Roy Lewallen wrote:
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.
Roy, thanks for this posting and it's immediate predecessors. They are
very helpful both to my understanding and toward understanding how you
are thinking of things.

Your comment about the power bill was particularly helpful because you
are exactly right about the sequence of power, first received, and then
given back. The standing wave is doing exactly that, with the only
difference being that power is moving from one side of the system to the
other, with one side containing "negative power" (negative only in the
sense that polarity is reversed) and the other positive power.

This does raise the question of the description of the traveling wave
used in an earlier posting.

The example was the open ended 1/2 wavelength transmission line, Zo = 50
ohms, with 1v p-p applied at the source end. The term wt is the phase
reference. At the center of the line, (using the source end as a
reference), you gave vr(t) =-.05*sin(wt-90 deg.) and ir(t) =
0.01*sin(wt-90 deg.)

By using the time (wt-90), I think you mean that the peak occurs 90
degrees behind the leading edge.

This posting was certainly correct if we consider only the first
reflected wave. However, I think we should consider that TWO reflected
waves may exist on the line under final stable conditions. This might
happen because the leading edge of the reflected wave will not reach the
source until the entire second half of the initial exciting wave has
been delivered. Thus we have a full wave delivered to the 1/2
wavelength line before the source ever "knows" that the transmission
line is not infinitely long. We need to consider the entire wave period
from (wt-0) to (wt-360.

If these things occur, then at the center of the line, final stable
vr(t) and ir(t) are composed of two parts, vr(wt-90) and vr(wt-270) and
corresponding ir(wt-90) and ir(wt-270). We should be describing vr(t)
as vrt(t) where vrt is the summed voltage of the two reflected waves.

Then vrt(t) = vr(wt-90) + vr(wt-270) = 0
and irt(t) = ir(wt-90) + ir(wt-270) = 0

Below I will use the following terms:

vf = 1v/2 = 0.5v
vr = vf = 0.5v (for open circuit line)
ir = 0.5/50 = 0.01a

At the ends, vrt(t) = 2*vr(wt-90) = -2*vr(wt-270) = 1v

The voltage is doubled because of reflection. vf(wt-270) is negative.

Now consider power at the 45 and 135 degree points under stable
conditions, 1/8 and 3/8 wavelength in from the source. This would be
the power in the standing wave at the 45 and 135 degree locations.

At the 45 and 135 degree transmission line points, we have two
components present because the peak of each wave will be in alignment
with the leading edge of the reflected part of the same wave.
(Reflected waves being reflected again) Peak current and peak voltage
will be matched with near the near zero leading edge voltage and current.

Pr(left side) = vr(wt-90)*ir(wt-90) + vr(wt-180)*ir(wt-180)
= Pr(right side)= vr(wt-270)*ir(wt-270) + vr(wt-360)*ir(wt-360)
= 0.005w

Notice that vr(wt-90) and ir(wt-90) are different polarity from
vr(wt-270) and ir(wt-270). but the products of v and i are positive.

Find total power in the standing wave at the 45 degree points from
(ignoring the zero leading edge component)

Prt = Pr(left side) + Pr(right side)
= vr(wt-90)*ir(wt-90)+ vr(wt-270)*ir(wt-270)
= 0.005w + 0.005w = 0.01w

which is the total power (rate of energy delivery) contained in the
standing wave at the points 45 degrees each side of center. If we want
to find the total energy contained in the standing wave, we would
integrate over the entire time period of 180 degrees.

So think I.

As you can see, by considering that a full wave or complete cycle may
exist on the 1/2 wave transmission line, we can explain power moving
from side to side over the transmission line. I could not have offered
this explanation without your help and guidance and Cecil's as well. I
hope you will agree that it is a correct analysis.

73, Roger, W7WKB