A Bird 43 directional wattmeter calibrated for 50+j0Ω responds to the combination of voltage and current in the very small region where the element couples to the line. At HF, the dimensions are such that that small region can be regarded as a point.
This worked example is for a 100H element (100W full scale deflection) at 14MHz.
If a 50+j0Ω load is connected to the instrument, and the element oriented to measure 'forward' power, the meter will read full scale (30ľA when 100W is delivered to the load. Under these conditions, the voltage V on the through line is (100*50)^0.5=70.71V and the current I is (100/50)^0.5=1.414A. The meter deflection is proportional to |V+50*I| and since it is 30ľA at 100W, we can write Im=|V+50*I|/4.71ľA (if we ignore the diode non-linearity which affects the low end of the scale) where I is the current in the direction of the arrow on the element.
Note that |V+50I| of itself cannot be used to indicate power as you will see as you read on.
For the grossly mismatched load, let us choose 5m of RG58C/U with an o/c a the far end.
From TLLC we expect that the impedance looking into the line will be 2.85+j38.11Ω, and the admittance is 0.001952-j0.026097S.
To simplify this analysis, let us assume that the distance from the element to the start of the RG58C/U is insignificant. At this frequency it is about λ/1000 in length and this assumption introduces very little error.
Let us apply RF at 14MHz, sufficient to cause 70V on the line at the element location.
We can calculate the line current into the 2.85+j38.11Ω load, it is 70/(2.85+j38.11)=0.137-j1.83A.
Let us calculate the 'forward' meter current using the formula above. Im=|V+50*I|/4.71=|70+50*(0.137-j1.83)|/4.71=25.3ľA which would indicate 71.22W on the meter scale.
Now, turning the element around, we can calculate the 'reflected' current by adjusting the sign of I for the changed direction. Im=|V+50*I|/4.71=|70+50*-(0.137-j1.83)|/4.71=23.6ľA which would indicate 61.66W on the meter scale.
Knowing the voltage and the load conductance (the real part of the load admittance) from TLLC we can calculate the power delivered to the stub, it is V^2*Y=70^2*0.001952=9.56W.
The 'forward' and 'reflected' meter readings respectively are 71.22W and 61.66W, yet we have calculated that there is only 9.56W delivered to and converted to heat in the stub. Clearly the 'forward' power reading on its own is rather meaningless, likewise for the 'reflected' power reading, but the astute reader will have noticed that 'forward' power minus 'reflected' power is exactly equal to the power delivered to the load.
It can be shown (Duffy 2007) that if the instrument is calibrated for an impedance that is purely real, then it is true in the general case that Power=Pf-Pr. Obviously, as Pr approaches Pf, confidence limits will become quite poor.
An interpretation often given to the meter readings in this scenario is to infer the Matched Line Loss. The inference being that if the 'reflected' power is XdB lower than the 'forward' power, then since the 'reflected' wave has travelled the length of line twice, the one-way or Matched Line Loss is half the dB loss quantity or X/2dB.
Let us calculate the Return Loss (as it is known), it is 10*log(71.22/61.66)=0.626dB, and half of it is 0.313dB. Is this the Matched Line Loss of this line? No, it isn't. The Matched Line Loss (from TLLC) is 0.265dB, not a lot different, but different.
Matched line loss may equal half the Return Loss with a s/c or o/c stub under some circumstances, and it may be close in others, but simply halving Return Loss in the general case does not accurately give Matched Line Loss.
Bird describe this method in their manual (Bird 2000) even though it is flawed:
Attenuation can also be determined by the open circuit method. The test set exhibits good equality between forward and reflected readings when the load connector is open or short circuited. When this is checked on an open circuit and an open-circuited length of line of unknown attenuation is connected to the load connector, the ratio shown is the attenuation in two passes along the line (down and back). To convert the reading to decibels, divide it in half or double the line length because twice the line length is being measured (down and back).
The fact is that the loss under mismatch conditions is not uniform along the line (Duffy 2008), and methods that depend on that assumption are flawed.
© Copyright: Owen Duffy 1995, 2021. All rights reserved. Disclaimer.