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Keith Dysart wrote:
When the two transmission lines are removed, how can
these imputed powers exist? What impedance did you
assign to the non-existant section of line to permit
you to compute these powers?

I have very deliberately chosen and limited the
scope of the examples to a 50 ohm environment.
Those imputed powers are therefore whatever an
ideal 50 ohm directional wattmeter indicates.

The distributed network model also works for circuits
as it is a superset of the lumped circuit model.
--
73, Cecil http://www.w5dxp.com

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On Apr 7, 8:27*am, Cecil Moore wrote:
Keith Dysart wrote:
Just as one does not expect the
partial values of volts and currents during superposition
to produce a power value that represents a real energy
flow, one should not expect it from Es and Hs which are
the partial values being superposed.

Given two coherent EM waves, W1 and W2, we know
that the power in W1 is E1xH1 and the power
in W2 is E2xH2. That is simple physics.

No matter what the results of superposition
of those two waves, the total energy in both
waves must be conserved. That is simple physics.

If you would think about what happens to two
coherent light waves in free space, you wouldn't
be able to justify your assertion above.

I did not expect you would be happy with the answer.

As long as you stick with simple assertions, followed
by sentences such as, "That is simple physics.",
spoken in a tone which says no further understanding
is necessary, you will be locked in the fruitless
search for the imputed reflected energy flow.

...Keith

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On Apr 7, 8:48*am, Cecil Moore wrote:
Keith Dysart wrote:
When the two transmission lines are removed, how can
these imputed powers exist? What impedance did you
assign to the non-existant section of line to permit
you to compute these powers?

I have very deliberately chosen and limited the
scope of the examples to a 50 ohm environment.
Those imputed powers are therefore whatever an
ideal 50 ohm directional wattmeter indicates.

You seem to be saying that the answers would be
completely different if you chose a different
impedance for the non-existant transmission line.

Or even for the one wavelength line.

That does not make a very robust explanation.

...Keith

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Keith Dysart wrote:
I did not expect you would be happy with the answer.

If you ever comprehend that those RF waves in a
transmission line could just as easily be light
waves in free space, you will realize that your
"answer" violates the existing laws of physics,
i.e. EM waves whose energy content varies
according to your whim.
--
73, Cecil http://www.w5dxp.com

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On Sun, 6 Apr 2008 19:21:00 -0700 (PDT)
Keith Dysart wrote:

On Apr 5, 10:06*am, Roger Sparks wrote:
On Sat, 5 Apr 2008 03:06:18 -0700 (PDT)

Keith Dysart wrote:
On Apr 4, 11:41*am, Roger Sparks wrote:
On Fri, 4 Apr 2008 06:30:08 -0700 (PDT)

Keith Dysart wrote:
On Apr 4, 1:29*am, Roger Sparks wrote:
[snip]
Another way to figure the power to the source would be by using the voltage and current through the source. *

This is how I did it.

Taking Esource.50[90..91] = 0.03046 J as an example ...

Psource.50[90] = V * I
* * * * * * * *= 0.000000 * 0.000000
* * * * * * * *= 0 W
Psource.50[91] = -2.468143 * -0.024681
* * * * * * * *= 0.060917 W

Using the trapezoid rule for numerical integration,
Esource.50[90..91] = ((Psource.50[90]+Psource.50[91])/2) * interval
* * * * * * * * * *= ((0+0.060917)/2)*1
* * * * * * * * * *= 0.030459 J

The other powers and energies in the spreadsheet are computed
similarly.

There was an error in the computation of the 'delta's; the sign was
wrong.
The spreadsheet available athttp://keith.dysart.googlepages.com/radio6
has
now been corrected. (And, for Cecil, this spreadsheet no longer has
macros
so it may be downloadable.)

To use this spreadsheet to compute my numbers above, set the formulae
for Vr.g and Ir.g in the rows for degrees 90 and 91 to zero.

Using current and voltage, the power at time 91 degrees of the reflected wave is 1.234v*1.43868a= 1.775w. *Over one second integrated, the energy should be 1.775 J.

I could not match to the above data to any rows, so I can't comment.

