Lots of explanations i've read on the internet written by ham radio operators and in some text books state that a half wave dipole is resonant when the inductive and capacitive reactances cancel out.

Exactly what is the cause for the zero reactance seen in the impedance at the center feed point of a resonant half wave dipole?

The image below is from Wikipedia and shows the voltage and current distribution for the standing wave which exists on a half wave dipole at resonance.

enter image description here

Resonance is determined by the fact that the length of each antenna element is exactly 1/4 of the wavelength of the applied signal, and so the wave reflected from the ends is exactly 360º behind in phase which is the same thing as being in phase with the applied signal and the two add together to produce a resultant waveform with larger amplitude.

Wikipedia describes this as "When an oscillating force is applied at a resonant frequency of a dynamical system, the system will oscillate at a higher amplitude than when the same force is applied at other, non-resonant frequencies."


My understanding is that at resonance the voltage of the standing wave which is 90 deg out of phase with the applied current is always zero at the feed point at resonance. For an antenna which is longer or shorter the zero crossing point no longer occurs at the feed point and then the feed point has a non-zero voltage which is out of phase with the feed point current.

So at resonance the voltage of the standing wave which is 90 deg out of phase with the current at the feed point is zero and it doesn't contribute any reactance to the feed point impedance.

Is this correct ?

See following some of the resources i used to try and find an answer to this question.

Chapter 5 of Practical Antenna Handbook by Joseph J. Carr Fourth Edition.







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    $\begingroup$ Related question with answers $\endgroup$ Nov 24, 2020 at 3:40
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    $\begingroup$ What's the question? $\endgroup$
    – Brian K1LI
    Nov 24, 2020 at 12:51
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    $\begingroup$ I suspect you're getting downvotes because it's unclear what you're asking. For the question in the title, this is a matter of definition: the frequencies where the impedance is nonreactive are defined as the resonant frequencies. But from all the text in the body it's clear you're asking something else, but it's not really clear what. $\endgroup$ Nov 24, 2020 at 14:58
  • $\begingroup$ @PhilFrost-W8II thanks for the comment, i am asking what am i missing ? Someone explain to me how inductive and capacitive reactances cancel at resonance because i cant see how this idea applies to a half wave dipole. I will happily accept that what i detail in my question is wrong if someone can explain why. The aim of the question is to determine exactly what causes the resonance and so zero reactance in the impedance of a half wave dipole. I changed the question title, maybe that will help. $\endgroup$
    – Andrew
    Nov 24, 2020 at 21:08
  • $\begingroup$ @PhilFrost-W8II The Original question asked "Is it true that the inductive and capacitive reactances cancel for a half wave dipole at resonance ?" Sorry but i don't see how your comment "this is a matter of definition: the frequencies where the impedance is nonreactive are defined as the resonant frequencies" relates to or answers that question. $\endgroup$
    – Andrew
    Nov 24, 2020 at 21:12

4 Answers 4


What you miss is that a resonant dipole doesn't need to be a half wave length long. You can add series inductors near the center or part way out from the middle, and/or capacitors (hats) near the tip ends, with L's and C's varying in value from small to large, to vary the dipole length from a just tiny bit shorter to significantly smaller (e.g. HT rubber duckies or 80M hamstick dipoles). And still be resonant at exactly the same frequency. (alas, with worse radiation efficiency and a narrower SWR bandwidth.)

Note that if you do the above mods, but keep the identical resonant frequency, the length can change drastically, but the LC ratio stays the same. So LC product is the constant for resonance, not the physical length.

Added: In fact, if you gradually change the L and C loading values as you shrink the dimensions of a dipole antenna from half wavelength down to the size of a few small lumped components, the resonant antenna will gradually become a non(or barely)-radiating lumped resonant LC circuit (with a center-tapped/split inductor at the feedpoint).

