I'm an amateur and looked to the calculators of TV/3G/WiFi antennas.

All articles state I can get away with all sizes divided by 2 or 4 (for biquad antenna & the range 400MHz - 2.1GHz dividing by 4 makes antennas compact enough to keep it indoor without ruining room aesthetic).

Why are fractions 1/2 or 1/4 of a wave length allowed and 1/3 or 1/8 are not suitable?

Following answers have related but not exact discussions:

  • \$\begingroup\$ Antenna length being 1/2 and 1/4 wavelength means that the actual electric and magnetic fields can satisfy suitable boundary conditions that allow effective radiation and reception, arising from the mathematical structure of the fields. 1/3 or 1/8 wavelength corresponds to neither the nodes nor the maxima of a sinusoid. \$\endgroup\$
    – nanofarad
    Feb 27, 2021 at 15:32
  • 2
    \$\begingroup\$ @nanofarad Minor quibble. A half-wave dipole is half an electrical wavelength long. It doesn't actually correspond to the a half wavelength in space. Regardless of what frequency it is fed, a half-wave dipole will always satisfy the boundary condition of the antenna, i.e. the current at the ends will be zero, and there will be a maximum at the feed point. But when fed with the wrong frequency, a half-wave antenna will not be oscillating at it's natural frequency, determined by the geometry, boundary conditions and (homogeneous) Differential Equations. \$\endgroup\$ Feb 27, 2021 at 16:18

2 Answers 2


Antennas of any length will radiate (or receive) electromagnetic waves. The efficiency of transmission depends upon 3 components: the transmitter, the feed-line and the antenna. To transmit efficiently, the loss in the feed-line should be small. To have small feed-line losses, one wants a power-factor in the feed-line to be near 100% (or the Voltage Standing Wave Ratio to be near 1:1). That is, one wants most of the apparent power to be true power, rather than reactive power. Reactive power increases the loss in the feed-line without adding anything to the radiated power.

Half-wave dipole and quarter-wave "monopole" antennas have purely resistive impedance. This means that the power-factor of the feed-line will theoretically be 100%. Short antennas, such as 1/8 wave-length, will have a strong capacitative reactance. This will cause the power-factor to be far from 100%. One can compensate for this capacitative reactance by adding a inductive "loading coil" at the base of the antenna. In this way, 1/8 wave-length or even 1/8.283947 (i.e. some random number) wave-length antennas are possible.

The following image from the Wikipedia article Dipole Antenna shows the dependence of reactance on normalized dipole length.

enter image description here

If one looks at approximately 0.25 wavelength, (remember this is a dipole), the resistive impedance looks to be less than 50\$\Omega\$ and the capacitative reactance looks to be about 500\$\Omega\$, i.e. about 10 times the resistive impedance.

  • \$\begingroup\$ Interesting note about capacitive reactance, I'm not sure about the correctness of my formulae interpretation but I see Xc=1/(2πfC) and capacity is proportional to the square or cube of linear size (depending on antenna's geometry): C=O(dim²) or C=O(dim³) so decreasing size two times may lead to 4-8 more reactance, with 1/8 it decreases 64-512 times and that might have an impact. \$\endgroup\$
    – gavenkoa
    Feb 27, 2021 at 17:02
  • \$\begingroup\$ @gavenkoa I added a diagram from Wikipedia that shows the rapid increase in capacitative reactance below 0.5 wavelength for a half-wave dipole. \$\endgroup\$ Feb 27, 2021 at 17:51
  • \$\begingroup\$ @gavenkoa I am not sure about your formulae either, but they seem to be in the right ball park. I don't know the formula used in the graph I added. I might research it later today. \$\endgroup\$ Feb 27, 2021 at 18:00
  • \$\begingroup\$ You should still use the 0.5 lambda position for a quarter wave monopole and it reveals an impedance of 37 ohms + j21 ohms (basically you half the impedance values for a monopole) when using the graph above. \$\endgroup\$
    – Andy aka
    Feb 27, 2021 at 19:55
  • \$\begingroup\$ @Andyaka it may be a terminological thing, but when I see the term "half-wave dipole", I understand it to be a "resonant half-wave dipole", which is half an electrical wavelength, rather than half a free-space wavelength. It is slightly shorter (than the half a space wavelength), and has 0 reactive impedance. Similarly for a "quarter-wave monopole". I understand that to be a "resonant quarter-wave monopole", which will be shorter than a quarter of a space wavelength, and will have 0 reactive impedance. I can't think of a good reason to make an antenna an exact multiple of 1/4 space wavelength. \$\endgroup\$ Feb 27, 2021 at 21:43

I think you should read this wiki article: -

enter image description here

Basically, because a sinewave can be broken in half in the time axis, you get a full internal coherent reflection with the forward applied voltage when you match the antenna length to the electrical half \$\lambda\$ of the applied voltage. This also works at a quarter \$\lambda\$ and multiples thereof (just like the pipe of an organ). These are called standing waves.

At other in-between frequencies you get asymmetrical reflections that have the effect of producing capacitive or inductive impedances as seen by the source voltage and, are not as effective as a means of emitting radio power. I'm not saying they don't work, just that they are not as effective.

It's exactly the same theory as the tube of a church organ. They are resonantly tuned in length to produce the correct fundamental pitch when excited with air. They can be open-ended or closed: -

enter image description here

Standing wave pictures from here.


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