Intuitively, we all understand that an antenna at fractional wavelengths will not radiate well even with perfect impedance matching systems.

From what I understand the efficiency of antenna has a few factors:

  • Conduction losses (due to finite conductivity of the metal that forms the antenna)
  • Dielectric losses (due to conductivity of a dielectric material near an antenna)
  • Impedance mismatch loss

This still does not explain why a 1-foot long dipole antenna on 160 meters does not radiate well.

If I have 100 Watts RF input, and a theoretical perfect matching system to transform my radios independence to the very high impedance of <1/10 fractional wavelength antenna such that there is no reflected power and no ohmic losses in the antenna, where does this 100 watts go, and why is it not all radiated in the form of RF energy as well as that of say a 1/4 wavelength vertical?

  • $\begingroup$ What do you mean by "perfect"? Do you mean just that it yields a perfect, 50 ohm match (like a dummy load?) Or do you mean it's made of ideal, lossless components? $\endgroup$ Commented May 6, 2018 at 23:19

2 Answers 2


In theory, if you had lossless conductors in the antenna and a lossless matching network, your shortened, 1 foot dipole would have a gain of only ~0.7 dB less on 160 meters than the gain of a full size 1/2 wavelength dipole. But the world is far from perfect.

The efficiency of an antenna is defined as:

$$\text {Efficiency}=\frac {R_r}{R_r+R_l} \tag 1$$

where $R_r$ is the radiation resistance of the antenna and $R_l$ is the resistive losses of the antenna.

As an antenna becomes shorter, its radiation resistance drops significantly. As you can see from formula 1, this causes the otherwise negligible resistive losses to significantly impact the efficiency of the antenna. Any inefficiency is simply dissipated as heat.

The gain of an antenna is defined as:

$$\text {Gain}=\text {Directivity}*\text {Efficiency} \tag 2$$

As the antenna is shortened, the directivity drops slightly from 1.64 for a 1/2 wavelength dipole to 1.50 for an infinitesimal dipole. This, combined with the lowered efficiency, results in an overall drop in gain.

So now let's plug your example into the above formulas.

The radiation resistance of a free space shortened dipole is given as:

$$R_r=197L_\lambda^2 \tag 3$$

where $L_\lambda$ is the length of the dipole expressed as a fraction of the wavelength (length in meters / 160 meters in this case).

Your 1 foot 160 meter dipole antenna example would therefore have an $R_r $ of only ~0.7 milliohms.

The RF resistance of 1 foot of 12 gauge wire at 1.8 MHz is ~17 milliohms (ref. http://chemandy.com/calculators/round-wire-ac-resistance-calculator.htm).

[Note: The following paragraph and the subsequent formula results are the result of Phil's critique of my original answer regarding resistive losses.]

The RMS current distribution in a very short dipole without end loading decays linearly from the maximum at the center point to zero at the ends of the antenna. As a result, the average RMS current of the short dipole is 1/2 that of the feedpoint current. Since resistive heat losses are the result of current squared, we must apply a 0.25 correction factor (0.5 squared) to the calculated calculated RF resistance of the wire in order to account for this RMS current slope along the length of the short dipole. The effective resistance of the wire in this shortened dipole is therefore ~4.3 milliohms.

Populating formula 1 tells us that the antenna will have a 14% efficiency. This means that 86 of your 100 watts of applied power goes to waste as heat. Formula 2 tells us that the antenna will have a free space gain of 0.21 or -6.8 dBi. For comparison, an efficient, full 1/2 wave dipole in free space has 2.2 dBi of gain.

Finally, a shortened antenna has a reactive component at its feedpoint that must be dealt with. This, in combination with a low resistive component, can be difficult to match to normal amateur transmission lines resulting in additional losses in the matching circuit and a corresponding reduction in antenna system gain. Of course if the matching circuit does not provide a match to the $Z_o$ of the transmission line, there will be additional losses in the transmission line and a possible reduction in the transmitter output power.

You mentioned dielectric losses. These are normally minimal at HF frequencies in a properly designed matching network.

  • 1
    $\begingroup$ @PeteNU9W Thank you. I edited my answer to define the term. $\endgroup$
    – Glenn W9IQ
    Commented May 3, 2018 at 11:38
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    $\begingroup$ Thanks for your answer! Where did you get the formula for radiation resistance of a fractional wavelength dipole, and where could I find similar equations for other antennas? $\endgroup$
    – Skyler 440
    Commented May 3, 2018 at 19:58
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    $\begingroup$ @Skyler440 You are welcome. I pulled the formula from memory but you will generally find these derivations in college text books. I recommend "Antennas" by Kraus if you are interested in the more advanced mathematics applicable to antennas. $\endgroup$
    – Glenn W9IQ
    Commented May 3, 2018 at 20:10
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    $\begingroup$ Can you explain how you obtained the 17 milliohm figure? $\endgroup$ Commented May 4, 2018 at 3:01
  • 1
    $\begingroup$ @PhilFrost-W8II I used W J Highton's calculator as I have found this in the past to correlate with my own calculations. I put a link in my answer. $\endgroup$
    – Glenn W9IQ
    Commented May 4, 2018 at 10:33

In short: if you had the perfect matching system you described and a lossless antenna, there would be no losses. But those pesky laws of physics prevent us from making anything perfect, and even getting anywhere close to perfect may not be so easy. Instead of the 1-foot dipole you suggested, let's try a 5-meter long dipole instead (i.e. a half-wave for 10m). Most people have room for one of those, right?

In this case, the radiation resistance goes all the way up to 192 milliohm, which is actually of the same order of magnitude as the resistive losses, which become 282 milliohm at this length. And the reactance becomes about 240kOhm capacitive. I don't know what kind of a network it takes to match 50R to 0.474-240000j at 1.8MHz and handle 100W, but seeing as how I can't just buy one at DXE and solve all of my problems, it stands to reason that such a box is either incredibly inefficient on its own, or else requires coils bigger than the 160m antenna I've avoided putting up.

  • $\begingroup$ If you could make the appropriate superconducting inductor that would cancel 240 kOhm capacitive so you would be left with the problem matching 0.47 ohms at 1.8 MHz, the bandwidth would be close to zero. Q would be something like 500000 and the bandwidth something like 3 Hz. I think, in real life, we want a bandwidt of at least $\endgroup$
    – sm5bsz
    Commented May 8, 2018 at 23:39

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