  • $\begingroup$ you're mixing up lumped constant circuit theory with traveling wave theory. fo = 1 / 2π x √LC applies for the lumped constant components and does not apply to the rest of the dipole. Saying that adding capacitors and inductors in series with the elements changes the reason that a half wave dipole is resonant is ridiculous. $\endgroup$
    – Andrew
    Nov 24, 2020 at 5:22
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    $\begingroup$ Of course I'm mixing them up. It's the mix of distributed wire L and C and lumped loading L and C that determines the resonant frequency of the combo. And traveling wave theory depends on that distributed L and C. $\endgroup$
    – hotpaw2
    Nov 24, 2020 at 6:07
  • $\begingroup$ thanks for the comments, to rephrase, a short dipole with coils or capacitors added to cancel out the reactance caused by the dipole not being half a wave length isn't the same thing as a true half wave dipole, it is a short dipole with inductors in series and capacitors at the ends, and the points in my question discuss a proper half wave dipole without added lumped constant components which is intrinsically or naturally resonant because the elements are 1/4 wave in length. We are talking about two different systems. $\endgroup$
    – Andrew
    Nov 24, 2020 at 6:39
  • $\begingroup$ Yes, but note that you can very gradually turn your "proper" dipole into LC elements, by slinky-twisting the middle by 0.0001% radius and flattening then tips to 1.0001% wire diameter. And continue from there into fully lumped LC elements. So there is no absolute difference between the two systems. $\endgroup$
    – hotpaw2
    Nov 24, 2020 at 6:48
  • $\begingroup$ And then as you do that traveling wave theory gradually starts to not apply to the antenna ... ? $\endgroup$
    – Andrew
    Nov 24, 2020 at 7:13

The original question was, "Why do ham radio operators insist that a half wave dipole is resonant when the inductive and capacitive reactances cancel out?" That question seems to be about psychology, and I will answer accordingly. I don't know that my answer will be very useful to others, but I'll proceed anyway.

People who are hams come from all walks of life, and their understanding of the technology of radio is imperfect, as it is for everyone. Many hams know a bit about antennas and a bit about RLC circuits. They observe that the impedance of a dipole that is shorter than resonant is capacitive, and the impedance of a dipole longer than resonant is inductive.

"Aha", someone must have said to himself, "a dipole must be like a series RLC circuit, and the capacitive reactance must go up and the inductive reactance must go down when the antenna gets shorter, and vice-versa." It's a gross oversimplification, as demonstrated by your observation that the dipole is also resonant for odd harmonics, but it worked for him at the time. The model was shared, and it made sense to other people, who passed it on and perpetuated it.

Fragile and incomplete mental models are how we humans make sense of the world; we need some sort of mental model to begin to understand just about anything, and we do the best we can with what we have available. In this case a better understanding is possible for some with a bit of work. At least the misunderstanding is understandable.

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    $\begingroup$ Thank you for that answer, i really like that actually and i think what you say is spot on, i include myself in those people that need the fragile mental models to begin to understand something, and those fragile models are often a very good start. The only bad thing is when the fragile models become folklore and end up all over the internet, trying to find the truth becomes increasingly difficult, especially for beginners who are blasted with large mounts of inaccurate and confusing information which never has any context. $\endgroup$
    – Andrew
    Nov 24, 2020 at 21:55
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    $\begingroup$ I'm in the same boat as you Andrew, and I agree that sometimes separating bogus folklore from more useful mental shortcuts can be difficult. At least we have some very-qualified folks on this site who can answer our questions, if only we can phrase them well enough. $\endgroup$
    – rclocher3
    Nov 24, 2020 at 22:44
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    $\begingroup$ I don't think it's a "fragile and incomplete" mental model. A dipole is not far from a transmission line stub. Such a stub can be capacitive, or inductive, right? And at just the right length, it's neither. Arguing about whether that's "no reactance" or "capacitive and inductive reactances cancelling each other" is like arguing about whether "0" and "1-1" are the same thing. $\endgroup$ Nov 25, 2020 at 19:52
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    $\begingroup$ Here's a counter-argument: if a dipole isn't "capacitive and inductive reactances cancelling each other", then why is a dipole different from a resistor? Or, how can you have a standing wave and resonance if there's no reactance? One could argue it is more complete to view a resonant dipole as one where the sum of reactance is zero, rather than one with no reactance at all. $\endgroup$ Nov 25, 2020 at 19:57
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    $\begingroup$ Hi @PhilFrost-W8II! I never said that a dipole isn't capacitive and inductive reactances canceling each other. $\endgroup$
    – rclocher3
    Nov 25, 2020 at 20:42

Zero reactance means voltage and current are in phase.

