NEC-LIST: RE: Modeling short dipoles

I'm going to take another shot at this efficiency calculation. It
turns out that the method I'd given previously does not work for NEC,
due to the way NEC calculates gain. I believe I have a good
correction factor this time. I hope that someone else out there has
looked at this and can confirm or correct it.

In simple MM programs that I've written, I feed the elements with
voltage sources in series with pure 50-Ohm resistors. The voltage
sources are sized such that the total power dissipated by the antenna
would be equal to one watt if the antenna consisted of pure 50-Ohm
loads. For a single source, the voltage would be sqrt(2*z0) =
14.14... V. With this voltage set, the reported E-Field will be
correct for any input impedance, assuming a 1-watt source with a
50-Ohm characteristic impedance. Absolute gain is then easily to
calculate. NEC, however, does not compute absolute gain this way. It
computes efficiency as the total radiated power divided by the total
input power, and multiplies all computed gain numbers by this
efficiency value. For a practical system, however, efficiency is
better represented by the total radiated power divided by the total
power that WOULD HAVE been input had the impedance been perfectly
matched.

I think that the best way to approach this is to first put a Z0
resistor in series with each feed. (This could be skipped for a
single element, but the resistors make for more-representative
patterns in the case of an array.) I believe that the ratio of these
two efficiency measures is equal to the sum of the true input powers
divided by the sum of the ideal input powers:

  (practical efficiency) / (NEC efficiency) = [Z0 + load resistance)] /
[sum(Vs^2/(4Z0))]

or for a single element with Z0 = 50 Ohms: (recalling that Zin will
include the 50-Ohm resistor)

  (practical efficiency) / (NEC efficiency) = [200*(real(Zin)] / (Vs^2)]

For example, I ran a tenth-wavelength dipole. With no series
resistor, NEC gave an input impedance of 2.4-j1234, a gain of 1.7 dBi
(1.5 in linear units), and an efficiency of 100%. With a 50-Ohm
resistor the gain showed up as -11.6 dBi and the efficiency was given
as 0.046. This is high, since the efficiency penalty due to mismatch
loss alone is:

  efficiency = 1-rho^2 = 1 - [(Zl-50) / (Zl+50)]^2 = 0.00315

which leads to a practical gain of approximately -23 dBi.

The equation above will give the ratio of these two values, which can
be used as a correction factor for NEC gain patterns when it is
desired to account for mismatch loss.

Comments would be appreciated.

                                Dave
Received on Wed Nov 26 1997 - 09:41:58 EST