K7ITM wrote:
On Feb 23, 2:32 pm, Cecil Moore wrote:
...
If you, like Roy and Tom, don't care where the power
goes,


I really do wish you'd learn to read. I did NOT say I don't care
where the power goes.

I care where the power goes, too. And I know where it goes: from my
transmitter through the transmission line to the antenna.

Roy Lewallen, W7EL

Roy Lewallen wrote:
I care where the power goes, too. And I know where it goes: from my
transmitter through the transmission line to the antenna.

Roy, here's a quote from your web page:
"I personally don't have a compulsion to understand
where this power 'goes'."

How can you care where the power goes when you don't care
to understand where the power goes?

You also scream out:
"THE REVERSE POWER IS NOT DISSIPATED IN OR ABSORBED BY THE
SOURCE RESISTANCE."

I just presented a special case example where all of the
reverse energy is dissipated (converted to heat) in the
source resistance. All it takes is one example to prove
your above statement to be false.

Your web page even contains one of those special case
examples. When your load is 0 +/- j50 at the end of
a 1/2WL piece of 50 ohm line, the forward wave and
reflected wave are 90 degrees out of phase at the
50 ohm source resistor. Under that special condition,
all of the reflected energy is dissipated in the
source resistor. Your own example proves your statement
to be false.
--
73, Cecil http://www.w5dxp.com

Cecil Moore wrote:


Maxwell and Bruene have been arguing about such for 20
years now. The need to discover "where the power goes"
apparently exists for them and others.

If you, like Roy and Tom, don't care where the power
goes, that OK, but please don't try to present yourselves
as experts on a subject you don't care enough about
enough to have ever studied it in detail.


Cecil,

I must have missed the messages in which you resolved this 20 year
battle through your pioneering analysis technique.

Who won?

8-)

73,
Gene
W4SZ

On Sun, 24 Feb 2008 16:23:49 GMT, Gene Fuller
wrote:

Who won?

Gad Gene,

Now we will get postings signed Cecil Balboa! (Older, pudgier, but
still swinging after all these years.)

73's
Richard Clark, KB7QHC

Gene Fuller wrote:
I must have missed the messages in which you resolved this 20 year
battle through your pioneering analysis technique.

I didn't say it had been resolved. Those guys argue
about real-world class-C amplifiers. I didn't want
to discuss sources at all but Keith led me down the
primrose path. :-)

The small part of the argument that I can resolve is
the ideal class-A voltage and current sources. But
perhaps that resolution will spill over into the
more complicated configurations. I don't recall
anyone else considering what effect interference
has on energy redistribution inside an RF source.

But I indeed do have it figured out for ideal linear
voltage and current sources. Not only does it agree
with the results of other valid analysis methods,
but it answers the question: What happens to the
reflected energy when it is incident upon an ideal
linear source?
--
73, Cecil http://www.w5dxp.com

On Feb 21, 9:53*pm, Cecil Moore wrote:
Keith Dysart wrote:

Hey Keith, how about this one?

* * * * * * * *Rs * * * * * * Pfor=50w--
* * * * *+----/\/\/-----+----------------------+
* * * * *| * *50 ohm * * * * * *--Pref * * * *|
* * * * *| * * * * * * * * * * * * * * * * * * |
* * * * Vs * * * * * * * * * 45 degrees * * RLoad+j0
* * * 100v RMS * * * * * * * 50 ohm line * * * |
* * * * *| * * * * * * * * * * * * * * * * * * |
* * * * *| * * * * * * * * * * * * * * * * * * |
* * * * *+--------------+----------------------+

The dissipation in the source resistor is:

P(Rs) = 50w + Pref

How can anyone possibly argue that reflected power
is *never* dissipated in the source resistor? :-)

It is time for a more complete analysis. For the sake of an
example let us set Rload to 150 ohms. That makes a nice
voltage RC at the load of 0.5.

Vs = 100 V rms

At the generator end
Vf.g = 50 V rms
Pf.g = 50 W avg

Vr.g = 25 V rms
Pr.g = 12.5 W avg

At the source resistor, before the reflection returns, the
dissipation in the source resistor is
Prs.br = 50 W avg
and after the reflection returns it is
Prs.ar = 62.5 W avg

Yup, it sure looks like that reflected wave is dissipated in the
source resistor since 62.5 = 50 + 12.5, does it not?

But let us examine in more detail...

