This page 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. |
I tend to agree that the D s should not exist-- at all! (not even at the step). That's because I believe that nothing flows from one conductor to the other in a capacitor or TL. Energy flows along the line -that is all. This question of infinitesemal rise time is not relevant to the problem. How does your view differ from this?-- Light current 23:21, 10 January 2006 (UTC)
What do you mean by the term displacement current?-- Light current 02:15, 11 January 2006 (UTC)
I tend to agree with all of your last paragraph Kevin-- Light current 02:15, 11 January 2006 (UTC)
Kevin: before the TEM step arrives at any part of the cable, there is zero voltage. As it arrives, the voltage increases and with it the current. Once it has peaked, it remains steady. The only period of interest is while the voltage is rising. Ivor ignores this, and also introduces other errors like the claim that displacement current flowing where the voltage is steady. Your sine wave radio signal versus square wave logic step argument is vacuous, since radio emission occurs so long as di/dt is not zero in any part of the aerial. Fourier analysis is useful to me for the purpose of translating a plot of current or field strength versus time into a frequency spectrum. According to Fourier analysis, you can represent a square wave as a series of sine waves! But that's not my main argument.
Consider the 2 conductor transmission lines as 2 parallel radio aerials. If you feed one with a signal (of any type) and leave the other passive, the first transmits energy to the passive one which receives energy only as a result of di/dt in the first one. This is indistinguishable from Maxwell's "displacement current" equation. Maxwell says vacuum "displacement current" i = e.dE/dt = e.dv/(x.dt) where e is permittivity and x is the width over which the step rises (definition: x = ct, where t is the rise-time). We see that if x = 0, then i = infinity. This disproves the idea of a truly abrupt step. Moreover, the current rises over the rise-time from 0 to its peak, and since radio emission occurs in proportion to di/dt, it becomes more intense as the step rise-time is made smaller.
Now here is the proof. Taking the 2 parellel aerials or transmission line conductors. Feed one with any signal, and feed the other with the inversion of that signal. While the signal strength rises, electrons accelerate and radio emission occurs in a perpendicular direction.
I've done this experiment and proved it experimentally. During the rise-time, each conductor transmits a radio signal that is the exact opposite of that emitted from the other conductor. At a long distance (several times the distance of the gap between the two conductors) there is no observable radio transmission at all, because each radio emission cancels out that of the other: perfect interference. (The same concept is often used as white noise to suppress sounds, but that is less effective.)
The point is that the entire radio energy emitted by each conductor during the step is transmitted to, and received by, the other conductor. This is the process by which the TEM wave is allowed to propagate. Catt, ironically, gives the conventional textbook slab of drivel on this point! See [1] (that web version misses out the formulae, but they are widely known) where Catt calculates the inductance of a single wire and finds: "The self inductance of a long straight conductor is infinite. This is a recurrence of Kirchhoff's First Law, that electric current cannot be sent from A to B. It can only be sent from A to B and back to A." I think it is unhelpful for Catt, having defined E and B in fixed ratio for a TEM wave (E=cB), then goes along with the unfruitful textbook treatment of inductance which considers inductance as a B field effect! The magnetic field loops around each conductor instead of going from one conductor to the other line "displacement current" or in fact radio energy. This is probably where the conventional theory went wrong! It is clear that the entire energy needed to propagate the TEM wave is transmitted as radio from one conductor to the other during the step. No loss occurs because the step in each is inverted with respect to the other in a TEM wave. I'm going to do the calculations to demonstrate how this solves the Catt anomaly. 172.209.113.91 20:52, 20 January 2006 (UTC)
Light current, radio is a fact. The Poynting vector is wrong in the way it is usually taken to suggest that nothing moves except in the propagation direction. The electric field vector, at right angles to the propagation (c labelled) vector also involves energy flow, hence the transverse nature of the wave. Catt interprets Heaviside and Poynting as a longitudinal wave, simply because they don't say anything about the transverse action.
The Poynting vector falsely omits transverse motion of energy: see the illustrated discussion on the blog here: [2]. 172.214.92.156 08:35, 21 January 2006 (UTC)
I agree with you here Kevin.- the electromagnetic disturbance ceases (to an external observer) when the cable is fully charged to the on load source voltage! But why should em waves suddenly stop- what is the mechanism of stopping them?- That is Catts question-- Light current 00:22, 21 January 2006 (UTC)
Im afraid I dont agree that charge carriers can travel at the speed of light- but of course an EM wave can!. The only question is, what is an em wave made of ?. -- Light current 00:22, 21 January 2006 (UTC)
My theory is that there is some sort of induced effect in two wires of a twin conductor TL but this is entirely due to the EM energy flowing between them down the dielectric! Poyntings vector again. Do you believe in it?-- Light current 00:22, 21 January 2006 (UTC)
My own view is that what's propagating at light speed is the electric and magnetic fields due to the applied voltage, with the electrons following on afterwards as fast as they can. However, I dont call this displacement current, I call it energy current or em energy. You see, there must be some coming together of em field theory at high freq and circuit theory at low freq/dc. This can only be achieved by assuming that whenever physical charges cannot move fast enough, the em wave takes over and makes up the difference to present the ultra fast rise times that weve all? witnessed. BTW can you go back to sequential posting rather than interleaved, as its very confusing when the posts get out of chron order (esp with 3 respondent).-- Light current 03:48, 21 January 2006 (UTC)
Which to you seems the most likely?:
If you say b), what is the proposed mechanism for the stopping of these travelling waves? -- Light current 19:58, 21 January 2006 (UTC)
I agree that "circuit theory" is merely a useful approximation which simplifies calculation at low frequencies and that the Telegrapher's equations which describe a TL at high frequencies are not "em field theory" but are derived from Kirchoff's Laws. However here we must depart from theory to look at the experimental evidence. Anyone who has worked with square pulses on TLs will tell you that the description of the TL in terms of impedance, characteristic velocity and "reflections" at impedance mismatches are in fact correct.(as far as we can see). THe answer to your apparent paradox here is that you do not allow yourself the "limit as x tends to 0" trick where all components become distributed. When all component values tend to zero in the TL equations, the HF cutoff disappears! There is no inconsistency I feel in the way Catt has interpresed the experimental truths and the theories.
