Colleagues, Pursuant to a completely different discussion a couple months ago , Mr. Keith Armstrong presented the following equation as one which predicts the electric field at 10 meters due to common mode currents on a cable. I now find myself in need of just such an equation, but I have some problems with this one which I am hoping the forum can address.
E (V/m) = 1.26 10^-4 (f*L*Icm) In this equation, f is frequency in MHz, L is cable length in meters, and Icm is current in milliamperes. Mr. Armstrong substituted values of 100 MHz, 1.5 meter cable length and 2 mA current to arrive at a predicted field intensity of 92 dBuV/m. Note that 1.5 meters is one half wavelength at 100 MHz, so this is a half-wave dipole. I could instead use: Pd = Pt G/4 pi r^2 and substitute E^2/120 pi = Pd. If I further assume I am driving 2 mA into the antenna which looks like 72 Ohms and has a numeric gain of 1.64, then I get 81 dBuV/m, which is 11 dB less than predicted by Mr. Armstrong's equation. I could account for 6 dB by assuming constructive interference from a ground bounce, but I see nothing in Mr. Armstrong's equation to lead me to believe that it includes a ground bounce factor. I attribute the total 11 dB difference to the fact that the current distribution is uneven on a half-wave dipole and perhaps Mr. Armstrong's equation is a low frequency approximation assuming uniform current distribution. Can anyone comment on this? Or offer other expressions and their derivations (or references)? Thank you in advance. Ken Javor
