Transient Emission Model

[idarney]

The paper ‘Modelling the transient emission from a twin-conductor cable’ describes a test where a step pulse is applied to one end of an open-circuit cable. -  http://ietdl.org/t/ZKt6U 
The waveform of the differential-mode current delivered to the line was recorded. This was different from the waveform of a lossless line, since there were radiation losses. A circuit model was created to simulate the hardware-under-review and was subjected to transient analysis. Adjusting component values of the model enabled close correlation to be achieved between test results and model simulation. The process of creating the model provides an improved understanding of the mechanisms involved.

The ‘References’ section of the paper includes a link to the Mathcad worksheet which performs the calculations.

Comments will be welcome.

[idarney]

Electromagnetic field can be visualised as a flow of charged, sub-atomic, entities. These entities are not electrons. When held in the atomic structure of the conducting material, they are spinning rapidly. When released from this state, they depart at a velocity similar to the speed of light, rather like a slingshot. The dynamic energy Vd involved in this change of state is associated with the relationship: Vd = L*(dI/dt), where L is an inductance and the current I is the rate of change of charge:

That is,  I = dQ/dt . Hence the dynamic energy of the released charge, Q, is

Vd = L*(d²Q/dt²).

The charge Q propagates at near-light velocity through the insulation surrounding the conducting wire. When it is captured by atoms in the material of an adjacent conductor, the energy level in that conductor increases. The static energy Vs acquired by this second conductor is  

Vs = Q/C, where C is a capacitance.

This voltage causes current to flow in the second conductor.

Effectively, there is current flow from one conductor to another, through the insulating material which separates those conductors. This should be no surprise. Every student of Electrical Engineering is taught that there is current flow between the two plates of a capacitor.

In an electrical system, not all of the charged, sub-atomic entities departing from the send conductor arrive at the designated return conductor. Those that arrive at another conductor are unwanted, and manifest themselves as electromagnetic interference.

Ian Darney

[-MFra-]

How to turn a two-wire line segment in a dipole....

With this ideal experiment, illustrated in the picture, I want to prove that the antenna radiates only an electromagnetic field and there is no expulsion of electric charges. The Hallén integral equation has as a solution, the electric current distribution on each antenna branch when it is powered by a sinusoidal source.

Experiment description: the electric line is constituted by two conductors separated by a dielectric that, although insulating, shows in fact a very high electrical resistance but finite. Whereby, when there is a voltage, constant or variable over time, between the two conductors flows a infinitesimal leakage current, the smaller, the smaller the conductance g existing between the two conductors. As the line ends move apart, rotating, finally each of 90 degrees, the conductance between the two ends tends to zero and so tends to zero the infinitesimal current that existed between the two conductors prior to the experiment. So, finally, the two antenna branches radiate an electromagnetic field but not an electric current that is all confined and trapped near the surface of each conductor, as a result to the skin effect

.

[AlvaroDíaz]

Ian Darney wrote:

Electromagnetic field can be visualised as a flow of charged, sub-atomic, entities. These entities are not electrons.

Nop.

Why? Because charged particles in movement generates the field, and loose some energy doing that. Recursively, you have not energy after few femtoseconds. This point of view can't preserve the system energy. Actually, that's the reason because Bhör atom's model is wrong.

EM field is generated by charged particles, but it's not composed by them. Uncharged particles are the EM field: photons.

Best regards.

Alvaro.

[idarney]

The posting of 25 June ‘How to turn a two-wire line segment in a dipole....’  describes a ‘thought experiment’ where a voltage is applied to a twin-conductor transmission line, the two conductors are rotated 90 degrees, and the distribution of the current flow in the conductance of the surrounding medium is analysed.
 
Of course the flow of current is miniscule. The ‘experiment’ defines steady state conditions. It ignores the existence of ‘displacement current’.

In a dipole which is operating at half-wave frequency, current and voltage change along the length of each conductor. The current distribultion along each monopole is sinusoidal, with maximum current at the termination with the feeder cable and zero current at the tip of the monopole. Reference:- [ Paul C. R. ‘Antennas’ in  ‘Introduction to Electromagnetic Compatibility’ First Edition, (John Wiley & Sons Inc. 1992). pp 181-192.] Alternatively,  the chapter on ‘Antennas’ in any book on Electromagnetic Theory will define the concept of ‘Radiation Resistance’ and derive the relevant formulae.  

