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Antenna properties:

  • Height x Diameter = 3.3 inches x 1.4 inches
  • Test Frequency: 900 MHz (Wavelength = 33 cm or 13 inches)

I see some conflicting formulas to calculate the near field distance for an antenna w.r.t. the frequency of interest.

Method 1: (multiple sources)

Reactive Field <= 0.63 x sqrt(Height^3/Wavelength) ===> 2.6 cm or 1.02 inches

Method 2:

Reactive Field <= Wavelength ===> 33 cm or 13 inches

Which one applies here accurately?

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2 Answers 2

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The general formula for reactive field does not incorporate height, but rather maximum dimension of the antenna:

$$R\approx 0.62\sqrt{\frac{D^3}{\lambda}} \tag 1$$

where D is the maximum dimension in meters and $\lambda$ is the wavelength in meters.

Note that this results in an approximation in meters, not an exact answer.

The radiative near field (between the reactive near field and the Fraunhofer region) ends at a distance from the antenna as:

$$R\approx \frac{2D^2}{\lambda} \tag 2$$

Your second formula is technically correct as it uses "<=". When you compare the more concise results of formula 1, it does not conflict with the broad prediction of your second formula.

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  • $\begingroup$ There are essentially 2 antenna dimensions, height and diameter. Based on the fact you provided, I have included height in my post as the maximum dimension instead of diameter. There seems to be 2 variants of the formula to calculate reactive field distance, and both yield very different distances. $\endgroup$
    – eecs
    Commented Aug 13, 2018 at 0:52
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    $\begingroup$ @eecs Your second formula is too broad due to its <= formulation. See the edit to my answer. $\endgroup$
    – Glenn W9IQ
    Commented Aug 13, 2018 at 1:14
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Below is the output result of a NEC4.2 analysis of an electrically short monopole, showing its near-field boundary radius to be located about 30 meters from the base of that monopole.

The groundwave field beyond about 30 meters decays at nearly a 1/r rate, but not exactly 1/r because of the surface wave propagation losses present for these conditions — even for the short path shown.

enter image description here

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    $\begingroup$ I am struck by the apparent difference in our answers. Any idea how to resolve this, Rich? $\endgroup$
    – Glenn W9IQ
    Commented Aug 17, 2018 at 1:50
  • $\begingroup$ Most likely because the distance to the radian sphere (r = λ/2pi) defines the boundary limit where the reactive power density from an "infinitesimal radiator" is greater than its radiated power density [Source: Antenna Theory, Analysis and Design, 2nd Edition - C. Balanis; §4.2.3(d)]. That radius for the NEC model posted above is about 29 meters. Thanks for asking a technical question. $\endgroup$
    – Richard Fry
    Commented Aug 17, 2018 at 11:47

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