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A half wave center-fed dipole has a resonant frequency where the input impedance appears to be purely resistive. How can this be, when the voltage and current distribution along the length of a half wave dipole when fed with an AC waveform of the resonant frequency are seemingly not in phase and in fact appear to be 90 degrees out of phase?

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  • $\begingroup$ I think the animation is not completely correct. 1. The voltage is not zero in the middle, but it is perhaps 10x smaller than the tips, so doesn't show. The current at the feedpoint is in phase, at resonance, by definition. This doesn't mean that the voltage at the tips is in phase with the current at the centre. $\endgroup$ – tomnexus Oct 27 '18 at 0:30
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Andrew, a dipole is a standing wave antenna. That means that the energy existing on the dipole that hasn't been radiated is in standing waves which do not change phase. The equation for a standing current wave is $I(x,t) = I_{\max} \sin(kx) \cos(\omega t)$. The distance $x$ determines the magnitude of the standing wave, not the phase. Only time determines the phase. At any instant in time, the phase of the standing wave is the same all up and down the length of the antenna. At resonance, the standing wave voltage and standing wave current are everywhere in phase. The forward current and reflected current are coherent phasors of close to equal magnitudes rotating in opposite directions. It's obvious that their sum would have constant phase. Same is true for the forward and reflected voltage phasors.

Phase has two different meanings for this context. The amplitudes of the voltage standing wave and current standing wave are out of phase in time but the phase of the voltage standing wave and current standing wave are in phase, i.e. equal in time.

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The phase and current of the r-f energy along each half of a dipole are the result of natural laws, which show that current must be at/near zero at the far ends of the dipole while voltage there must be maximum.

The practical concern to the user of that dipole is its self-impedance at its feedpoint terminals.

A thin-wire, center-fed dipole that physically is 1/2-wavelength long has a feedpoint self-impedance of 73 +j43.5 Ω in free space (source: Kraus' Antennas, 3rd Edition, p. 446).

To achieve resonance (where the voltage and current at its feedpoint are in phase and the "j" term of its impedance is zero), that dipole must be shortened a few percent, which somewhat reduces that 73 Ω radiation resistance — as a function of the O.D. of the radiating conductors of that dipole.

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  • $\begingroup$ Hi Richard thanks for the reply. But the part I don't understand is that if the dipole is shortened, and the reactive component is then zero, the current is still maximum in the center and zero at the ends, and the voltage is zero in the center and maximum at the ends, and the two are not in phase. I think i'm getting phase in the time domain and voltage and current distribution along the length of the wire mixed up ... $\endgroup$ – Andrew Oct 26 '18 at 21:56
  • $\begingroup$ I think the answer is this : The AC voltage and current in the time domain are actually in phase at resonance at the feed point, so if you drew a graph of time versus voltage and current amplitude when measured at the feed point you would see that they are in phase. However the voltage and current distribution along the wire are actually standing waves which are set up because the length of the antenna is such that the applied AC voltage reinforces that which is reflected from each end, and these have nothing to with the phase of the voltage and current.over time, Is that correct ? $\endgroup$ – Andrew Oct 26 '18 at 22:42
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    $\begingroup$ One certainty is that as far as its characteristics as a dissipator of r-f power, a resonant antenna can be replaced by a pure, physical resistor of the same ohmic value, wherein current and voltage always are in phase (unity power factor). The mechanism whereby the e-m fields radiated by the antenna into space produce that radiation resistance is complex. I'll do some research on that, but maybe other readers here have a good, short explanation to post as to how/why that occurs. $\endgroup$ – Richard Fry Oct 27 '18 at 16:01
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nice to get back to communicate with you.

In my file "Displacement of Resonance Frequency", I wrote that theoretical statement, but it was already corrected to avoid confusing readers.

Reality is how you explain it, when the physical length of the antenna is in resonance with the frequency, the current wave distributions in the antenna are in phase and the RF voltage waves are also in phase.

This condition in the antenna, determines its best radiation resistance and maximum efficiency of it.

To this answer, I added the updated link of the file "Displacement of Resonance Frequency.pdf", which unfortunately has not yet updated its complement file "Coaxial Cable Length and Node.pdf".

This file was written with the purpose of clarifying a reality, typically considered as an old myth and the scarce information that links themes of resonance lengths in optimization processes of antenna systems.

I know that should not include links, because they probably will not work, but if required in the future, it is possible to locate it from the web.

Greetings.

Displacement of Resonance Frequency.pd: https://files.acrobat.com/a/preview/7021823f-839c-4122-bdb7-c41eede5891d

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Let me try and answer this myself as simply and directly as possible.

This picture from Wikipedia shows the voltage and current distribution along a resonant half wave dipole.

https://en.wikipedia.org/wiki/Dipole_antenna#/media/File:Dipole_antenna_standing_waves_animation_461x217x150ms.gif

For transmit, the picture is showing standing waves which are present on the dipole which are caused by the interaction of the applied voltage at the feed point and the resultant voltage reflected from the ends. The standing waves are just like the ones which exist on coax when the antenna impedance doesn't match. The amplitudes of the voltage and resulting current standing waves are fixed in space along the length of the dipole. If you could connect an AC meter to the dipole without affecting it's operation (which you can't), you could actually measure these voltages and currents because they are fixed and don't change. So the standing waves have a varying amplitude in space along the length of the dipole, and the particular amplitude at each point doesn't change..

In contrast, the phase relationship in time which exists between the amplitudes of the existing voltage and resultant current wave forms at any point along a resonant half wave dipole is zero. To re-phrase, the AC voltage and current are in phase (in time) everywhere along the dipole because the dipole has zero reactance which makes it resonant.

It's easy to mix up the idea of standing voltage and resultant current waves in space along the length of the dipole, and the phase relationship in time between the applied voltage and resultant current which is zero for a resonant antenna.

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