electromagnetic transmission power relativistic effects

If we're talking "traditional" antennas that can be modeled with regular old R's, L's, and C's, I believe the answer is, "yes, absolutely."

If you add in things like anisotropic or non-linear (e.g., saturating) or semiconductor materials, I'd speculate that you might be able to get a "no" because it might be impossible to construct a scenario where they're "equivalently excited." E.g., if you have some lossy and saturable material, but the loss is a function of position within the antenna, if you illuminate that antenna externally with an EM field, you might be able to saturate it and while it will produce some output voltage/current, feeding that same voltage/current back into it might not necessarily saturate it (since the losses are heavier closer to the electrical terminals) and hence might produce a different resulting EM field.

This is probably not that great of an example -- and it's just speculation on my part anyway. But for something simpler -- which isn't an antenna -- if you take a transformer with a different number of windings on the primary vs. secondary, you might find that exciting the primary saturates it, and while there will still be *some* voltage on the secondary, if that secondary has a greater number of turns, feeding that same voltage back into the secondary won't necessarily saturate it and hence the voltage at the primary won't necessarily match the initiation excitation.

The concept we're talking about here -- "reciprocity" -- actually only says that, in a circuit, if you swap the position of a voltage source and ammeter, the ammeter will read the same and the voltage source will provide the same current as in the original circuit. However, the various node voltages and branch currents typically are *not* the same in both cases. In an antenna system, those node voltages/branch currents would presumably represent the EM field in space inbetween the two antennas, so I wonder if that implies that swapping Tx and Rx antennas on a system only means that, while the receiver antenna does still end up with the same signal strength, is the EM field itself perturbed in the process?

BTW, one interesting aspect of antennas is that they don't have to obey Foster's reactance theorem, that is, despite being passive, the impedance of an antenna does not have to monotonically increase with frequency. On a network analyzer, antennas are one of the few (maybe only?) passive devices where the trace on a Smith chart with respect to frequency can sprial in a counterclockwise direction!

The electrons don't leave the active element. Well, OK, maybe in a really high-powered system a few of them do due to the very high field strengths involved, but in general you just need to move electrons around to generate radiation; no removal of them from a conductor is required.

(And when you hook an antenna up to, e.g., a transmission line, since we're talking AC here anyway, the electrons just sit there and wiggle back and forth in sync with the RF anyway; energy is transferred, but there's no net motion of the electrons.)

An (AC) current will begin to flow. When one says that Tx and Rx antenna behave the same, an implicit assumption is that the terminal conditions are the same as well.

You get coupling and the results can rapidly become difficult to predict. However, as you're probably aware, antennas such as log periodics are specifically designed to take advantage of this coupling so as to generate currents in the reflector and director elements, thereby forming a phased array that focuses antenna pattern.

BTW, the guys over at rec.radio.amateur.antenna have some *very* savvy antenna guys within their ranks -- guys like Cecil Moore and Ray Lewallen; you might want to post your question there as well. (Wim Telkamp, who posts here at times, also posts there.)

---Joel

t.net ---

You need to understand the concepts of near field and far field in radio transmission.

To quote the wikipedia

"The "far-field", which extends from about two wavelengths distance from the antenna to infinity, is the region in which the field acts as "normal" electromagnetic radiation. The power of this radiation decreases as the square of distance from the antenna, and absorption of the radiation has no effect on the transmitter. By contrast, the "near-field", which is inside about one wavelength distance from the antenna, is a region in which there are strong inductive and capacitative effects from the currents and charges in the antenna, which do not behave like far-field radiation. These effects decrease in power far more quickly with distance, than does the far-field radiation power. Also, absorption of radiated power in this region has effects which feed-back to the transmitter, increasing the load on the transmitter that feeds the antenna by decreasing the antenna impedance that the transmitter sees. Thus, the transmitter can sense that power has been absorbed from the near-field zone, and if this power is not absorbed, the transmitter does not draw as much power."

I am afraid I do not possess the tools to refute your post. All I can say is that there are many books out there that support what I have posted. I did not dream this up by myself.

Please look up John D. Krause (deceased) to examine his credentials.

Cheers, John

I don't think that this is important to your statement, but according to the book I referenced earlier, page 36:

"Although the radiation resistance, effective aperture, and directivity are the same for both receiving and transmitting, the current distribution is, in general, not the same. Thus, a plane wave incident on a receiving antenna excites a different current distribution than a localized voltage applied to a pair of terminals for transmitting."

Cecil is still a contributor but Roy Lewallen no longer posts there. Wim Telkamp is an excellent source of knowledge.

John

Sorry, I mispelled his last name. It is Kraus.

John

[deletia]

I'll definitely defer to your expertise here, Phil, but tell us... would you consider Kraus's quote up there Just Plain Wrong, or more like... unclear/unqualified/etc.?

(You can be honest with us; Kraus unfortunately is dead so he won't be offended!)

What would that physically look like if you were trying to build a more-efficient-than-a-dipole antenna for, e.g., some GHzish range radio receiver antenna?

Thanks,

---Joel

P.S. -- Kraus's semi-autobiographical book, Big Ear Two, is definitely worth a read! I liked it so much that, years ago, I bought a copy for my then-fields-and-waves professor at school as well.

Let us know when you get around to writing your autobiography... :-)

There are a lot of wrongheaded generalizations made by experts that get misunderstood and then passed on uncritically. A couple of my love-to-hate examples are folks saying that you should run photodiodes at zero bias and that op amp TIAs are the best you can do, which are both so ludicrously wrong that they (almost) deserve the Phil Allison treatment.

