Hi Walt,

I'm a little confused here. I hope you can straighten me out.

Walter Maxwell wrote:

It appears to me that even with all the successive posts on the subject of power in the standing wave, you all
seem to be missing the ingredient that proves why there is no useable power in the standing wave. It is
because the current and voltage in the standing wave are 90° out of phase. Multiplying E x I under this
condition results in zero power.

I've always regarded a "standing wave" as being a description of the
envelope caused by the interference between forward and reverse
traveling waves. But you're saying there are currents and voltages "in"
the standing wave. Are you referring to the total current and voltage at
any point along the line? If so, why are they always in quadrature?
Certainly, the total V and I are in quadrature if the line is terminated
by an open, short, or purely reactive load. But not in any other case.
Or do you regard a line as having a "standing wave" with its own voltage
and current which are different from the total V and I? If so, how do
you define a "standing wave"? Are there separate equations for "standing
wave" V and I that are different than for total V and I?

. . .

Roy Lewallen, W7EL

Dave wrote:
you can measure the 'standing' wave voltage,
that has been known for a long time... but the effects are NOT due to power
in standing waves.

Are the effects due to energy in the standing waves?
--
73, Cecil http://www.w5dxp.com

Roger wrote:
Haste makes waste, and errors as well. The standing wave power
equation is incorrect. It should read "Power = V^2 / Zo + I^2 * Zo"

I'm afraid you will find that those are the equations for
power associated with a traveling wave. Actually should be
Power = V^2/Z0 = I^2*A0. There is no net power transfer
associated with a standing wave. For a pure standing wave,
V*I*cos(A) = 0 watts.
--
73, Cecil http://www.w5dxp.com

Walter Maxwell wrote in
:

....
It appears to me that even with all the successive posts on the
subject of power in the standing wave, you all seem to be missing the
ingredient that proves why there is no useable power in the standing
wave. It is because the current and voltage in the standing wave are
90° out of phase. Multiplying E x I under this condition results in
zero power.

Walt,

I am trying to make sense of this and the first issue is what you mean by
the term "standing wave".

The only meaning that seems possible is that it is the magnitude of the
time alternating voltage or current at some displacement along the
transmission line.

If that is the meaning, then the situation you describe of 90° phase
difference between E and I is rather specific, it can only occur with a
distortionless line AND a load that is (s/c OR o/c OR purely reactive).

Is that the case?

If so, should you have stated the assumptions and how does the case you
discuss help in explanation of general principles?


In addition to another comment above that implies that reflected power
is reactive power, this is not true--reflected power is as real as
forward power. The only differences are that they are traversing in
opposite directions, and that while the voltage and current travel in
phase in the forward direction, they are traveling 180° out of phase
in the rearward direction. Multiplying voltage and current while 180°
different in phase results in the same power as when they are in
phase.

You seem to be inferring that it is legitimate (in a general sense) to
calculate the power of forward and reflected waves as voltage times
current (eg Vf*If).

Isn't the instantaneous power at a point a function of time, and it is p
(t)=v(t)*i(t) and the expansion of that equals Vf*If-Vr*Ir ONLY when the
other two terms of the expansion cancel, and that is the special case of
a distortionless line.

Are you illustrating general principles with a special case without
stating the underlying assumptions.

Why is it that so many attempts to explain transmission line behaviour,
particularly regarding real and imaginary components of power at a point,
aren't consistent with basic AC circuit theory?

Owen

Gene Fuller wrote:
Where do you get so many goofy ideas? Do you have any references at all
that support your contention that standing wave energy does not meet the
definition of EM energy? I have been in the wave business professionally
for about 40 years, and I have read many technical papers, reference
books, and text books. I have yet to encounter anything that indicated
the inferior nature of standing waves in the energy community.

I guess the authors of the textbooks never thought anyone
would be so ignorant as to believe that EM waves can stand
still. :-)

EM waves are photonic in nature must travel at the speed of
light in the medium. A standing wave stands still and oscillates
in place. Therefore, A standing wave is not an EM wave - It is
something else, by definition.
--
73, Cecil http://www.w5dxp.com

"Power waves" is a standing joke around here. The last person to
seriously consider such things is Cecil, and he now denies ever saying
such.

