Phase diff between tx and rx antennas spaced by whole multiples of wavelength - Page 1
Forum section: electronics / rf-microwave

author: msat
author_rank: Regular Contributor
timestamp: September 07, 2021, 03:33:43 pm
content: I don't know if this is common knowledge or not, but I have been unable to find relevant information on this. Hopefully someone here knows. Here's the question:Assuming two identical antennas, with one transmitting [a sine wave] and one receiving, spaced apart by some whole number multiple of the wavelength, and both probed identically and fed to an oscilloscope, would the waves be in phase, or 180 degrees out (or something else?!)?I'm currently assuming they would be 180 degrees out of phase.Definitive answers would be appreciated. Otherwise, please provide your supporting argument.Thanks!-Mark

author: TimFox
author_rank: Super Contributor
timestamp: September 07, 2021, 03:59:56 pm
content: The time delay between two antennas spaced one wavelength apart equals the (time) period of the sine wave. Therefore, the two signals (carefully probed) will be 360 degrees out of phase (receive antenna lagging behind transmit antenna), which is equivalent to zero phase angle. This assumes antenna construction such that if the two antennas coincide in space (zero wavelengths apart), the signals are in phase.Similarly for other integer values of spacing/wavelength."Careful" probing requires equal transmission lines from the two antenna feedpoints to the two oscilloscope inputs to equalize the probe delays.

author: msat
author_rank: Regular Contributor
timestamp: September 07, 2021, 04:30:50 pm
content: @TimFoxI understand the effects of time delay between transmission and reception based on the distance between antennas, as well as the need to consider the way the antennas are probed to maintain equal transmission line delays. I also get that if we were to plot an EM sine wave as is commonly done, we would see that they are in phase diagrammatically. What I'm interested in is not what happens diagrammatically, but what happenselectricallyat the antenna.So I guess I need to clarify what I'm seeking to understand by adding some detail to my hypothetical experiment. Lets say the antennas being used are dipoles oriented vertically. In our phase diagram, the upper half of the sine wave corresponds to the upper half of the dipole having a positive charge relative to the lower half, and vice-versa for the lower half in the diagram.In other words, for a given point in a transmitted wave, are the induced charge distributions on a receiving antenna the opposite of the antenna which created it?

author: msat
author_rank: Regular Contributor
timestamp: September 07, 2021, 04:46:36 pm
content: I suppose I should give my line of reasoning which is what I'm basing my assumption off of.We know that like charges repel and opposites attract. If some point of an EM wave was generated by a downward pointing e-field (vectors pointing from positive charges towards negative), then when that point reaches the receiving antenna, the charge carriers (electrons) would be forced to move in the direction opposite of the vector.To my understanding, this at least holds true strictly for e-fields, but I don't know if it does for EM radiation.

author: radiolistener
author_rank: Super Contributor
timestamp: September 08, 2021, 05:02:22 am
content: Quote from: msat on September 07, 2021, 04:30:50 pmNo. If both dipoles have the same orientation (the same polarization) and placed at N*lambda distance (multiple of wavelength), then the charge density distribution will be the same for both at any given moment of time. And phase offset will be 0 degree (folded 360 degree).Phase offset 180 degree will be at N*lambda + lambda/2 distance.Note that there is phase delay between two antennas and resonant antenna bandwidth is limited due to high Q. So if you change sine amplitude on transmitting antenna it will affect second antenna after some period of time, which is include time delay due to antenna bandwidth limitations and time delay for wave traveling across space.Antenna cannot change amplitude immediately and needs at least Q cycles to radiate some carrier change. For half-wavelength dipole Q is about 10-12, so it needs to wait for about 10-12 sine cycles to radiate a new sine amplitude + time delay for wave traveling at specific distance.Quote from: msat on September 07, 2021, 04:46:36 pmNo. For example, let's name up side of vertically oriented half-wavelength dipole as U and down side as D.When you apply some electric potential to the antenna A, let's say + for U and - for D. The antenna B at N*lambda distance will sense this electric potential with reduced level due to distance.And since both antennas A and B have exactly the same orientation, antenna B will sense the same + on U and - on D side.In other words, when you charge some thing with positive charge, you will measure positive charge around that thing, NOT negative.

author: Rod
author_rank: Contributor
timestamp: September 08, 2021, 06:11:58 am
content: Quote from: msat on September 07, 2021, 04:46:36 pmTrue, and true. But don't confuse any changing quantity with theforcedriving it, it's 2nd derivative with respect to time.Imagine a pendulum, at one extreme of its swing.- Its displacement x is at a maximum.- Its velocity dx/dt is zero.- Its acceleration d2x/dt2is proportional to the force on it, and is at a maximum, but is in theoppositedirection to x.It isnotstatic. What now happens?An antenna is also a resonant system, just like a pendulum, only the charges are "swinging" along it. Imagine the transmitting antenna at the instant:- The charge (voltage) on one end is at a maximum.- The rate of change of the charge dQ/dt = I, the current, is zero.- The force of the electric field on the electrons is also at a maximum, and again is in theoppositedirection to the charges on the antenna.What's now happens?The force (electric field) experienced at the receiving antenna (n lambda away) is identical to that on the transmitting antenna, and its instantaneous charge distribution is identical, too: they are in-phase.There's also a magnetic field, proportional to the instantaneous current. Same story, it's just 90 degrees out of phase with the electric field, proportional to the voltage.Hope this helps?

author: msat
author_rank: Regular Contributor
timestamp: September 09, 2021, 04:29:01 pm
content: I want to thank everyone for taking the time to respond, and I truly mean that even though I have yet to be compelled. I'm not trying to come off as ungrateful or hard headed, but there's a reason I want to get this right and have as concrete proof as possible even if that leads me towards creating an experiment to do it.My question actually stems from what I consider to be a completely unsatisfactory classical description of "simple" linear polarizing filters. So while this isn't necessarily about RF communication systems per-se, it's relevant in that I think these systems provide the best means for testing any hypothesis on the matter. It also has implications for the interaction of EM radiation with conductors in general (and I suppose insulators as well, but I haven't given it any meaningful thought). I know this might already start to sound absurd, but hear me out.Full disclaimer: I have pretty limited math literacy, and have little hope of solving the "simplified" Maxwell Equations, nevermind Maxwell's actual equations. In light of that, what I'm positing might already be fully explained by them, but the explanations that I've read and videos I've watched on the topic don't make it clear.The premise is pretty simple: conductors interacting with external EM radiation (that is, radiation emitted from something else), re-radiate the energy (whatever hasn't been damped), but with the wave inverted. I think there's pretty good evidence for this, which I'll elaborate on shortly. At first glance, it would seem to require that the charge distributions on the receiving antenna at a given point in time be the opposite of those which created that point at the original radiator. That was the thing I hoped to illuminate in my OP. That said, I don't know if the charges would actually have to be the opposite to create an inverted field, but I can't help but to suspect it would.So far, I don't think I've said anything particularly controversial. We know that multiple antennas in relatively close proximity, especially when in parallel, can affect one another, even if they're both only receivers. I don't know much on the topic, but I assume re-radiation is a well-established explanation. We can further use this to explain the establishment of standing waves when EM radiation interacts with a conductive surface, with one of the nodes being at that surface. Like I said before, I'm incapable of solving Maxwell's Equations, but it's my understanding that they require that such an interaction require the establishment of a standing wave with a node terminating at the conducting surface. I was able to model this by simulating an emitted EM wave interacting with a surface, inverting, and re-emitting (reflecting). When we take the superposition of the incoming and inverted reflected wave, we get the standing wave as described.https://www.desmos.com/calculator/8spopwnkh2In the above link, when you press the play button next to the variableh, you'll see 3 sine waves. In this model, the vertical line at the origin represents the conductive surface. If you want to imagine the wave interacting with the left side of the surface, then the red wave represents the transmitted wave, and the green the reflected. It's the opposite if you see the transmitted wave coming from the right. Of course, in both instances, the blue wave is the established standing wave. The equations used are really just hacked together, so I wouldn't look into them describing any sort of physical reality, but to the best of my knowledge, the plotted results are accurate.Out of curiosity, I wanted to see what the plot would look like if the reflection wasn't inverted:https://www.desmos.com/calculator/doym28qua7The main takeaway is that the standing wave node is no longer at the surface, which again, to the best of my understanding, the Maxwell Equations require. So the reflection has to be inverted. Actually, I watched a video that stated as much, but it merely stated that's what happens when a pulse of EM radiation interacted with a conductive surface.Going back full circle, despite not yet having attempted to perform a mathematical analysis (didn't I already say I suck at math?) of this concept on the behavior of linear polarizing filters, I suspect this will sufficiently explain those behaviors, including the "three polarizer puzzle". All without having to resort to the insanity of QM probability distributions etc etc.

author: Rod
author_rank: Contributor
timestamp: September 09, 2021, 09:56:52 pm
content: Your illustration of standing wave is very nice.Quote from: msat on September 09, 2021, 04:29:01 pmCorrect... butonlyif the conductor has the same voltage (electric potential) at all points along it. That can occur only if currents are flowing in the wire which create an EM field which exactly cancel the external EM field. This is how a mirror works. This is why a ground plane acts as a reflector, a mirror. This is also why aninfinitely longwire (one far longer than the wavelength) doesnotact as a receiving antenna: it would reflect everything and receivenothing. It would create a standing wave, exactly as you illustrate. How does it do this?A key fact you've failed to consider: current doesnotflow at infinite speed in a conductor, but only at the speed of light. This is true no matter how long the conductor is.If you accept this reality, then youcanfigure out how antennas work. Antennas work as antennas only if they are 1/4 wavelength long (or an odd multiple thereof, as described herehttps://en.wikipedia.org/wiki/Dipole_antenna. This is what I attempted to suggest in the previous posting, but you didn't get: antennas resonate in-phase with the EM wave.)And if you accept this reality, you can also come to understand why an infinitely long wire willnotact as an antenna, but as a reflector. Each 1/4 wavelength along it, at each instant in time, the external EM fieldattemptsto induce anoppositevoltage, and anoppositecurrent on it. The is exactly what you state in the quote above. But as time continues, these currents and resulting voltages will continue propagate down the wire inbothdirections, and at the speed of light, exactlycancellingone another at every point along the wire at every instant in time.A quarter-wave antenna is different,only because no current can flow in it beyond it's ends.It is as simple as that!As a consequence, a voltage develops across the antenna (and, 90 degrees out of phase with it) a current flows along it. The ends of an antenna arecurrent nodes, and voltageanti-nodes. It isnotout of phase with the external EM field, it isin phasewith it.Let's stop here. If you understand how quarter-wave antennas work, then you will be able to predict how two joined together to create a half-wave long antenna will work... as a reflector, not as a receiving antenna... and why... current can flow through their joined ends. If this is not yet obvious to you, then you do not yet understand how antennas work. Please go back and try again to do so before proceeding further, as this misunderstanding will only continue to mislead you.Now, let's turn to linear polarizers. A simple quarter-wave dipole transmitting antenna emits a linearly polarized EM wave. A receiving antenna parallel to it receives it, and one perpendicular to it does not. Same for a second receiving antenna behind the first. No "three polarizer paradox" occurs. The receiving antennas do not alter the polarization of the transmitted EM wave.A passive receiving antenna does also emit an EM wave, in phase with the one it receives. The two EM waves do interfere, constructively on-axis, and destructively off-axis, together creating a directional antenna. The elements arenotacting as reflectors, but as in-phase re-radiators. But one mayappearto act as a reflector if it is spaced 1/2 wavelength behind the transmitting antenna. (At a receiving antenna, they mayappearout-of-phase depending only on the total difference in path length from the transmitting element to each passive element and then to the receiving antenna. That's why such an array is directional. But most directional antennas have multiple driven elements, with phase differences between them. Lots of examples herehttps://en.wikipedia.org/wiki/Category:Radio_frequency_antenna_types)The fact that all these work is proof your intuition"At first glance, it would seem to require that the charge distributions on the receiving antenna at a given point in time be the opposite of those which created that point at the original radiator."is incorrect. We observe exactly the opposite. (This fact is so fundamental it even has a name: Lorentz reciprocityhttps://en.wikipedia.org/wiki/Reciprocity_(electromagnetism), reflecting its consequences.)Also sorry you feel math is difficult. You're clearly smart enough to master and use it. Please don't give up on it. It would help you a whole lot. But this is not a math problem, but conceptual one.

