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Why do high-voltage transmission lines buzz?
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)
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)
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)
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
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)
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)
Yes. Confusing , isn't it? Can you enlighten us?-- Light current 02:21, 13 October 2005 (UTC)
(moved from Talk:capacitor no 17/10/05)
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] :-)
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)
No, they are not. How are they similar? — Omegatron 22:16, September 11, 2005 (UTC)
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)
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)
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)
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)
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)
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)
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)
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."
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!.
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.
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)
If the charging resistor is much larger than the cable impedance, the rise would be logarithmic.;-) -- Light current 16:14, 12 September 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)
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)
1 - *--------* | | 0 - | *-------- | -1 - ------------*
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)
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)
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)
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)
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)
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)
I believe that 'O' is essentialy totally correct on this point also
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)
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)
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)
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)
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)
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)
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)
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.
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)
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?)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
-- Light current 00:09, 12 October 2005 (UTC)
sounds good.. Pfalstad 04:59, 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 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)
BTW How is your house looking now? Do you think you'll get a buyer soon?-- Light current 01:43, 13 October 2005 (UTC)
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)
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)
near end line length = T seconds(say) of line ---> --------------------------------------------------------- ********************************************************* --> * * V * --> *** **** ********************************************************* <---* ^ * <-- * V * *** ***
---------------------------------------------------------
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)
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's the problem with just "thinking" and not doing the math, isn't it? Pfalstad 00:13, 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)
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.
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)
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)
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)
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?
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)
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?
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)
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)
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)
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)
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)
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)
That's interesting. FYI, only TM and TE modes exist in a dielectric t-line. Alfred Centauri 12:34, 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)
-- Light current 00:37, 14 September 2005 (UTC)
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 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)
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)
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)
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)
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)
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)
Still awaited
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)
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)
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)
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)
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)
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)
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)
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)
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.
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)
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)
I see two significant gaps in the information here...
Anyone want to take a stab at those, or should I give it a start?-- ssd 15:48, 30 December 2005 (UTC)
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)
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)
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)
![]() | This is an archive of past discussions. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page. |
Archive 1 | Archive 2 |
Why do high-voltage transmission lines buzz?
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)
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)
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)
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
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)
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)
Yes. Confusing , isn't it? Can you enlighten us?-- Light current 02:21, 13 October 2005 (UTC)
(moved from Talk:capacitor no 17/10/05)
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] :-)
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)
No, they are not. How are they similar? — Omegatron 22:16, September 11, 2005 (UTC)
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)
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)
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)
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)
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)
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)
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)
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."
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!.
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.
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)
If the charging resistor is much larger than the cable impedance, the rise would be logarithmic.;-) -- Light current 16:14, 12 September 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)
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)
1 - *--------* | | 0 - | *-------- | -1 - ------------*
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)
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)
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)
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)
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)
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)
I believe that 'O' is essentialy totally correct on this point also
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)
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)
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)
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)
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)
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)
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)
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.
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)
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?)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
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)
-- Light current 00:09, 12 October 2005 (UTC)
sounds good.. Pfalstad 04:59, 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 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)
BTW How is your house looking now? Do you think you'll get a buyer soon?-- Light current 01:43, 13 October 2005 (UTC)
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)
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)
near end line length = T seconds(say) of line ---> --------------------------------------------------------- ********************************************************* --> * * V * --> *** **** ********************************************************* <---* ^ * <-- * V * *** ***
---------------------------------------------------------
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)
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's the problem with just "thinking" and not doing the math, isn't it? Pfalstad 00:13, 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)
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.
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)
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)
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)
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?
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)
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?
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)
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)
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)
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)
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)
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)
That's interesting. FYI, only TM and TE modes exist in a dielectric t-line. Alfred Centauri 12:34, 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)
-- Light current 00:37, 14 September 2005 (UTC)
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 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)
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)
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)
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)
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)
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)
Still awaited
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)
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)
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)
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)
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)
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)
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)
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)
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.
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)
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)
I see two significant gaps in the information here...
Anyone want to take a stab at those, or should I give it a start?-- ssd 15:48, 30 December 2005 (UTC)
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)
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)
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)