Roy Lewallen wrote in
:

For all the fluff about
photons, optics, non-dissipative sources, and the like, I have yet to
see an equation that relates the dissipation in the resistance in one
of those painfully simple circuits to the "reflected power" in the
transmission line it's connected to.


I saw the challenge and note the lack of response.

Let me offer a steady state solution.

In the case of a simple source being an ideal AC voltage generator of Vs
and an ideal series resistance Rs of Ro, and that Zo=Ro, for any
arbitrary load, at the source terminals, Vf=Vs/2, Vl=Vf+Vr=Vs/2+Vr, and
the voltage difference across Rs is Vs/2-Vr (noting that Vr is a complex
quantity and can have a magnitude from 0 to Vs/2 at any phase angle),
therfore dissipation in Rs is given by:

Prs=(Vs/2-Vr)^2/Rs where Vs is the o/c source voltage, Vr is the complex
reflected wave voltage equivalent, Rs is the source resistance.

Clearly, dissipation in Rs is related to Vr, but it is not simply
proportional to the square of Vr as believed by many who lack the basics
of linear circuit theory to come to a correct understanding.

Roy, is that a solution?

Owen

On Jun 15, 6:36*pm, Roy Lewallen wrote:
Owen Duffy wrote:
lu6etj wrote in news:da3e5147-cad8-47f9-9784-
:

...
OK. Thank you very much. This clarify so much the issue to me. Please,
another question: On the same system-example, who does not agree with
the notion that the reflected power is never dissipated in Thevenin
Rs? (I am referring to habitual posters in these threads, of course)

Thevenin's theorem says nothing of what happens inside the source (eg
dissipation), or how the source may be implemented.
. . .

Cecil has used this fact as a convenient way of avoiding confrontation
with the illustrations given in my "food for thought" essays. However,
those models aren't claimed to be Thevenin equivalents of anything. They
are just simple models consisting of an ideal source and a perfect
resistance, as used in may circuit analysis textbooks to illustrate
basic electrical circuit operation. The dissipation in the resistance is
clearly not related to "reflected power", and the reflected power
"theories" being promoted here fail to explain the relationship between
the dissipation in the resistor and "reflected power". I contend that if
an analytical method fails to correctly predict the dissipation in such
a simple case, it can't be trusted to predict the dissipation in other
cases, and has underlying logical flaws. For all the fluff about
photons, optics, non-dissipative sources, and the like, I have yet to
see an equation that relates the dissipation in the resistance in one of
those painfully simple circuits to the "reflected power" in the
transmission line it's connected to.

Roy Lewallen, W7EL

obviously its not the 'reflected power'... that can be easily
disproved by showing that the length of the line changes the impedance
seen at the source terminals without changing the power that was
reflected from the load. since it is the impedance at the source
terminals that determines the performance of the amp the power that is
reflected is irrelevant.

however, it should be relatively easy to derive such a relation for a
simple thevenin source with a lossless line and a given load
impedance... just transform the impedance along the length of the line
back to the source then calculate the resulting current or voltage
from the source. that would give you the power dissipated in the
source. then you could also calculate the reflection coefficient and
separate the forward and reflected waves... and if you did it all
correctly and kept everything in terms of RL, Z0, and the length of
the line you could come up with a family of parametric curves relating
the power dissipated in the source resistance to the reflected power
over a range of load impedances for a given line length, or for
varying line length for a given load. obviously a purely academic
exercise that should be left for a rainy day.

On Jun 15, 7:06*pm, K1TTT wrote:
On Jun 15, 6:36*pm, Roy Lewallen wrote:



Owen Duffy wrote:
lu6etj wrote in news:da3e5147-cad8-47f9-9784-
:

...
OK. Thank you very much. This clarify so much the issue to me. Please,
another question: On the same system-example, who does not agree with
the notion that the reflected power is never dissipated in Thevenin
Rs? (I am referring to habitual posters in these threads, of course)

Thevenin's theorem says nothing of what happens inside the source (eg
dissipation), or how the source may be implemented.
. . .

Cecil has used this fact as a convenient way of avoiding confrontation
with the illustrations given in my "food for thought" essays. However,
those models aren't claimed to be Thevenin equivalents of anything. They
are just simple models consisting of an ideal source and a perfect
resistance, as used in may circuit analysis textbooks to illustrate
basic electrical circuit operation. The dissipation in the resistance is
clearly not related to "reflected power", and the reflected power
"theories" being promoted here fail to explain the relationship between
the dissipation in the resistor and "reflected power". I contend that if
an analytical method fails to correctly predict the dissipation in such
a simple case, it can't be trusted to predict the dissipation in other
cases, and has underlying logical flaws. For all the fluff about
photons, optics, non-dissipative sources, and the like, I have yet to
see an equation that relates the dissipation in the resistance in one of
those painfully simple circuits to the "reflected power" in the
transmission line it's connected to.

