Cecil Moore wrote:
wrote:
... the quantities being measured are line volts and line amps
and neither is proportional to incident or reflected watts.
line volts = Vfwd + Vref, line amps = Ifwd + Iref
load power = Vfwd*Ifwd - Vref*Iref at the load
Not with my math....
power = volts * amps
= (Vfwd + Vref) * (Ifwd + Iref)
= Vf*If + Vf*Ir + Vr*If + Vr*Ir
Seems you've lost a couple of terms in there.
This is why, in general (using my definition of general), superposition
does not hold for power. Those extra terms get in the way.
Much happens to the measured voltage and current before the result is
inaccurately interpreted as forward and reverse power.
If you wade through the math you will see why it works for low-loss
feedlines.
You have started to state this caveat.
Does this mean that forward and reverse waves only have power in
the special case of low-loss feedlines?
Where is that example that proves your assertion that reflected waves
don't exist? How do standing waves occur when there are only forward-
traveling waves?
With my model, incident and reflected VOLTAGE waves and CURRENT waves
do exist. This is in common, I think, with most authors on the subject.
And this all works fine since superposition holds for voltage
and current.
What does not exist in general (using my definition of general) is
POWER in these incident and reflected voltage and current waves.
While a power term can be computed, and correct results will be
obtained in some special cases, in general (using my definition
of general), this is an invalid operation because superposition
does not hold for power.
This, I suspect, is the reason that at least one author has
carefully chosen fuzzy words when describing the 'power' in
the waves.
To recap. Everything with the incident and reflected wave model
works as long as you stick to voltage and current waves. It is only
when extended to include power (as done by Bird and others), that
the model starts to deteriorate.
So to get the right answers in the general case (using my definition
of general), compute your forward and reverse voltages and currents.
Use superposition to derive the resultant voltages and currents at
any point on the line and then use p(t) = v(t) * i(t) to compute
the power, which you may then average if you desire.
In the special case of low loss RF, the shortcut of 'regarding'
the waves as having power will produce an answer that is good
enough. But remembering that you have taken a shortcut and
by only 'regarding' the waves as having power, you will indicate
an enhanced understanding of the subject.
....Keith