That's a false statement. I supported power waves back in
the 1990s. I changed my mind around 1998, almost ten years
ago. I have not supported power waves in this 21st century.
Some other posters on this news still support power waves.
However, I do support EM energy waves.
--
73, Cecil http://www.w5dxp.com

John Smith wrote:
Roy Lewallen wrote:

...
This is correct. Cecil and others have often muddled things by
considering only average power, and by doing this, important
information is lost. (As was the case of the statistician who drowned
crossing a creek whose average depth was only two feet.)
...


This:
2. Microwave ovens use standing waves to cook food. This means that
nodes, where the amplitude is zero (where the wave crosses the x-axis),
remain at nearly fixed locations in the oven, and cooking won't occur at
those locations.

From he
http://faculty.fortlewis.edu/tyler_c..._microwave.htm

Now, you can argue that any damn way you wish, but "standing waves of no
power" is a myth for idiots!

Oddly, "John", another source tells us:

Another hazard is the resonance of the magnetron tube itself. If the
microwave is run without an object to absorb the radiation, a standing
wave will form. The energy is reflected back and forth between the tube
and the cooking chamber.

http://en.wikipedia.org/wiki/Microwave_oven

Dave K8MN

John Smith wrote:
Roy Lewallen wrote:

...
This is correct. Cecil and others have often muddled things by
considering only average power, and by doing this, important
information is lost. (As was the case of the statistician who drowned
crossing a creek whose average depth was only two feet.)
...


This:
2. Microwave ovens use standing waves to cook food. This means that
nodes, where the amplitude is zero (where the wave crosses the x-axis),
remain at nearly fixed locations in the oven, and cooking won't occur at
those locations.

From he
http://faculty.fortlewis.edu/tyler_c..._microwave.htm

Now, you can argue that any damn way you wish, but "standing waves of no
power" is a myth for idiots!

Another source, "John", says:

Why is food cooked in a microwave oven sometimes not cooked uniformly?

Inside the microwave oven, the microwaves bounce off the metal internal
walls and set up complex 'standing wave' patterns. As with any wave,
microwaves have peaks and troughs and the intensity of the microwaves is
greatest in the peaks and troughs and lowest at points in between.

So if some food is near one of the peaks it will absorb lots of
microwaves and get really hot, while if it is midway between peaks and
troughs it may receive hardly any microwaves and so not get very hot at
all.

http://www.bbc.co.uk/food/tv_and_rad...icrowave.shtml

So you're telling us that what cooks food in a microwave oven is the
standing waves and that it isn't because the food itself is the load for
the output of the magnetron? You'd have us believe that standing waves
which result from operating a microwave oven without such a load are
present and actually cooking food?

Dave K8MN

Richard Harrison wrote:
Keith Dsart wrote:
"Therefore, the forward and reverse waves can not be transferring energy
across these points."

Waves in motion are transporting energy no matter how their constituents
seem to add at a particular point.

We can make Keith's assertion true by the addition of one
word.

"Therefore, the forward and reverse waves cannot be transferring
*net* energy across these points. As Ramo and Whinnery say about
the forward and reflected Poynting vectors:

If Pz+ = Pz- then Pz+ - Pz- = 0
--
73, Cecil http://www.w5dxp.com

Walter Maxwell wrote:
It appears to me that even with all the successive posts on the subject of power in the standing wave, you all
seem to be missing the ingredient that proves why there is no useable power in the standing wave. It is
because the current and voltage in the standing wave are 90° out of phase. Multiplying E x I under this
condition results in zero power.

In addition to another comment above that implies that reflected power is reactive power, this is not
true--reflected power is as real as forward power. The only differences are that they are traversing in
opposite directions, and that while the voltage and current travel in phase in the forward direction, they are
traveling 180° out of phase in the rearward direction. Multiplying voltage and current while 180° different in
phase results in the same power as when they are in phase.

Seems to me everything would be made clear by the addition of the
word "net". There is no net power in pure standing waves. There is
no net energy transfer in pure standing waves.
--
73, Cecil http://www.w5dxp.com