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Keith Dysart wrote:
As long as you stick with simple assertions, followed
by sentences such as, "That is simple physics.",
spoken in a tone which says no further understanding
is necessary, you will be locked in the fruitless
search for the imputed reflected energy flow.

My vision is returning and could turn out to be
the best vision that I've ever had in my life. :-)

You have been asking for the mechanism for storage
and return of the interference energy in the system.
That mechanism is standing waves. Are you aware that
standing waves store energy and return it to the
system every 90 degrees? In the examples being discussed,
there are standing waves inside the source. For the 1/8WL
shorted line, there appears to be 125 watts of forward
power and 25 watts of reflected power at points on each
side of the source.

With powers given in average values, the circuit that
you should be using for your instantaneous power
equations is:

50 ohm
----50-ohm----/\/\/\/\----50-ohm----
125w-- 100w 50w--
--25w --50w

I will be very surprised if the instantaneous
powers don't balance.

Your previous problem is that you were using net
power values on one side of Rs and component power
values on the other side of Rs. That's apples and
oranges, a known no-no.
--
73, Cecil http://www.w5dxp.com

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Roger Sparks wrote:
. . .
If we remove the transmission line from the circuit, we have an open circuit with no current. Without current, there can be no power. How can power arrive at Rs if there is no power coming through the transmission line?
. . .

I regret that I haven't been able to take the time to contribute to or
even follow this very interesting thread. So please pardon me if the
following comment isn't relevant.

The quoted statement reminded me of an earlier thread where Keith asked
a similar question, and proposed some interesting alternative
explanations. The point in question was a current node in a lossless
line which had infinite SWR, e.g., an open circuited line. A look at the
energy flow in the vicinity of the node reveals a plausible explanation.
It turns out that energy is being stored at the node, in the line's
capacitance, to the tune of C*V^2/2. For half the cycle, energy is
flowing into the node equally from both directions, so the power
measured at that node is zero. During the other half of the cycle,
energy is flowing out of the node equally in both directions. So the
power remains zero at the node for the entire cycle. This is, of course,
consistent with the zero current at the node. The node's energy
increases to a maximum and down to zero periodically in step with the
square of the node voltage.

Now, I don't know of any way to assign "ownership" to bundles of energy.
But let's suppose that the energy which flows into the node from the
left side during the "inhalation" part of the cycle is the energy which
flows out to the right during the "exhalation" part of the cycle, and
the energy flowing into the node from the right exits on the left. So
now we've managed to get energy past the node going in both directions
while maintaining zero power and current at the node and conserving
energy as we must.

This can be seen graphically with the little TLVis1 program I made up a
while back -- see the thread of that name for more info. The phenomenon
I'm describing can be seen in demo 4, one cycle from the output (right)
end of the line.

Roy Lewallen, W7EL

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Roy Lewallen wrote:
Now, I don't know of any way to assign "ownership" to bundles of energy.

One way is to add a unique bit of modulation to each
bundle of wave energy. I am fond of using a TV signal
and observing ghosting on the screen. This, of course,
assumes that the modulation stays with the same component
wave to which it was originally associated.

But let's suppose that the energy which flows into the node from the
left side during the "inhalation" part of the cycle is the energy which
flows out to the right during the "exhalation" part of the cycle, and
the energy flowing into the node from the right exits on the left. So
now we've managed to get energy past the node going in both directions
while maintaining zero power and current at the node and conserving
energy as we must.

This agrees with the distributed network model. Since
there is no impedance discontinuity and no impedor at
the node, there can be no reflections at the node. The
forward wave flows unimpeded through the node as does
the equal magnitude reflected wave. The net energy flow
is zero. The average energy flow is zero.

Anyone who believes there is zero energy at a standing-
wave current node should touch that point on a transmission
line (which just happens to be the same point as the
maximum voltage anti-node).

One must be careful not to confuse the net signal with the
component signals. One must be careful not to confuse the
average values with the instantaneous values.

This can best be visualized using light waves in free
space. Unimpeded EM waves do not bounce off of each other.
--
73, Cecil http://www.w5dxp.com

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Keith Dysart wrote:
You seem to be saying that the answers would be
completely different if you chose a different
impedance for the non-existant transmission line.

