Roger wrote:
Cecil Moore wrote:
Roy Lewallen wrote:

The existence of both voltage and current at any point along the line
tells us that there is instantaneous power at that point, ...

Not if the voltage and current are always 90 degrees out of
phase which is a fact of physics for pure standing waves. There
is no power, instantaneous or otherwise, in pure standing waves.
The cosine of 90 degrees is *always* zero.

These comments in total are very interesting, by both authors. Thank
you for them.

It is clear that there is not a standard way of describing energy that
is in the process of being stored in either an inductor or capacitor.
Clearly it is stored, not used (converted) as in a resistor.

Sure there is. The energy stored in a capacitor is 1/2 * C * V^2, and in
an inductor, 1/2 * L * I^2. The rate of energy flow into or out of an
inductor or capacitor is called "power". It's exactly the same stuff as
energy moving into or through a resistor, and it's measured and
described in exactly the same way.

Storage of energy in a capacitor (for instance) occurs over time and
requires power to complete.

Yes, power is the rate of energy transfer.

So how do we describe that power if it is
always zero because voltage and current are 90 degrees out of phase?

See the following example which shows just that situation.

Should we recognize that only the peak voltage and peak current is 90
degrees out of phase, with the entire charging time occurring within
those two time extremes? Between those two time extremes, the voltage
and current are in phase but at changing impedance, with power flowing
into the capacitor. So think I.

The whole concept of impedance gets a bit flaky when working in the time
domain. I recommend sticking to voltage, current, power, and energy when
doing so. Go through the following example, and I believe you'll get a
much better idea of what's happening on a basic level.

But you've made some good observations. The answer is that, as I
mentioned, the instantaneous power (p(t)) is not zero except at the ends
and middle of the example line. Some of the posters are confusing
average and instantaneous power, which is leading to incorrect
statements and conclusions. Instantaneous power can easily be non-zero
while maintaining zero average power, as I'll illustrate.

Let's look, for example, at the half wavelength open circuited line
described by "Dave", being driven by v = 0.5 sin(wt), 1/8 wavelength
from the end. At that point,

vf(t) = 0.5 sin(wt - 135 deg.)
vr(t) = 0.5 sin(wt - 225 deg.)
if(t) = 0.01 sin(wt - 135 deg.)
ir(t) = -0.01 sin(wt - 225 deg.)

considering the positive direction of ir(t) as the same as the positive
direction of if(t) so we can add if(t) and ir(t) to get i(t).

Using elementary trig identities for the addition, and dropping the
explicit identification of degrees for simplicity:

vf(t) + vr(t) = v(t) = sin(wt - 180) * cos(45) = 0.7172 * sin(wt - 180)
if(t) + ir(t) = i(t) = 0.02 * cos(wt - 180) * sin(45) = 0.01414 *
cos(wt - 180)

One important thing to notice is that, unlike the ends or center of the
line, neither the voltage nor the current is zero at this point.
Consequently, their product is also non-zero.

The voltage and current are, of course, 90 degrees out of phase, as they
are everywhere along an open circuited or short circuited line, or one
terminated in a pure reactance. And when the average power is computed
from the instantaneous power, the result will be zero. But that doesn't
mean that the instantaneous power is zero as some are claiming. Let's
see what it does mean. Using another trig identity,

p(t) = v(t) * i(t) = 0.01 * sin(wt - 180) * cos(wt - 180)
= 0.005 * sin(2wt)

This is the instantaneous power which, as I've described before, is a
sinusoidal function with rotational frequency 2w. It has no offset, so
the average value is zero. But it shows that energy moves back and forth
past this point every cycle. Equal amounts move each direction each
cycle, so no net power flows either direction.

There are simpler ways to get the same answer, but this one shows very
basically where the energy is going at every instant. If instead you
start out by throwing away the time information and just looking at
averages, you lose a lot of information along with the basic
understanding of energy movement in the system. I'm afraid some people
have taken this simpler approach without realizing what information and
insight they've lost by doing so. The average value can always be
determined from the instantaneous function, but never the other way around.

I think of the "standing wave" as being equivalent to the graph on my
power bill that shows power used daily over a month. Granted that the
standing wave is recorded in V or I, but we only need to know the R that
the standing wave is acting through to determine the power that it
represents.

Not quite. The power bill graph is a graph of the power, much like the
p(t) function I calculated. The energy, which you pay for, is the
integral of that graph. If you used a constant power of 1 kW for half
the month, then sent 1 kW back into the power grid for the other half of
the month, you'd technically owe nothing, since you'd use no net power.
You might even have stored it somewhere and given the very same energy
back rather than using some and generating an equal amount later. That's
exactly what's happening at the point on the line 1/8 wavelength from
the end.

Roy Lewallen, W7EL

-- I've had many opportunities when doing this to make arithmetic or
trigonometric errors, and I haven't taken the time to thoroughly check
my work -- although the conclusion is correct. I'd appreciate anyone
pointing out any mathematical errors he's found.