in a dielectric t-line?

While you ponder these questions, I'll ponder the best way to answer your question above. Alfred Centauri 23:13, 13 September 2005 (UTC)

This may take a little time (as Deep Thought [9] once said)--Light current 23:33, 13 September 2005 (UTC)

I'll answer your questions in your order:

1) The device is a capacitor (it has capacitance between the 2 leads)

2) I will decline to answer this part until I have done more thinking. Im sure the answer is yes but I can't justify my answer just yet.

Im now prepared to answer this part. To analyse this situation, we need to apply the principle of superpostion. Firstly, lets assume the line is a balanced twin 300 R feeder cable with wires A,B, fed from a transformer. Initially, make plate/wire B the ref. THe situation is one now of an UNbalanced transmission line with the energy flowing 'forward'(lets say) with a voltage step of +V. Now reverse the situation, and common/ground etc plate/wire A. In this case we have a similar unbalanced line with the energy flowing in the 'reverse' direction (a voltage step of -V). THe polarity of the voltages/ currents flowing in each direction are of course opposite to each other. These two independent waves each travel along the line and get reflected in phase at the open ends. Therefore, after the transient period, there is a steady voltage difference between the 2 wires/plates etc of +V-(-V) = 2V which was, by definition the voltage that was applied to the device. SO, it would appear that this device does indeed act as both a transmission line and a capacitor. Hey, do you realise the device you described is how they construct extended foil low inductance (and maybe other types of) capacitors?--Light current 14:20, 14 September 2005 (UTC)

I understand superposition fairly well but I don't understand what you have done here. Specifically, there is only once source so how do you use superposition for one source? If you split the source in two, where does the common node between the sources connect? How exactly have you made the t-line unbalanced? Alfred Centauri 15:12, 14 September 2005 (UTC)

OK Lets look at it another way. The source is differential with an o/p voltage of 2V (o/c). So, it can be considered that a positive pulse + 2V is applied to line A wrt line B. This results in a pulse of 1V travelling down the line due to the matching loss. At the same time, a negative pulse -2V is applied to line B wrt line A.(This is of course exactly the same thing said in a different way). This results in a pulse of -1V (again due to matching loss) travelling the other way down the line. When the pulses reach the ends they reflect in phase. So, after the transient period, there is always a 2V difference in potential between the lines and the line has fully 'charged'. There are no longer any travelling waves. We know experimentally that this is the case. Do you agree?--Light current 16:26, 14 September 2005 (UTC)

I'm still not sure I understand exactly what you are saying here so let me give you my thoughts on how this 'thing' must react Let the positive terminal of the source be connected to the left end of the upper conductor of the TL. Let the negative terminal be connected to the right end of the lower conductor. An instant after these connections are made, what is the voltage between the upper and lower conductors on each end of the TL? My intuition tells me that the voltage at either end is essentially zero. At this point, each conductor is acting like a long wire antenna. At some point in time later, the positive charge density wave moving to the right on the upper conductor 'meets' the negative charge density wave moving to the left on the other conductor. Where the waves overlap, there is a voltage between the conductors. What is this voltage? It's hard to say. To calculate this, one would need to determine the characteristic impedance looking into just one of the conductors (think of driving one of the conductors as an antenna). At any rate, if the source resistance is twice this impedance, the voltage between the conductors where the charge density waves overlap will be 1V I suspect. Thus, I picture a 1V voltage pulse starting in the middle of this thing sometime after the source is connected and propagating in both directions to the ends. Here the individual charge density waves are reflected. When these reflected pulses 'meet' in the center, the voltage between the conductors there becomes 2V and this pulse propagates outward towards the ends. At this point the charge density waves have returned to their starting points and the charge densities there should now match so this thing is now in steady state. This should be a fairly easy experiment to setup and test. Alfred Centauri 17:19, 15 September 2005 (UTC)

NB the following 2 answers assume that the radiation is fed in from one end only as I cant see how you could do it as you describe above at the moment.

3) Assuming that a fibre optic cable exhibits total reflection of the light (EM radiation) at each end and the propagation mode is TEM (I'm not sure if it is), then, yes, I believe it does act as a capacitor -- it is storing energy in the same? way as a capacitor does by bouncing the energy continuously from one end to the other.

