Electronics > RF, Microwave, Ham Radio
Phase diff between tx and rx antennas spaced by whole multiples of wavelength
msat:
--- Quote from: radiolistener on October 04, 2021, 11:15:48 am ---
--- Quote from: msat on October 04, 2021, 04:26:53 am ---Uhhhh.... Isn't this EXACTLY what I've been saying throughout this entire thread which you've been disagreeing with me about?!?!
--- End quote ---
What 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.
--- End quote ---
--- Quote from: radiolistener ---The wave re-emitted by filter is 180 phase degree with original incident wave, so it cancel original wave due to wave superposition principle.
--- End quote ---
So which one is it?
--- Quote from: radiolistener ---
--- Quote from: msat on October 04, 2021, 05:03:38 am ---Notice that when an atom absorbs photons, it always must emit other photons as part of the process. This is absolutely required by the physics, but rarely mentioned in QM courses.
--- End quote ---
No. 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 :)
--- End quote ---
What 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.
radiolistener:
--- Quote from: msat on October 05, 2021, 02:04:49 am ---So which one is it?
--- End quote ---
First 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 am ---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.
--- End quote ---
What 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 am ---As best we know, any absorbed "photon" will eventually be re-emitted.
--- End quote ---
Before 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?
cbutlera:
--- Quote from: msat on October 04, 2021, 05:03:38 am ---
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 here
--- Quote ---Usually the intuitive, non-math explanation of receiving antennas is avoided in introductory texts. Instead, they steer around the problem by only explaining transmitters and the emission of EM waves. Yet the absorption of EM waves by receiving antennas is quite fascinating. It's also a minor 'hole in physics' which needs some serious filling. (Phd engineering students take note!)
...
That's how EM absorption works. The receiving antenna has launched a wave which "sucks in" energy from the incoming wave. The reception of EM waves is a wave-cancellation process.
...
The majority of introductory textbooks do us a great disservice: they insist that we only learn about wave-emission. Then we're supposed to automatically understand everything about wave-absorption ...because reciprocity?
Uh, no.
--- End quote ---
--- End quote ---
There 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 a band strip on 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.
cbutlera:
--- Quote from: Rod on October 02, 2021, 08:54:38 am ---Pondering the experiment proposed by the original poster...
--- Quote from: Rod on September 15, 2021, 10:23:24 pm ---Quote from: msat on Yesterday at 07:47:48 pm
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.
--- End quote ---
Realize 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...
--- End quote ---
...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.
--- End quote ---
Perhaps I haven't understood what you are saying.
Equation 16-4-3 on page 718 of Kraus shows 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 R21 and X21 will both approach infinity zero*. 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 Kraus shows 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 R21 and X21 would be approximately halved doubled*. 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.
Navigation
[0] Message Index
[*] Previous page
Go to full version