author: TimFox
author_rank: Super Contributor
timestamp: September 10, 2021, 05:39:11 pm
content: One of my "conservative" engineer co-workers stated that he didn't "believe in quantum mechanics". I told him that he would have to give up using solid-state electronics, since quantum mechanics is essential to their operation, and go back to vacuum tubes, which canalmostbe described with classical physics.

author: Marco
author_rank: Super Contributor
timestamp: September 11, 2021, 09:56:22 am
content: If they overlap the displacement current will be clearly opposed simply from capacitive effect, I don't see any reason why it would suddenly change at multiple wavelengths. So yeah 360*x+180 phase delay, but still just 360*x group delay of course.

author: radiolistener
author_rank: Super Contributor
timestamp: September 12, 2021, 06:29:09 am
content: Quote from: msat on September 09, 2021, 04:29:01 pmNo, it doesn't have to be an inversion. There is phase delay, because wave propagation in any environment has limited speed also phase delay depends on material properties. Depending on exact phase delay value it can be in-phase, inverted, or have any other difference. It depends on exact phase delay value.Since EM wave propagation has a limited speed (for vacuum this is a speed of light) you will see some phase delay between two EM wave sensors placed at some distance in the space.You will get inversion between two sensors if you place them at half wavelength distance.But if you place them at wavelength (or multiple wavelength) distance they will be in-phase and you will get the same polarity on the sensors terminals.But note, if you use some cable to attach your sensors to the measurement device, the cable has it's own time delay. So you're needs to use two cables with exactly the same electrical length to have exactly the same time delay. Otherwise you will see phase difference due to phase delay in the cable.Quote from: msat on September 09, 2021, 04:29:01 pmre-radiation is not immediate process. It takes some time delay between EM wave consumption and radiation. This time delay exists between two events. And this time delay depends on matter type. This is why wave propagation in the matter is smaller than the speed of light in the vacuum. Each particle of matter store energy for some period of time and re-radiate it after some delay. It leads to lower wave propagation speed.Quote from: msat on September 09, 2021, 04:29:01 pmI don't think that there is needs to involve QM, photons and other complicated stuff to explain "three polarizer" effect, in my opinion it can be done within classical (non quantum limit) theory with pure wave electrodynamics.In my opinion the secret of "three polarizer" is that polarized film doesn't "filter photons", it consume and re-emits EM wave. And when you place 45 degree third polarized film between two other, it can consume half of EM wave passed through first polarized filmand re-emit a new EM wave which will be 45 degree for the third polarized film. So, the third polarized film will pass a part of that EM wave. So, there is no quantum effect, this is a classic wave behavior.And let's involve a little math. For example polarized film A consume light on the input and re-emits it with horizontal polarization. Now let's see what will happens with other polarized films:1) When you place polarized films B with 90 degree to A, the output on B will be:cos(90°) = 02) When you place polarized film B with 45 degree to A and polarized film C with 45 degree to B, the output of polarized film C will be:cos(45°) * cos(45°) = 0.7071 * 0.7071 = 0.5As you can see, it's pure waveform effect. And there is no needs to involve quantum effects to describe that.

author: Rod
author_rank: Contributor
timestamp: September 13, 2021, 12:22:22 pm
content: Quote from: radiolistener on September 12, 2021, 06:29:09 amReflection, whether off a mirror or off the shorted end of a transmission line, occursinstantaneously. There isnotime delay.If there were any fixed time delay, it would result in a shift in phase. If there were any random time delay, it would result in a loss of phase coherence. Neither of these occur during reflection.There is no "absorption" or "consumption" of the EM wave, its energy is not "stored for some period of time" and is not "re-radiated" later.(You're describing a process calledfluorescence. The wave energy absorbed and stored, and after some delay is re-emitted: but in all directions, and with random phase. It is not magically re-emitted in the same direction with the same phase!)Maxwell's classical wave equations describe perfectly well "why wave propagation in the matter is smaller than the speed of light in the vacuum." Any medium (matter or vacuum) propagating electromagnetic waves is described by only two intrinsic properties: itspermittivityandpermeability(or, equivalently for waveguides or cables, their capacitance and inductance per unit length). Theirproductdetermines the speed of propagation. Theirratiodetermines the impedance, the ratio of electric field to magnetic field of the EM wave. If, as is commonly the case, the wave propagates into a medium with different permittivity, both the speed (refractive index) and impedance change. Two separate waves are produced: a refracted wave is transmitted, and a wave is reflected, both instantaneously withnotime delay.In 1887, Heinrich Hertz first demonstrated the polarization, reflection and refraction of radio waves, measured their speed in air to be the same as light, measured their refraction through pitch resin, all as predicted by Maxwell's wave equations.The permittivity and permeability of any medium can be measured independently. The capacitance and inductance per unit length of any waveguide or cable can be calculated directly from their dimensions, and/or measured independently. These predict the wave speed and impedance that is observed.No "absorption, delay, and re-emission" occurs. Reflection and refraction are instantaneous and are predicted by classical wave equations.Quote from: radiolistener on September 12, 2021, 06:29:09 amDirac first posed the "three polarizer paradox" in 1930. Here's a discussionhttps://www.informationphilosopher.com/solutions/experiments/dirac_3-polarizers/

author: ejeffrey
author_rank: Super Contributor
timestamp: September 13, 2021, 04:13:02 pm
content: Quote from: Rod on September 10, 2021, 05:29:03 pmActually we *can* observe single microwave photons although it is much harder than observing single optical photons. For instance MKIDshttps://web.physics.ucsb.edu/~bmazin/mkids.html.

author: ejeffrey
author_rank: Super Contributor
timestamp: September 13, 2021, 04:33:00 pm
content: Quote from: Rod on September 13, 2021, 12:22:22 pmYou definitely can. I don't have a reference of anyone doing this explicitly, but part of the reason is that it wouldn't surprise anyone. It's just classical E&M. The weird part about the three polarizer experiment is that it still works when you do it with single photons.QuoteLight doesn't exactly act like a classical wave but it is extremely close.

author: radiolistener
author_rank: Super Contributor
timestamp: September 14, 2021, 09:44:04 am
content: Quote from: Rod on September 13, 2021, 12:22:22 pmTime delay between EM wave energy absorbtion and re-emission exists and is present in any environment which consists of a charged particles (like electrons, protons, etc). And you can easily find the proof of this as a slower wave propagation speed in the matter.For exampel, the wave propagation speed for a water is about 1.333 times slower in comparison with vacuum. This slowdown is due to the fact that it takes some time delay between absorption and re-emission by charged particles in a water.Quote from: Rod on September 13, 2021, 12:22:22 pmYes, and any mirror reflects light with some phase delay.Quote from: Rod on September 13, 2021, 12:22:22 pmThe time delay is not random, it is fixed for almost the same value and depends on material properties. So all reflected waves have fixed phase delay and they are coherent to the source. But you're right reflection add some very small amount of phase delay dispersion, but usually you cannot notice it due to a very small difference.Quote from: Rod on September 13, 2021, 12:22:22 pmThe time delay between absorbtion and re-emission is exist and leads to add a reactive component in a medium impedance. This reactive component appears because energy is stored for some time period in particles of the medium.For comparison, vacuum impedance is 376.73 + j0 Ω, and as you can see it doesn't have reactive component. This is because vacuum doesn't have charged particles, so there is no needs to absorb and re-emit wave and as result no time delay.The phase delay for a vacuum depends on the distance only (we don't take into account gravity relativistic corrections here, because they are very small).But the phase delay of a medium which consists of charged particles will depends on two factors: 1) distance and 2) particle properties.Quote from: Rod on September 13, 2021, 12:22:22 pmNo. Reflection is NOT instantaneous and requires some period of time between energy absorption and re-emission.And this time delay leads to a refraction which happens between two medium with different time delay between absorption and re-emission.The refraction happens the same as a vehicle starts to skid when the wheels on one side get stuck in the soil more than the other side. When we're talking about EM wave it starts to skid because time delay between absorption and re-emission is different for two mediums. When one side of EM wave slows down, while other side still moving at faster speed, the vector of wave is changed and refraction happens.But phase delay doesn't matters for three polarizer behavior. The secret here is that second filter rotate polarization of light at 45 degree and now it can pass through the last filter.

author: msat
author_rank: Regular Contributor
timestamp: September 14, 2021, 05:50:39 pm
content: I want to respond to many parts of your reply because you provided a lot of good information here and to clarify any misunderstandings I may have caused. It's also evident that I should have attempted to be more careful with some of the terminology I used.Quote from: Rod on September 09, 2021, 09:56:52 pmIt is perfectly sensible to me that charges in a metal conductor, whether it's an antenna or not, would redistribute themselves in an attempt to cancel an external e-field induced on it. This is a major behavior I was trying to get at, though perhaps not very articulately on my part. Let me use an image to help make my point.(modified image taken fromhttps://em.groups.et.byu.net/embook/ch9/mod9.2.html)Lets assume that in the above example the distance between the antennas are spaced by some whole multiple of wavelength so that they are in phase with one another. Lets also say that quarter of a cycle after the start of every new cycle, the transmitter is at peak amplitude with point A being positive and point B being negative. Since we are driving the antenna, it's logical that the upper leg of the dipole is positively charged while the lower leg is negatively charged. Measurements of point A and B confirm that. But what happens when we measure the voltage at points C and D at the same point in the cycle? The polarity will presumably be the same (though the voltage will be reduced due to distance unless we're perfectly directional). That seems obvious enough. Yet, as I believe you're saying, which is also in agreement with what I'm saying, is that the actual charge distribution between the two antennas will actually be the opposite. Why? Because since the external e-field induces itself on the receiving antenna, the charges want to distribute themselves to cancel it out. So when we take a measurement with a voltmeter, we see that point C is positive with respect to D, and because of this, current in the form of electrons will flow from the lower leg of the antenna towards the upper leg. Again, which is the opposite charge distribution of the transmitting antenna.As far as how much those charges could generate their own EM radiation I suppose depends on how much of their own fields are "locked in" cancelling out the induced e-field. I really have no idea. This I think is of prime importance to what I'm proposing, but I don't want to dwell on it at this moment.I want to make a point that while nobody should consult me to design their antenna, I do have at least a little more than a superficial understanding of how they work. I've looked at various sources of info on antenna theory before coming in to this discussion. Likewise, I'm cognizant of fundamental limits such as the speed of light, and by extension, propagation delays. I may unintentionally neglect to make those facts clear, and I may also sometimes choose my terminology poorly.QuoteAs far as I'm aware, I wasn't in disagreement of you on this.QuoteTo be fair, I never said a certain antenna arrangement created a three polarizer paradox. All I was saying is that inverted charge distributions in a conductor, whether in an antenna or not, could possibly re-emit EM radiation that's effectively 180 degrees out of phase with the incoming radiation. Such re-radiation may explain (I say may because I didn't do any mathematical analysis to validate) the 3 polarizer paradox.QuoteAll receiving antennas would experience the same effect I'm trying to describe, and thus all other effect still hold true best I can tell.QuoteIt does occur with light, and by extension, all EM radiation. Because of the wavelengths involved with light, it's hard for there to not be at least a bit of a mystery when it comes to optical polarizers. That's why I liked the following polarizer experiment which makes it a whole lot more macroscopichttps://sciencedemonstrations.fas.harvard.edu/presentations/microwave-propertiesPSHow the heck do I embed the uploaded images!?

