Author Topic: Phase diff between tx and rx antennas spaced by whole multiples of wavelength  (Read 3612 times)

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Offline msat

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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
 

Offline TimFox

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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.
 

Offline msat

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@TimFox

I 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 happens electrically at 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?
 

Offline msat

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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.
 

Offline radiolistener

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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?

No. 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.

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.

No. 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. :)
« Last Edit: September 08, 2021, 05:59:11 am by radiolistener »
 
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Offline Rod

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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.

True, and true.  But don't confuse any changing quantity with the force driving 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/dt2 is proportional to the force on it, and is at a maximum, but is in the opposite direction to x.
It is not static.  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 the opposite direction 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?
« Last Edit: September 08, 2021, 06:29:28 am by Rod »
 
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Offline msat

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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/8spopwnkh2
In the above link, when you press the play button next to the variable h, 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/doym28qua7
The 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.
 

Offline Rod

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Your illustration of standing wave is very nice. 

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.

Correct... but only if 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 an infinitely long wire (one far longer than the wavelength) does not act as a receiving antenna: it would reflect everything and receive nothing.  It would create a standing wave, exactly as you illustrate.  How does it do this?

A key fact you've failed to consider: current does not flow 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 you can figure out how antennas work.  Antennas work as antennas only if they are 1/4 wavelength long (or an odd multiple thereof, as described here https://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 will not act as an antenna, but as a reflector.  Each 1/4 wavelength along it, at each instant in time, the external EM field attempts to induce an opposite voltage, and an opposite current 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 in both directions, and at the speed of light, exactly cancelling one 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 are current nodes, and voltage anti-nodes.  It is not out of phase with the external EM field, it is in phase with 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 are not acting as reflectors, but as in-phase re-radiators.  But one may appear to act as a reflector if it is spaced 1/2 wavelength behind the transmitting antenna.  (At a receiving antenna, they may appear out-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 here https://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 reciprocity https://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. 
« Last Edit: September 21, 2021, 06:18:13 am by Rod »
 
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Offline TimFox

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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 can almost be described with classical physics.
 
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Offline Marco

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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.
« Last Edit: September 11, 2021, 09:59:00 am by Marco »
 
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Offline radiolistener

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Re: Phase diff between tx and rx antennas spaced by whole multiples of wavelength
« Reply #10 on: September 12, 2021, 06:29:09 am »
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.

No, 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.

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.

re-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.

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.

I 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 film and 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°) = 0

2) 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 = 0.5

As you can see, it's pure waveform effect. And there is no needs to involve quantum effects to describe that.  :)

« Last Edit: September 12, 2021, 08:21:26 am by radiolistener »
 

Offline Rod

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Re: Phase diff between tx and rx antennas spaced by whole multiples of wavelength
« Reply #11 on: September 13, 2021, 12:22:22 pm »
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.

re-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.

Reflection, whether off a mirror or off the shorted end of a transmission line, occurs instantaneously.  There is no time 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 called fluorescence.  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!   :palm:)


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: its permittivity and permeability (or, equivalently for waveguides or cables, their capacitance and inductance per unit length).  Their product determines the speed of propagation.  Their ratio determines 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 with no time 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.

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.

I 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 film and 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.

Dirac first posed the "three polarizer paradox" in 1930.  Here's a discussion https://www.informationphilosopher.com/solutions/experiments/dirac_3-polarizers/   
« Last Edit: September 21, 2021, 06:21:30 am by Rod »
 

Offline ejeffrey

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Re: Phase diff between tx and rx antennas spaced by whole multiples of wavelength
« Reply #12 on: September 13, 2021, 04:13:02 pm »
We can't observe single radio frequency photons.  Or even microwave photons.  The reason is our antennas are utterly swamped beneath a sea of them generated within the antenna itself by Johnson noise, also known as blackbody radiation.  We can only observe the superposition of enormous numbers of them, veritable tidal waves of them, large enough to overwealm this thermal noise.  They exhibit wave behavior.