That's interesting. FYI, only TM and TE modes exist in a dielectric t-line. Alfred Centauri 12:34, 14 September 2005 (UTC)

4)I'm not sure of your defn of charge current here. If you mean displacement curent when the EM radiation is first applied, then, no for the same reason that in a TEM wave there is no source of current transverse to the direction of propagation.--Light current 05:39, 14 September 2005 (UTC)

By charge current, I mean flow of electric charge (dQ/dt). Your last statement - that there is no displacement current in a TEM wave is equivalent to stating that there is no time rate of change of the electric field in a TEM wave. Is this what your are saying? Alfred Centauri 12:34, 14 September 2005 (UTC)

No. I refer the honorable gentleman to the reply I gave on 'Maxwells version of Displacement current'--Light current 18:23, 14 September 2005 (UTC)

--Light current 00:37, 14 September 2005 (UTC)

Transmission Line 'Capacitor' --- Experimental setup and Results

Suggested setup

Copied from prev posts

Yes. probaly best to use 300R twin feeder? I have to go out v. shortly, but I'll see if I can find some bits when I get home--Light current 17:49, 15 September 2005 (UTC)

I do think twin-lead is the best choice as it should be quite easy to probe the voltage at points on the line by using insulation piercing probe leads. I do think it will be important to use a true differential probe when measuring the voltage across the line - grounding one side with the O'scope just wont do! Alfred Centauri 18:59, 15 September 2005 (UTC)

I think the setup would need to be matched to properly observe this effect, so this means a 50R to 300R balun. I have some 75R/300R baluns but im not sure of the effect of mismatch from my 50R gen to the balun i/p. Also a fiarly fast edge will be needed if a short length of cable is used. THe fastest rise I can produce/measure is about 1ns (at least thats what it looks like on my scope). I dont know the velocity factor of twin lead either but I guess were going to need quite a few feet to see much happening.--Light current 00:09, 16 September 2005 (UTC)

Light travels 1 meter in about 3.3ns so 2 or 3 meters should be plenty. If my intuition is correct, the matching balun isn't going to do what you might think. Alfred Centauri 00:44, 16 September 2005 (UTC)

You could be right- we dont know whats going to happen but I was suggesting the balun partly to convert my unbalanced gen o/p to balanced. As to the matching, maybe its not important in the transient period. I dont think we can drive this 'thing' unbalanced--Light current 00:55, 16 September 2005 (UTC)

Ahh! Good point! Alfred Centauri 01:14, 16 September 2005 (UTC)

VF twin 300R is 0.82--Light current 01:08, 16 September 2005 (UTC)

Im just stuck for a back to back 75R connector now!(I normally work with 50R) Ive got everything else ready , (having destroyed my FM radio aerial!) You havent got a spare you could lend me, have you?--Light current 16:55, 17 September 2005 (UTC)

Experimental results from Light Current

1. BTW I have now completed the set up for the contra fed capacitor/TL. The results as yet are inconclusive. I may have a (low end) BW limitation on the balun that is causing some ringing. Also I am not able to match my 50R gen properly to the 75R balun i/p. Ive tried a 12dB 75 pad but thats no good. I need a 50 R pad but I ve got problems with the connectors not fitting at the moment!

end of copied text

2. I found my standard X10 scope probes have too much capacitance and distort the pulse waveform applied to the 300R feeder. Experimenting with low cap 500R home made probe. (altho' the Zin is a bit low, at least its lower capacitance doesnt completely foul up the waveform!) Have you had any success yet on the experiment? BTW dont forget to use matched length cables to your scope to get the right delay data!--Light current 19:56, 18 September 2005 (UTC)

3. I am usig a 1MHz square wave with 1ns rise to drive the balun. The voltage step wrt common (source 0v) at each o/p of the balun was 1 div (giving 2 div diff to the cable).

4. I found three (maybe four, its hard to tell) distinct voltage levels at the open ends of the line. For one of the lines they were approximately -0.5, +1.25, +1.75, +2.0 divisions all occupying about 6ns each. 6ns is the approx length of my cable( ~1.5m @ 0.82 vf). The other line had the inverse of these levels ie +0.5. -1.25 etc. There was no delay measured (well not detectable on the scope) between the generator o/p rising edge and the rising edge of the differential signal appearing at the cable input.

5. Heres the interesting one: there appears to be no delay between the input signal and the first 6ns excursion at the open ends!! How can this be??. BTW I arranged the cable in an approximate circle to try to avoid any signal shortcuts/pickup. It does make a difference to the waveform amplitudes you see. Do you have any thoughts on these strange (to me) results?--Light current 20:43, 18 September 2005 (UTC)

6. That first transient period of 6ns must be due to direct capacitive coupling at the near end of the cable. There's no other way it can happen. I shall ignore the first 6ns fluctuations as an artifact. Is this due to ACs postulated packets of charge coupling to the other wire of the cable??

