On Dec 22, 8:43 am, Denny wrote:
Nice graphic, Cecil.. But the thread has drifted beyond recognition..
Part of the original dispute across a couple of threads as I
remember it, was the contention that there is no energy contained
within the reflected wave and therefore no energy contained within the
standing wave, i.e. a mere artifact...

I'd suggest this is a mischaracterization of the contention. I have
seen no disagreement with the notion that the line contains energy.

Assertions about a lack of energy in reflected waves is not
inconsistent with the line containing energy.

I simply wanted to point out that the standing wave on a line does
contain energy and it is a childishly simple exercise to prove it,
therefore the reflected wave must contain energy...

Prepare yourself to rethink this connection.

As far as the questioner, where does the energy go between the
standing wave peaks - oy vey....
If it is a real question - as opposed to a rhetorical device which I
hope was the intent -

It was not rhetorical, but an educational question that followed
from the claim. With the claim that a lit flourescent bulb
demonstrates the presence of energy, it is entirely reasonable
to question what a dark lamp means and the original post did
not suggest this understanding.

then the profound ignorance

There is no need to descend to the level of insult commonly
used by some of the more prolific posters.

of basic physics is
vastly beyond the limited space I have to go over it... See ANY
introductory level, physics textbook for details...

--------------------------------------------------------------

Let us consider a transmission line....

There IS a voltage and current distribution on this line. For
the moment attempt to forget standing waves, travelling waves,
forward waves, reflected waves, .... Just that:

There IS a voltage and current distribution on this line.
These distributions can be expressed as functions of distance
along the line and time:

V(x,t)
I(x,t)

These are the instantaneous real voltage and current
at a particular location (x) and time (t). They can be
measured with a voltmeter and ammeter, though this gets
more challenging at higher frequencies.

Now we know from basic electricity that Power is Volts
times Amps, so we have:

P(x,t) = V(x,t) * I(x,t)

P(x,t) is the instantaneous power at any point and time
on the line. Power being the rate of energy flow, P(x,t)
is the instantaneous energy flow at that point and time
on the line.

If you disagree with any of the above please read no
further and post any objections now.

Good! Agreement.

So let's consider the specific example of sinusoidal
signal applied to a transmission line that is open
at the end. After settling, there is a voltage and
current distribution on this line, but how can we
describe it? Now some of you are immediately
thinking "standing wave", and you'd be right. Its
an excellent description, but we need to look at
the details.

So V(x,t) = A cos(x) cos(wt)
where w is radians/second and x is measured in
degrees back from the open end.

Consider t=0. The spatial voltage distribution
is a sinusoid with a maximum at the open end. As
time advances, this spatial sinusoid drops in
amplitude until the voltage everywhere on the
line is 0, then the amplitude heads towards
minus max. Noting that the zero crossings are
always in the same place and the shape is
sinusoidal leading to the name "standing wave".

From a time perspective, every point on the line
has a sinusoidal voltage, but the amplitude
changes with position. The peaks and zero crossings
occur at the same time everywhere, thus the
claim that there is no phase shift as one
moves down the line.

The current is also a sinusoid, but shifted 90
degrees from the voltage sinusoid, thus there
is a current zero where-ever there is a voltage
maximum.

Now power is really interesting. Recall that

P(x,t) = V(x,t) * I(x,t)

At certain values of t, the voltage everywhere
on the line is 0, so at these times, no energy
is flowing anywhere on the line. Similarly for
current.

And at certain positions on the line (n*180+90)
the voltage is always 0, so the power is always
0 at these points. No energy is ever flowing
at these points. Similarly for current at points
(n*180).

The sumary, being as pedantic as I can is that
"standing wave" is "merely" a description of the
voltage and the current on the line. And
"merely" is in quotes, because it is a very
powerful and useful description, but be careful
not to ascribe too much to it. Always keep in
mind that the "standing wave" is a description
of the conditions on the line, not the creator
of those conditions.

Now some of you will be saying "Yes!", it IS
the forward and reverse waves that create the
conditions on the line. Not so fast. These
too are "merely" partial descriptions that
when summed, describe the conditions on the
line. Very powerful and useful, but again
descriptions, not creators.

Some readers overuse forward and reverse waves
and start ascribing power to them. These readers
think that the forward wave transports energy
to the end of the line which is reflected
back in the reverse wave. To those readers I
offer the following counter-proof....

In the setup above used for "standing waves"
it can be seen that there is zero power in
the line every 90 degrees back from the open
end. At a zero power point, no energy is
being transferred. Therefore, the forward
and reverse waves can not be transferring
energy across these points. Conclusion:
forward and reverse waves do not always
transport energy.

....Keith