author: msat
author_rank: Regular Contributor
timestamp: September 14, 2021, 06:36:37 pm
content: Quote from: Rod on September 10, 2021, 05:29:03 pmThere's a few things that I find unsatisfactory. Firstly, It's that it would be one of the few macroscopic phenomenon that only has a QM solution.The common explanation for the operation of linear polarizers is that they filter out the component of EM radiation that's parallel with the filtering elements (seehttps://sciencedemonstrations.fas.harvard.edu/presentations/microwave-properties) and convert that energy into heat. This would seem to imply that if those filters in the link I just posted were superconductors, then they would no longer act as polarizers. Now, I don't think any such experiment has ever been demonstrated, I doubt that would actually be the case. At any rate, the action of filtering out a component of the light inherently means there's no classical solution to the 3 polarizer paradox. Maybe that's truly the case, but I doubt it, and that's why I'm trying to figure out an explanation. That said, this post, and perhaps even this forum isn't the place to debate it. Understanding the the mechanics of antennas may give me the tools to either prove or disprove my theory, and perhaps build some electronics test equipment to do it (which *is* appropriate for this forum).QuoteFor the time being, I respectfully disagreeQuoteThe true nature of photons is something I find extremely interesting. Not even Einstein felt he understood exactly what they were, so with all due respect, I doubt you nor I ever will. I think most QM physicists feel the same way.QuoteWhat I'm proposing is actually strictly a matter of the superposition of various waves with various polarities and phases. So I suspect you're misunderstanding me. I'm trying to figure out if conductors will re-radiate incident EM radiation with its phase inverted. And if so, then why, and what are the implications of that?QuoteI'm not actually arguing against QM. My point is that pretty much every macroscopic phenomenon also has a classical explanation, yet the 3 polarizer paradox does not.

author: msat
author_rank: Regular Contributor
timestamp: September 14, 2021, 06:40:56 pm
content: Quote from: Marco on September 11, 2021, 09:56:22 amCould you elaborate on this a bit more? I'm pretty sure we're talking the same thing here.

author: msat
author_rank: Regular Contributor
timestamp: September 14, 2021, 06:58:07 pm
content: Quote from: radiolistener on September 12, 2021, 06:29:09 amBy and large, I think we've sort of been talking past one another. I'm not in disagreement with pretty much anything you've said, rather I just think I was misunderstood. Therefore I clipped most of your post to just reply to the bit above.My explanation for the "paradox" is along the lines of what you stated, but I think you have a misunderstanding of how polarizers work. So once again, I'll link a microwave polarizer demonstration:https://sciencedemonstrations.fas.harvard.edu/presentations/microwave-properties. In the experiment, the wave is nulled if the transmitted wave is vertically polarized and the polarizer's wire element are vertically oriented. Therefore, the only way to make sense of this classically is if the "filter" inversely re-radiates the absorbed energy (effectively 180 degrees out of phase), which cancels out at least part of the original wave. You can then rotate the polarizers and calculate the results using superposition, which is basically what you state.

author: msat
author_rank: Regular Contributor
timestamp: September 14, 2021, 07:47:48 pm
content: So I've been pondering ways to experiment on this. I didn't want to measure at the feed points of a dipole (particularly the receiver) since probing there might obscure what I'm trying to find. It also seems reasonable to probe the transmitter and receiver the same way. Here's a high level diagram of what I'm thinkingIt should be pretty self-explanatory what's going on there. The black boxes are high-impedance, low-ish capacitance amplifiers. I don't particularly care about the accuracy of the measurements, other than knowing the polarity of the potentials.I'm thinking of working in the ~6MHz ISM band due to only having a 25MHz scope. This also makes locating the antennas kind of easy at the expense of having somewhat unwieldy distances to work with. The two amp to scope coaxes should be of equal length for obvious reasons. I'm not sure if I need the amps to reference the neutral points on the antennas, especially since I can occasionally ground them to eliminate potentially any built up charges.Any thoughts on circuit topologies? Currently leaning towards having the antenna ends feeding the non-inverting input on an op-amp with a battery powered double-ended power supply. I "know" I need to match the output impedance of the amp to the transmission line impedance. Since I've never worked with RF, thoughts and tips would be appreciated.

author: Marco
author_rank: Super Contributor
timestamp: September 15, 2021, 09:37:19 am
content: Why no resistor in the middle of the receiver?

author: msat
author_rank: Regular Contributor
timestamp: September 15, 2021, 11:08:25 am
content: Quote from: Marco on September 15, 2021, 09:37:19 amOriginally I considered that, or actually, 3 series resistors and I'd take a measurement across the two terminals of the center resistor (the reasoning I couldn't fully wrap my head around), but eventually considered it unnecessary if I probed the end. I loosely based the logic off the fact that the director elements of a Yagi also don't have any resistive elements between their halves. The induced e-field should still redistribute charges along the antenna's length, which is what I'm trying to detect, specifically at the ends.

author: Marco
author_rank: Super Contributor
timestamp: September 15, 2021, 01:10:56 pm
content: Shrug, you are not measuring what you asked about in the original post then. A receiver antenna is terminated, not a resonator.

author: msat
author_rank: Regular Contributor
timestamp: September 15, 2021, 04:12:47 pm
content: Quote from: Marco on September 15, 2021, 01:10:56 pmWhat I've been trying to ask about is the charge distribution on a conductor (such as an antenna) when an e-field is induced on it by external EM radiation.

author: Rod
author_rank: Contributor
timestamp: September 15, 2021, 10:23:24 pm
content: Quote from: msat onYesterdayat 07:47:48 pmRealize the wavelength at 6 MHz is 50 meters. Your proposed antennas will be 25 meters long, and have to be suspended above the ground plane (the earth) to work. You propose to space them an integer number of wavelengths apart. This experiment will span a football field. The long coaxial cables from your receivers to the oscilloscope will be expensive...Heinrich Hertz first demonstrated the electromagnetic waves predicted by Maxwellhttps://en.wikipedia.org/wiki/Heinrich_Hertz#Electromagnetic_wavesusing a detectors tuned to ~350 MHz (85 cm) or ~450 MHz (66 cm). These experiments filled a large room. It appears you wish to reproduce this at a much longer wavelength, on the scale of a football field.The Harvard demonstrations of Hertz' experiments which you recommend chose 3 GHz (10 cm). You might also enjoy Bragg's demonstration of these same experiments, including polarization; he chose 30 GHz (1 cm). The shorter wavelength allows the size of the experiment to be reduced to a table top.

author: radiolistener
author_rank: Super Contributor
timestamp: September 16, 2021, 07:46:42 am
content: Quote from: msat on September 14, 2021, 06:58:07 pmNo. The phase delay doesn't matter here at all. Phase delay and polarization are completely different things.The secret is that polarizer re-emit wave with polarizationwhich is differentfrom input wave. This is clearly explained in the text from your link:QuoteAnd this is what I talked about.There is no quantum effect and no needs to involve photons or phase delay to explain three polarizers behavior.Regarding to your schematic for testing, it is invalid. Because your measurements will be affected by your probes.The correct way is to use two identical half wavelength dipoles connected to coax cables with RF chokes. Both cables should have exactly the same electrical length. Connect cables to two channels of oscilloscope with using pass-through terminators.And keep dipoles at least half-wavelength distance from each other to avoid inductive or capacitive coupling.See fixed circuit in attachment.Quote from: msat on September 15, 2021, 04:12:47 pmAs said before, at wavelength distance it will be exactly the same as on transmitting antenna.

author: msat
author_rank: Regular Contributor
timestamp: September 16, 2021, 11:17:49 am
content: @RodI'm aware of those drawbacks related to using 6MHz (I forgot to add that I was also considering 27MHz - which isn't a whole lot different). Of course I'd prefer to use much higher frequencies, but I want to hook up the experiment to an o-scope, and the best I have access to is 100MHz (and my personal scope being 25MHz).@radiolistenerWith all due respect, I think you're misunderstanding me. I did not mention "phase delay" - I talked about inverted re-radiation which, when considering a pure sine wave, would essentially appear as being 180 degrees out of phase, but not that it actually is! I also never said or implied that it is somehow related to polarization!As far as my experiment, I can see no other way to verify what I'm looking for other than what I propose. Your experiment would not work due to reasons I already outlines. Of course the probes will have an effect on the "antenna" under test, but I have come across other experiment where antennas were probed along their lengths. The effects of probing is another reason why this experiment is better performed at the low-ish frequencies I'm considering."As said before, at wavelength distance it will be exactly the same as on transmitting antenna."I'll accept that if that's the case, but no one has yet replied to this thread with hard evidence. I know what the "conventional wisdom" is, and I posited a reason for why conventional wisdom might not reflect reality. At this point, it appears this discussion is at a stalemate, so it seems appropriate to settle this with an experiment.

author: msat
author_rank: Regular Contributor
timestamp: September 16, 2021, 02:08:51 pm
content: Attached is a graphic I made to describe the effect I'm positing

author: radiolistener
author_rank: Super Contributor
timestamp: September 17, 2021, 07:35:07 am
content: Quote from: msat on September 16, 2021, 11:17:49 am180 degree out of phase is phase delay = 180 degree.Quote from: msat on September 16, 2021, 11:17:49 amAre you trolling me?!I proposed you working experiment. If you do it in such way you will not be confused with questions "why it doesn't works?". It takes into account RF specific issues. There is very deep background behind it which you don't know (it's pretty clear from your picture). If you're interesting to learn why, you can read about transmission lines.Quote from: msat on September 16, 2021, 02:08:51 pmyour picture has mistake. Since both antennas placed at the same E field they both will have the same polarity on their terminals. But your picture shows inverted polarity. This is incorrect.Inverted polarity will be if you move antenna into center of your image (where phase delay is 180 degree).

author: msat
author_rank: Regular Contributor
timestamp: September 17, 2021, 01:05:30 pm
content: Quote from: radiolistener on September 17, 2021, 07:35:07 amThis is theentirepoint of this thread. If you look closely, you will see that, as far as voltage measurements go, the polarities WILL be the same. And it's because of these polarities that cause the redistribution of charge carriers to try to cancel these fields. By what other mechanism would there be a current in a receiving antenna?!I posited a hypothesis of why I think this is correct, and due to its perhaps counterintuitiveness, I'm trying to design an experiment to test it. You say I'm wrong, which may very well be true, but you have not provided supporting evidence. Just telling me something is some way does not constitute proof.