7. Now the situation is this: After 6ns (one way travel time) the voltage on the open ends has magnitude 1 div. After another 6ns, the magnitudes go up to 2 div and stay there for about 15ns. After that the voltages return to zero in about 10ns and then show a back swing of about 0.5 div. This backswing lasts for about 20ns.. There is then what appears to be a small (0.1 div) positive reflection lasting about 20ns. The voltage on each line end is stable for about 20ns at +2 div and -2 div. giving a diff voltage of 4 div. NB the drive signal is only +/- 1 div from the balun! I need to think more about it!

8. AHH! hold on. The o/c o/p voltage from the balun is almost +/- 2div, so initially this gets reduced to +/- 1 div by the load (after 6ns of course) then after the transient, no further loading occurs and the line is charged to +/- 2 div!!(until the balun o/p cant hold it up any longer)--Light current 23:06, 18 September 2005 (UTC)

9. At the centre point of the cable things are harder to see. Nothing happens here until 3ns after the applied pulse from the balun, then there are a number of steps of rising magnitude but with more rounded wave shapes, and not quite as great in magnitudes at those at the cable ends. - Its difficult to do proper measurements at the cable centre. I need to get the proper 50R/75R adaptor so I can use a 50R pad. Im using a 75R 12db pad at the moment which is reducing my applied signal too much.--Light current 23:38, 18 September 2005 (UTC)

The first three to six nanoseconds

During this period, it is not clear what is happening on line. Until the pulses of charge meet at the center of the line, each wire may be acting as an aerial as there is no 'return path' for the flow of charge. If these 2 charge packets are travelling, what guides them and what is their speed of propagation? Where is the other 'conductor' to guide these 'waves'?.

The magnitude of the pulse at the open ends of the cable during the first 6 ns does appear to be highly dependent upon the physical form of the loop of cable. If the cable is folded so it consists of 2 closely coupled strips of equal length, then the magnitude of the initial 6 ns pulse approaches zero. later caharcteristics of the waveform remain mainly unchanged. If the cable is formed into a circle, the magnitude of the initial pulse is maximised. This would tend to indicate that a higher capcitance to the return conductor is absorbing the charge and giving rise to a lower voltage at the cable ends during the first 6ns. One conclusion that can be drawn is that the system does not appear properly act as a balanced transmisson line until the signals have reached the far end of the cable. That is, a transmssion line needs a return conductor in which to induce charge current. If this return conductor is not present then transmission line characteristics do not obtain. From the foregoing, it would appear that the (contra fed) capacitor construction is not the ideal arrangement in that radiation may be emitted during the first one way travel time of the capacitor due to antenna effects. For a capacitor of length 1.5cm this would mean an initial radiating period of 60ps with poyethylene dielectric. Of course this assumes an open wire construction of which the extended foil capacitor is certainly not an example. Also the majority of capacitors are not operated in the balanced mode-- sometimes one end is connected to common. With extended foil capacitors, the outer foil is usually connected to common and this tends to screen the inner foil creating an unbalanced transmission line. In this case, radiation should not occur.--Light current 13:52, 19 September 2005 (UTC)

General conclusions drawn by LC

It would appear wise, therefore, when using transmssion lines as fast capacitors, that connections should be made at the same end, leaving the far end o/c. It is not wise to try to use balanced TLs as capacitors. Extended foil capacitors do approximate, to a large extent, transmission lines as long as they are used in the unbalanced mode(ie one end grounded/commoned).--Light current 14:04, 19 September 2005 (UTC)

Results from Alfred Centuari

Still awaited

Relation between voltage and current

Actually, I did use a different functions g1() and g2() for the current and this is still correct if you want to work out the current at any point on the line. Of course then the voltage would be constrained by the current values in th e ratio of Z. Its not quite what I intended, but on the whole I think what you have put may be preferable to avoid any confusion.--Light current 09:06, 19 October 2005 (UTC)

Trapped EM waves in transmission lines Part II

I have carefully considered ACs analysis of the waves in an oc TL. He has shown that in the SS, there are no waves detectable but only dc due to the constant value Ao. This is correct. After all, Ive always said that the travelling waves cancel out to give steady dc.