Pages 189 to 197 of the book ‘Circuit Modeling for Electromagnetic Compatibility’ describe an experiment where a sinusoidal signal of constant voltage and varying frequency is injected into the centre section of a length of isolated cable and the current in one monopole is measured using a current transformer. A circuit model of the setup is created, and analysed using a Mathcad worksheet. Good correlation is achieved between test results and model simulation. Using such a setup, it is possible to measure the value of the radiation resistor.

The set of Mathcad worksheets used in the book http://digital-library.theiet.org/content/books/ew/sbew502e is available for download from   www.designemc.info/WorksheetsMCD.zip  This download includes the worksheet ‘worksheet 7.4.xmcd’, which records both the test results and the simulation provided by the model.

[-MFra-]

Hi Ian Darney

Now it seems to me that we agree. In the previously reported diagram, ( sketch), (argument known for nearly a century) I have considered an infinitesimal element of line, length dz, and of it, I drew a lumped equivalent circuit. I have taken into account of the transversal currents (considering all the variables involved, defined in the domain of the Laplace transform), the one that flows in the conductance g, and the one that flows in the admittance sc (due to  the displacement current),  that are  both very small respect to the longitudinal one. Furthermore the same  transversal currents flowing among the branches  of the dipole (as drawn in the sketch) (gap excluded) can be considered practically inexistent.

All the interested in the subject  electromagnetism in general, and in particular the antennas,  they possess one or more text, including myself.

So, I am opposed to the saying of the article (radiated currents), from you mentioned at the beginning of the whole discussion and for this reason I did not read the rest of that article.

It seems only a waste of time arguing about things long known.

Cheers

FM.

[idarney]

Response to Alvaro Diaz

The Introduction to the paper ‘Modelling the Transient Emission from a Twin Conductor Cable’ includes the words:- 

‘This led to the conclusion that, as the step pulse progressed, half the charge was transferred to the return conductor, creating a current flow back to the near end. In turn, this transient flow delivered charge back to the send conductor.

There was a continuous flow of charges into and out of, the surface of each conductor. If the current flow in one direction is equal to a current flow in the opposite direction, the nett current flow is zero, but the voltage between the conductors is doubled. This continuous interchange of charges between the conductors ensures that the nett flow of energy along the cable follows the path of the cable.’

This paper was submitted to the Editor of the Journal of Engineering at the Institution of Engineering and Technology and received the response ‘Thank you for submitting your paper to The Journal of Engineering. The referral process is now complete. I am pleased to say that the referees found your paper interesting and worthy of immediate publication. My decision is therefore to accept your paper as it stands.’ Such a decision would not have been made if there had been any concern about the scientific validity of any of the statements made in the paper.

Ian Darney

[RichardJ]

Then the reviewers didn't do their job. It's unfortunate, but sometimes the reviewers only skim read, or don't read at all, the papers they agree to review. It's very irresponsible, but it happens. A big red flag for this is when the reviewers come back with no comments or suggestions at all, because when a reviewer does do their job it's unlikely they can not find any mistake, or think of anything that might be better. As I said before, your statements are not just incorrect at some detailed level, they violate the basic laws of physics.

[idarney]

Hi Richard 

In response to the comment:  ‘Then the reviewers didn't do their job.’

The Journal of Engineering claims to be a ‘pure gold open access journal’. Please see
http://digital-library.theiet.org/journals/joe/about .

The task of submitting a paper for review is definitely non-trivial. Please see
http://digital-library.theiet.org/journals/joe/author-guide

I can testify to the fact that the editorial team carry out their task rigorously. My first submission was rejected, but the reviewers were kind enough to identify the errors in my reasoning which led to their decision. The next submission gave rise to several critical comments. I was able to modify the text in the light of those comments and re-submit the paper. This revision was accepted for publication.

Best regards,

Ian.