author: radiolistener
author_rank: Super Contributor
timestamp: September 17, 2021, 04:35:15 pm
content: Quote from: msat on September 17, 2021, 01:05:30 pmWhen two antennas are placed in the same condition (the same electric and magnetic field gradients in the space) they both will have exactly the same output. It doesn't matter if antenna is receiver or transmitter, because antenna works in both directions exactly the same. It doesn't matter who applying energy to create electric field around antenna.Current flows through antenna terminals because electric field in the space is not static and changing over the time. The Voltage over space is changing and as result it leads to change voltage on antenna terminals. Antenna just collects this Voltage from the space. If you short antenna terminals, the antenna will re-emit consumed RF energy back to the environment (with some loss on heating). When you attach some load (for example receiver), your load will consume that energy from antenna and this consumed energy will not be re-emited back to the environment, it will be absorbed by your load (receiver for example).In reality antenna works a little more complicated, because not all applied RF energy are radiated, part of it remains in antenna in form of standing waves. But it doesn't matter to understand overall picture.Quote from: msat on September 17, 2021, 01:05:30 pmI already provided you with simple and easy picture on how to build test setup and make sure of everything in practice. So, I don't understand what you're expecting more?Instead of build that setup and check it, you're trying to talk about QM, photons and paradoxes. All these things (antennas and three polarizers behavior) are well described with classic electromagnetic fields and Maxwell's equations which don't needs to involve QM and photons at all.Quote from: msat on September 17, 2021, 01:05:30 pmIf you're needs proof, just build test setup and get it. It's not hard.

author: msat
author_rank: Regular Contributor
timestamp: September 17, 2021, 06:22:19 pm
content: @radiolistenerI appreciate that you've taken the effort to respond to my posts, but to me it seems that our discussion is counterproductive for a few reasons. First off, you're attributing things other people have said to me. Secondly, I don't think you've put much effort in understanding my hypothesis. Maybe we can get on track, or maybe it's pointless, but at least I'll try.I'm not the one who brought up photons. I *responded* to the person who did, but I didn't use them to explain anything. So we can drop that.I didn't invoke QM as an explanation. If you look back, I argued AGAINST QM being the only viable explanation for the 3 polarizer paradox. So we can drop the QM thing too. My entire intent is to provide a CLASSICAL explanation for the paradox. You say it already exists, but I have yet to see one that addresses experimental results. What I'm proposing I believe WOULD explain experimental results.I'm basically in agreement with you regarding how antennas work, up to the point where you neglect to mention anything regarding the movement (current) of charge carriers (electrons). So I implore you to think about it for a little while, because that's one of the most critical aspects of my hypothesis. If you refer back to my last diagram, think about the effect the E-field has on the direction of the current. Since electrons don't just magically appear and disappear, they have to be displaced from somewhere and redistribute themselves in a way where they attempt to sum the external and internal forces to 0.Your experiment basically tells me to measure the voltage across the resistor. That voltage should be as you suggest. I'm not in disagreement there. But that voltage will cause current to flow, right? Which way will that current flow? And because of that current flow, where will those charges finally end up? Do you see what I'm saying?

author: radiolistener
author_rank: Super Contributor
timestamp: September 19, 2021, 10:43:31 am
content: Quote from: msat on September 17, 2021, 06:22:19 pmYou cannot argue against QM, becausethere is no paradoxand no needs to involve QM to explain such behavior, because this is classic wave behavior.First polarizer absorb wave and re-emit only vertical polarization. When vertically polarized wave falls on second 90 degree polarizer, it will not be re-emited and will be dissipated as heat, because cos(90°) = 0. When it falls on 45 degree polarizer, it will be absorbed and re-emited at 45 degree, and this is why it will be passed through third polarizer, because cos(45°) * cos(45°) = 0.5.This is why light didn't passed through 2 polarizers and passed through 3 polarizers.You're just needs to understand how polarizer works.It rotate polarization.That's it.Exactly the same thing happens with RF polarizer grid. When you place vertical grid, electrons can move only in vertical direction (along grid wires) and cannot move in horizontal direction. This is why re-emited wave will have only vertical polarization.Quote from: msat on September 17, 2021, 06:22:19 pmJust open the link from harvard.edu, which you posted above. It include classic wave explanation for 3 polarizers behavior. And description how it can be tested.Quote from: msat on September 17, 2021, 06:22:19 pmFor both antennas current flow direction will be identical at one lambda distance. If it's not, you can be sure that the distance between antennas is not multiple of wavelength.And this is known for at least 100 years.

author: msat
author_rank: Regular Contributor
timestamp: September 21, 2021, 12:45:15 am
content: Quote from: radiolistener on September 19, 2021, 10:43:31 amI only use the name "3 polarizer paradox" as a way of referencing the experiment. I did not come up with the name, nor am I actually suggesting there really is a paradox. You're paying too much attention to the wrong aspects of my posts.QuoteOnce again, you prove that you don't actually understand the experiment or how it works. Take a look again at the Harvard page. If the microwave source and receiver is vertically polarized, and then the "polarizing filter" is placed in between with it's wire elements also oriented vertically, no microwaves reach the receiver. The explanation given in that page (and ALL others that I've read) is that the energy is absorbed and converted to heat. If that's true, the rest of the experiment CANNOT be explained that way. What *could* explain it, however, is if the polarizer does indeed re-radiate the energy parallel with the filter wire elements but only if it is INVERTED (basically 180 degrees out of phase for a symmetrical wave).QuoteOnce again, your explanation breaks down under scrutiny.In a transmitter, the amplifier forces the different charges in a dipole, and these charges are responsible for creating the electric field around the antenna and the magnetic field around the antenna due to the flow of charges.A receiving antenna CANNOT be explained the same way! In a receiver, there is not an amplifier moving the charges around to each leg of a dipole. It must be the induced E-field from the incident EM wave which causes the charges in the antenna to move. So, if we take your example, that would mean that the electrons move from the positive side of the E-field to the negative side. If that was true, then that goes against all our knowledge of the behavior of electricity.So, what exactly is it that "is know for at least 100 years"??Honestly, the point of my original post wasn't to try to convince anybody of anything. I sought to get evidence of charge distributions in a receiving antenna relative to the direction of the E-field of the incident RF wave. No one has provided any such compelling evidence. In all explanations so far, the charges have been completely ignored. And if the charges are ignored, then the explanation is incomplete.

author: Rod
author_rank: Contributor
timestamp: September 21, 2021, 10:08:14 am
content: Very much appreciate your diagrams, msat. "A picture is worth a thousand words" and communicate what you're thinking very clearly.Question: How fast can a voltage, or a current, or any electrical signal, travel down a wire: instantly, or only at the speed of light?Your diagrams suggest you do agree that electric field waves travel through space at a finite speed, the speed of light, not instantly. But your diagrams also suggest you assume voltages travel down a wire (an antenna) instantly, at an infinite speed. Specifically, your diagram shows a receiving antenna having the same voltage along its entire length, and the voltage at the end of a receiving antenna wire arriving instantly a quarter-wavelength away at its terminal. That can only occur if the voltage at its end propagates instantly down the wire. Do you believe it does? Or if it does not, how would your diagram change?Paradoxically, your diagram also shows a transmitting antenna emitting an electric field which looks (correctly) like a sine wave along its length, highest near the ends of the antenna and zero at its center terminals. This (correctly) implies it was created by an alternating voltage on the transmitting antenna wire which is greatest at its ends, and is near zero at its terminals. This can only occur if (and because) the voltage along the length of the transmitting antenna wire doesnottravel at infinite speed, but only at the speed of light. (If the voltage travelled instantly at infinite speed down the transmitting antenna wire, the entire wire would be at the same voltage at every instant in time, and would emit along its length asquarewave, not a sine wave.)Question: How fast can a voltage, or a current, or any electrical signal, travel down a wire: instantly, or only at the speed of light?

author: radiolistener
author_rank: Super Contributor
timestamp: September 21, 2021, 06:09:58 pm
content: Quote from: msat on September 21, 2021, 12:45:15 amNo. No. And again No. Phase doesn't matter here at all. If wave will be re-emited at 180 degree it will be passed through all polarizer the same as for phase delay 0 degree, 45 degree, 90 degree and any other phase delay. It can be re-emited at any phase delay, because phase delay doesn't affect result at all.The polarizer deal with polarization of wave. And output of polarizer is polarized according to polarizer orientation. You're just don't understand what polarization is.Phase delay and polarization are completely different thing. You're needs to learn what polarization is before drawing any conclusionsQuote from: msat on September 21, 2021, 12:45:15 amReceiver antenna works exactly the same as transmitting antenna. It just works in opposite direction, but all things exactly the same.

author: msat
author_rank: Regular Contributor
timestamp: September 21, 2021, 09:55:18 pm
content: Quote from: Rod on September 21, 2021, 10:08:14 amThanks, Rod. I agree that an associated visual depiction can often be much more useful at conveying an idea than just words alone.I want to state that, to the best of our understanding, the speed of light is the ultimate velocity limit of everything (particles, waves, etc) in the universe, and in no way am I trying to argue against it. So our explanations of any physical phenomenon needs to inherently apply that constraint.I think there's an error in your interpretation of what I'm positing, as it doesn't require any action that exceeds the speed of light to the best of my knowledge. Any notable changes at the receiver (essentially the movement of charges) change at a rate either equal to or less than that of the transmitter. Since there's a finite rate of change at the transmitter (at the most fundamental level, it would seem to be -in a superficial equation sort of way- that the velocity due to the "amplitude" or physical displacement of a charge multiplied by it's frequency must equal less than C), there is similarly a finite rate of change in the receiver. Any action that charge movements in the receiver cannot sufficiently react to will exhibit some other effect, which I imagine for a polarizer means the "unreacted" energy will be passed through, though I don't know what it means for a conductive plate.However it's worth considering that, especially for perfectly aligned transmitter/receiver pairs, a hypothetical "slice" of EM radiation will be incident on the receiver's entire length simultaneously. Though as mentioned above, the difference between the current "slice" and the previous one cannot have been created by something that exceeded C.I hope that makes sense. I think it addresses your points.

author: radiolistener
author_rank: Super Contributor
timestamp: September 22, 2021, 04:44:05 am
content: Quote from: msat on September 21, 2021, 09:55:18 pmNo. Your knowledge is wrong. According to your picture the voltage at the end of dipole is the same as voltage on antenna terminals. It means that wave propagation speed between antenna terminals and the end of wires is infinite and breaks speed of light limitation.In reality the dipole doesn't have constant voltage across dipole length. There is always fixed time delay between you apply voltage to the terminals and when that voltage appears at the end of dipole wires. This phase delay will be added between electric field change and transmitter voltage change for transmit antenna. And the same this phase delay will be added between electric field change and voltage change for receiving antenna.You didn't take this time delay into account. That's your mistake. You're needs to understand that charges cannot move across antenna length immediately it needs some time delay. This is very important to take this into account, otherwise your model for charge distribution will be completely incorrect.