In order to show the travelling waves, the analysis should be done before the refelected wave reaches the near end. In that case, the input current is not zero but still V/Z. If this analysis is done, it will show counter propagtinig waves right up until the exact time the reflected wave reaches the near end. At this time all waves will seem to disappear.

Is this because they have been absorbed? If so, by what?. My suggestion is that they are reflected, continue travelling from end to end in a reciprocating manner. The net visible effect is one of complete cancellation. So one could say both that the waves ARE there, and at the same time, the waves ARE NOT there. At this point were are entering the world of philosophy. Do we want to go there?--Light current 14:27, 20 October 2005 (UTC)

Further exposition of the subject of counter propagating waves.

Because the consideration of counterpropagating waves in a single open ended transmission line is rather confusing and tends to obscure the true nature of what is happening, I propose to offer an alternative but, I hope equivalent expalnation.

In this case, we take not one, but two TLs of equal length and lay them side by side. At each end ot the pair is a change over switch that can be used to connect the cable inners together or to a voltage source (I am using coax). So The LH end of line A is initially connected to a matched generator with a voltage V on it. The RH end of line B is connected to a similar generator again with voltage V. At t=0, both generators are switched on and remain connected for T seconds, where T is the length of each line. At this time both switches are simultaneously operated.

Now I think no one can deny that, just before the switches are operated, we have a square travelling pulse flowing from left to right in line A, and a square travelling pulse flowing from right to left in line B. The pulses are of equal magnitude. When the switches are operated what happens to the pulses: do they stop, or do they carry on flowing into the other line?

If we accept the obvious answer,(that they carry on) we have a situation akin to the single line case of counter propagating pulses, with a net current of zero on the pair of wires, but non zero currents in each individual wire. Yet there is a constant voltage at all points of each cable (equal to half the generator o/c voltage) In this argument, no mathematics is needed, an no mathematics should be needed to argue against it. Can anyone argue against it?--Light current 16:49, 20 October 2005 (UTC)

And now... just for fun, lets open one of the switches ( but not connect it to the generator which has been removed from the experiment). What now happens to the pulse? Is it destroyed or absorbed< I dont think either. The energy is still there. The pulse now simply starts to reflect off the open ends again. Close the switch, energy proceeds to travel in a circle. Open the switch.... well Im sure you get the idea!--Light current 17:11, 20 October 2005 (UTC)

I don't get it. How are line A and B connected? Anyway, if individual wires have nonzero current in them, I don't see how that applies to the case where they add up to DC, because in that case, there is no current. Pfalstad 20:52, 20 October 2005 (UTC)

When switches are open, each line is connected to its own gen with far end o/c. Lines A,B are connected in a loop when both switches are closed. You miss the point about about the 2 lines. When the two lines are connected by the switches, there is net zero current (flowing from left to right or right to left) BUT current does flow in the individual lines.Its the voltage that remains, the current cancels out in the parallel configuration.

        switch                    Line A 
         |----------------------------------------------------------------|
        /                                                                 |
 GEN---/                                                                    /----GEN
         |                        Line B                                   /
         |----------------------------------------------------------------| switch 

Setup of cables gens and switches. Only line inners shown. Switches now spco types--Light current 22:15, 20 October 2005 (UTC)

A situation with constant nonzero current (in a circle) but constant nonzero voltage is clearly unrealizable. Since the lines have finite resistance, there is no way this can happen. If we assume perfect conductors, I am still suspicious of what happens at the edges. You're using those perfectly square edges again and I think that misleads us. I don't think the waves would combine into a nice little loop. I also still miss the point about how the 2 lines compare to the 1 line situation. Pfalstad 02:33, 21 October 2005 (UTC)

Honestly I cant see why its clearly unrealisable. Can you explain? Im assuming the lines are ideal but with Zo =50 ohm say. So V=IZo surely?? Lets leave the edges for the moment. I'll mention them later. They're not reall important to the argument. AS regards the comparison, in this situation, the counter propagating waves have been split up an put on their own lines. Instead of the waves refelecting off the end of the single line, here they go back on the other line. Simple? --Light current 02:39, 21 October 2005 (UTC)

Well it's realizable in a superconductor I guess, except for what happens at the edges which I'm still not sure about. Pfalstad 16:00, 21 October 2005 (UTC)

Remember the voltage is on the line wrt the outer conductor which is not shown (its all coax cable!)--Light current 03:09, 21 October 2005 (UTC)