If you restrict the discussion to small wire antennas illuminated by plane waves, there is bound to be a lot of reflection, but the physics of that is the mode mismatch. No mismatch, no reflection.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal
ElectroOptical Innovations
55 Orchard Rd
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845-480-2058

email: hobbs (atsign) electrooptical (period) net
http://electrooptical.net

Well, let's see. You admit that there is reradiation taking place. Is it possible to reradiate power if there is no radiation resistance (which, by the way is not normal resistance. It is the apparent resistance caused by radiating energy away from the wire)?

You're ignoring the phase relationships between the scattered and re-radiated fields. They don't have to add, they can also cancel.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal
ElectroOptical Innovations
55 Orchard Rd
Briarcliff Manor NY 10510
845-480-2058

email: hobbs (atsign) electrooptical (period) net
http://electrooptical.net

I don't like arguing with dead people--it looks too much like a cheap shot.

Isolated fields like antennas and lens design tend to develop a lot of conventional wisdom that's mostly true within some limited universe of possibilities, but that falls apart when you try applying it more generally. I have absolutely no doubt that Kraus and Balanis and folks like that would run rings round me when it comes to designing wire antennas.

A single pair of wires can interrogate only one mode of the electromagnetic field, which is a phase space volume of

A Omega = lambda**2/2.

where A is the intercepted area and Omega is the projected solid angle. Small wire antennas have equivalent areas much larger than their geometric cross-sections, which leads to some fairly strange properties.

A simple example is a Yagi. You put a reflector element behind the driven element, phased so that its scattered field cancels out the driven element's in the backward direction.

You can also make a mode-matching structure, such as a large paraboloidal reflector, that matches the incoming mode to the dipole or feedhorn. Another example is an Agilent waveguide-to-coax transition, which is basically a 1/4 wave whip antenna stuck down into the waveguide, 1/4 wave from the shorted end. Waveguides only have one propagating mode in each direction, so mode matching is much simpler.

Electrically large antennas are also easier to understand.

Cheers

Phil Hobbs

Now _that_ would be boring...I'd probably go to sleep writing it. ;)

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal
ElectroOptical Innovations
55 Orchard Rd
Briarcliff Manor NY 10510
845-480-2058

email: hobbs (atsign) electrooptical (period) net
http://electrooptical.net

[snip]

Sort of like arguing with Larkin ?:-)

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What mechanism would scatter the field from a 1/2W dipole? It is reradiating half of the watts/m^2 but that is not considered scattering.

If there is no scattering, there is no phase relationship. Since you do not accept Kraus and Balanis, I have no backing on this statement. Is there some author you would accept as gospel?

Cheers, John

Thanks Phil, that makes things a lot clearer.

Hmm... and apparently the scattering pattern off of a small dipole in the center of a dish is somewhat different than the radiation pattern of that same dipole, since otherwise the energy scattered would just head back towards wherever the initially incident radiation was to begin with, right (and we'd be back to ~50% efficiency)?

Better you than someone with nothing better to do who writes you up in Wikipedia some fine day!

---Joel

By whom?

Good luck making an antenna that doesn't scatter, at least with a mode-mismatched incident wave. The phase relationships are key.

Since you do

It isn't Kraus and Balanis I'm arguing with, it's you. I'm just pointing out a flawed generalization. Anybody can make them--certainly including me.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal
ElectroOptical Innovations
55 Orchard Rd
Briarcliff Manor NY 10510
845-480-2058

email: hobbs (atsign) electrooptical (period) net
http://electrooptical.net

BTW, Phil, I think we have drifted off the question anyway. I think Jamie wants to know if there would be a delay that would effect the transmitter *assuming there was reradiation back to the source* however minscule.

Jamie is calling into question quantum effects. Can you help us with that?

Cheers, John

But, Phil, I copied verbatim from Kraus' book! Do I need to scan and post it?

Cheers, John

We've probably flogged the cover off this dead horse. You and Kraus are quite right, a small wire antenna will scatter and (if you like) reradiate a lot.

The _argument_ based on the radiation resistance is pure nonsense, because it applies in exactly the same form to two coax cables, or a waveguide-to-coax transition, and neither of those reflects anything like 50%. (The same argument would tell you that half the radiated power was dissipated in the transmitter. 'Tain't so.)

There are a lot of those sorts of silly notions in electromagnetism. Another one is the idea that the Poynting vector (E cross H) always corresponds to energy flow.

Poynting's theorem is true, so the surface integral of (E cross H) dot n-hat over some closed surface does indeed give you the total power leaving the enclosed volume. However, it isn't always true locally.

As a counterexample, you can charge up a high voltage capacitor and put it next to a rare-earth magnet--you'll have a huge E cross H, but no power flow whatsoever.

I'm much readier to believe the math than the accompanying arguments.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal
ElectroOptical Innovations
55 Orchard Rd
Briarcliff Manor NY 10510
845-480-2058

email: hobbs (atsign) electrooptical (period) net
http://electrooptical.net

No. He assumed that a distant receiver would *absorb* power, and asked how the transmitter power supply would make it up. The answer is, it wouldn't.

As I pointed out, power that is absorbed isn't reflected.

John

I disagree that transmission lines (coax cables) and a half-wave dipole are the same. The characteristic resistance of a lossless transmission line is non-dissipative and energy can circulate in the line forever under appropriate conditions. The radiation resistance of a half-wave dipole is the resistance that represents emission from the dipole (that is, watts radiated and leaving the scene). There is no actual resistance in a perfect antenna. The process of having a wave move from the injection point to the end of the antenna and be reflected to return to the injection point gives rise to radiation that appears to be resistance.

They are not the same.

Cheers, John