Dave wrote:
Roy, try a real simple case.

50 ohm line, 1/2 wavelength long, open circuit, lossless, sinusoidal steady
state, fed with a 1v p-p voltage source.

Everyone will agree there is a standing wave on this line of course.

now, to make everyone happy... in the middle of the line calculate v(t),
i(t), and p(t), these are the standing wave voltage, current, and power.

No problem.

Let's use the voltage source as the phase reference, so it will be
(using w to represent the rotational frequency omega):

v(t) at the line input = 0.5 * sin(wt)

At the center of the line, i(t) = 0.02 * sin(wt - 90 deg.) and v(t) = 0.

p(t) = v(t) * i(t) = 0

Note that this isn't the average power (although the average power is
also zero), but the instantaneous power -- meaning that the power at
that point is zero at all times.

These aren't "standing wave" voltage, current, and power, but simply the
total voltage and current, and the power, at that point.

now, calculate vf(t), if(t),pf(t), vr(t), ir(t), pr(t) at that point, where
the 'f' terms are the forward wave, the 'r' terms are the reflected wave.

At the center of the line,

vf(t) = 0.5 * sin(wt - 90 deg.)
if(t) = 0.01 * sin(wt - 90 deg.)
vr(t) = -0.5 * sin(wt - 90 deg.)
ir(t) = 0.01 * sin(wt - 90 deg.)

where I've taken the positive direction of ir to be forward, the same as if.

compare and comment on the relation between pf, pr, and p.

Calculation of P(t) is as follows:

p(t) = v(t) * i(t) = [vf(t) + vr(t)] * [if(t) + ir(t)]

= 0 * 0.02 sin(wt - 90 deg.) = 0 as calculated before.

I haven't seen a definition of pr and pf, but they're not relevant to
the discussion. If you get a different result for power than zero by
using whatever you take them to mean, then the concept is invalid. There
is no average power leaving the source and no average power being
dissipated in the load(*). So there had better be no average power
anywhere in the line. There will be non-zero instantaneous power
everywhere along the line except at the input, far end, and midway, but
its average value will be zero, indicating the movement of energy back
and forth but no net energy flow.

extra credit:
repeat calculations at the far end of the line. again compare pf, pr, and
p.

Sure.

At the far end of the line,

v(t) = sin(wt - 180 deg.)
i(t) = 0

p(t) = v(t) * i(t) = 0 at all times.

vf, if, vr, and ir, are the same everywhere on the line, so see the
previously calculated values. Likewise, the calculation of p(t) from the
forward and reverse traveling waves is the same as before, with the same
result.

I've shown that I can calculate the correct power at two points along
the line by using p(t) = v(t) * i(t) as I can for any point on any line
with any termination. Your question implies that the results I got (zero
at all times at both points) are incorrect. What power (instantaneous
and average) do you calculate for those two points?

(*) In fact, calculation of the power at the source will show that the
power there is also zero at all times. The energy moving back and forth
in the line at all points except the quarter wavelength nulls was put
into the line during the turn-on process. After steady state has been
reached, the source can be turned off (converted to a short circuit)
with no change in what's happening on the line.

Roy Lewallen, W7EL