This long discussion is too much for me to wade through. It would help if you recapped the problem you are posing with all assumptions (lossy vs. loss-less, for example). A couple of observations though. First of all, it is common to induce a current in a super-conducting ring that goes on for ever. Second, if you take a piece of transmission line with a given characteristic impedance and open-circuit it, it no longer has a single impedance. Rather the impedance varies with frequency, dropping to zero at odd multiples of quarter wavelength, infinite at even multiples. I believe the sort of stuff you are proposing will generate RF at the resonant frequencies, much like a spark-gap transmitter. --agr 03:29, 21 October 2005 (UTC)

Im trying to prove that equal amplitude counter-propagating square waves on an 'open circuit at both ends' lossless transmission line sum to zero current but give twice the voltage of each wave and the waves then effectively disappear leaving only dc on the line. Now read on!--Light current 03:37, 21 October 2005 (UTC)

First of all (unless I am missing something), when you close the switches, the length of the line becomes 2T, so in your model, at best, you have two width-T pulses circulating on a 2T line. Second, a square wave in voltage does not imply a square wave in current. I believe in current, to first order, you will get a spike that starts at the maximum current your generator can deliver and decays exponentially as local capacitance is charged. The two spikes will will keep traveling along the looped line by the principle of superposition. --agr 09:47, 21 October 2005 (UTC)

Your first part - I agree. Second part - I disagree. I=V/Zo so current is constant. Remember, the initial waves in the individual line are square cos they are provided from a matched generator. Max current from matched load is I= Vo/2Zo where Vo is the o/c output of the generator. There is no exp decay in the waves. They are all square because the system is matched. Try it at home with a sq wave gen matched to the line & scope. Your last comment: yes the two square waves will keep travelling (no space between them either) around the looped line.--Light current 14:16, 21 October 2005 (UTC)

Lecher lines (moved from my talk page)

Do you have any thoughts on the subject of tuned circuits for UHF/SHF, I am planning on expanding the page a little to make it more useful.

Yes, But I have to gather my thoughts. The trouble is that you have reverted some of my corrections to this page in your addition. Can you pleae be more careful when adding material that you dont destroy peoples improvements? BTW can you please sign your posts by typing four tildes so we know who to reply to. Thanks.--Light current 21:13, 27 December 2005 (UTC)

I think that a clear need exists for the division between balanced and unbalanced transmission lines. Many aerials are balanced such as the centre fed half wave dipole, these aerials can be fed using a balanced line which is connected directly to the some old equipment or via a balun in the shack. The other option is to use a balun near the feed point of the aerial and an unbalanced feeder (This arrangement normally leads to higher losses, but it is more EMC friendly). While the use of long runs of balanced feeder is something which many people outside of the radio community are unaware of, it is a concept which is very important and WP will be better if such things are included.

Sorry I overwrote any improvements which another editor made, but I was reverting some changes which had made the lecher line section into dire state where it was not in the correct order.Cadmium 23:00, 29 December 2005 (UTC)

Yes but we dont want to turn it into page on Ham radio solely- its about transmission lines. Sounds like we need a page on aerial matching however, aerial feeders (bal and unbal) are I think are appropriate on this page.--Light current 23:51, 29 December 2005 (UTC)

I have already been working on the subject of Antenna tuner, while the term tuner is a misnomer (a true ATU is a hacksaw/wire cutters for changing the length of an element) I reason that most people who have had some exposure to radio electronics will know a aerial matcher as a ATU so I think it should keep its slightly bad name.

While in the ham community the use of balanced feeders for HF systems is common, I would say that the use of balanced feeders is not confined to the HAM community. I am sure that with regards to (nuclear electromagnetic pulse) NEMP that a signal line which is a balanced feeder fitted with a good balun at each end would be a far better choice than a simple length of coax cable. The reasoning is that most of the RF energy of NEMP is below 50 MHz, and that it induces a common mode current on any long conductive object. I can get you a reference (A IEEE monograph on the protection of electronic equipment) if you want to read about this set of ideas for designers. Even while the cold war is over, I suspect that NEMP is still an interesting topic and it is similar to lightning strikes.Cadmium 10:25, 30 December 2005 (UTC)

missing info

I see two significant gaps in the information here...