[RichardJ]

The fact is, your paper makes factually, and grossly, incorrect statements. Photons do not carry charge. Any theory based on this is based on an incorrect premise, and is therefore wrong. It's not hard to check this. Just Google "do photons have charge".

[nas0k]

Richard, interesting results from the search; an excerpt from one:

Also, the system cannot be static, so is it feasible (relative to non-static system)  not to have one (photons) without the other (charge), or vise-versa?

[idarney]

Thank you for the advice to   Google "do photons have charge".

The reasoning in my paper was based on the fact that the response of the model correlated closely with that of the test setup. The model itself was then used to deduce the mechanics of signal propagation. Had I been aware of the information provided by the Q & A session at https://van.physics.illinois.edu/qa/listing.php?id=2348  , I would have provided a different description of the mechanism.

I note that the Q & A session contains the statement ‘ Photons, real and virtual, are emitted and absorbed by charged particles, even though they are not charged themselves. They only interact with charged particles, and not with each other.’ Also, Norm Schutzkus asks: ‘Also, the system cannot be static, so is it feasible (relative to non-static system) not to have one (photons) without the other (charge), or vise-versa?’ 

My reasoning now is:

If there is a surplus of electrons on the surface of a conductor, then that surface is negatively charged. If there is a deficit of electrons, then the effect of the protons is to create a positive charge on that surface.

If a free electron moves to another location in the atomic structure of the conducting material, it releases a photon. Photons move in a straight line at the velocity of light. If a photon impacts on another free electron, it causes that electron to move. The movement of electrons can be likened to the balls in a Newton’s Cradle. If the movement is axial, then charges propagate along the conductor. If the movement is radial, then charges propagate to the surface. This results in a preponderance of electrons on the surface. The surface becomes negatively charged.

If a photon does not impact on an electron in the structure of the conductor, it propagates out into the environment. If it impacts on an electron on the surface of another conductor, it causes that electron to move into the material of the receiving conductor. This causes a positive charge to appear on the surface of that conductor.

Effectively, a charge is propagated from one conductor to another.

Ian Darney

[RichardJ]

No, you are closer but what you are saying is still not correct. However, I have lost interest in discussing it. As I already said, this is physics 101, and I don't have much inclination to give an impromptu online class in the subject in a thread on the Mathcad forums.

One final tip: electromagnetic radiation can be thought of as either a particle or a wave (or field). Either description is valid, and if you can prove a theory by treating it as a wave, then you can also prove it by treating it as a particle (and vice versa). That does not mean that the two proofs are equally easy to derive though, or equally intuitive to understand. If you want to understand the mechanism whereby current in one conductor induces current in another conductor, it's much easier to think of the electromagnetic radiation as a wave, or if you prefer a field. If you really want to treat it as a particle then you had better read up on quantum mechanics.

[LucMeekes]

As Richard wrote:

"if you can prove a theory by treating it (EM radiation) as a wave, then you can also prove it by treating it as a particle (and vice versa)".

I would add: This does by no means imply that EM radiation consists of particles, or of waves.

Keep in mind that the particle or the wave approach to deal with EM are models, no more, no less.

They tell you nothing about the physical reality.

Now

if all you have done in your paper is to demonstrate that the outflow of EM energy from an antenna, or a set of wires, can be modelled using a bunch of resistors with current flowing through them, to yield a simple approach of the subject, you may have a point.

But then present it as,

and keep reminding yourself and the reader that it is,

a model, no more, no less.

Success!
Luc

[idarney]

In response to:   But then present it as, and keep reminding yourself and the reader that it is, a model, no more, no less:-

The widespread belief in the concepts of the ‘single-point ground’ and the ‘equipotential ground, together with the stricture to ‘avoid earth loops’, attest to the fact that many graduates leave university with the belief that the circuit diagram represents physical reality. The idea that the circuit network is ‘just a model’ is not even mentioned in books on Circuit Theory, let alone emphasised. (At least, in the books I’ve read.)

In response to: Keep in mind that the particle or the wave approach to deal with EM are models, no more, no less. They tell you nothing about the physical reality.