Vs = 100 V rms
Vs(t) = 141 cos(wt)

Vf.g = 50 V rms
Vf.g(t) = 70.7 cos(wt)

Pf.g = 50 W avg
Pf.g(t) = 50 cos(2wt) + 50

Vr.g = 25 V rms
Vr.t(t) = 35.35 cos(wt-90degrees)

Pr.g = 12.5 W avg
Pr.g(t) = 12.5 cos(2wt) + 12.5

The power in the source resistor, before the reflection returns is
Prs.br = 50 W avg
Prs.br(t) = 50 cos(2wt) + 50

and after the reflection returns
Prs.ar = 62.5 W avg
Prs.ar(t) = 62.5 cos(2wt-53.13degrees) + 62.5

So, while the average powers seem to sum nicely to support the
claim, when the actual power as a function of time is examined
it can be seen that the power in the source resistor is NOT the
sum of its dissipation pre-reflection plus the power from the
reflection.

This solidly disproves the claim that the reflected power is
dissipated in the source resistor.

You may find the spreadsheet at
http://keith.dysart.googlepages.com/...oad,reflection
useful in taking some of the tedium out of the calculations.
And yes, if your anti-virus tools do not like Excel macros, you may
have to invoke your authority over your tools. (They do know who is
boss, do they not?)

...Keith

Keith Dysart wrote:
This solidly disproves the claim that the reflected power is
dissipated in the source resistor.

I will respond to your posted data after awhile. In the
meantime, let's correct the misconception contained in
your above statement.

No one is asserting that reflected power is *always*
dissipated in the source resistor so that discussion
is a dead horse. The assertion that I am making is:

Reflected RF energy is *sometimes* dissipated (converted
to heat) in the source resistor and sometimes it is not.
With an ideal voltage or current source, I can tell you
when it happens and when it doesn't happen along with
the quantitative values involved.

What is being questioned is Roy's assertion in his
"Food for Thought" paper. Quoting:

"THE REVERSE POWER IS NOT DISSIPATED IN OR ABSORBED BY
THE SOURCE RESISTANCE."

It seems clear that he is saying that reverse power is
*never* dissipated in the source resistance. That is
simply not true. Sometimes, there is nowhere else for
the reflected energy to go.

And yes, if your anti-virus tools do not like Excel macros, you may
have to invoke your authority over your tools. (They do know who is
boss, do they not?)

I am reluctant to turn off my firewall and virus protection
while I download the EXCEL file. I once had a virus/worm
that even survived a reformatting of my hard drive. Apparently,
it was stored somewhere besides the hard drive.
--
73, Cecil http://www.w5dxp.com

Keith Dysart wrote:
So, while the average powers seem to sum nicely to support the
claim, when the actual power as a function of time is examined
it can be seen that the power in the source resistor is NOT the
sum of its dissipation pre-reflection plus the power from the
reflection.

Whoa Keith, Dr. Best's power equation that is being used
does not work for instantaneous powers. It is adapted from
the irradiance equation from optical physics which is a
*time averaged power density*. The irradiance equation
does NOT work for instantaneous powers even in the
field of optical physics and it is not supposed to.

Trying to use Dr. Best's power equation on instantaneous
powers is akin to trying to measure the feedpoint impedance
of an antenna with a DC ohm-meter. It is the misuse of a
tool. All powers or power densities appearing in the
power equation *must* be integrated over at least one
complete cycle. Instantaneous powers are simply excluded
from this energy model that we have been discussing.

Note that every voltage, current, and power in Dr. Best's
QEX article is an average value.

I have made absolutely no assertions about instantaneous
powers so your instantaneous power data is irrelevant to
my assertions. When I say "power", I am always talking about
time averaged power - never about instantaneous power.

You are free to make assertions about instantaneous power
but those assertions do not apply to my statements
about average power. They may be interesting to you but
have nothing to do with anything that I have been saying.
--
73, Cecil http://www.w5dxp.com

Keith Dysart wrote:
So, while the average powers seem to sum nicely to support the
claim, when the actual power as a function of time is examined
it can be seen that the power in the source resistor is NOT the
sum of its dissipation pre-reflection plus the power from the
reflection.

Here's a quote from "Optics", by Hecht, concerning power
density (irradiance). "Furthermore, since the power
arriving cannot be measured instantaneously, the detector
must integrate the energy flux over some finite time, 'T'.
If the quantity to be measured is the net energy per unit
area received, it depends on 'T' and is therefore of limited
utility. If however, the 'T' is now divided out, a highly
practical quantity results, one that corresponds to the
average energy per unit area per unit time, namely 'I'."
'I' is the irradiance (*AVERAGE* power density).

i.e. The irradiance/interference equation does not work
for instantaneous powers which are "of limited utility".
--
73, Cecil http://www.w5dxp.com