A dipole which is exactly one half wave length allows reflections within the antenna to be in phase with the applied signal. To me it seems that this situation gives an antenna which will produce the most output for a given input signal.

If the antenna is not exactly one half wave length then the reflections won't be in phase exactly with the applied signal and the antenna output will drop because the reflections are subtracting from the applied signal for some duration of the waveform cycle.

So, in order to make the antenna resonant, wouldn't it be a better idea to not shorten a half wave dipole to remove the inherent inductive reactance but rather leave it at exactly half a wave length and cancel out the reactance with series capacitance ?


1 Answer 1


A dipole can be physically a half wave, or electrically a half wave.

A dipole that's physically a half wave has a nonzero reactance at the feedpoint. This means it is not resonant, and that the reflections within the antenna are not precisely in phase with the signal. (These are two ways of saying the same thing.)

The electrical length is determined by the time taken for the wave to traverse the antenna. Making the dipole a little shorter until the reactance is zero makes the dipole electrically a half wave. Now it is resonant.

Making the antenna resonant doesn't make it better per se, but since any low-loss feedline (like, any one you'd use in practice) has a negligible reactance to its characteristic impedance, an antenna that is not resonant (which means a non-zero reactance at the feedpoint) must necessarily be mismatched to some extent. A mismatch means additional SWR loss.

An antenna that is resonant is not necessarily matched to the feedline, or even at minimum SWR, unless it just so happens the feedpoint impedance and feedline's characteristic impedance happen to be the same. Resonant dipoles typically have an impedance around 75 ohms; bending the dipole into a "V" decreases the feedpoint impedance and it's usually possible to get it down to 50 ohms. So it's usually possible to match a dipole to 50 or 75 ohm coax without any additional matching devices, though you'd still want a balun in there.

Of course you could always match the antenna some other way, but most ways you might do it (like adding discrete capacitors or inductors) would introduce additional loss, usually more than is saved by reducing the mismatch losses.

In practice on HF, all this is more theoretical than practical because the feedline losses, and thus the mismatch losses, are low. Any SWR under 1.6:1 is within the specifications of most transmitters and "plenty good enough" for most HF applications. The advantage to twiddling with the dipole shape to get closer to 50 ohms is really just that you then have a wider band where you're under 1.6:1.

  • $\begingroup$ What is the speed at which an electromagnetic wave travels down an antenna? Inside a conductor, electromagnetic waves propagate very, very slowly relative to c (en.wikipedia.org/wiki/Speed_of_electricity), but since the EM wave travels along the surface near the conductor, it's much faster. Typical free space half wave dipole is 5% shorter than a physical half-wavelength to bring it to resonance, so does that mean light travels 5% slower around the conductor? $\endgroup$ Dec 17, 2019 at 19:58
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    $\begingroup$ @SterlingN0SSC Yep. But don't fall into the trap of thinking the wave is traveling just immediately around the conductor. The "wave" we're talking about is a wave in the electromagnetic fields, and those exist in all the space between the arms of the dipole: a volume of space approximately one or two wavelengths in radius around the antenna (with the exact value depending on just how you define the transition to "free space") $\endgroup$ Dec 17, 2019 at 21:33
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    $\begingroup$ @Andrew Resonance is usually easier to match with low loss, yes. If the antenna isn't resonant that means reactance is non-zero, which means there must be some amount of circulating reactive power. With real components, dealing with that reactive power must involve some loss, because although that reactive power doesn't go to radiation, it still involves current and voltage, and is thus subject to ohmic and dielectric losses, among other things. And also you have to spend money to make a matching network. $\endgroup$ Dec 17, 2019 at 21:37
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    $\begingroup$ @Andrew anyone can downvote, and I have no control over them. I'm just an ordinary user, just like you. $\endgroup$ Dec 17, 2019 at 22:29
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    $\begingroup$ @Antonio51 I'm not really sure what you're asking, but perhaps it is best to do it as a new question so you can get proper replies? You can always link back to this one, for context. $\endgroup$ Jul 10, 2021 at 19:22

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