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Let's assume we have a transmitter, a feed line, and a standing wave antenna (dipole, etc.). The transmitter is matched with the feed line, and the feed line is matched with the antenna, which is why the traveling wave in the feed line does not reflect from the antenna's feed point and passes without any obstacle. That is, there is no standing wave in the feed line, and this is good (radio amateurs appreciate this).

Now our antenna creates a standing wave. But if the wave doesn't see an obstacle when it moves towards the antenna, why doesn't the same happen in the reverse direction, when the wave goes from the tip of the antenna and reaches its base? Why does the wave bounce multiple times inside the antenna and why it doesn't move back into the feed line if there is a perfect impedance match at the base? Yet a standing wave is still created in the antenna. The question is, how?

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  • $\begingroup$ "the wave bounce multiple times inside the antenna" Why do you say that? Does it matter? When excited by a voltage, at some frequency, the antenna does its thing, consistent with the laws of physics, and at its terminals It looks like a 50 ohm resistor. You could ask the same question about a bandpass filter or resonant RLC circuit - the circulating currents don't matter when you draw a box around it, it just looks like a resistor. $\endgroup$
    – tomnexus
    Nov 6, 2023 at 14:08
  • $\begingroup$ @tomnexus that's fair, but it just begs the question "OK, so now what's happening inside the box, exactly"? $\endgroup$ Nov 6, 2023 at 14:14
  • $\begingroup$ @tomnexus ok the only thing I was curious about was why the impedance matching works in one direction only so to speak and why doesn't the wave travel back into the feed line the same way it got into the antenna in the first place... $\endgroup$
    – elgroovy
    Nov 7, 2023 at 20:23
  • $\begingroup$ @tomnexus I used the words "bounces around multiple times" to emphasize the fact that the wave is somehow trapped inside the antenna and cannot travel back into the feedline for some reason, which is what I'm trying to figure out... $\endgroup$
    – elgroovy
    Nov 7, 2023 at 20:33
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    $\begingroup$ As a transmission-line-and-antenna question, it's equivalent to think about a parallel RLC circuit, and easier to analyse. Your question could be: why with all the current bouncing around between L and C, does it just look like a resistor, and none of it comes out of the box or reflects back into the line? The answer is simple: At resonance (by definition) its impedance is purely resistive, so it "feels" from the outside like it's just the R. The L and C are invisible. There is still circulating current, perhaps a lot more than in the resistor, but it doesn't matter to the transmission line. $\endgroup$
    – tomnexus
    Nov 7, 2023 at 23:50

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Keep in mind that when you are talking about standing waves, you are talking about voltage. Also, a systems' feedline is containing the EM energy, and the antenna is shedding that energy.

With regard to your question: "Why does the wave bounce multiple times inside the antenna and why it doesn't move back into the feed line if there is a perfect impedance match at the base?", The wave doesn't bounce multiple times on the radiator, what is reflected is voltage, which is reduced by the material resistance; and in the case of a radiating element that is a 1/4 wavelength or odd multiple thereof, what returns to the feed-point is 180 degrees out of phase with the incoming voltage, and is cancelled out, and is what make the feed-point low impedance. In the case of a radiator that is a 1/2 wave, or multiple thereof, the reflected voltage which is now reduced, is in phase with the incoming voltage, and is what makes the feed-point high impedance, and again, the reflected voltage is reduced due to resistance losses.

A standing wave in the feedline should only occur if there is an impedance mismatch between the feed-line and the antenna, or if the radiating element(s) are not an electrically resonant length, i.e., some multiple of 1/4 wavelength. -At least that's the way I see it...

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Now our antenna creates a standing wave.

Nope, not a wave in the "propagating wave" sense. That's the fallacy here: there's not a wave propagating forth and back.

What is standing is indeed points maximum and minimum current; but that's not because there's a wave that gets reflected; it's due to the boundary conditions on the underlying current, E-field and H-field dictating that.

Consider "propagating, and reflecting EM waves" to be a phenomenon that arises from Maxwell's equations being solved by it under the given boundary conditions (i.e., material, geometry and excitation); but they're not what's happening here. The same equations are at play here, but the wave only exists as a result in the far field of the antenna; it's not identical thing as the current on the conductors of your dipole.

