On Apr 4, 1:29*am, Roger Sparks wrote:
On Wed, 2 Apr 2008 02:41:24 -0700 (PDT)

Keith Dysart wrote:
On Apr 1, 8:08 am, Cecil Moore wrote:
Keith Dysart wrote:
Cecil Moore wrote:
If the average interference is zero, the average
reflected power is dissipated in the source resistor.
All of your unethical lies, innuendo, and hand-waving
will not change that fact of physics.

More precisely, the average value of the imputed reflected
power is numerically equal to the increase in the average
power dissipated in the source resistor.

More precisely, at the *exact* time of arrival of the
reflected wave, the average power dissipated in the
source resistor increases by *exactly* the magnitude
of the average reflected power. Do you think that is
just a coincidence?

Not quite, but close. And averages can not change instantly.

But your phraseology suggests another way to approach
the problem. Let us consider what happens just after
the arrival of the reflected wave. For computational
convenience, we'll use a frequency of 1/360 hertz so
that 1 degree of the wave is one second long.

The reflected wave arrives at 90 seconds (or 90 degrees)
after the source is turned on. If the line is
terminated in 50 ohms, then no reflection arrives and
during the one second between degree 90 and degree 91,
the source resistor dissipates 0.01523 J of energy. The
line also receives 0.01523 J of energy and the source
provides 0.03046 J.
* Esource.50[90..91] = 0.03046 J
* Ers.50[90..91] * * = 0.01523 J
* Eline.50[90..91] * = 0.01523 J
* Esource = Ers + Eline
as expected.
* Efor = Eline
since Eref is 0.

OK, Power at the source is found from (V^2)/50 = 0.03046w. *

Not quite.

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 at http://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.
But
perhaps the explanation above will correct the discrepancy.

Now let us examine the case with a 12.5 ohm load and
a non-zero reflected wave.

This was an ooooopppps. The spreadsheet actually calculates
for a shorted load.

During the one second between degree 90 and degree 91,
the source resistor dissipation with no reflection
is still 0.01523 J and the imputed reflected wave
provides 99.98477 J for a total of 100.00000 J.

But the source resistor actually absorbs 98.25503 J
in this interval.

100 - 1.775 = 98.225 J

I computed 98.25503 J from applying the trapezoid rule for
numerical integration of the power in the source resistor
at 90 degrees and 91 degrees.

I am not sure where you obtained 1.775 J.

The apparent impedance of the reflect wave has changed from 50 ohms to 0.8577 ohms.

There is still a confusion between what power comes from where at time 91 degrees, at least partly coming from how Vg is used. *Or maybe the confusion is because we are mixing a circuit with a shorted line with calculations from a line with a 12.5 ohm load.

I do not think the latter is happening. If you want to see
the results for a 12.5 ohm load, set the Reflection Coefficient
cell in row 1 to -0.6.

Ooooopppps. It does not add up. So the dissipation
in the source resistor went up, but not enough to
account for all of the imputed energy in the
reflected wave. This does not satisfy conservation
of energy, which should be sufficient to kill the
hypothesis.

But Esource = Ers + Eline as expected.
* Esource.12.5[90..91] = -1.71451 J
* Ers.12.5[90..91] * * = 98.25503 J
* Eline.12.5[90..91] * = -99.96954 J
Note that in the interval 90 to 91 degrees, the source
is absorbing energy. This is quite different than what
happens when there is no reflected wave.

clip
To recap, it is when a total flow is
broken into multiple non-zero constituent flows
that energy flow imputed to the constituents
is a dubious concept.

...Keith

The power seems pretty well accounted for degree by degree so far as I can see right now. *

I think your spread sheet would be easier to follow if the reflected power beginning at 91 degrees was positive, like adding batteries in series. *That is what happens and there is no reversal of voltage when examined from the perspective of the resistor. *There is only a delay in time of connection to the voltage source on the line side of the resistor.

When I wrote my original spreadsheet I had to decide which convention
to
use and settled on positive flow meant towards the load. Care is
certainly
needed in some of the computations. Choosing the other convention
would
have meant that care would be needed in a different set of
computations.
So I am not sure it would be less confusing, though it would be
different.

When using superposition, one has to pick a reference voltage and
current
direction for each component and then add or subtract the contributing
voltage or current depending on whether it is in the reference
direction
or against the reference direction. Mistakes and confusion with
relation
to the signs are not uncommon. Careful choice of reference directions
will
sometimes help, but mostly it simply leads to some other voltage or
current that needs to be subtracted instead of added.

...Keith