Engineers depend on the use of modelling techniques during the design process, whatever the hardware under review. Circuit models are used to good effect to predict the functional performance of complex electronic systems.

In response to:- If all you have done in your paper is to demonstrate that the outflow of EM energy from an antenna, or a set of wires, can be modelled using a bunch of resistors with current flowing through them, to yield a simple approach of the subject, you may have a point.

The use of circuit models can be extended to ensure that the equipment under development will meet its EMC requirements and that it will not present any electromagnetic or electrostatic hazards. But first, it is necessary to update Circuit Theory to ‘include the concepts of partial currents, partial voltages, time delays, distributed parameters, and virtual components’. Please see the conclusion to the paper http://ietdl.org/t/ZKt6U  .

[AlvaroDíaz]

Hi Ian.

Congratulations for the approbation. But still some conjectures in the paper, that are not proven, and, eventually, refused by classical theory. In you introduction stats: "The starting point of this paper is a brief summary of the analysis of the half-wave dipole. When this antenna is driven at its resonant frequency, it delivers current into the environment. That is, it emits a stream of charged particles from every point on the conducting surface. These charges radiate away at near-light velocity."

What I can calculate is that there are two things that contradicts this affirmations. One: charges radiated at relativistic speeds emits a very high energy, in the form of X or gamma rays. The other is that there are not current from an antenna to the environment. As you unground your system, charges emission must to charge the wires. With enough time and current, all the free electrons must to be emitted, and the metal in the conductor go to be ionized. As I know, that don't happen.

Your model is ok, what isn't is the concept that " ... The first observation was that the first trailing edge seemed to follow an exponential decay. This was because the current in the send conductor was radiating away into the environment. " What is radiate is an EM field, by the electrons moving very slow (at cm/sec). And the EM field is composed by photons, without any charge, so there are not any current to the environment. This is the true explanation about why there are the exponential decay, with some thermal dissipation by Joule-Thomsom effect too, but, again, in the form of photons, in the range of infra-red waves.

Other impressions came from some assertions like: "In textbooks on electromagnetic theory, the final section on transmission lines usually deals with the transient behaviour when a step pulse is applied. Currents and voltages propagate in both directions along each conductor, and are reflected at both terminations." Here, some considerations. There is not only usual apply a step, there are some mathematical background to do that. But this is ok. What's it is not ok, is the concept about that the voltage can travel along a conductor. The voltage is punctual property: each space point have some potential. Is a scalar property too. Is the equivalent of U gravitational potential or T, the thermodynamic temperature. What is propagated isn't V, U or T, is the EM fiel, the gravitational field, and the heat.

About the avoiding the concept of grounding electrical system in the process of ensuring the EM compatibility, in the appendix of your referenced book, we can read that the 'hybrid equations' are taken from "a set of lecture notes copied from a blackboard at Glasgow University in 1959." as can read in http://digital-library.theiet.org/docserver/fulltext/books/ew/sbew502e/SBEW502E_appendixc.pdf?expires=1467034754&id=id&a… .

But in the develop of those set of equations, the author use a ground system, just between equations (C.10) and (C11) (curiously, don't numerate it), when establish the boundary conditions, with V=Vr for x=l. That's what a ground is: an arbitrary voltage for some point.

As you stay at the beginning, "Comments will be welcome.", those are mine.

Best regards.

Alvaro.

[idarney]

Hi Alvaro.

Many thanks for the link to the analysis of electron velocity.

I agree that: ‘charges radiated at relativistic speeds emits a very high energy..’ where those charges are electrons. It is not suggested that electrons are emitted. The paper goes on to state: ‘Such a velocity precludes the concept that this flow of current is carried by electrons or protons. So it is reasonable to conclude that sub-atomic particles are involved. The name given by the scientific community to such entities is ‘photons’. A photon is an elementary particle, the quantum of light and all other forms of electromagnetic radiation.’

