[Antennaware] dBi conversion to Antenna Factor

Terry Conboy n6ry at arrl.net
Tue Nov 20 21:57:02 EST 2007

At 03:12 PM 2007-11-20, Andy Ikin G8LUG wrote:
>Guy, thank you very much for you reply and clarifying that the EZNEC
>modelling calculates the dBi assuming a 50 Ohm system.

I'm not sure that's what Guy was implying.  I don't think that there 
isn't any particular impedance assumed by EZNEC, but it does normally 
assume that all the power from the source is delivered to the antenna 
(which implies a perfect conjugate match).  The sources in NEC can be 
either voltage (zero impedance) or current (infinite impedance), so 
they have no internal generator impedance and thus no possible 
internal losses and always deliver full power to the load (the antenna system).

>I have done a bit more digging into resolving the relationship of dBi to
>Antenna Factor. In sixth edition of the RSGB Radio Communication Handbook,
>Chp.11.3 Propagation. f ig.11.4 shows the relationship for field strength of
>a half-wave dipole to voltage at the Rx end of a correctly matched feeder
>connected to it. i.e. 1uV at Rx relative to 1uV/m. At 50MHz 1uV at Rx equals
>1uV/m (0dB), at approx. 1.8MHz Rx it is +30dBuV for 1uV/m. Therefore
>assuming that the above dipole is in free-space we can subtract approx.
>2.2dB to arrive at the dipole dBi? Hence, approx. +28dBi = an Antenna Factor
>of +30dB (1uV/m to +30dBuV). Assumming that my prognosis is correct,  the
>K9AY with -26dBi gain at 1.8MHz has an antenna Factor of +2dB.

Check out this web page 
which has a detailed discussion of antenna factor.  Note that their 
definition of antenna factor (AF) is the reciprocal of yours (which 
they define as the "effective length", h-sub-e).

This states: "An antenna with 0 dBi gain possesses a 0 dB antenna 
factor at approximately 30.81 MHz. Thus a half-wavelength dipole has 
a zero dB antenna factor at 39.46 MHz. At 19.73 MHz, the antenna 
factor of a half wave dipole is -6 dB."

They also say that for an antenna with constant gain, the antenna 
factor increases 6 dB per octave with frequency.
By extrapolation, at 1.8 MHz, the AF would be -24.67 dB for an 
isotropic (0 dBi).

For a K9AY, the AF would be 26 dB higher or 1.33 dB.  (Or the 
"effective length" is -1.33 dBi, which should compare to your result).

You could also use the formulas:
AF = 9.73 * SQRT(Gain) / (wavelength, meters) in a 50 ohm system.

In terms of frequency, this is
AF = fMHz / ( 30.81 * SQRT (Gain) )

or in decibels
AF = 20*LOG10(fMHZ/(30.81*sqrt(Gain)) = 20*LOG10(fMHz) - 29.77 - GdBi

To check: for the K9AY at 1.8 MHZ, AF = 5.11 - 29.77 + 26 = 1.34 dBi

The discrepancy may be in part due to the impedance of a half-wave 
dipole being ~72 ohms instead of 50 (a 3.2 dB voltage difference for 
the same power).  Also the ratio of 50 MHz to 1.8 MHz represents a 
change of 28.87 dB, so there is some approximation involved in 
reading the RSGB chart.

73, Terry N6RY

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