Roger wrote:
. . .
You give a good example Keith. It would be correct for measurement at
the load and at every point 1/2 wavelength back to the source from the
load, because the standing wave has the same measurements at these
points. At the 1/4 wavelength point back from the load and every
successive 1/2 wave point back to the source, the equation would also be
correct as demonstrated in Roy's example earlier today.

Excepting for these points, we would also be measuring a reactive
component that could be described as the charging and discharging of the
capacity or inductive component of the transmission line. (Imagine that
we are measuring the mismatched load through a 1/8 wave length long
transmission line, using an Autek RX VECTOR ANALYST instrument) The
inclusion of this reactive component would invalidate the power reading
if we were assuming that the measured power was all going to the load.
. . .

Well, let's look at that problem. Make the line 1/8 wavelength long
instead of 1/4 wavelength. The ratio of V to I at the source can be
calculated directly with a single formula or by separately calculating
the forward and reverse traveling voltage and current waves and summing
them. The result, for my 50 ohm transmission line terminated with 25
ohms, is 40 + j30 ohms. V at the input is fixed by the source at 100
volts RMS. I = V / Z = 1.6 - j1.2 amps = 2.0 amps magnitude with a phase
angle of -36.87 degrees.

Now I'll translate V and I into time domain quantities. (I could have
calculated I directly in the time domain, but this was simpler.)

Using w for omega, the rotational frequency,

If V(t) = 141.4*sin(wt) [100 volts RMS at zero degrees] then
I(t) = 2.828*sin(wt-36.87 deg.) [2 amps RMS at -36.87 degrees]

Multiplying V * I we get:

V(t) * I(t) = 400 * sin(wt) * sin(wt - 36.87 deg.)

By means of a trig identity, this can be converted to:

= 200 * [cos(36.87 deg) - cos(2wt - 36.87 deg.)]

cos(36.87 deg) = 0.80, so

V(t) * I(t) = 160 - 200 * cos(2wt - 36.87 deg.)

This is a waveform I described in my previous posting. The cosine term
is a sinusoidal waveform at twice the frequency of V and I. The 160 is a
constant ("DC") term which offsets this waveform. The fact that the
waveform is offset means that the power is positive for a larger part of
each cycle than it is negative, so during each cycle, more energy is
moved in one direction than the other. In fact, the offset value of 160
is, as I also explained earlier, the average power. It should be
apparent that the average of the first term, 160, is 160 and that the
average of the second term, the cosine term, is zero.

Let's see how this all squares with the impedance I calculated earlier.

Average power is Irms^2 * R. The R at the line input is 40 ohms and the
magnitude of I is 2.0 amps RMS, so the power from the source and
entering the line is 2^2 * 40 = 160 watts. Whaddaya know. Let's try
Vrms^2 / R. In this case, R is the shunt R. The line input impedance of
40 + j30 ohms can be represented by the parallel combination of 62.50
ohms resistive and a reactance of 83.33 ohms. The 62.50 ohms is what we
use for the R in the V^2 / R calculation. Vrms^2 / R = 100^2 / 62.50 =
160 watts.

We can also calculate the power in the load from its voltage and current
and, with the assumption of a lossless line I've been using, it will
also equal 160 watts.

P = V(t) * I(t) always works. You don't need power factor or reactive
power "corrections", or to have a purely resistive impedance.

This is really awfully basic stuff. Some of the posters here would come
away with a lot more useful knowledge by spending their time reading a
basic electric circuit text rather than making uninformed arguments.

Roy Lewallen, W7EL