"Energy current" is helpful in thinking about these problems in that one no longer has to think about current and voltages on the wires (which probably do not exist at high frequencies anyway). Energy current will always give the correct answer. How do you think time domain reflectometry works if you dont believe in the Heaviside step of energy current?
Catt's "energy current" is actually the vector cross product of the electric and magnetic field (ie the Poynting vector that no one disputes)-- Light current 19:12, 21 January 2006 (UTC)
OK Kevin. But you see what I mean about signing - otherwise we don't know who the hell were talking to do we?-- Light current 23:24, 21 January 2006 (UTC)
If the HF cutoff is real, at what frequency does it occur in real coax cables for example? -- Light current 22:26, 21 January 2006 (UTC)
No you are wrong here. The only cutoff freq is when the cable goes into waveguide mode. This occurs at 20GHz for a 5mm dia cable with polyethylene dielectric. (ref Some questions and answers on fundamentals of coaxial cables - Tektronix UK Ltd c 1977) Losses are different and Im not talking about lossy cables. -- Light current 23:58, 21 January 2006 (UTC) The explanation is the at the Ls counteract the Cs and the Cs counteract the Ls giving a purely resistive impedance that is not frequency dependent. (assuming L/C=R/G -- the so called lossless- or Heaviside condition) See Transmission line-- Light current 00:41, 22 January 2006 (UTC)
No, sorry. try another argument-- Light current 00:59, 22 January 2006 (UTC)
This page 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. |
I tend to agree that the D s should not exist-- at all! (not even at the step). That's because I believe that nothing flows from one conductor to the other in a capacitor or TL. Energy flows along the line -that is all. This question of infinitesemal rise time is not relevant to the problem. How does your view differ from this?-- Light current 23:21, 10 January 2006 (UTC)
What do you mean by the term displacement current?-- Light current 02:15, 11 January 2006 (UTC)
I tend to agree with all of your last paragraph Kevin-- Light current 02:15, 11 January 2006 (UTC)
Kevin: before the TEM step arrives at any part of the cable, there is zero voltage. As it arrives, the voltage increases and with it the current. Once it has peaked, it remains steady. The only period of interest is while the voltage is rising. Ivor ignores this, and also introduces other errors like the claim that displacement current flowing where the voltage is steady. Your sine wave radio signal versus square wave logic step argument is vacuous, since radio emission occurs so long as di/dt is not zero in any part of the aerial. Fourier analysis is useful to me for the purpose of translating a plot of current or field strength versus time into a frequency spectrum. According to Fourier analysis, you can represent a square wave as a series of sine waves! But that's not my main argument.
Consider the 2 conductor transmission lines as 2 parallel radio aerials. If you feed one with a signal (of any type) and leave the other passive, the first transmits energy to the passive one which receives energy only as a result of di/dt in the first one. This is indistinguishable from Maxwell's "displacement current" equation. Maxwell says vacuum "displacement current" i = e.dE/dt = e.dv/(x.dt) where e is permittivity and x is the width over which the step rises (definition: x = ct, where t is the rise-time). We see that if x = 0, then i = infinity. This disproves the idea of a truly abrupt step. Moreover, the current rises over the rise-time from 0 to its peak, and since radio emission occurs in proportion to di/dt, it becomes more intense as the step rise-time is made smaller.
Now here is the proof. Taking the 2 parellel aerials or transmission line conductors. Feed one with any signal, and feed the other with the inversion of that signal. While the signal strength rises, electrons accelerate and radio emission occurs in a perpendicular direction.
I've done this experiment and proved it experimentally. During the rise-time, each conductor transmits a radio signal that is the exact opposite of that emitted from the other conductor. At a long distance (several times the distance of the gap between the two conductors) there is no observable radio transmission at all, because each radio emission cancels out that of the other: perfect interference. (The same concept is often used as white noise to suppress sounds, but that is less effective.)