There you go again, trying to shove your words
into my mouth. (Pattooieee!) Please don't do that.

You have apparently done the math and found it
to be valid so, once again, you have to change
the specified conditions in order to try to make
your point.

A Z0 of 50 ohms is the *only* characteristic
impedance that will meet the specified precondition
of zero average interference. Choosing any other
characteristic impedance will move the example
outside of the scope of my Part 1 article. I
understand why you want to do such a thing but
obfuscation, diversions, and straw men are not
part of the scientific method. My Part 1 article
has a very narrow scope. Please abide by it.

Killing a snake and using it for the transmission
line would likewise violate the same specified
preconditions. Please feel free to do that and
report the results.
--
73, Cecil http://www.w5dxp.com

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On Apr 7, 12:14*pm, Roger Sparks wrote:
On Sun, 6 Apr 2008 19:21:00 -0700 (PDT)

Keith Dysart wrote:
On Apr 5, 10:06*am, Roger Sparks wrote:
Pg(t) is the result of a standing wave, containing *power from Pf(x) and Pr(x+90). *

This is one way of thinking of it, but it is less misleading to
consider
that Pg(t) describes the actual energy flow, just as Vg(t) describes
the
actual voltage and Ig(t) describes the actual current. Using
superposition
Vf, If, Vr and Ir can be derived and from these Pf and Pr.

Your argument is correct to the extent that the power you describe is passing point Pg(t) at the instant (t). *It is the equivalent statement that an observer watching cars pass on the freeway would make, saying "one blue car moving left and one red car moving right, so two cars are passing". *Not wrong, just "how is the information useful"?

Pg(t) is the actual power at that point in the circuit. It can be
derived by simply multiplying the direct measurement of the actual
voltages and currents at that point in the circuit. One measures
the same voltages and currents regardless of whether it is a
transmission line to the right of point g, or the equivalent
lumped circuit element.

While Vf, Vr, etc. can be used to derive the same information and,
therefore is arguably just a different point of view, Vf and Vr,
If and Ir, etc., must always be used in pairs to arrive at the
actual circuit conditions. It is when one starts to look at them
separately, as if they individually represent some part of reality,
that confusion awaits.

Thus I strongly suggest that Vg, Ig, Pg, represent reality. The
others are a convenient alternative view for the purposes of
solving problems.

Typically we see Vg split into Vf and Vr, but why stop at two.
Why not 3, or 4? Analyzing a two wire telephone line will use
four or more, forward to the east, forward to the west, reflected
to east, reflected to the west, and sometimes many different
reflections. How do we choose how many? Depends on what is
convenient for solving the problem. The power of superposition.
But assigning too much reality to the individual contributors
can be misleading.

If we can't account for the power, it is because we are doing the accounting incorrectly.

And the error in the accounting may be the expectation that the
particular set of powers chosen should balance. Attempting to
account for Pr fails when Pr is the imputed power from a partial
voltage and current because such computations do not yield powers
which exist.

If we remove the transmission line from the circuit, we have an open circuit with no current. *Without current, there can be no power. How can power arrive at Rs if there is no power coming through the transmission line? *

There is power coming from the transmission line. Looking at Pg(t),
some of the time energy flows into the line, later in the cycle
it flows out. The energy transfer would be exactly the same if the
transmission line was replaced by a lumped circuit element. And
we don't need Pf and Pr for an inductor.

But this flow is quite different than the flow suggested by Pf and
Pr. These suggest a continuous flow in each direction. It is only
when they are summed that it becomes clear that flow is first in
one direction and then other.

Would it help to consider that before the "reflection from the short" arrives, power arrives via the transmission line path but the impedance is 100 ohms for our example, composed of Rs = 50 ohms and transmission line = 50 ohms? *After the "reflection from the short" arrives, the impedance drops to 70.7 ohms so the power to the circuit goes up (assuming a constant voltage source). *How can this happen if power is not carried via the "reflection from the short"?

It goes up because the impedance presented by the transmission
changes when the reflection returns. This change in impedance
alters the circuit conditions and the power in the various
elements change. Depending on the details of the circuit,
these powers may go up, or they may go down when the reflection
arrives.

...Keith