Roy Lewallen, W7EL

obviously its not the 'reflected power'... that can be easily
disproved by showing that the length of the line changes the impedance
seen at the source terminals without changing the power that was
reflected from the load. *since it is the impedance at the source
terminals that determines the performance of the amp the power that is
reflected is irrelevant.

however, it should be relatively easy to derive such a relation for a
simple thevenin source with a lossless line and a given load
impedance... just transform the impedance along the length of the line
back to the source then calculate the resulting current or voltage
from the source. *that would give you the power dissipated in the
source. *then you could also calculate the reflection coefficient and
separate the forward and reflected waves... and if you did it all
correctly and kept everything in terms of RL, Z0, and the length of
the line you could come up with a family of parametric curves relating
the power dissipated in the source resistance to the reflected power
over a range of load impedances for a given line length, or for
varying line length for a given load. *obviously a purely academic
exercise that should be left for a rainy day.

Tom, I understand the variation of the line-input impedance resulting
from the change in length of the trombone line with the reflection
coefficient of 0.01. However, I'm totally unaware of how the change in
line-input impedance relates to a change of frequency of the forward
wave of 100 Hz. Please explain.

Further, will you please also explain how the effect of the trombone
exercise determines the value of the source impedance, which, for
example, you stated is 56+j16 ohms? In other words, what mechanism
produced that value ? Sorry, Tom, I'm a little dense on this issue.

Walt

On Jun 15, 1:23*pm, Roy Lewallen wrote:
As I've said many times, and
you've continually disagreed with, increased dissipation and/or damage
at the transmitter is due to the impedance it sees, and not by
"reflected power".

Roy, unfortunately for your argument, the impedance seen by the
transmitter is:

Z = (Vfor + Vref) / (Ifor + Iref)

where the math is phasor math. ANY DEVIATION AWAY FROM THE Z0-MATCHED
LINE VALUE OF IMPEDANCE = Vfor/Ifor IS *CAUSED* BY THE REFLECTED WAVE!
Given that fact of physics, how can you possibly argue that the
reflected wave doesn't affect dissipation/damage at the transmitter???
--
73, Cecil, w5dxp.com

On Jun 15, 1:25*pm, Roy Lewallen wrote:
Thanks, but I've designed TDR systems and prepared and given classes on
the topic.

Given your strange concepts that violate the laws of EM wave physics
and your ignorance of how a redistribution of energy is often
associated with the presence of interference, I feel sorry for your
students.
--
73, Cecil, w5dxp.com

On Jun 15, 1:36*pm, Roy Lewallen wrote:
For all the fluff about
photons, optics, non-dissipative sources, and the like, I have yet to
see an equation that relates the dissipation in the resistance in one of
those painfully simple circuits to the "reflected power" in the
transmission line it's connected to.

The power density equation containing an interference term is what you
need to use and I seriously doubt that, after 5+ years, you are
ignorant of that equation. In your food-for-thought, forward/reflected
power example, all you have to do is figure out the power in the
forward wave and the power in the reflected wave at the source
resistor and plug them into the following equation:

Ptot = P1 + P2 + 2*SQRT(P1*P2)*cos(A)

where 'A' is the angle between the forward voltage and reflected
voltage. For instance, if the reflected voltage arrives back at the
source resistor in phase with the forward voltage, cos(A) = cos(0) = 1
and there is constructive interference which increases the dissipation
in the source resistor. If the reflected voltage arrives back at the
source resistor 180 degrees out of phase with the forward voltage,
cos(A) = cos(180) = -1 and there is destructive interference which
decreases the dissipation in the source resistor. If the reflected
voltage arrives back at the source resistor 90 degrees out of phase,
cos(A) = cos(90) = 0, and there is no interference and all of the
reflected power is dissipated in the source resistor. If you had ever
read my energy article, published many years ago, you would know what
effect superposition accompanied by interference can have on the
redistribution of energy. But you instead said, "Gobbleygook" (sic)
and plonked me. Time to pull your head out of the sand.

The above power density equation not only agrees with all of your
power calculations, it tells anyone who desires to acquire the
knowledge, exactly where the reflected energy goes and why it is not
always dissipated in the source resistor.
--
73, Cecil, w5dxp.com

On Jun 15, 5:59*pm, Owen Duffy wrote:
I saw the challenge and note the lack of response.

You seem to always note the lack of response while normal people are
trying to sleep. If Roy would simply use the power density equation
published by Dr. Best in his QEX article, he wouldn't still be
ignorant of where the reflected energy goes.
--
73, Cecil, w5dxp.com

On 16 jun, 00:21, Cecil Moore wrote:
On Jun 15, 1:36*pm, Roy Lewallen wrote:

For all the fluff about
photons, optics, non-dissipative sources, and the like, I have yet to
see an equation that relates the dissipation in the resistance in one of
those painfully simple circuits to the "reflected power" in the
transmission line it's connected to.