The point is that the entire radio energy emitted by each conductor during the step is transmitted to, and received by, the other conductor. This is the process by which the TEM wave is allowed to propagate. Catt, ironically, gives the conventional textbook slab of drivel on this point! See [1] (that web version misses out the formulae, but they are widely known) where Catt calculates the inductance of a single wire and finds: "The self inductance of a long straight conductor is infinite. This is a recurrence of Kirchhoff's First Law, that electric current cannot be sent from A to B. It can only be sent from A to B and back to A." I think it is unhelpful for Catt, having defined E and B in fixed ratio for a TEM wave (E=cB), then goes along with the unfruitful textbook treatment of inductance which considers inductance as a B field effect! The magnetic field loops around each conductor instead of going from one conductor to the other line "displacement current" or in fact radio energy. This is probably where the conventional theory went wrong! It is clear that the entire energy needed to propagate the TEM wave is transmitted as radio from one conductor to the other during the step. No loss occurs because the step in each is inverted with respect to the other in a TEM wave. I'm going to do the calculations to demonstrate how this solves the Catt anomaly. 172.209.113.91 20:52, 20 January 2006 (UTC)
Light current, radio is a fact. The Poynting vector is wrong in the way it is usually taken to suggest that nothing moves except in the propagation direction. The electric field vector, at right angles to the propagation (c labelled) vector also involves energy flow, hence the transverse nature of the wave. Catt interprets Heaviside and Poynting as a longitudinal wave, simply because they don't say anything about the transverse action.
The Poynting vector falsely omits transverse motion of energy: see the illustrated discussion on the blog here: [2]. 172.214.92.156 08:35, 21 January 2006 (UTC)
I agree with you here Kevin.- the electromagnetic disturbance ceases (to an external observer) when the cable is fully charged to the on load source voltage! But why should em waves suddenly stop- what is the mechanism of stopping them?- That is Catts question-- Light current 00:22, 21 January 2006 (UTC)
Im afraid I dont agree that charge carriers can travel at the speed of light- but of course an EM wave can!. The only question is, what is an em wave made of ?. -- Light current 00:22, 21 January 2006 (UTC)
My theory is that there is some sort of induced effect in two wires of a twin conductor TL but this is entirely due to the EM energy flowing between them down the dielectric! Poyntings vector again. Do you believe in it?-- Light current 00:22, 21 January 2006 (UTC)
My own view is that what's propagating at light speed is the electric and magnetic fields due to the applied voltage, with the electrons following on afterwards as fast as they can. However, I dont call this displacement current, I call it energy current or em energy. You see, there must be some coming together of em field theory at high freq and circuit theory at low freq/dc. This can only be achieved by assuming that whenever physical charges cannot move fast enough, the em wave takes over and makes up the difference to present the ultra fast rise times that weve all? witnessed. BTW can you go back to sequential posting rather than interleaved, as its very confusing when the posts get out of chron order (esp with 3 respondent).-- Light current 03:48, 21 January 2006 (UTC)
Which to you seems the most likely?:
If you say b), what is the proposed mechanism for the stopping of these travelling waves? -- Light current 19:58, 21 January 2006 (UTC)
I agree that "circuit theory" is merely a useful approximation which simplifies calculation at low frequencies and that the Telegrapher's equations which describe a TL at high frequencies are not "em field theory" but are derived from Kirchoff's Laws. However here we must depart from theory to look at the experimental evidence. Anyone who has worked with square pulses on TLs will tell you that the description of the TL in terms of impedance, characteristic velocity and "reflections" at impedance mismatches are in fact correct.(as far as we can see). THe answer to your apparent paradox here is that you do not allow yourself the "limit as x tends to 0" trick where all components become distributed. When all component values tend to zero in the TL equations, the HF cutoff disappears! There is no inconsistency I feel in the way Catt has interpresed the experimental truths and the theories.
"Energy current" is helpful in thinking about these problems in that one no longer has to think about current and voltages on the wires (which probably do not exist at high frequencies anyway). Energy current will always give the correct answer. How do you think time domain reflectometry works if you dont believe in the Heaviside step of energy current?
Catt's "energy current" is actually the vector cross product of the electric and magnetic field (ie the Poynting vector that no one disputes)-- Light current 19:12, 21 January 2006 (UTC)
OK Kevin. But you see what I mean about signing - otherwise we don't know who the hell were talking to do we?-- Light current 23:24, 21 January 2006 (UTC)
If the HF cutoff is real, at what frequency does it occur in real coax cables for example? -- Light current 22:26, 21 January 2006 (UTC)
No you are wrong here. The only cutoff freq is when the cable goes into waveguide mode. This occurs at 20GHz for a 5mm dia cable with polyethylene dielectric. (ref Some questions and answers on fundamentals of coaxial cables - Tektronix UK Ltd c 1977) Losses are different and Im not talking about lossy cables. -- Light current 23:58, 21 January 2006 (UTC) The explanation is the at the Ls counteract the Cs and the Cs counteract the Ls giving a purely resistive impedance that is not frequency dependent. (assuming L/C=R/G -- the so called lossless- or Heaviside condition) See Transmission line-- Light current 00:41, 22 January 2006 (UTC)
No, sorry. try another argument-- Light current 00:59, 22 January 2006 (UTC)