author: msat
author_rank: Regular Contributor
timestamp: September 22, 2021, 07:38:55 am
content: Quote from: radiolistener on September 22, 2021, 04:44:05 amYou are making too many assumptions from my simple diagram. I drew it to make my point apparent, not for complete accuracy. I NEVER said there's a constant voltage along the length of the dipole, or that anything happens with 0 delay! Therefore, there's no need for me to refute the rest of the things you just said, because it's based off of your misinterpretations.Look, I'm more than willing to try to support my hypothesis in good faith, and also attempt to provide clarifications to those asking for it. However, I find it tiring having to spell out every little detail because someone can't, or worse, is unwilling to make reasonable inferences. More frustrating still is if I have to refute arguments based off of words someone else put in my mouth.When you stop and think about it, my point is pretty simple. Apparently many people seem to think that the distribution of charges (i.e. which leg of a dipole is dominated by which charge type) are the same for a transmit and receive antenna that are in phase. I'm merely positing that they are opposite - and I think there's pretty good evidence that it's true because it possibly explains other phenomenon. It's that simple. There's no other significant departure from antenna theory and everything we understand about how antennas work still applies!

author: radiolistener
author_rank: Super Contributor
timestamp: September 22, 2021, 01:16:07 pm
content: Quote from: msat on September 22, 2021, 07:38:55 amyour mistake is that you ignore wave propagation delay in the antenna wire. For half wavelength dipole it leads to a phase delay about 90 degree. If you sum phase delay for transmitter and receiver antenna phase delay will be about 90+90 = 180 degree. If you add 180 + 180, you will get 0 degree phase offset on receiving antenna. You can also add 360 degree phase delay for wave propagation in the space between antennas and you will get 0+360 = 0 degree phase offset.Since terminals on both antennas are in phase and frequency is the same, this is just impossible to get different charge distribution on both antennas.

author: Rod
author_rank: Contributor
timestamp: September 24, 2021, 05:44:13 am
content: Quote from: radiolistener on September 22, 2021, 01:16:07 pmAn incorrect statement in this posting is highlighted. Please allow me to clarify.We are discussing two half-wave dipole antennas, spaced N wavelengths apart, N = 0,1,2,3...The voltage at the ends of both antennas, and thecurrentthrough the terminals of both antennas, are in-phase with one another. A standing wave exists on both antennas; these standing waves are in-phase with one another.However, thevoltageacross the terminals of the transmitting and receiving antennas are anti-phase, 180 degrees out of phase, with one another. Why is this?1) Ohm's Law: any resistor R develops a voltage V across it proportional to the current I through it, V = IR.2) Kirchkoff's law: any power source (DC battery or AC signal source) in series with a resistor creates that voltage V, and is the source of that current I. However, the sign of V across the source isoppositethe sign of I through it. Any power source attached to a resistor R exhibits anegativeresistance -R. (More generally, any power source attached to a load of any impedance Z exhibits a negative impedance -Z.)A half-wave dipole antenna may play different roles, simply depending on what resistance R we place across its terminals:1) if the terminals are attached to a positive resistor R>0, the voltage across the terminals is in-phase with the current through the terminals, this voltage resists the current through flowing the terminals, removing power from the antenna. This is a receiving antenna.2) if the terminals are shorted together, R=0, the half-wave dipole is a half-wave monopole antenna.3) if the terminals are tied to a transmitter, R<0, the voltage across the terminals is anti-phase with the current through the terminals, adds to the standing wave current through them, adding power to the antenna. It is a transmitting antenna.At the top of the screen is one guiding principle: "Be excellent to each other." Not merely polite, buthelpful. The original poster of this thread asked for help designing an experiment to test a hypothesis. (We know this hypothesis is incorrect, but he rejects any attempt to merely tell him that. He needs to think things through for himself to believe them. That's OK, even admirable. If he wishes, we can help him. That requires developing step-by-step the concepts that conservation of energy requires reflection at the ends of the antenna wires, allowing him to realize that two travelling waves in opposite directions in a half-wavelength wire add up to a standing wave, that a resistor placed in the voltage node and current anti-node at its center create a dipole, that the receiving antenna is capacitively coupled to the transmitting antenna, etc etc. That's not what he asked for, hasn't done it, and is pretty resistant to doing it!)So, can we help him? He posed two experiments:1) probe the voltage on the ends of the transmitting and receiving antennas (which we know will be in-phase), or2) probe the voltage on their terminals (we know will be anti-phase, but he doesn't, so he will be misled).Can we help him understand this? I don't know. But any help we offer has to be encouraging,verysimplistic and involve no math, or it simply isn't helpful. By this standard, neither the posting I quoted above, nor I fear my response to it, have been helpful.

author: radiolistener
author_rank: Super Contributor
timestamp: September 27, 2021, 08:09:13 pm
content: Quote from: Rod on September 24, 2021, 05:44:13 amyes, exactly. And phase offset due to wave propagation delay in the space between these antennas will be N * 360 degree = 0 degree.Quote from: Rod on September 24, 2021, 05:44:13 amNow add 90 degree phase offset (1/4 wavelength = 360/4=90) due to delay in transmitter antenna wires and 90 degree due to delay in receiver antenna wires:180 + 90 + 90 = 0 degree phase offset.Quote from: Rod on September 24, 2021, 05:44:13 amTransmitter impedance doesn't affect antenna impedance. Since energy loss in the transmitter and feeder line doesn't matter here, we can just assume that transmitter Zout = 0 + j0 Ω for simplicity.If we're talking about resonant half-wave dipole it's reactance on the terminals will depends on the dipole length and conductor thickness.Actually resonant frequency of classic half-wave dipole a little bit shifted. Let's see it, the impedance on a half-wavelength dipole terminals (with a feeding point at center) can be estimated in the following way:Code:R = (Zm / (2*pi)) * ((EulerGamma + log(2 * k * l) - Ci(2 * k * l) +cos(2 * k * l) / 2 * (EulerGamma + log(k * l) + Ci(4 * k * l) - 2 * Ci(2 * k * l)) + sin(2 * k * l) / 2 * (Si(4 * k * l) -2 * Si(2 * k * l))));X = (Zm / (2 * pi)) * (Si(2 * k * l) + sin(2 * k * l) / 2 * (EulerGamma + log(k * l) + Ci(4 * k * l) -2 * Ci(2 * k * l) - 2 * log(l / r)) + cos(2 * k * l) / 2 * (2 * Si(2 * k * l) - Si(4 * k * l)));whereR - active part of impedance (resistance)X - reactive part of impedance (reactance)Zm = environment impedance (Zm=376.73 + j0 Ω for a free space)l = dipole arm lengthr = conductor radiusk = wave number (k = 2 * pi / lambda)Ci(x) - integral cosineSi(x) - integral sineEulerGamma - Euler’s constant = 0.57721566490153286060651209008240243...Note: this formula doesn't take into account energy loss on wire heating due to Ohmic resistance of a wires.If we take dipole length = 0.5*lambda and r=0.001, the impedance on dipole terminals will be Z = 73.079 + j42.515 Ω.You can see it on this chart:As you can see, the actual resonant frequency of a half-wavelength dipole is a little bit shifted, so we needs to use a little bit shorter dipole to fit with it's exact resonant frequency. If we use dipole length = 0.4775072*lambda and r=0.001, the dipole impedance will be Z = 63.665 + j0.000 Ω.And since reactance on antenna terminals is X=0.000 Ω, the current and voltage on antenna terminals will be in phase. Isn't it?Quote from: Rod on September 24, 2021, 05:44:13 amIn order to get anti-phase, there is needs at least some difference in wave propagation speed or distance, how such difference can happens for a two antennas with identical sizes placed into exactly the same environment for exactly the same frequency and distance = N wavelengths?

author: Rod
author_rank: Contributor
timestamp: September 28, 2021, 08:59:32 am
content: radiolistener, you have entirely missed avery basicconcept. Please review Kirchkoff's lawhttps://en.wikipedia.org/wiki/Negative_resistance#Negative_static_or_%22absolute%22_resistancePlease think carefully about the very simple figure shown there, of a battery in series with a resistor, captioned "FromKVL, the static resistance of a power source (RS), such as a battery, is always equal to the negative of the static resistance of its load (RL)."Repeating, thecurrentacross the terminals of the transmitting and receiving antennas are in-phase. However, the transmitter (R<0) has an output voltage thatincreases that current, while the receiver input resistance (R>0) produces a voltage whichdecreases that current. Thevoltageacross their terminals are 180 degrees out of phase.This is KVL. It is true ofanypower source and load. It is not unique to antennas. It is much simpler than you think.

author: radiolistener
author_rank: Super Contributor
timestamp: September 28, 2021, 01:13:06 pm
content: Quote from: Rod on September 28, 2021, 08:59:32 amIf you measure voltage on antenna terminals, you will be unable to detect energy flow direction and unable to detect current source - antenna or feeder.Don't confuse current of standing waves within antenna (this current is 90 degree phase offset with voltage) with current from transmitter at antenna terminals (it is 0 degree phase offset with voltage).Since transmitter working at resonant frequency of antenna, reactance on antenna terminals will be X = 0 Ω. It means that RF energy is not returned back to the transmitter and remains in antenna (antenna holds it in form of reactive field oscillations).Here is animation of standing wave and feeding current:As you can see it's very different from static picture which is provided by msat, it needs to consider wave propagation delays in antenna wires.Also note that amplitude voltage on dipole end is Q times higher than amplitude voltage applied to antenna terminals. Where Q is a Q factor of antenna which affects antenna bandwidth. The voltage is higher because antenna accumulate amplitude of last Q wave cycles. For transmitting antenna it helps to radiate power from transmitter (higher voltage leads to higher radiation loss). For receiving antenna it helps to get voltage gain (amplitude of last Q cycles are summed together).

author: Rod
author_rank: Contributor
timestamp: September 28, 2021, 10:53:34 pm
content: That's a very nice picture of the voltages and currents on a half-wave antenna. Thank you!In return, I have a very simple picture for you, a battery in series with a resistor.Observe identical current is flowing thesamedirection in both the battery and the resistor. However, the voltages across the battery and the resistor are also identical but are inoppositedirections.This is very simple. This is very fundamental. This is very important. This is Kirchkoff's voltage law: the sum of voltages around a closed loop is zero. That requires that the voltage across the battery is exactly theoppositeof the voltage across the resistor.This has implications. If we define a property called resistance R as the ratio of the voltage to the current through a device, R = V/I, what resistance is the battery exhibiting? We observe it is -V/I = -R.How much power is each device dissipating? The resistor is dissipating P = VI. The battery is dissipating -VI = -P. That is, it is dissipatingnegativepower; it is a powersource!This remains true if we reverse the battery. This remains true if we replace the battery with an AC power source. The voltage across any power source is in theoppositedirection as the voltage across a resistor with the identical current flowing in the identical direction through it.If there isanything above you disagree with, please say so!If not, we can apply it toanycircuit. In fact, itmustapply in any closed circuit (any circuit in which the total energy is conserved).