Actually we *can* observe single microwave photons although it is much harder than observing single optical photons.  For instance MKIDs https://web.physics.ucsb.edu/~bmazin/mkids.html.
 

Offline ejeffrey

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Re: Phase diff between tx and rx antennas spaced by whole multiples of wavelength
« Reply #13 on: September 13, 2021, 04:33:00 pm »
:wtf:  Well, then, why has this "three polarizer effect" never, in the 134 years since Hertz, been seen by anyone working with radio or microwaves?

You 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.

Quote
Might it be... :-//  because polarizers simply don't "consume and re-emit a new EM wave"?   It appears not!

Dirac first posed the "three polarizer paradox" of light in 1930.  Here's a discussion https://www.informationphilosopher.com/solutions/experiments/dirac_3-polarizers/   Light simply does not act like a classical wave.

Light doesn't exactly act like a classical wave but it is extremely close.
 

Offline radiolistener

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Re: Phase diff between tx and rx antennas spaced by whole multiples of wavelength
« Reply #14 on: September 14, 2021, 09:44:04 am »
Reflection, whether off a mirror or off the shorted end of a transmission line, occurs instantaneously.  There is no time delay. 

Time 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.

If there were any fixed time delay, it would result in a shift in phase.

Yes, and any mirror reflects light with some phase delay.

If there were any random time delay, it would result in a loss of phase coherence.

The 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.

No "absorption, delay, and re-emission" occurs.

The 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.

Reflection and refraction are instantaneous and are predicted by classical wave equations.

No. 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. :)
« Last Edit: September 14, 2021, 02:02:32 pm by radiolistener »
 

Offline msat

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Re: Phase diff between tx and rx antennas spaced by whole multiples of wavelength
« Reply #15 on: September 14, 2021, 05:50:39 pm »
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.

Your illustration of standing wave is very nice. 

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.

Correct... but only if 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 an infinitely long wire (one far longer than the wavelength) does not act as a receiving antenna: it would reflect everything and receive nothing.  It would create a standing wave, exactly as you illustrate.  How does it do this?

A key fact you've failed to consider: current does not flow 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 you can figure out how antennas work.  Antennas work as antennas only if they are 1/4 wavelength long (or an odd multiple thereof, as described here https://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.)

It 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.

[attachimg=1]
(modified image taken from https://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.

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And if you accept this reality, you can also come to understand why an infinitely long wire will not act as an antenna, but as a reflector.  Each 1/4 wavelength along it, at each instant in time, the external EM field attempts to induce an opposite voltage, and an opposite current 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 in both directions, and at the speed of light, exactly cancelling one 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 are current nodes, and voltage anti-nodes.  It is not out of phase with the external EM field, it is in phase with it.

As far as I'm aware, I wasn't in disagreement of you on this.


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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. 

To 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.

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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 are not acting as reflectors, but as in-phase re-radiators.  But one may appear to act as a reflector if it is spaced 1/2 wavelength behind the transmitting antenna.  (At a receiving antenna, they may appear out-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 here https://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 reciprocity https://en.wikipedia.org/wiki/Reciprocity_(electromagnetism), reflecting its consequences.)

All receiving antennas would experience the same effect I'm trying to describe, and thus all other effect still hold true best I can tell.

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The "three polarizer paradox" does occur with light.  It is one proof that light does not propagate purely as a wave.  It is somehow localized... as if it were propagating as a particle, not as a wave.  Sorry, but that cannot be rationalized by misunderstanding waves, nor how antennas work. nor by confusing antennas with their opposite: reflectors.


It 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 macroscopic https://sciencedemonstrations.fas.harvard.edu/presentations/microwave-properties

PS
How the heck do I embed the uploaded images!?
« Last Edit: September 14, 2021, 05:54:41 pm by msat »
 

Offline msat

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Re: Phase diff between tx and rx antennas spaced by whole multiples of wavelength
« Reply #16 on: September 14, 2021, 06:36:37 pm »
I have to apologize.  I was so distracted into attempting to clarify misunderstandings on how antennas, reflectors etc work that I missed your underlying goal:

My question actually stems from what I consider to be a completely unsatisfactory classical description of "simple" linear polarizing filters.   ..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.