Photons carry energy. As far as electrical circuitry is concerned, energy is defined by the voltage, and voltage is directly related to charge : Q = C*V. Photons also have energy due to the rate of change of velocity of that charge along the conductor, and that energy is measured by the voltage developed across the inductance. E-field is defined in terms of Volts per metre and H-field in terms of Amperes per metre

In response to the statement  ‘there are not current from an antena to the enviroment.’ :- The picture below is an extract from a book by C. R. Paul. It shows that power is delivered to the environment. This flow of power can be represented by a voltage source delivering current into a ‘Radiation Resistance’.

In response to  ‘What's it is not ok, is the concept about that the voltage can travel along a conductor.’

If a step voltage of V Volt is delivered to the input terminals of a zero-loss transmission line, a step current I is caused to flow in the line, where I = V/Ro and Ro is the characteristic resistance. The voltage and current steps propagate in synchronism along the line at near-light velocity. When the step arrives at the output terminals, a voltage V in series with a resistance Ro is applied to the load. Please see www.designemc.info/31TravellingWave.pdf

In response to ‘But in the develop of those set of equations, the author use a ground system, just between equations (C.10) and (C11) (couriosly, don't numerate it), when stablish the boundary conditions, with V=Vr for x=l. That's what a ground is: an arbitrary voltage for some point.’

The model used to formulate the hybrid equations assumes that current flows in a loop, along the send conductor and back via the return conductor. Although voltage is developed along the inductance of the return conductor, this does not affect the relationship between currents and voltages at the near and far ends of the line.

Best regards

Ian

[AlvaroDíaz]

Hi Ian.

If I run against a wall, my kinetic energy was transferred to the wall. But that's don't mean that the wall was to move. There are some complicated process involved, and a bed hospital waiting for me.  If some I = dQ/dt disappear, that's mean only that some Q goes static, like some Alvaro's mass in the hospital bed. As I don't need to make a mass transfer to the wall for transfer my kinetic energy, electrons don't need to propagate to the space to make I to decrease. What shows the surface integral from the book is the calculus about how many energy (power, in that case [1]). In my crash against the wall, I can see me as a fictitious hot radiant source with Q = 1/2*m*v^2. Fortunately, my v is very small, but the problem could be my m. And god thanks, 1/2 (and not an other big constant) came from the Einstein develop in series therms, taking only the first derivative as the approximation.

So, again: notice the word "fictitious" resistance: this R isn't real, just because there are not charged particles going out from the conductor.

About the unsigned article Traveling Waves, there are some lost precisions. First, that Impedance, phasors, reactances are well defined only for sinusoidal signals, which can be expressed as phasors. That what's mean Vred = V*exp(j*w*t), and the similar eq for I. Those are eqs (1) and (2) in the paper. There V, I isn't the instantaneous value of V nor I, as in the paper is says !. Are the peak values.

Near the end, it is stated that " Since this relationship is true for all frequencies, it becomes possible to apply it to the Heaviside Step Function, which is the sum of frequency components ranging from zero to infinity. Figure 3 illustrates this. The propagation velocity is defined by equation (25) and the characteristic resistance is defined by equation (26)."

I say a lot of times here that there are some "math background" about linear systems, linear diff equations, and linear algebra. First, you confuse Heaviside Step Function (one step: OFF-ON) with a rectangular function (OFF-ON-OFF: see Rectangular function - Wikipedia, the free encyclopedia). The math background is related to the analysis of infinite dimension space vectors, functions as vectors, the use sin, cos as an infinite base space, and things like this. As I say before, those systems can be determined knowing the answer to a pulse and a step. So, maybe the conclusion that can apply the results for a set of finite sinusoidal waves to an infinite dense ones could be correct, don't shows how could be done this transfer from finite vectorial spaces to infinite ones. Just saying that there are some theorems anywhere could be enough, but not calling that as sums of freqs from zero to infinity. Or with complicated phrases like: "For a linear time-invariant black box, the step response can be obtained by convolution of the Heaviside step function control and the impulse response h(t) of the system". Or better: Oops, there are some complicated math background involved, just take this conclusion as true.

There are some other impressions in the paper, like "Since voltage is due to the energy difference between different charges" and things related.

And the worst was pointed by Richard: photons are uncharged, so, can't carry current.

Best regards.

Alvaro.

