Let's say I have three dipoles, one is resonant at some frequency and is matched at the antenna, the second is shortened with loading coals and is also matched at the antenna, the third is shortened but has no loading coils and is also matched at the antenna.

Supposing that all three are matched to 50 ohms and I use a short (ish) piece of coax to the antenna, how would they perform with regard to efficiency, gain, etcetera?

Assuming at least a half wavelength above ground at the same location, only one antenna installed at a time.

  • $\begingroup$ If the third antenna is shortened and has no loading coils, then how is it still matched? $\endgroup$ Mar 28, 2014 at 13:23
  • $\begingroup$ Assuming all three antennas have some soft of matching network between the antenna and the feedline. $\endgroup$
    – s3c
    Mar 28, 2014 at 13:27
  • $\begingroup$ A loading coil is a matching network. If you take the loading coils out, then the "matching network" you need is exactly the inductance you just took out. We can't exactly compare the performance of antennas that are specified so vaguely. $\endgroup$ Mar 28, 2014 at 13:30

1 Answer 1


Antenna efficiency can be defined as the ratio of the radiation resistance to the total resistance of the antenna (radiation resistance, plus ohmic resistance of the antenna, plus ground losses, etc), normalized to the feedpoint impedance. If radiation resistance is the only resistance in the antenna, then it's 100% efficient.

$$ \text{efficiency} = \frac{R_\text{radiation}}{R_\text{radiation} + R_\text{ohmic} + R_\text{ground}+R_\text{any other kind}} $$

Performing this calculation can be tricky since there are so many potential sources of loss which may not be thought of as a "resistance", but can be expressed as such. Dielectric losses are a good example.

Loading coils and matching networks are significant only to the extent that they introduce additional loss. Sources of loss might include:

  • resistive losses in the coils
  • magnetic hysteresis losses in the core material of inductors
  • dielectric losses in capacitors
  • unintentional near-field coupling to nearby lossy materials (mast, trees, etc)

Shortening the antenna tends to decrease the radiation resistance. This doesn't make the antenna less efficient itself: it just makes other losses more significant, since they are now a greater fraction of the effective total resistance of the antenna.

It doesn't matter much if you use loading coils or "some sort of matching network". Loading coils are a matching network. If you take the loading coils out, your matching network, if at the feedpoint near where the loading coils would have been, will be an inductance of similar value. What matters is the losses you introduce with whatever matching technique you use. You can make very good loading coils that introduce no significant losses (fat, low resistance wire around a low-loss core like air), or you can make very poor coils that turn your antenna into a heater (very thin wire around a lossy iron core).

Shortening a dipole has a very small effect on the radiation pattern. As the dipole becomes shorter, it becomes less like a half-wave dipole and more like a Hertzian dipole. Wikipedia provides a nice graphical comparison of radiation patterns, with the half-wave dipole as a solid line, and the Hertzian dipole as a dashed line:

comparison of dipole radiation patterns

As you can see, the half-wave dipole has a marginally higher directivity, but they are close enough that for most practical concerns they can be considered identical.

  • $\begingroup$ Thanks Phil, that helped. Basically what I want to know is does the matching done to get the antenna to 50 Ohm (or whatever your feedline is) make it less efficient? If I read your response correct you are saying that a good design using good components for the matching won't make a big difference? $\endgroup$
    – s3c
    Mar 28, 2014 at 19:26
  • $\begingroup$ @s3c yes, that's right. If you had ideal, lossless components to make the matching network, efficiency would not be reduced, no matter how much you shortened the antenna. Of course, we don't have ideal components, so there's always necessarily some loss, but we can approach the ideal of no loss through better engineering (and sometimes, more expensive components). $\endgroup$ Mar 28, 2014 at 23:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .