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Chuck wrote in
:

On Sun, 9 Mar 2008 15:07:26 -0700 (PDT), K7ITM wrote:

snip


Note that, as far as I've been able to determine, Michelson did not
have a coherent light source to shine into his interferometer, but
still he saw interference patterns. Perhaps he had invented lasers

snip


It is said he used sodium vapor gas light (~589 nm). Coherent enough.

I have a wonderful old book, named "A Treatise on Optics", written sometime
in the late 1800's. I have to look it up here, Perhaps Cecil would be
interested in a pdf copy if I locate it again.

Those old boys could do an awful lot that we do today, and I was amazed at
the precise measurements that they could do with almost entirely mechanical
devices. Coherent light is not the half of it!

We are both more and less advanced than we might think.

- 73 de Mike -

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K7ITM wrote:
Cecil Moore wrote:
Tom, you are the one who implied that coherency or
incoherency doesn't matter.

No, you're the one who inferred that from what I actually did say,
which is (in assorted different wordings) that we're dealing with a
linear system, and as such we can fully analyze it and indeed get MORE
insight into what's going on by using instantaneous voltages and
currents than we can in looking only at averaged powers.

Tom, I have apologized, eaten crow, and said Mea Culpa.
What more do you want from me?
--
73, Cecil http://www.w5dxp.com

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Roger Sparks wrote:
For the resistor Rs, it will have power applied from two sources, the source and the reflected power from the transmission line, i.e., the source power and reflected power are in series when considered in relationship with the resistor Rs. The problem is that 'what looks like two sources to resistor Rs, is really only one source, Vs'.

Actually, it will have energy applied from three sources, the
source, the reflected energy, and the reactive energy stored
in the transmission line. The energy that Keith is missing
comes from the reactance in the transmission line.
--
73, Cecil http://www.w5dxp.com

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On Mon, 10 Mar 2008 02:38:05 GMT
Cecil Moore wrote:

Roger Sparks wrote:
For the resistor Rs, it will have power applied from two sources, the source and the reflected power from the transmission line, i.e., the source power and reflected power are in series when considered in relationship with the resistor Rs. The problem is that 'what looks like two sources to resistor Rs, is really only one source, Vs'.

Actually, it will have energy applied from three sources, the
source, the reflected energy, and the reactive energy stored
in the transmission line. The energy that Keith is missing
comes from the reactance in the transmission line.
--
73, Cecil http://www.w5dxp.com

Here is a link to a discussion of AC power. http://en.wikipedia.org/wiki/AC_power

How do you propose to seperate reactive power from reflected power?

--
73, Roger, W7WKB

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Cecil Moore wrote:
K7ITM wrote:
Cecil Moore wrote:
Tom, you are the one who implied that coherency or
incoherency doesn't matter.

No, you're the one who inferred that from what I actually did say,
which is (in assorted different wordings) that we're dealing with a
linear system, and as such we can fully analyze it and indeed get MORE
insight into what's going on by using instantaneous voltages and
currents than we can in looking only at averaged powers.

Tom, I have apologized, eaten crow, and said Mea Culpa.
What more do you want from me?

Rending of hair and wearing of sackcloth? 8^)

- 73 de Mike N3LI -

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Roger Sparks wrote:
How do you propose to seperate reactive power from reflected power?

For average power, it's easy using phasors.
real power = V*I*cos(theta)
reactive power = V*I*sin(theta)

I know very little about instantaneous power having
avoided thinking about it for most of my life.
--
73, Cecil http://www.w5dxp.com

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Cecil Moore wrote:
Dave wrote:
"Keith Dysart" wrote in message
Cecil Moore wrote:
So I stopped talking about "reflected power" and
started talking about "reflected energy". Now you object
to the use of the term "reflected energy". Would you and
the rest of the guru attack gang please get together on
what term you would like for me to use?

NEITHER! they are both confusing. use the most fundamental things
that you can measure, either voltage or current. either one is
completely defined in the basic maxwell equations, and either one is
completely sufficient to describe ALL effect on a cable or in any
circuit.

If Maxwell's equations could be used to answer the
questions that we are asking, why haven't they been
answered a long time ago?

How can Maxwell's equations be used to track the path
and fate of the energy in a reflected wave?

Cecil,

There may be some terminology confusion here.

Maxwell's equations are really the only relevant physical equations
there are to work with, at least in the classical regime. The discussion
about constructive, destructive, superposition, linearity, etc.
represents merely mathematical manipulation of the basic physical
entities embodied in Maxwell's equations.

All of this *math* is of course very important and very useful. However,
it is not a replacement for the *physical* laws known as the Maxwell
equations.

(And we all know that these endless threads are closely parallel to the
blind men describing an elephant puzzle.)

73,
Gene
W4SZ

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On Mar 9, 10:35 am, Cecil Moore wrote:
Keith Dysart wrote:
My issue was that you seemed, in that sentence, to be saying
that the reflected energy was dissipated in the source
resistor. But earlier you had stated that was not your claim.

My earlier claim was that the average power in a reflected
wave is dissipated in the source resistor when the forward
wave is 90 degrees out of phase with the reflected wave at
the source resistor. In that earlier claim, I didn't care
to discuss instantaneous power and thus excluded instantaneous
power from that claim.

For instantaneous values, it will be helpful to change the
example while leaving the conditions at the source resistor
unchanged. Here's the earlier example:

Rs Vg Vl
+----/\/\/-----+----------------------+
| 50 ohm |
| 1/8 WL |
Vs 45 degrees 12.5 ohm
100v RMS 50 ohm line Load
| |
| |
+--------------+----------------------+
gnd

Here's the present example:

Rs Vg Vl
+----/\/\/-----+----------------------+
| 50 ohm |
| 1 WL |
Vs 360 degrees 23.5+j44.1
100v RMS 50 ohm line ohm Load
| |
| |
+--------------+----------------------+
gnd

If I haven't made some stupid mistake, the conditions at
the source resistor are identical in both examples.

No silly mistakes. This number is computed by the spreadsheet at
http://keith.dysart.googlepages.com/...oad,reflection
as 23.529411764706+44.1176470588235j, so you are pretty close.

But
in the second example, it is obvious that energy can
be stored in the transmission line during part of a cycle
(thus avoiding dissipation at that instant in time) and be
delivered back to the source resistor during another part
of the cycle (to be dissipated at a later instant in time).
That is the nature of interference energy and is exactly
equal to the difference between the two powers that you
calculated. You neglected to take into account the ability
of the network reactance to temporarily store energy and
dissipate it later in time.

I don't think I did.

The power delivered by the generator to the line is
Pg(t) = 32 + 68 cos(2wt)
The average power delivered to the line is 32W while
the peak power is 100W towards the load and 36W from
the line to the generator.
The exact same function describes the power delivered
to the line for 12.5 ohm load 45 degree line example,
and the 23.5+44.1j ohm load 360 degree line example.

And if the load was connected directly to the generator,
the same power would be delived directly to the load.

In all cases the power dissipated in the source resistor
is
Prs(t) = 68 + 68 cos(2wt-61.92751306degrees)
This power varies between 0.0W and 136W.

This is true even when the load is connected directly to
the generator without a line to created reflected power.

The power provided by the source is
Ps(t) = 100 + 116.6190379 cos(2wt-30.96375653degrees)
= Prs(t) + Pg(t)
so all of the energy delivered by source is nicely
accounted for. All of the energy dissipated in the
source resistor and delivered to the line originates
in the voltage source.

Looking at the line where it connects to the generator,
we find that the forward power is
Pf.g(t) = 50 + 50cos(2wt)
and the reflected power is
Pr.g(t) = -18 + 18cos(2wt)

Pf.g(t) + Pr.g(t) = 32 + 68cos(2wt)
= Pg(t)
As expected, the sum of the forward and reflected power
at the generator terminals is exactly the power delivered
by the generator to the line.

So you are correct, energy is stored in and returned from
the line on each cycle, but this is the net energy, not
the forward or reflected power since these are both
computed into non-reactive impedances, their respective
voltage and current are always in phase, so their powers
always flow in only one direction.

All of the reflected energy is dissipated in the source
resistor, just not at the time you thought it should be.

So as yet, there is no mechanism to explain storage of
reflected power so it can be dissipated at a different
time in the source resistor.

I stand by my contention that the reflected power is not
being dissipated in the source resistor because the
dissipation does not occur at the correct time.

You could convince me to change my contention by locating
the entity that is storing the energy, and showing some
analysis that explains why the amount of the energy
time shift is a function of the magnitude of the
reflected power.

....Keith

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Keith Dysart wrote:
So as yet, there is no mechanism to explain storage of
reflected power so it can be dissipated at a different
time in the source resistor.

Of course there is, Keith. That's what reactances do.
A reactance stores energy during part of the cycle
and gives it back during a different part of the cycle.
The energy stored during part of the cycle is the
destructive interference energy that you are missing
from you equation. It is delivered back during the
next part of the cycle and dissipated 90 degrees
later.

When the signs of the two superposed voltages are opposite,
there is "excess" energy available which is stored in the
transmission line. 90 degrees later, when the voltages
have the same sign thus requiring constructive interference
energy, that "excess" energy is delivered back to the source
resistor to be dissipated.

Since we already know that the interference energy averages
out to zero, the energy imbalance that you discovered is
obviously energy being displaced in time by the reactance
whether it is from a coil or from the transmission line.
--
73, Cecil http://www.w5dxp.com

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On Mar 11, 7:18*pm, Cecil Moore wrote:
Keith Dysart wrote:
So as yet, there is no mechanism to explain storage of
reflected power so it can be dissipated at a different
time in the source resistor.

Of course there is, Keith. That's what reactances do.
A reactance stores energy during part of the cycle
and gives it back during a different part of the cycle.
The energy stored during part of the cycle is the
destructive interference energy that you are missing
from you equation. It is delivered back during the
next part of the cycle and dissipated 90 degrees
later.

When the signs of the two superposed voltages are opposite,
there is "excess" energy available which is stored in the
transmission line. 90 degrees later, when the voltages
have the same sign thus requiring constructive interference
energy, that "excess" energy is delivered back to the source
resistor to be dissipated.

Since we already know that the interference energy averages
out to zero, the energy imbalance that you discovered is
obviously energy being displaced in time by the reactance
whether it is from a coil or from the transmission line.

I can see that words describe a somewhat plausable conjecture,
but to be convincing, a mathematical exposition is needed.

When is the energy stored where? Energy stored in a capacitor,
for example, should correlate with the voltage on the
capacitor. Similarly for current in an inductor.

To be convincing, the various functions of time need to
align appropriately.

So far, the only convincing expressions I have seen
a
Ps(t) = Prs(t) + Pg(t)
and
Pg(t) = Pf.g(t) + Pr.g(t)
and, of course, all the corresponding voltage and
current functions line up as expected.

To show that the reflected power is dissipated in
the resistor will require the derivation of some
X(t) such that
Prs(t) = 50cos(2wt) + Pr.g(t) + X(t)
and X(t) needs to be in terms of some known
physical quantities of the circuit.

...Keith