The article Antennas – disturbing the thing being measured – open wire lines illustrated the pitfalls of a simplistic model of an antenna presented on two terminals of an open wire line.

A more complete representation of the antenna can be formed by making three impedance measurements (Schmidt nd).

Running the same NEC-4.2 model as used in the earlier article, we can find the values of the three measurements by changing the source configurations.

Entering the values into Find three terminal equivalent circuit for an antenna system we get:

It is the existence of the third arm to ground that completes the model and given two grounded sources driving terminals 1 and 2 the currents in each of the three legs can be calculated.

It can be seen that if terminal 3 is disconnected, the impedance from terminal 1 to 2 reconciles with the measurement of the isolated line end.

It can also be seen that if a high impedance common mode choke is inserted lose to the measurement plane, it is approximated by inserting it in series with terminal 3 to ground, and if Zchoke is sufficiently high, the differential impedance seen by the choke is simply Z1+Z2, I3≅0 and ∴ I1≅-I2 (this condition is current balance).

We can solve this simple network with 1V of differential drive with a little complex maths, here in Python.

import math import cmath z1=33.59-190.8j z2=38.09-28.43j z3=118.6-109.2j i1=1/(z1+1/(1/z2+1/z3)) i2=-1/(z2+1/(1/z1+1/z3)) i3=i1+i2 id=(i1-i2)/2 ic=(i1+i2)/2 cmath.polar(i1) cmath.polar(i2) cmath.polar(i3) cmath.polar(id) cmath.polar(ic) 20*math.log10(cmath.polar(ic)[0]/cmath.polar(id)[0])

… and with the results…

>>> import math >>> import cmath >>> >>> z1=33.59-190.8j >>> z2=38.09-28.43j >>> z3=118.6-109.2j >>> i1=1/(z1+1/(1/z2+1/z3)) >>> i2=-1/(z2+1/(1/z1+1/z3)) >>> i3=i1+i2 >>> id=(i1-i2)/2 >>> ic=(i1+i2)/2 >>> cmath.polar(i1) (0.004495880078213788, 1.2859135100313297) >>> cmath.polar(i2) (0.00725188807638577, -2.236275910455074) >>> cmath.polar(i3) (0.003501657720876183, -2.7334538372765103) >>> cmath.polar(id) (0.0057737342748441335, 1.0504553253528246) >>> cmath.polar(ic) (0.0017508288604380915, -2.7334538372765103) >>> 20*math.log10(cmath.polar(ic)[0]/cmath.polar(id)[0]) -10.364261911425348

So, the common mode component of current at this point in each conductor is 10.4dB less than the differential component, not very good current balance at all.