For example, consider a voltage source connected to a nonreactive load. This means the peaks of the applied voltage should coincide with the peaks of the resulting current.

It's equally valid to consider a current source connected to a nonreactive load. In this case the peaks of the applied current should coincide with the peaks of the resulting voltage.

Either way, reactance is part of impedance, and impedance is the relationship between voltage and current.

A dipole is just a bit of balanced transmission line that's been pulled apart. What happens when a DC voltage step is applied to the end of a section of transmission line that's open on the end opposite the voltage source?


simulate this circuit – Schematic created using CircuitLab

If we want to know the impedance of this "load" (the transmission line) we need to know how much current flows. We know that eventually the current must be zero, because the circuit is open at the end. But how can the voltage step know that, not yet having seen the open end?

So what happens is initially some current flows, in an amount defined by the surge impedance (also known as characteristic impedance) of the transmission line. But the current is constrained to zero at the open end, so a reflected wave is superimposed on the initial wave, propagating from the open end and back to the voltage source. It may help to play with a time domain transmission line simulator to get an intuition for this process.

What happens when the reflected wave gets back to the source is key. In the case of a DC step, the source will see too much voltage, and so it will reduce current. And this sets off another round of wave propagation, with each iteration getting closer to what we know the DC solution must be: zero current, that is infinite impedance.

But in the case of AC, the voltage source is not a step but rather a sinusoid. We must consider both the phase of the reflected wave, and the additional phase delay introduced by the propagation of the forward wave and then the reflected wave.

Exactly what is the cause for the zero reactance seen in the impedance at the center feed point of a resonant half wave dipole?

When the transmission line is open, the current of the reflected wave will always be equal but opposite the forward wave, because the open end always wants to cancel the current to make it zero. In other words, the reflection adds 180 degrees of phase delay.

When the length of the transmission line is 90 degrees, it is resonant. This is due to the 90 degrees of delay for the forward wave, plus 90 degrees for the reflected wave, plus 180 degrees for the phase of the reflection equals 360 or 0 degrees. Current is in phase with voltage, which means zero reactance, which means resonance.

I don't understand yet how the radiation resistance fits into all of this.

In the case of an ideal 1/4 wave transmission line, the impedance seen by the voltage source is exactly 0+0j ohms. This is because the current from each reflected wave reinforces each forward wave, and there's no loss in the system, so the current builds to infinity. But in an ideal resonant dipole some energy is lost to radiation (represented by a resistance), and so the current builds to a high but finite quantity, resulting in the low but non-zero impedance of about 70+0j ohms.

Now, what about this graphic:

enter image description here

At a glance, it looks like the red and blue curves, labeled "voltage" and "current" respectively, are not in phase, but quadrature. How is this reconciled with the above explanation, where voltage and current are in phase?

More confusing, but perhaps more helpful is the old version of the image which shows only the standing wave, but does not include the influence of the voltage source (perhaps it would be better if the illustration did not include a voltage source, since its effects are not illustrated):

enter image description here

Here, the red and blue curves are exactly in quadrature. And this is no mistake, since the standing wave is purely reactive.

I think the confusing thing about this image is it just says "V" (for voltage) without really explaining what that means. Anything measured in volts could be called voltage. That's not very specific or helpful.

If we are concerned about the feedpoint impedance, the voltage we are concerned about is more specifically the electric potential difference between the two feedpoint terminals.

If we are concerned about the electromagnetic fields around the dipole, we are probably more concerned about the electric field intensity, which is a vector quantity for some point in space around the antenna, measured in volts per meter.