author: radiolistener
author_rank: Super Contributor
timestamp: September 29, 2021, 12:32:59 pm
content: Quote from: Rod on September 28, 2021, 10:53:34 pmThe resistor is not voltage source.If you know second Kirchhoff's rule, you know that the sum of voltage drop will be zero if there is no EMF source in circuit. So, the source for voltage drop is EMF source (the voltage source inside the battery), not resistor.In other words, the polarity on resistor terminals is determined by external voltage source, not by resistor itself.The equivalent circuit of a battery consists of two elements: voltage source and internal resistor. The resistor doesn't contains voltage source and don't generate voltage on it's terminals. The voltage drop on resistor terminals and internal resistor is generated by voltage source inside battery.It means that there are at least two heat loss points for your circuit - the battery (it's internal resistance) and the resistor. Even if you remove resistor from your circuit, the internal battery resistance is still there. So if you short circuit the battery it will produce heat on it's internal resistance and the battery temperature will start to grow.Quote from: Rod on September 28, 2021, 10:53:34 pmWhen we're talking about antenna with transmitter. They both are AC voltage source, because antenna stores AC energy. And these two AC source have it's own phase which depends on antenna size and frequency (due to wave propagation delay in antenna wires). This is what we're talking about - about phase offset between AC sources.On your circuit there is no phase offset between voltage sources, because it consist of a single voltage source and this is DC source with constant phase.So, if you want to deal with DC circuit and Kirchhoff's law, the circuit will be the following:But DC equivalent cannot explain things which happens in antenna, because it doesn't take into account wave propagation delays and polarity alternating over time on voltage sources.Also note that you will be unable to detect which battery has higher voltage in this circuit by using DMM to measure voltage on battery terminals. The same you will be unable to detect if antenna transmitting or receiving when you measure voltage on antenna terminals. Exactly the same voltage can be for transmitting antenna and for receiving.

author: Rod
author_rank: Contributor
timestamp: September 29, 2021, 09:55:12 pm
content: Quote from: radiolistener on September 29, 2021, 12:32:59 pmIs that what KVL means? Does KVL only apply to circuits which donotcontain an EMF source? Does KVLnotapply to a circuit loop containing a battery? Please explain your understanding of KVL to me. (It appears to be different than that in any textbook?) Thank you!Quote from: radiolistener on September 29, 2021, 12:32:59 pmReally?! That's quite remarkable. But... you animationdoesshow a voltage across the antenna terminals. (That voltage appears to be in phase with the current through the antenna terminals. The voltage on each terminal appears to beleadingthe phase of the voltage on the end of each dipole by 90 degrees. It appears to bedrivingthe current... adding power to the antenna. I wonder if this is also true if we attach a receiver withpositiveimpedance to the terminals...removingpower from the antenna... or might theoppositethen have to be true? Hmm?)But before we dive deeper into this world of Tesla's secret zero-point energy antenna, I confess I am simple-minded and still live in a world in which energy is conserved, and ask you to please explain KVL to me first.

author: radiolistener
author_rank: Super Contributor
timestamp: September 30, 2021, 02:14:41 am
content: Quote from: Rod on September 29, 2021, 09:55:12 pmNo. It apply to any circuit loop. If there is no EMF source, all voltage drops will be zero and the sum of all zero will be also zero. As you can see it works.Quote from: Rod on September 29, 2021, 09:55:12 pmNo. It apply to any circuit loop. When circuit consists of battery with some voltage, it will produce voltage drops on resistor. But the source of voltage drop on resistor is not resistor, the source of that voltage is the battery (voltage source inside the battery).When you connect voltmeter to the battery terminals it shows + voltage on + terminal of battery and - voltage on - terminal of battery. But it doesn't means that voltmeter generates voltage "which is opposite to the battery". The source of this voltage is the battery, not voltmeter.Quote from: Rod on September 29, 2021, 09:55:12 pmMy understanding is that algebraic sum of all voltages in a loop must be equals to zero. Is your understanding is different?Quote from: Rod on September 29, 2021, 09:55:12 pmYes, really. In order to detect energy flow direction through antenna terminals, you're needs to connect antenna through directional coupler. Such approach is used for SWR meters to measure direct and backward energy flow. Measuring voltage on antenna terminals is not enough to detect RF energy flow direction.Just because you will be unable to distinguish voltage from forward wave and voltage from backward wave, they both will be summed on antenna terminals due to wave superposition principle. Even if there is just a single direction wave flow, you can see wave on terminals, but you will be unable to detect direction of that wave flow through terminals.So, if you measure just voltage on antenna terminal, you will be unable to distinguish where is the source of that voltage. You will be unable to detect if wave flows from antenna to transceiver or from transceiver to antenna.You can distinguish it if you know wave flow direction. But you will be unable to detect wave flow direction by measuring voltage on antenna terminals. Both wave flow directions will looks exactly the same for your voltmeter on antenna terminals.I can say more. You can just disconnect antenna from antenna terminals and your voltmeter will show you the same wave on terminals, but that wave doesn't flow into antenna because it is completely disconnected.Quote from: Rod on September 29, 2021, 09:55:12 pmDon't confuse voltage of standing waves inside antenna and voltage on antenna terminals. These are different voltages. Note that voltage on antenna terminals is a superposition of a standing waves traveling in the antenna body. Also note that these voltages on antenna end have very different amplitude from voltage on antenna terminals (for half-wave dipole the difference is about 10-15 times).You will be unable to simulate processes in a half-wave dipole with DC circuit equivalent, just because it doesn't take into account time and space dependent processes across antenna wires.

author: cbutlera
author_rank: Regular Contributor
timestamp: September 30, 2021, 09:48:12 pm
content: Quote from: msat on September 07, 2021, 03:33:43 pmThis is actually a very interesting question, which gets right to the heart of how dipoles work.The short answer is that if they are thin, parallel, side-by-side, half-wave dipoles, spaced a whole number of wavelengths apart, and with the receive dipole connected to a conjugate matched load, then the voltage across the receive dipole terminals will lead the voltage across the transmit dipole terminals by about 45o.This in itself is perhaps not so surprising, because of the phase shift introduced by the reactive part of the dipole impedances. It is rather more surprising if we look at the phase relationship between the currents going in to the terminals of the two dipoles. It turns out that the current going in to the receive dipole will lag the current going in to the transmit dipole by 90o, unless the dipoles are less than a couple of wavelengths or so apart. Or to put it another way, compared to a straight length of hypothetical, light-speed, coaxial cable. The electromagnetic wave will appear to have covered the distance a quarter of a cycle quicker!One way to analyse this system is to look at the mutual impedance Z21between the parallel side-by-side half-wave dipoles for any given separation.Figure 10-12 on page 426 of "Antennas" (second edition) by John D. Krausshows almost what we need. R21and X21are the real and imaginary components of Z21. Unfortunately, the graph in Kraus only goes up to two wavelengths, where there are still significant near-field effects. So I wrote a few lines of code to carry out the numerical integration ofequation 10-5-3 on page 424 of Kraus, so that I could extend the graph to five wavelengths. I have attached the mutual impedance graph to the end of this post.This graph shows that beyond the near-field, the mutual impedance is purely reactive when the dipoles are separated by a whole number of wavelengths. As expected, at zero separation, the mutual impedance is equal to the self impedance of the dipoles (Z11and Z22).If this two-port network is conjugate matched terminated, then the current going in to port 2 will be in-phase with the current going in to port 1 when the mutual impedance is negative real. As shown on the mutual impedance graph, for larger distances, this occurs at (N-0.25) wavelengths where N is an integer. So at N wavelengths, the current going in to the receive dipole will lag the current going in to the transmit dipole by 90o.The relevant solutions to Maxwell's equations for the E and H fields, in the vicinity of a dipole at a particular frequency, contain three coupled modes. These modes are akin to the vibrational modes of a bell.One mode, whose amplitude falls off with distance as 1/r3, that is coupled directly to the charge density along the dipole.One mode, whose amplitude falls off with distance as 1/r2, that is coupled directly to the current density along the dipole. For a short electric dipole, this mode leads the 1/r3mode by 90o.One mode, whose amplitude falls off with distance as 1/r, that is not coupled directly to either the charge or current densities along the dipole. This 1/r mode is the only one that extends into the far-field.In terms of Maxwell's equations, the 1/r2mode couples to the 1/r mode through the action of additional terms in the curl partial differentials, which arise because of the extreme rate of change of amplitude with distance close to the dipole.The consequence of this coupling is that the 1/r mode leads the 1/r2mode by 90o. There is a diagram showing these couplings at the end of this post.At a receiving dipole, the 1/r mode from the transmitter can directly couple to the charge and current densities along the dipole, with no inherent change of phase (other than the current being of opposite sign). However, if the dipole is resonant at a different frequency, the current could lag or lead by up to an additional +/- 90o.I have redone my comment on parasitic elements, because my first attempt was misleading. The lesson being that maths is good, maths plus intuition is better, but intuition on its own can easily lead one astray. To get anywhere near to an additional +/- 90owould require almost complete detuning of the parasitic element, rendering it worthless.Section 11-9 starting on page 476 of Krausanalyses parasitic elements in terms of their mutual and self impedances. Based on the equations presented by Kraus, the most that can be achieved when detuning a half wavelength dipole is about +/- 45o, which would result in a 3dB reduction in the reflected/directed power. If a 6dB reduction is acceptable, then about +/- 60ocould be achieved. So with respect to its 1/r mode, a practical director element (shorter than the resonant length) could get to within -30oto -45oof being in phase with the received signal. Similarly, a practical reflector element could get to within +30oto +45oof being in anti-phase with the received signal.

author: Rod
author_rank: Contributor
timestamp: September 30, 2021, 11:10:20 pm
content: Quote from: radiolistener on September 30, 2021, 02:14:41 amOf course you can: the simplest way to measure current flowing in any wire is to insert a small current shunt resistor and measure the phase of the voltage across it. Insert the shunt resistor across the antenna terminals in series with either the transmitter or receiver, and compare the phase of the voltage across the shunt resistor (the phase of the current) to the phase of the voltage across the terminals with an oscilloscope.Allanydirectional coupler does is measure the relative phase of the voltage and current. This is particularly obvious in directional couplers we use at HF and VHF frequencies, which use discrete broadband RF transformers. "The relative sign of the induced voltage and current determines the direction of the outgoing signal."https://en.wikipedia.org/wiki/Power_dividers_and_directional_couplers#Cross-connected_transformers(For a more detailed discussion, see Figure 2 inhttps://michaelgellis.tripod.com/direct.html)Any directional coupler relies on precisely the principle we are discussing: by KVL, the relative sign of the voltage and current change (their relative phase changes by 180 degrees) if the direction of the power flowing through it is reversed.Quote from: radiolistener on September 30, 2021, 02:14:41 amWhat??? Please explain. Are you trying to explain a quarter-wave monopole antenna? Or did you intend to write "You can justshortthe antenna terminals. The same current will flow between the terminals, but no power will flow into it from the shorted transmitter output impedance, nor will any power be delivered from the antenna into the shorted receiver input impedance."This is key to understanding what we've been talking about.

author: radiolistener
author_rank: Super Contributor
timestamp: October 01, 2021, 02:05:11 pm
content: Quote from: Rod on September 30, 2021, 11:10:20 pmNo. If you insert shunt resistor, the standing wave mode will be changed and your measurement will be not relevant. Because when you add shunt, you will measure different circuit with different wave modes.And you will be unable to use resistor shunt to measure amplitude of two waves moving in opposite direction in the same wire. Which is needed to distinguish forward power flow from a backward power flow.Quote from: Rod on September 30, 2021, 11:10:20 pmdirectional coupler is not intended to measure sign and phase. It measure amplitude of two waves which are flowing through cable simultaneously in two opposite directions - forward and backward wave. And these measurements are always positive, because amplitude cannot be negative.There is no sense to measure phase between two waves in the cable because such phase difference is always changed in time from 0 to 360 degree in a loop, just because waves are moving towards each other.You cannot simulate wave interference and other wave properties and wave effects of two waves with using DC circuit.