The power density equation containing an interference term is what you
need to use and I seriously doubt that, after 5+ years, you are
ignorant of that equation. In your food-for-thought, forward/reflected
power example, all you have to do is figure out the power in the
forward wave and the power in the reflected wave at the source
resistor and plug them into the following equation:

Ptot = P1 + P2 + 2*SQRT(P1*P2)*cos(A)

where 'A' is the angle between the forward voltage and reflected
voltage. For instance, if the reflected voltage arrives back at the
source resistor in phase with the forward voltage, cos(A) = cos(0) = 1
and there is constructive interference which increases the dissipation
in the source resistor. If the reflected voltage arrives back at the
source resistor 180 degrees out of phase with the forward voltage,
cos(A) = cos(180) = -1 and there is destructive interference which
decreases the dissipation in the source resistor. If the reflected
voltage arrives back at the source resistor 90 degrees out of phase,
cos(A) = cos(90) = 0, and there is no interference and all of the
reflected power is dissipated in the source resistor. If you had ever
read my energy article, published many years ago, you would know what
effect superposition accompanied by interference can have on the
redistribution of energy. But you instead said, "Gobbleygook" (sic)
and plonked me. Time to pull your head out of the sand.

The above power density equation not only agrees with all of your
power calculations, it tells anyone who desires to acquire the
knowledge, exactly where the reflected energy goes and why it is not
always dissipated in the source resistor.
--
73, Cecil, w5dxp.com

Hello boys... good day to you. You are make me study so hard forgotten
stories, dusting off old books... First, I sorry for I lost some posts
(I miss free news servers, ISPs here, nones!). Thanks to Roy, Owen,
K1TTT, etc. I read them today.

(: Please do not quarrel! :). Someone said (I think it was Nikita
Kruschev to the Pope, I am not sure) = "If we can not agree on
heaven's things, let us at least agree on the earth's things...." :)

Why not make a "truce" for a few hours with the "why's" to verify if
the proposed methods arrives at the same numerical results in terms of
PRs and Pl first with our minimalistic Vg and Rs? (for the sake of
novice readers)

For not to work too much, calculations could be reduced to three
resistive Zls, and three lengths of TL. Let me suggest 25, 50 and 100
ohms RLs and 0.5, 0.25 and 0.125 of lambda (to honor Cecil and Roy
papers), Vg voltage = 100 V RMS, Rs 50 ohms, Zo obviously 50 ohms too
(lossless line). Zl=50 ohms and 1/2 lambda TL are for beginners, as
me ;D

73 - Miguel - LU6ETJ

lu6etj wrote in
:

....
For not to work too much, calculations could be reduced to three
resistive Zls, and three lengths of TL. Let me suggest 25, 50 and 100
ohms RLs and 0.5, 0.25 and 0.125 of lambda (to honor Cecil and Roy
papers), Vg voltage = 100 V RMS, Rs 50 ohms, Zo obviously 50 ohms too
(lossless line). Zl=50 ohms and 1/2 lambda TL are for beginners, as
me ;D

Miguel, I gave an expression for power in your simple source circuit in a
recent post, did you see it. Well here it is again:

Prs=(Vs/2-Vr)^2/Rs where Vs is the o/c source voltage, Vr is the complex
reflected wave voltage equivalent, Rs is the source resistance.

If the reflected wave was always, and entirely dissipated in Rs in that
type of source, then Prs=Vr^2/Zo=Vr^2/Rs... but it isn't, see the above
expression.

You can work one case, three cases, three x three cases, but they are not
as complete as the expression above that shows that in the steady state,
Prs is not simply equal to the 'reflected power'.

But if you want, work the cases and report the results here, it is a
trivial exercise. You will accept the results more readily if you work it
our yourself.

Owen

On 16 jun, 03:48, Owen Duffy wrote:
lu6etj wrote :

...

For not to work too much, calculations could be reduced to three
resistive Zls, and three lengths of TL. Let me suggest 25, 50 and 100
ohms RLs and 0.5, 0.25 and 0.125 of lambda (to honor Cecil and Roy
papers), Vg voltage = 100 V RMS, Rs 50 ohms, Zo obviously 50 ohms too
(lossless line). Zl=50 ohms and 1/2 lambda TL are for beginners, as
me ;D

Miguel, I gave an expression for power in your simple source circuit in a
recent post, did you see it. Well here it is again:

Prs=(Vs/2-Vr)^2/Rs where Vs is the o/c source voltage, Vr is the complex
reflected wave voltage equivalent, Rs is the source resistance.

If the reflected wave was always, and entirely dissipated in Rs in that
type of source, then Prs=Vr^2/Zo=Vr^2/Rs... but it isn't, see the above
expression.

You can work one case, three cases, three x three cases, but they are not
as complete as the expression above that shows that in the steady state,
Prs is not simply equal to the 'reflected power'.

But if you want, work the cases and report the results here, it is a
trivial exercise. You will accept the results more readily if you work it
our yourself.

Owen

Do not argue with me Owen... If you do not want put your numbers you
are free, Roy played theirs, Cecil too, it is a simple and loving
exercise of numeric agreement. No arguments, no algebra, no calculus,
no photons, no Quantum, no Maxwell, no clever and slippery words...
simple and crude numbers... Not more than five minutes. :)

73. Here 04:26 I am go to dream with the little angels... Thanks,
Miguel LU6ETJ