What's "unsatisfactory" about it, exactly?

There'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 (see https://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).

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msat, it appears you don't understand why the "three polarizer paradox" is a paradox: it is irreconcilable with how waves work.  This paradox only appears when we start observing single photons.  It doesn't appear when we're observing waves.

For the time being, I respectfully disagree  :)

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We can't observe single radio frequency photons.  Or even microwave photons.  The reason is our antennas are utterly swamped beneath a sea of them generated within the antenna itself by Johnson noise, also known as blackbody radiation.  We can only observe the superposition of enormous numbers of them, veritable tidal waves of them, large enough to overwealm this thermal noise.  They exhibit wave behavior.

The 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.


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That you'd propose to devise an experiment in which waves do not act like waves, well, reflects a fundamental misunderstanding of how waves work.  I encourage you to continue to pursue understanding waves; it's very useful.

What 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?

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(By the way, QM is not insane, nor is it sane.  It is counterintuitive, but it's simply the way mother nature works.  It proves our intuition is flat wrong.  Physicists were forced to this realization a century ago.  We can learn from them.  Attempting to cling to intuition in the face of proof it's wrong... well, that wouldn't be sane, but is perquisite for the Flat Earth Society, "free energy" suppression and countless more conspiracy theories.  No one in the EEVblog community is going there.  Not because of any need for conformity here.  Simply because: it don't work.)

I'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.
 

Offline msat

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Re: Phase diff between tx and rx antennas spaced by whole multiples of wavelength
« Reply #17 on: September 14, 2021, 06:40:56 pm »
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.

Could you elaborate on this a bit more? I'm pretty sure we're talking the same thing here.
 

Offline msat

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Re: Phase diff between tx and rx antennas spaced by whole multiples of wavelength
« Reply #18 on: September 14, 2021, 06:58:07 pm »


I 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 film and 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°) = 0

2) 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 = 0.5

As you can see, it's pure waveform effect. And there is no needs to involve quantum effects to describe that.  :)

By 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.
 

Offline msat

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Re: Phase diff between tx and rx antennas spaced by whole multiples of wavelength
« Reply #19 on: September 14, 2021, 07:47:48 pm »
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 thinking



It 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.
 

Offline Marco

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Re: Phase diff between tx and rx antennas spaced by whole multiples of wavelength
« Reply #20 on: September 15, 2021, 09:37:19 am »
Why no resistor in the middle of the receiver?
 

Offline msat

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Re: Phase diff between tx and rx antennas spaced by whole multiples of wavelength
« Reply #21 on: September 15, 2021, 11:08:25 am »
Why no resistor in the middle of the receiver?

Originally 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.
« Last Edit: September 15, 2021, 11:12:09 am by msat »
 

Offline Marco

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Re: Phase diff between tx and rx antennas spaced by whole multiples of wavelength
« Reply #22 on: September 15, 2021, 01:10:56 pm »
Shrug, you are not measuring what you asked about in the original post then. A receiver antenna is terminated, not a resonator.
 

Offline msat

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Re: Phase diff between tx and rx antennas spaced by whole multiples of wavelength
« Reply #23 on: September 15, 2021, 04:12:47 pm »
Shrug, you are not measuring what you asked about in the original post then. A receiver antenna is terminated, not a resonator.

What 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.
 

Offline Rod

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Re: Phase diff between tx and rx antennas spaced by whole multiples of wavelength
« Reply #24 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.


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...

Heinrich Hertz first demonstrated the electromagnetic waves predicted by Maxwell https://en.wikipedia.org/wiki/Heinrich_Hertz#Electromagnetic_waves using 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. 
« Last Edit: September 15, 2021, 10:40:55 pm by Rod »
 


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