Note: [1] I'm always remark the difference between power and energy calculations just because in a lot of situations power and energy words are interchanged without any regarding. Fortunately, mathcad and the use of units prevent a lot of damage. For example, in aspen one, or aspen hysys "energy" streams aren't, are power streams. Maybe this seems like a minor problem, but, believe me, it is not. Can carry big mistakes confusing thinks like this in excel.

[idarney]

Hi Alvaro

In response to the statement:  ‘If I run against a wall, my kinetic energy was transferred to the wall. But that's don't mean that the wall was to move. There are some complicated process involved, and a bed hospital waiting for me.  If some I = dQ/dt disappear, that's mean only that some Q goes static, like some Alvaro's mass in the hospital bed

As reasoned in my response to Richard Jackson, photons provide the means whereby charge is transported axially along a conductor, radially out of the surface of that conductor, through the insulation, and radially onto the surface of an adjacent conductor. They do not need to carry charge themselves. 

Photons do not have mass. So there is no need to invoke the expression 1/2*m*v^2  anywhere in any equation involving that entity or in any equation involving charge propagation. 

So the prognosis is that your alter-ego will make a full recovery.

In response to the comment:  ‘in a lot of situations power and energy words are interchanged without any regarding.’ 

There is a clear correlation between the equations defining the response of an electrical circuit and the equations of motion of a mechanical system, even though different units are involved. Please see Figure 3 of the article  www.designemc.info/11Concepts.pdf . It is also possible to establish correlation between the equations of Electromagnetic Theory and those derived from a representative circuit model.

In response to the comment:  About the unsigned article Traveling Waves, there are some lost precisions:

Thank you very much for reviewing my article. I will revise the text in the light of your comments, at the first opportunity.

Ian

[RichardJ]

As reasoned in my response to Richard Jackson, photons provide the means whereby charge is transported axially along a conductor, radially out of the surface of that conductor, through the insulation, and radially onto the surface of an adjacent conductor. They do not need to carry charge themselves. 

Charge is not transferred through an insulator. That is the whole point of, indeed the definition of, an insulator.

Photons do not have mass.

Incorrect. Photons do not have a rest mass, but that is something of a moot point because they are never at rest. The mass of a photon is E/c^2, where E is the energy of the photon and c is the velocity of light in a vacuum.

[AlvaroDíaz]

Ian Darney wrote:

...

Photons do not have mass. So there is no need to invoke the expression 1/2*m*v^2  anywhere in any equation involving that entity or in any equation involving charge propagation. 

So the prognosis is that your alter-ego will make a full recovery.

... or in any equation involving charge propagation. (!!!)

Absolutely false. Charge propagation always have a moving mass involved, with their correspondent kinetic energy associated. That's include holes, because must to have paired electrons.

Alvaro.

[AlvaroDíaz]

About electron speed: They are very slow.

Best regards.

Alvaro.

[RichardJ]

Your point about relativistic electrons moving in a curve is a very good one. It is possible to do that, in which case they emit large quantities of EM radiation tangentially to the path, with frequencies all the way up into the deep X-rays. Enough radiation to kill you almost instantly if you were exposed to it, even though the current is typically less than 1A. It's called a synchrotron. The electrons travel in an ultrahigh vacuum tube though, not a conductor.

[AlvaroDíaz]

With some nice applications: BBC NEWS | Science/Nature | Secret 'dino bugs' revealed . Other apps are important, but maybe some boring.

Notice that Copper in previous calculus is an ideal conductor, providing a path and electrons, but without any resistance. For first approach can take m.r = m.e/sqrt(1-(v/c)^2) as the relativistic mass of the electron.

For energy considerations, can use Ek = 1/2*m*v^2, then P = Ek/dt = 1/2*m.e*v^3*A*n for get the kinetic energy (by time, actually is a power) that can be obtained from the n electrons across a section A at velocity v. Actually, this eq is the same for a wind turbine.

Some of this kinetic energy (by time, power) can be transformed into radiation. I don't remember right now the equation of that conversion for relativistic speeds. The longitude of the wave, lambda, is inverse-proportional to the energy. For a big energy, a low wave, like X-rays, obtained with high speeds.

Best regards Richard.

Alvaro.