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  • $\begingroup$ ok that's interesting, thanks. But that also means the accepted answer for this other question is completely misleading: electronics.stackexchange.com/a/531835/205482 It basically says there are wave reflections inside antenna and standing wave as a direct result of that (wave interference pattern). So where is the truth? It's surprising that the answers are so different on such a fundamental topic of antenna theory... $\endgroup$
    – elgroovy
    Nov 7, 2023 at 20:53
  • $\begingroup$ From that answer "To be accurate, this image represents an antenna storing energy, not radiating". When the energy is radiated, it leaves the antenna. Resistive losses + radiated + reflected power = total power. $\endgroup$
    – webmarc
    Nov 8, 2023 at 13:46
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    $\begingroup$ yep, and that's why I used the term 'in the sense of "propagating wave"'; this is a voltage wave, alright, but it's not a propagating EM wave. $\endgroup$ Nov 8, 2023 at 14:45
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    $\begingroup$ In case you guys are curious: link to the question on Quora. You will need to change from "All Related" to "Answers" in the dropdown list to see all answers there. $\endgroup$
    – elgroovy
    Nov 9, 2023 at 12:47
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    $\begingroup$ @elgroovy what I would do: read the relevant standard literature that a friendly answerer linked to already; that would enable me to understand the answers.. In any case, if I would ask a new question somewhere else I'd very clearly link back to the question you asked in the first place, so that people would not duplicate efforts. $\endgroup$ Nov 12, 2023 at 14:51
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Don't forget: The antenna is a 2 port device:

  • power transferred from the transmission line to the antenna (#1), and
  • power further transferred from the antenna to free-space (#2).

The power transferred to free-space cannot then ALSO be kept to recirculate on the antenna... this would violate conservation laws and end the universe. The mods will appreciate that you do not do this.

When power is reflected back to the transmitter, that is indicative of an impedance mismatch somewhere in the system, preventing transfer of at least some energy thru the antenna system to free-space.

Hope this helps!

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  • $\begingroup$ Thanks for your answer. My point was merely to emphasize the fact that the wave is somehow trapped inside the antenna and cannot travel back into the feedline the same way it got into the antenna in the first place.. The perfect impedance match at the antenna feedpoint should work for wave travelling in both directions. This is what I'm trying to figure out $\endgroup$
    – elgroovy
    Nov 7, 2023 at 20:38
  • $\begingroup$ and I think that’s where the misconception is: it is not “trapped inside the antenna”! The energy sent in to the antenna is transferred to free space; it leaves the antenna. $\endgroup$
    – webmarc
    Nov 7, 2023 at 21:26
  • $\begingroup$ I understand that but I think what confuses me is the accepted answer to this other question electronics.stackexchange.com/a/531835/205482 It basically says there are wave reflections inside antenna and standing wave as a direct result of that (wave interference pattern). This also means it accumulates the energy besides working as a radiator and loses the accumulated energy over time by radiating it. Is this correct? $\endgroup$
    – elgroovy
    Nov 7, 2023 at 21:31
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    $\begingroup$ It's not correct. See this answer, especially, "Many radio antenna books show how the radiation field of a dipole is derived by assuming a standing wave like sinusoidal current distribution along the rods. That's ok engineering approximation for many practical purposes, but the exact current distribution along the dipole rods is a little different and it cannot be derived properly without solving the wave equations in the space around the rods." $\endgroup$
    – webmarc
    Nov 8, 2023 at 4:59
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Consider a single finite length wire, bent in an L shape, 1/4 wavelength of some given frequency F0 from the end. A current impulse, say driven by a single spark from a spark transmitter, will indeed be seen traveling in the reverse direction at the corner of the L at time roughly half the period of F0 later. However this "bounced" current impulse will be reduced, due to a portion "lost" due to wire resistance plus EM energy loss due to coupling into free space (or any nearby or distant conductors). This loss is usually less than 100% for typical wires of typical lengths in free space.

Now add another L shaped conductor nearby in a mirrored arrangement. A portion of the "bounced" current might flow as displacement current across the two L corners to the other 1/4 wavelength L stub, due to the mutual inductive field.

So some of the current (the portion not "lost" and not sent a displacement current) indeed flows back down the feedline, which, summed with any current send up the feed line, adds up to a total current to voltage ratio, which we can call the impedance of the feed point.

Now consider a sinusoidal drive of frequency F0 to be the sum of a bunch of these impulses (infinite in number, infinitesimal in size), modulated in amplitude by sin(F0).

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