author: Rod
author_rank: Contributor
timestamp: October 02, 2021, 08:54:38 am
content: cbutlera, thank you!Your contribution here is a gift, and is spot-on. I'm extremely grateful. Your second attachment, depicting the origins of the mutual coupling between the antennas, is deeply insightful and simply marvelous.Please accept my gratitude and admiration, sir!Pondering the experiment proposed by the original poster...Quote from: Rod on September 15, 2021, 10:23:24 pm...we might (ahem) reflect (intentional pun, shamefully admitted) on the effect of the Earth ground plane on the numerical values of R21 and X21 you've calculated between two isolated half-wave dipoles. I think it reduces the numerical values of both R21 and X21 by a factor approaching 2, but has no effect on your conclusions as to their spacing.This reduction in both R21 and X21 arises from two causes:1) elevation of these horizontal dipoles above the ground plane. This is likely to be very low. This is included in equation 11-7-11 on Kraus page 468, and approaches 2 as the elevation approaches zero if the ground is perfectly conductive or lossless.2) at 7 MHz, the ground is likely to be low-loss dielectric, its relative permittivity given by equation 16-4-4 on Kraus page 718 and illustrated in Figure 16-6, so will do little to reduce this factor from 2. But even if it is not (in some regions of the desert southwest US, ground is dry to depths >50 m), it still has negligible effect on your conclusions.I was idly pondering this, conclude it has no effect on your conclusions as to antenna spacing, but would appreciate your comments either confirming or correcting my thinking. Thank you again!

author: cbutlera
author_rank: Regular Contributor
timestamp: October 02, 2021, 01:41:15 pm
content: Quote from: Rod on October 02, 2021, 08:54:38 amThanks for your comments, I'm glad to hear that my diagram was helpful.Unfortunately no one is infallible, least of all me. I wasn't paying proper attention to the sign of the induced current in my closing comments on the receive antenna. So that didn't agree with what I had already concluded by looking at the mutual impedance graph. I've now corrected it. I should have paid more heed to what you have been saying in this regard.

author: Rod
author_rank: Contributor
timestamp: October 02, 2021, 03:01:50 pm
content: Quote from: radiolistener on October 01, 2021, 02:05:11 pmThe key word in my phrase "small current shunt resistor" is the word "small". "Small" compared to what? Small (that is,negligible) compared to the antenna impedance. This is, in fact, the impedance any VSWR meter or directional coupler also imposes in series with the antenna terminals, as well. This has small, granted not zero, but negligible effect on the antenna circuit or on any standing wave on it. It must have a small positive impedance, as it directs a small part of the energy on the line to the coupled port. So by energy conservation (which is the principle underlying, and expressed in, KVL), it must have a small positive impedance. As does my small shunt resistor.Quote from: radiolistener on October 01, 2021, 02:05:11 pmAh, radiolistener, you are the gift that keeps on giving! An endless wellspring of internal inconsistencies, of which you areutterlyoblivious. I stand in sheer awe, sir.Beg, may I remind you of a very fundamental concept:forward==backward* cosine(180 degrees) =backward* -1Pray take heed: it's very helpful, indeed essential, when applying KVL to a circuit to be consistent in direction, to avoid this sign error.A directional coupler or VSWR meter is built on and works on this very principle, sir! Again, please do review how directional couplers we use at HF and VHF frequencies, which use discrete broadband RF transformers, work. As their components are discrete, their principles are particularly obvious and simple to understand. A transformer is used to sample the net current through the directional coupler, which is simply thesumof the forward and backward (reflected) "waves" in your words, or thevector sumof the forward and reflected "power*" in mine. If the phase of both the forward and backward "waves" are each stable, then thephase of their sumis also stable. It is readily measured; this is precisely what a VSWR meter does. There is no magic here; there is no need for such confusion.So, let us consider three cases:1) the transmitter is on, and its impedance perfectly matched with the antenna impedance. This is the trivial case:- the small shunt resistor develops a small voltage across it which haszerodegrees phase difference from the voltage on the line,- the directional coupler, there is 100% forward and 0% reflected power, or- the VSWR meter reads zero.2) the transmitter is on, and its impedance is completely mismatched with the antenna. (You imagined such a case in previous postings**, but obviously had no clue that's what you were doing!)- the small shunt resistor has zero voltage across it, as the forward and reflected waves are identical but opposite in phase (that cos(180) above!) so perfectly cancel. No net current or power flows down the line.- the directional coupler shows identical forward and reflected power, and- the VSWR meter readsplusinfinity.3) the transmitter is off, but the receiver is connected, and is perfectly matched with the antenna, and it is receiving a signal.- the small shunt resistor has a tinynegativevoltage across it; i.e. its phase is 180 degrees, and cos(180) = -1.- the directional coupler can be read if we attached two identical receivers to its forward and reflected ports; the forward port has zero signal, while the reflected port does receive the signal.- the VSWR meter, if similarly equipped with two receivers, will now need a circular scale centered at the top on 0, so it can correctly turn CCW to read VSWR =minusinfinity. (This is distinguishable from case 2, in which it rotated CW to indicate +infinity, although +-infinity are the same point at 6 o'clock, opposite 0, on its scale! Interesting, isn't it?! That's what happens when you approach dividing by zero from two different directions!)(More physically meaningfully, we should place a second VSWR meter in series with the first, and label its scale "* -1". It will then correctly read 0 during reception, which we can interpret as -0. That avoids the need to have a bidirectional negative scale VSWR meter!)I painstakingly go through these cases in the vain hope they might encourage you to think more clearly. What we're talking about is really very simple. As Obi-Wan Kenobi might say, the force runs deep in you, Luke...but that force is self-contradiction, utter confusion, merely imaginary!It remains possible to dispel it.*yes, power is a vector, not a scalar, quantity. It is the dot product of the voltage times current. It has a definite phase, as well-defined as the phase of the voltage and current are.**your postings to which I refer is on page 2:Quote from: radiolistener on September 27, 2021, 08:09:13 pmThis is a marvel of internal inconsistencies!First, what voltage? If X=0.000 Ω, then V = 0 and its phase is... simply undefined! Your point is meaningless.Second, what transmitter? If X=0.000 Ω, then either you have shorted its output (...perhaps with a superconducting shunt, may we imagine?) or the transmitter produces zero voltage and infinite current, or both. But in any case, this transmitter is perfectly mismatched to the finite input impedance of your antenna, so any power the transmitter puts out is reflected right back into it, and none is coupled to the antenna. VSWR = +infinity.So indeed your assumption"we can just assume that transmitter Zout = 0 + j0 Ω for simplicity..."is indeed simple: it means nothing happens at all! What can be simpler than that?Umm, nothing!Yet you persist in it:Quote from: radiolistener on September 28, 2021, 01:13:06 pmTrue, if X = 0 Ω, there is zero voltage on the antenna terminals, but that means something: we will be able to detect a definite result: zero power flow between antenna and feeder, just as you state. We transmit nothing. The transmitter is not "working" at all - it literally does no work, produces no power - if X = 0 Ω. This is meaningless.A working transmitter will have a nonzero negative impedance, and if matched to your antenna, it will be Z = -63.665 Ω.Quote from: radiolistener on September 29, 2021, 12:32:59 pmReally! Well, I guess that senseless statement makes sense if you do attach the antenna to a X = 0 Ω transmitter or receiver, nothing will be transmitted or received, and 0 == 0. But that's not a veryusefulcase, now is it?I suggest you abandon the X = 0 Ω fallacy, and perhaps use loads matched to your finite antenna impedance. Because a vector of zero length may point forward or backwards or we know not where, anywhere. But if you choose a finite impedance, with a sign which points in a definite direction, then we may have some faint hope of discerning whether you are pointing in the right direction! (This is unclear, as other internal inconsistencies in your postings suggest a rather mystical view of reality, at least as I know it, but are certainly entertaining!) So, onward, radiolistener, and we may(?) yet sort this out, sir!

author: gcewing
author_rank: Regular Contributor
timestamp: October 03, 2021, 12:22:53 am
content: Quote from: TimFox on September 10, 2021, 05:39:11 pmHe'll also have to give up believing in the stability of the atoms making up his body. According to classical physics, their electrons should radiate all their energy away and spiral into the nucleus in a very small fraction of a second.

author: gcewing
author_rank: Regular Contributor
timestamp: October 03, 2021, 06:22:31 am
content: Quote from: msat on September 14, 2021, 05:50:39 pmIn that case they would reflect all of the energy they didn't pass through.The page you linked to says that dissipating the energy as heat is "one way to do it", not that it's theonlyway to do it. It doesn't matter whether the wires absorb the energy or reflect it or some combination of both, the important thing is that they don't let through the component of the electric field parallel to the wires.The remaining component is perpendicular to the wires and reduced by a factor of cos(a), where a is the angle between the incoming E field and the wires. If you do that twice with a = 45 each time, the result is cos(45) * cos(45), which is not zero, therefore some energy gets through.I don't think most of the stuff that's been said here about transmitting and receiving antennas is all that relevant, because the wires in that setup arenotacting as resonant dipoles. They're more like the "infinitely long wires" that were mentioned in another post. So figuring out the phase relationships in dipoles at resonance isn't going to help you much.

