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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
No. 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 polarization which is different from input wave. This is clearly explained in the text from your link:Quoteone can show that the radiation transmitted by the filter is polarized in a new direction which depends on the orientation of the grid
And 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.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.
As said before, at wavelength distance it will be exactly the same as on transmitting antenna.
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@Rod
I'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).
@radiolistener
With 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. -
Attached is a graphic I made to describe the effect I'm positing
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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!
180 degree out of phase is phase delay = 180 degree.Your experiment would not work due to reasons I already outlines.
Are 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.Attached is a graphic I made to describe the effect I'm positing
your 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). -
your 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).
This is the entire point 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. -
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?!
When 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.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.
I 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.Just telling me something is some way does not constitute proof.
If you're needs proof, just build test setup and get it. It's not hard. -
@radiolistener
I 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?
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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 cannot argue against QM, because there is no paradox and 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.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.
Just 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.But that voltage will cause current to flow, right? Which way will that current flow?
For 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.
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You cannot argue against QM, because there is no paradox and no needs to involve QM to explain such behavior, because this is classic wave behavior.
I 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.QuoteFirst 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.
Once 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).Quote
For 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.
Once 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. -
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 does not travel 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 a square wave, 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? -
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).
No. 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 conclusionsIn 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.
Receiver antenna works exactly the same as transmitting antenna. It just works in opposite direction, but all things exactly the same. -
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 does not travel 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 a square wave, 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?
Thanks, 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. -
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.
No. 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. -
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.
No. 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.
You 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! -
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.
your 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. -
your 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.
An 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 the current through 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, the voltage across 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 is opposite the sign of I through it. Any power source attached to a resistor R exhibits a negative resistance -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, but helpful. 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), or
2) 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, very simplistic 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. -
We are discussing two half-wave dipole antennas, spaced N wavelengths apart, N = 0,1,2,3...
yes, exactly. And phase offset due to wave propagation delay in the space between these antennas will be N * 360 degree = 0 degree.However, the voltage across the terminals of the transmitting and receiving antennas are anti-phase, 180 degrees out of phase, with one another.
Now 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.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.
Transmitter 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: [Select]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)));
where
R - 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 length
r = conductor radius
k = wave number (k = 2 * pi / lambda)
Ci(x) - integral cosine
Si(x) - integral sine
EulerGamma - 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?
1) probe the voltage on the ends of the transmitting and receiving antennas (which we know will be in-phase), or
2) probe the voltage on their terminals (we know will be anti-phase, but he doesn't, so he will be misled).
In 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? -
radiolistener, you have entirely missed a very basic concept. Please review Kirchkoff's law https://en.wikipedia.org/wiki/Negative_resistance#Negative_static_or_%22absolute%22_resistance Please think carefully about the very simple figure shown there, of a battery in series with a resistor, captioned "From KVL, 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, the current across the terminals of the transmitting and receiving antennas are in-phase. However, the transmitter (R<0) has an output voltage that increases that current, while the receiver input resistance (R>0) produces a voltage which decreases that current. The voltage across their terminals are 180 degrees out of phase.
This is KVL. It is true of any power source and load. It is not unique to antennas. It is much simpler than you think.
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Repeating, the current across the terminals of the transmitting and receiving antennas are in-phase. However, the transmitter (R<0) has an output voltage that increases that current, while the receiver input resistance (R>0) produces a voltage which decreases that current. The voltage across their terminals are 180 degrees out of phase.
If 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). -
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 the same direction in both the battery and the resistor. However, the voltages across the battery and the resistor are also identical but are in opposite directions.
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 the opposite of 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 dissipating negative power; it is a power source!
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 the opposite direction as the voltage across a resistor with the identical current flowing in the identical direction through it.
If there is anything above you disagree with, please say so!
If not, we can apply it to any circuit. In fact, it must apply in any closed circuit (any circuit in which the total energy is conserved).
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Observe identical current is flowing the same direction in both the battery and the resistor. However, the voltages across the battery and the resistor are also identical but are in opposite directions.
The 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.If there is anything above you disagree with, please say so!
When 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.
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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.
Is that what KVL means? Does KVL only apply to circuits which do not contain an EMF source? Does KVL not apply 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!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.
Really?! That's quite remarkable. But... you animation does show 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 be leading the phase of the voltage on the end of each dipole by 90 degrees. It appears to be driving the current... adding power to the antenna. I wonder if this is also true if we attach a receiver with positive impedance to the terminals... removing power from the antenna... or might the opposite then 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. -
Does KVL only apply to circuits which do not contain an EMF source?
No. 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.Does KVL not apply to a circuit loop containing a battery?
No. 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.Please explain your understanding of KVL to me. (It appears to be different than that in any textbook?) Thank you!
My understanding is that algebraic sum of all voltages in a loop must be equals to zero. Is your understanding is different?Really?! That's quite remarkable.
Yes, 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.
But... you animation does show a voltage across the antenna terminals.
Don'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. -
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?!)?
This 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 Z21 between 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. Kraus shows almost what we need. R21 and X21 are 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 of equation 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 (Z11 and 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/r3 mode 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.
The consequence of this coupling is that the 1/r mode leads the 1/r2 mode 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 +/- 90o would require almost complete detuning of the parasitic element, rendering it worthless.
Section 11-9 starting on page 476 of Kraus analyses 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 +/- 60o could be achieved. So with respect to its 1/r mode, a practical director element (shorter than the resonant length) could get to within -30o to -45o of being in phase with the received signal. Similarly, a practical reflector element could get to within +30o to +45o of being in anti-phase with the received signal.
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Really?! That's quite remarkable.
Yes, 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.
Of 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.
All any directional 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 in https://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.
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.
What??? Please explain. Are you trying to explain a quarter-wave monopole antenna? Or did you intend to write "You can just short the 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.