The "voltage" in the graphic shows the electric potential for each point along the length of the antenna. Electric potential is the electric potential difference between the measured point, and a theoretical point infinitely far away, which is 0 volts by definition. In the case of a dipole, the electric potential right at the center is also 0 volts.

Now the question is: how can the electric potential difference between the feedpoint terminals be in phase with current when on the graphic the blue curve is clearly not in phase with the red curve?

The answer is quite simple: theoretically the feedpoint terminals are separated by only an infinitesimal distance. It doesn't actually matter what the blue curve is doing, because the electric potential difference between two points approaches zero as the two points approach zero separation.

Put another way, electric potential difference between two points in a uniform electric field is the electric field intensity (volts/meter) multiplied by the distance between the points (meter). If the distance is small, the electric potential difference can be neglected.

The full picture of what happens on the dipole is the superposition of:

  • the standing waves, shown in the image above, where the electric and magnetic fields are in quadrature, and
  • the influence of the voltage (or current) source driving the antenna, where voltage and current are in phase.

My understanding which is becoming more and more confused as time goes on is that at resonance the voltage of the standing wave which is 90 deg out of phase with the applied current is always zero at the feed point at resonance.

Although it is true the standing waves are associated with a high electric field intensity around the feedpoint terminals, as long as the terminals are not far apart this has negligible significance to the electric potential difference between the terminals.

  • $\begingroup$ Thanks very much Phil for that. You have touched on numerous ideas in your answer. There isn't enough room in comments for me to respond. Is it appropriate if i create an answer for this question which discusses your answer ? $\endgroup$
    – Andrew
    Nov 28, 2020 at 22:17
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    $\begingroup$ @Andrew how about chat? chat.stackexchange.com/rooms/11162/ham-shack $\endgroup$ Nov 28, 2020 at 23:13
  • $\begingroup$ excellent answer, as usual! $\endgroup$ Dec 1, 2020 at 18:44

To add to what Hotpaw stated ...

  1. There's the theoretical half-wave dipole, which is 492/f.

  2. However, a practical resonant dipole is shorter. When making one, we must account for the end effect, by cutting its length to 468/f.

The end effect results from the fact that the antenna is normally operating surrounded by air, and the signal is travelling in a conductor which is of finite length. More specifically, the antenna end effect results from a decrease in inductance and an increase in capacitance towards the end of the antenna conductor.

Quote from https://www.electronics-notes.com/articles/antennas-propagation/dipole-antenna/length-calculations-equation-formula.php

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    $\begingroup$ I've always wondered whether the self inductance of the dipole wires in free space, plus the tiny bit of distributed capacitance between the distant tips of each dipole half, was near the right amount of loading required to account for the practical shortening in length. $\endgroup$
    – hotpaw2
    Nov 24, 2020 at 4:36
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    $\begingroup$ An antenna does not suffer from "self-inductance of the wire." Self-inductance of a wire is that electromagnetic effect which we don't like to think about when using lumped elements, so we pretend it's a lumped inductor. Antennas grasp the whole field solution, including current that's not equal along the wire... they have to. Capacitance from the ends is different. $\endgroup$
    – tomnexus
    Nov 24, 2020 at 4:52
  • $\begingroup$ A traveling wave moves slower in metal than the emf wave it produces in air, so the length of one half wave is shorter inside a half wave dipole than in air for that reason. The larger the cross sectional area of the elements the slower the wave travels. This is the main reason the electrical length is shorter. $\endgroup$
    – Andrew
    Nov 24, 2020 at 7:38
  • $\begingroup$ By definition a half wave dipole is always exactly resonant. Ham operators like to use the free space length and then apply that to a metallic dipole and say that a half wave dipole should be 5 % shorter to be resonant and then everyone gets confused because they think that a half wave dipole has an impedance of 73+j73 ohms which is inaccurate and misleading. $\endgroup$
    – Andrew
    Nov 24, 2020 at 7:39
  • $\begingroup$ It's like saying the length of a piece of wood changes when you immerse it in water and all you have to do is chop a bit off the end to make it the right length again :/ $\endgroup$
    – Andrew
    Nov 24, 2020 at 7:59

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