I've read that antennas work best when their length equals 1/2λ, so how does one make an antenna that is 3/2λ, 2λ, 3λ, etc. work? How do I calculate how much energy I am losing compared to the 1/2λ length antenna?


2 Answers 2


The issue is not that a long dipole is inefficient, but that it may have an odd radiation pattern. For example, from Radiation Patterns of Dipole and Loop Antennas by Bhowmik, the theoretical polar plot of a four-wavelength dipole is shown below.

4-wavelength dipole radiation pattern, Bhowmik

As the length increases in proportion to a wavelength, the number of major lobes or points of maximum radiation increases, and they move outwards aligning further with the axis of the antenna.

Depending on your needs, a multi-wavelength dipole may work well... or not. Test empirically, and note that a small change in orientation can cause a large change in signal. Also consider that near-field and far-field effects may differ.


How do I calculate how much energy I am losing compared to the 1/2λ length antenna?

The "energy" is not lost in radiator, it's radiation pattern ( as noted above) changes.

Also its "feed point impedance" varies , it is no longer around 70 Ohms as in center fed 1/2 wavelength radiator - and assuming the radiator is primarily used for transmitting. It varies with actual length of the radiator and position of the feed point.

The "energy" (loss) or better term woudl be the system (radiator) efficiency is therefore affected by mismatch between feed point impedance of the radiator and feed line impedance.

Knowing such mismatch woudl be the key to calculate the efficiency , or your quest for energy loss calculation, of the system.


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