Everyone knows that a 1/2 λ dipole antenna has on it a standing wave of RF energy the voltage and current of which are about 90° out of phase with each other. The standing wave is circulating reactive stored energy present due to the fact that the antenna is a resonant system.

Apparently there is a departure of phase difference away from 90° which is the in-phase component of the standing wave responsible for radiation, the out of phase energy of the standing wave remains in the antenna.

At the ends of the antenna where there is a complete open circuit, how can the phase difference between voltage and current of the standing wave be anything else other than exactly 90° ? But there must be some non-reactive component otherwise there will be no radiation !

So, does the phase difference between voltage and current of the standing wave start off as exactly 90° at the outer ends of the elements, and then gradually change to a lesser value along the antenna elements with a minimum value at the current maximum at the feed points, in which case the standing wave must not be a perfect sine wave ? or is the phase difference the same everywhere on the antenna, and so the reflection at the ends must not be exact ?

Moderator Note: The same question was asked here at Amateur Radio.SE so interested readers may want to visit that page too. To avoid any appearance that this situation is the norm, duplicating questions across different SE sites is generally strongly discouraged.

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    \$\begingroup\$ Nice question. There is also a further question: if there is a 90 deg phase difference, how come the TEM wave has its E and H back in phase? \$\endgroup\$ Aug 24, 2021 at 9:47
  • \$\begingroup\$ @MarkoBuršič Is that the question no one knows the answer to ? \$\endgroup\$
    – Andrew
    Aug 24, 2021 at 9:49
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    \$\begingroup\$ Not me, I have asked myself this question many times. You have a generator, transmission line, antenna - all perfectly matched. The generator and line have V,I in phase, then antenna has them out of phase generating EH filed out of phase, after they become perfectly in phase, kind of black magic. The real knowledge would be a method to keep the EM wave out of phase for a long distance. \$\endgroup\$ Aug 24, 2021 at 10:15
  • \$\begingroup\$ The standing waves represent the amplitudes of voltage and current at points along the antenna, they are not the voltage and current signals as functions of time. So referring to a phase angle between voltage and current standing waves is not correct. \$\endgroup\$
    – Chu
    Aug 24, 2021 at 15:25
  • \$\begingroup\$ @chu Thanks for your comment but what you just said is completely wrong, there is a absolutely a phase difference of a bit less than 90 deg between voltage and current of the standing wave on a dipole, this is explained in just about every antenna book in existence. [Edited by a moderator.] \$\endgroup\$
    – Andrew
    Aug 24, 2021 at 21:58

1 Answer 1


The whole concept to think the branches of the dipole together as a transmission line is a big simplification because there's the radiation. That does not happen in a transmission line. But even that simplification should show the direction to you. The radiation can be considered as loss. If you calculate the standing wave (or actually total voltage and total current magnitudes as functions of the distance from the end) on a lossy open ended line you see that the current and voltage are 90 degrees out of phase only at the open end, the differece decreases towards the feeding point because the reflected wave weakens and the incident wave becomes stronger.

  • \$\begingroup\$ Thanks for that answer, this is my opinion also, i just wanted to get a few second opinions. I like the idea of referring to radiation as a loss because that's what it actually seems to be ! \$\endgroup\$
    – Andrew
    Aug 24, 2021 at 9:14
  • \$\begingroup\$ The next question is, why ? \$\endgroup\$
    – Andrew
    Aug 24, 2021 at 9:19
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    \$\begingroup\$ why what? The phase shift deviation from 90 degrees doesn't cause radiation nor other losses, the phase shift other than 90 degrees between current and voltage is the consequence of how the electric and magnetic field happen to settle into the given geometry from the given excitation. This metal geometry is used because from there the wave jumps partially also into the free space, it doesn't only reflect forth and back in a limited area. \$\endgroup\$
    – user136077
    Aug 24, 2021 at 9:48
  • \$\begingroup\$ So if the departure in phase from 90 deg doesn't cause radiation what does ? Everyone seems to think that this in-phase non-reactive component is that which results in the radiation. And the why what is why the electric and magnetic field happen to settle into the given geometry from the given excitation ? \$\endgroup\$
    – Andrew
    Aug 24, 2021 at 9:53
  • \$\begingroup\$ The phase shift deviation from 90 degrees is caused by loss, the loss is not caused by the deviation. The electromagnetic wave occurs as soon as someone causes with some method an accelerating change to electric or magnetic field somewhere. Sinusoidal voltage source or a sinusoidally oscillating circuit is a good method because the generated wave has the simplest possible spectral content. The signal which arrives along the transmission line is already a wave in the space around the wires, the parallel wires only direct its propagation along the line. Antenna lets it spread to the space. \$\endgroup\$
    – user136077
    Aug 24, 2021 at 10:47

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