• In the practical transmission line section, there should probably be a few sentences on wave guides, which are a very special form of transmission line.
• There is an equation that relates diameter of conductors (and dielectric material) to complex impedance of the line. This could potentially lead to a large section, as this has implications for both balanced line and coax and limits minimum and maximum impedance of both, but I think it would be OK to keep it small. At a glance, I can't tell if this equation is related to the transmission line equation that is already there, but I don't see any practical application of that equation in the article anyway, and certainly no mention of relation of wire construction to impedance.

Anyone want to take a stab at those, or should I give it a start?--ssd 15:48, 30 December 2005 (UTC)

You start- I'll watch! BTW Im a bit uneasy about lumping WG with TLS. They work on diffrernt principles that Im not sure about--Light current 17:03, 30 December 2005 (UTC)

Please see page on Waveguides--Light current 01:13, 24 January 2006 (UTC)

First, they work on identical principles: solutions to Maxwell's equations under boundary conditions determined by the physical structure.

Second, since the intro declares that waveguides are a type of transmission line (this text comes from Federal Standard 1037C, so it probably is a correct definition in some community, probably US government contractors)it would be a good idea to include some description and linke to the waveguide article. Or we could have a long discussion about whether transmission lines are a kind of waveguide, or waveguides are a kind of transmission line, or they are two distinct categories with no overlap (See Talk:Waveguide).

Since an encyclopedia is about ideas, not words, that would be futile. It's better to let the article itself define what idea it is going to be about, and then describe that idea. So if this article is about all things that transmit signals or power along a line, then lets just make it the best possible article about that idea, and not waste energy fighting about the word. If you want to add a note about the use of the nomenclature in different contexts or countries, that would be a reasonable thing to do. But if Ssd wants to add a brief description of waveguides, that's reasonable too.-- The Photon 04:54, 11 March 2006 (UTC)

Displacement current in TLs - more questions

If displacement current (ie current orthogonal to direction of energy flow) is not needed in a vacuum TL, why should it be needed in a TL with a dielectric? Does it exist in a TL with a dielectric? Has this been confirmed experimentally? --Light current 07:29, 1 January 2006 (UTC)

Page splitting

Does anyone else think it may soon be time to start moving the matl on different sorts of TLs to their own pages? THe detail on some sorts of lines is now getting rather large!--Light current 01:11, 24 January 2006 (UTC)

Yes, this page definitely needs to be split up. I think that there should be a separate page for the Telegraphers Equations, and a separate page for lossy transmission lines because these require more advanced treatment and the current page is confusing for the bulk of people who don't care about lossy lines or partial differential equations. The part on variations of cable impedance with frequency, is only relevant for lossy transmission lines of large distances (e.g. cables > 60km at 50Hz) so it should be moved to the lossy transmission line section. I'm going to just go ahead and change it today as much as I have time for. -- Perceptual Chaos 00:46, 16 May 2006 (UTC)

After making significant changes to the page I don't think there needs to be a separate page for lossy transmission lines, just that the variation of impedance with frequency section should be moved to the article on characteristic impedance. I will do that soon. -- Perceptual Chaos 02:44, 16 May 2006 (UTC)

I really think the telegraphers equations are absolutely essential to this article. Thats why I merged them originally. Removal just rips the guts out of the article and remove the whole basis of TL theory. I ask you to consider replacing this material. 8-(--Light current 03:03, 16 May 2006 (UTC)

I agree that not mentioning the telegraph equations and what they mean would remove the whole basis of TL theory, but I think that having a big link saying "main article : telegraph equations" which interested readers can go to is adequate. If anything, it makes the equations more important by giving them their own article. I'm open to argument but I think it was just cluttering up the article... think of the majority of people who don't really know what a transmission line is and aren't comfortable with PDE's - this just confuses the majority of readers and clutters the page. People with a more advanced background can read the main article, people without that background can just read a summary of what they are and what they mean. Don't you think??? Btw, I'm finished my editing spree for the most part, anything else you like or don't like about what I've done? --Perceptual Chaos 03:37, 16 May 2006 (UTC)

On second thought, I think that putting the second order wave equations for a lossless transmission line would be useful, there is a nice simple representation in my stage 2 E & M text book that I will use --Perceptual Chaos 03:55, 16 May 2006 (UTC)

Done, are you happy with this now Light current? I think its the simplest form, although the main article uses the form with the time derivative term still included which may be confusing to some readers. I should probably put my form as an alternative representation into the Telegrapher's Equations main article and perhaps a note and how to get there from the other representation. -- Perceptual Chaos 04:48, 16 May 2006 (UTC)