author: radiolistener
author_rank: Super Contributor
timestamp: October 03, 2021, 02:30:08 pm
content: Quote from: Rod on October 02, 2021, 03:01:50 pmYour first mistake here is that you didn't take into account that the small shunt resistor will change transmission line impedance and it will leads to change reflection coefficient. The standing wave mode will be changed. It means that adding shunt resistor will leads to amplitude change for forward and reflected wave. And your measurement will be invalid (amplitudes with resistor will be different than amplitudes without resistor).Your second mistake is that you didn't take into account phase delay inside shunt resistor and transmission line between shunt resistor and antenna terminals. The shunt resistor has a physical length, so it has phase delay for the wave propagation between resistor terminals. Also there is a short transmission line between shunt resistor and antenna terminals and it also has it's own phase delay. These phase delays are summed together and voltage on antenna terminals and on resistor terminals will be different voltage and usually is out of phase (it will depends on the shunt resistor electrical length and the distance between shunt resistor and antenna terminals).Regarding to the directional coupler and current shunt resistor - these are different things. And you cannot replace directional coupler with a shunt resistors.You insert port 1 and port 2 into cross section of the feeder and then you measure forward wave power on port 4 and backward wave power on port 3 (with take into account coupling coefficient).Quote from: Rod on October 02, 2021, 03:01:50 pmNo. Don't confuse reactive part of impedance X with active part R.X=0.000 Ω just means that there is no reflected wave from antenna to transmitter. But there is still present R=63.665 Ω which means that antenna consumes active energy from transmitter and amplitude voltage on antenna terminals is not zero.Note that the half-wave dipole impedance is a complex value and at resonant frequency will be:Z = 63.665 + j0.000 Ω(when wire radius = 0.001*lambda, the feeding point is at the center and there is no reflectors in the near field region)Quote from: Rod on October 02, 2021, 03:01:50 pmNo. Transmitter impedance doesn't matters here. Just read more about transmission lines. The input impedance of the transmission line will depends on transmission line impedance, it's electrical length and the load on the transmission line output. It doesn't depends on transmitter impedance.The output power on transmitter terminals is defined in it's specification datasheet for a specific load. The transmitter goal is to keep specific voltage amplitude on it's output terminals which are connected to the feeder with specific input impedance (usually coax cable 50 Ω). So the actual transmitter impedance doesn't matters here. It matters for RX mode or when there is a high SWR and significant amplitude of refelcted wave.Quote from: Rod on October 02, 2021, 03:01:50 pmNo. Z = 0 + j0 Ω just means that there is no internal power loss inside transmitter also it means that transmitter works as a short circuit load for reflected wave.Just look at this circuit:As you can see, if X = 0.000 Ω and R = 0.000 Ω, it doesn't means that transmitter output is zero.Quote from: gcewing on October 03, 2021, 06:22:31 amThese wires in the filter allows to move charges along it. It leads to re-emit a new wave and thatre-emitted wave will be polarized according to filter wires orientation.The wave re-emitted by filter is 180 phase degree with original incident wave, so it cancel original wave due to wave superposition principle. This is why horizontally oriented incident wave doesn't passed through filter with horizontally oriented wires. Actually it is passed, but there is appears second wave (re-emitted by wires) which is 180 phase degree and when you receive both waves on receiver after the filter they are just summed together and produce zero voltage on the receiver antenna.When you add third filter in the middle, which wires oriented at 45 degree to the wave re-emitted from the first filter, these wires now re-emit waves with 45 degree polarization against incident wave from the first filter. Now there are two waves falling into third filter. One wave has 90 degree polarization (from the first filter) and second wave is re-emitted on second filter and it's polarization is 45 degree. The first wave with 90 degree polarization will be canceled by third filter (the same as it happens for two filters at 90 degree). But the second wave which polarization is 45 degree now can pass through third filter.PS: note that phase delay degree and polarization degree are different things. And there is no quantum effects, this is just a pure wave behavior.

author: msat
author_rank: Regular Contributor
timestamp: October 04, 2021, 04:26:53 am
content: FWIW, I'm still following this thread, but just haven't felt compelled to post as I really didn't have anything worthwhile to add. Plenty of interesting comments since my last response however, and I want thank cbutlera for the post that I'm still parsing due to the richness of information.I've also been discussing this outside of the forum. In those conversations was a suggestion of using a VNA rather than an o-scope in the experiment. I realized the added benefits of using such a tool so I pivoted towards that. Because of that, a few aspects of my proposed experiment have changed.1) I want to raise the frequency to ~600MHz which makes the physical dimensions of the experiment a lot more manageable and convenient.2) I've considered additional variants of the experiment such as the receive dipole being a hollow cylinder with the tx antenna at it's center. The idea being that the received power is much closer to transmit power than if it was a simple wire dipole. Such power levels should be able to drive something like an LED in an optocoupler attached at the feed directly which would also indicate the direction of current. That said, cbutlera's post indicates to me that a proper experiment would require that such a rx antenna would need to have a radius greater than 2 wavelengths as there could still be significant near-field interactions.Quote from: radiolistener on October 03, 2021, 02:30:08 pmUhhhh.... Isn't this EXACTLY what I've been saying throughout this entire thread which you've been disagreeing with me about?!?!It was this thought that has been driving me to attempt to determine what the underlying physical process is that causes the re-emission to be an inversion of the incident EM wave. I've already stated what I believe that process is, and now I want to either prove or disprove it experimentally. That said, the hypothesis also needs to corroborate all existing relevant experiments, or at least justify any claims of experiment/logic/result errors.

author: msat
author_rank: Regular Contributor
timestamp: October 04, 2021, 05:03:38 am
content: Quote from: gcewing on October 03, 2021, 06:22:31 amBut then it stands to reason that, assuming the total area of the gaps is larger than the total area of the filtering elements (the wires in the case of the microwave filters), then more energy passes through than is reflected, but of course we know that's not what happens. So what's going on?I somewhat unintentionally came across a stackexchange post a few weeks back (https://electronics.stackexchange.com/questions/132325/how-does-a-receiving-antenna-generate-current-voltage/146766), which I wanted to repost here sooner but decided not to. However, I do think there's a relevant response by user wbeaty which is worthwhile to quote hereQuote

author: radiolistener
author_rank: Super Contributor
timestamp: October 04, 2021, 11:15:48 am
content: Quote from: msat on October 04, 2021, 04:26:53 amWhat I disagree is that polarizer filter cancels wave with specific phase offset. The polarizer doesn't deal with phase offset. It deals with polarization of wave.Quote from: msat on October 04, 2021, 05:03:38 amNo. When electron absorbs a photon, it doesn't emit a new photon. On the contrary, original photon is collapsed and it's energy is transfered into movement of electron. Otherwise it will break conservation of energy principle.I suggest you to forgot about photons and quantum physics and don't use it all, because it confuses you, there is no needs and no sense to involve such things as photons and quantum effects to explain these things. Quantum effects do not have a noticeable effect on the discussed things.People knew how antenna and polarizer work many years before concepts such as photon and quantum physics were introduced, and these knowledge still works, QM doesn't cancel it. So there is no needs to involve photons to describe EM field behavior at non quanum limit

author: msat
author_rank: Regular Contributor
timestamp: October 05, 2021, 02:04:49 am
content: Quote from: radiolistener on October 04, 2021, 11:15:48 amQuote from: radiolistenerSo which one is it?Quote from: radiolistenerWhat are you going on about? What you're responding to isn't something I said, but a quote from a thread I linked. And for some reason, in that entire quote, what you extracted from it is the bit about QM. And even still, the author wasn't wrong. As best we know, any absorbed "photon" will eventually be re-emitted. That was my take from what he said. I still don't understand your focus on that when in fact it's probably one of the least interesting things he said.

author: radiolistener
author_rank: Super Contributor
timestamp: October 05, 2021, 12:02:05 pm
content: Quote from: msat on October 05, 2021, 02:04:49 amFirst one talking about polarization and second one talking about absorption. Note that the result on polarizer output depends on polarization, but doesn't depends on phase offset of input wave, because re-emitted wave is coherent to input wave.Here is more interesting effect than "three polarizer". This is anisotropic crystals between two polarizers. When background is black both polarizers are 90 degree and light didn't passed. But as you can see light can pass it if there is anisotropic crystal between polarizers, because crytal rotate polarization of light after first polarizer. The effect here is more interesting, because anisotropic crystal has double refraction (birefringence) and produce two waves with different polarization simultaneously (ordinary wave and extraordinary wave have different polarization). Also it has dispersion, so refraction for different wavelength is different. And dispersion for ordinary wave and for extraordinary wave is different. As result it produce nice colorful picture. And colors are changed when you rotate polarizer.There is no paradox, this is just a wave behavior.Quote from: msat on October 05, 2021, 02:04:49 amWhat your author wrote is complete bullshit, include these "punches a hole", "bulls-eye pattern", "dark disks", photons and QM. It looks that he just don't understand what is reactive field and what is going on in the near-field region of antenna. He also misinterpret wave as photon.I recommend you to forgot all that bullshit and read classic literature about classic antenna design theory, it already has pretty detailed explanation without these "punches a hole", "dark disks", "bulls-eye", photons, QM and other stuff. This theory works well and based on classic physics, it is used in practice for antenna design and already verified with a lot of experiments.For example, I can suggest you the following book: Constantine A. Balanis, ANTENNA THEORY ANALYSIS AND DESIGN. See chapters 1 and 2, it already consists detailed explanation.Quote from: msat on October 05, 2021, 02:04:49 amBefore talking about photons you're needs to understand what is photon, where it is applicable and where it is useless.For example, do you realize why you cannot attach inertial system to photon? Do you realize why photon doesn't have size and mass?

author: cbutlera
author_rank: Regular Contributor
timestamp: October 05, 2021, 08:09:29 pm
content: Quote from: msat on October 04, 2021, 05:03:38 amThere is some misunderstanding here about what an electric or a magnetic field is.Classical electrodynamics is essentially the study of how stationary or moving charges at one place apply forces to stationary or moving charges at another, and how those charges respond to the forces. The definition of the electric field at a particular point is the force per unit charge that would be experienced by a test charge placed at that point. The definition of the magnetic flux density at a particular point is the force per unit current per unit length normal to the magnetic field that would be experienced by a current element placed at that point.It makes no sense to question how or why a charge or current element experiences a particular force when in the presence of an electric or magnetic field, when that force is the very definition of those fields. It is like asking why an object experiences a force of 9.8 N/kg when subect to a gravitational field of 9.8 N/kg.Energy transfer will occur if a charge moves in response to an applied electrical force, in the direction of that force. There is no mysterious 'sucking in' of energy, it is just simple Newtonian mechanics. Work done is equal to the force applied to the charge multiplied by the distance the charge moves in the direction of that force.All the receiving dipole arms (of a thin dipole) are doing is to rearrange their surface charge in such a way as to generate an additional electric field, that when added to the incident electric field, leaves no net tangential (to the dipole surface) component at their surface. An electric field can only be normal to the surface of a conductor where it meets that surface (or almost normal, due to the skin depth of an imperfect conductor). In doing this there will be side effects, and the net electric and magnetic fields at other points around the receiving dipole will be modified from what they would have been in the absence of the receiving dipole.The complexity of this process simply arises from the problem of calculating exactly how this surface charge rearrangement occurs. There is no mystery about it.The missing piece of the puzzle with regard to the polariser is the surface charge distribution around the circumference of the wires. Considering the case where the polariser wires are aligned with the incident electric field. The negative and positive surface charge distribution on the wires will be almost entirely confined to abandstripon the leading side of the wires (with respect to the incident electromagnetic waves, and except for the wires at the edges), because that is the only place where it needs to be to cancel the tangential component of the incident electric field. On the trailing side of the wires, there is no significant electric field remaining either tangential to or normal to the wires(again ignoring edge effects). Without this electric field there can be no electromagnetic wave generated by the wires in this direction. The surface charge distribution on the leading side of the wires, with its accompanying electric field that is normal to the surface of the wires at their surface, will generate a reflected electromagnetic wave back towards the source. Depending on the lengths of the wire elements with respect to the wavelength, there may also be an additional phase change in this reflected signal. Some of the incident energy will also of course be lost to heat in the polariser.

author: cbutlera
author_rank: Regular Contributor
timestamp: October 06, 2021, 12:38:21 am
content: Quote from: Rod on October 02, 2021, 08:54:38 amPerhaps I haven't understood what you are saying.Equation 16-4-3 on page 718 of Krausshows that for a horizontal dipole, the relative electric field E||will approach zero at low elevation angles and heights of the horizontal dipoles above the ground, both for the perfectly conducting ground case, and the high permittivity ground case. The Reflection coefficient in both cases being close to -1. So R21and X21will both approachinfinityzero*. Which is what I would have expected, given the close proximity of the dipoles to their reflected images.Conversely,equation 16-4-8 on page 719 of Krausshows that for a vertical dipole, the relative electric field will approach 2 under the same conditions, the reflection coefficient now being close to +1. In which case R21and X21would be approximatelyhalveddoubled*. Which again is what I would have expected, as the real antenna and its image are now reinforcing one another.* Sorry, I was thinking of a large impedance between the antennas, which is of course not the same as mutual impedance.