Where does it go? (mismatched power)
 

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

Where does it go? (mismatched power)
 

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.

Where does it go? (mismatched power)
 

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

Where does it go? (mismatched power)
 

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

Where does it go? (mismatched power)
 

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

Where does it go? (mismatched power)
 

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

Where does it go? (mismatched power)
 

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

Where does it go? (mismatched power)
 

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

Where does it go? (mismatched power)
 

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

Where does it go? (mismatched power)
 

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

Where does it go? (mismatched power)
 

On Jun 16, 1:48*am, Owen Duffy wrote:
Prs is not simply equal to the 'reflected power'.

I don't remember anyone saying that Prs is equal to the reflected
power and of course it is not. What you guys are missing are the
effects accompanying interference from superposition. There is another
mechanism besides the reflection mechanism that can redistribute the
reflected power. The power density equation includes an interference
term that indicates what happens to the energy.

Prs = Pfor + Pref + 2*SQRT(Pfor*Pref)*cos(A)

This equation gives the same answer as your equation but it also
indicates what happens to the energy components. A is the phase angle
between Vfor and Vref. The last term is the *interference* term.

If the interference term is zero, there is no interference and all of
the reflected power is dissipated in Rs.

If the interference term is negative, there exists destructive
interference at Rs and power is redistributed toward the load as
constructive interference. This is technically not a reflection
although the results are the same as a reflection.

If the interference term is positive, there exists constructive
interference at Rs and excess power is dissipated in Rs.

Let's look at your equation, Prs=(Vs/2-Vr)^2/Rs

Vfor = Vs/2, so Prs = (Vfor+Vref)^2/Rs (phasor addition)

Prs = Vfor^2/Rs + Vref^2/Rs + 2*Vfor*Vref*cos(A)/Rs

Prs = Pfor + Pref + 2*SQRT(Pfor*Pref)*cos(A)

I just derived the power density equation (with its interference term)
from your equation. The power density equation reveals the
interference term which tells us exactly where the reflected power
goes.

I learned about destructive and constructive interference at Texas A&M
in the 1950s.
--
73, Cecil, w5dxp.com

Where does it go? (mismatched power)
 

On Jun 16, 1:29*am, 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, Owen's equation, Prs=(Vs/2-Vr)^2/Rs, and the following power
density equation are equivalent and yield the same answers for Prs.
However, the power density equation indicates the magnitude of
interference present at Rs and the sign of the interference indicates
whether the interference is destructive or constructive. This general
equation is how optical physicists track the power density in their
light and laser beams.

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

where A is the phase angle between the electric fields of wave1 and
wave2. In our case, it will be the phase angle between the forward
wave and the reflected wave at the source resistor.

For our purposes, based on your specifications, we can rewrite the
equation:

Prs = Pfor + Pref + 2*SQRT(Pfor*Pref)*cos(A)

Pfor is fixed at 50 watts. For a 50 ohm load, Pref=0.
For a 25 ohm or 100 ohm load, Pref=5.556 watts.
Angle A is either 0, 90, or 180 degrees depending upon which example
is being examined.

For all three of the RL=50 ohm examples, Pref=0, the line is matched
and Prs = Prl = 50 watts. There is no question of where the reflected
power goes because Pref=0.

For all three of the 0.125WL examples, since A=90 deg, the
interference term in the power density equation is zero so for the 25
ohm and 100 ohm loads:

Prs = Pfor + Pref = 55.556 watts

When there is no interference, all the reflected power is obviously
dissipated in the source resistor. This matches my "zero interference"
article.

For all four of the other examples, the SWR is 2:1.
Pfor = 50w, Pref = 5.556w, Pload = 44.444w

If the reflected wave arrives with the electric field in phase with
the forward wave,
angle A = 0 degrees so cos(A) = +1.0
The positive sign tells us that the interference is constructive.

Prs = Pfor + Pref + 2*SQRT(Pfor*Pref)*cos(A)

Prs = 50 + 5.556 + 2*SQRT(50*5.556)

Prs = 55.556 + 33.333 = 88.89 watts

In addition to the sum of the forward power and reflected power being
dissipated in Rs, the source is forced to supply the additional power
contained in the 33.333 watt constructive interference term.

So Psource = 100w + 33.333w = 133.33w

Because of the constructive interference, the source not only must
supply the 100 watts that it supplies during matched line conditions,
but it must also supply the constructive interference power of 33.333
watts.

When the reflected wave arrives at Rs 180 degrees out of phase with
the forward wave,
cos(A) = cos(180) = -1.0. The negative sign indicates that the
interference is destructive.

Prs = 50 + 5.556 - 2*SQRT(50*5.556)

Prs = 55.556 - 33.333 = 22.223 watts

The source is forced to throttle back on its power output by an amount
equal to the destructive interference power of 33.33 watts. It is
supplying the 22.223w dissipated in Rs plus the difference in the
forward power and the reflected power.

Psource = Prs + Pfor - Pref = 66.667w

Note that the source is no longer supplying the reflected power. The
destructive interference has apparently redistributed the incident
reflected energy back toward the load as part of the forward wave.

In attempted chart form, here are the Prs values:

line length, 0.5WL, 0.25WL, 0.125WL

25 ohm load, 88.889w, 22.223w, 55.556w

50 ohm load, 50w, 50w, 50w

100 ohm load, 22.223w, 88.889w, 55.556w
--
73, Cecil, w5dxp.com

Where does it go? (mismatched power)
 

Owen Duffy wrote in news:Xns9D995B6088AFDnonenowhere@
61.9.191.5:

....
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.

Thinking about this some more, it should say:

Clearly, dissipation in Rs is related to Vr, but the contribution due to Vr
is not simply Vr^2/Rs as believed by many who lack the basics of linear
circuit theory to come to a correct understanding.

Owen

Where does it go? (mismatched power)
 

Owen Duffy wrote in
:

....
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.

Thinking about this some more, the last paragraph should say:

If the reflected wave was always, and entirely dissipated in Rs in that
type of source, then Prs=a+Vr^2/Zo=a+Vr^2/Rs (where a is some constant
value)... but it isn't, see the above expression.

Owen

Where does it go? (mismatched power)
 

Owen Duffy wrote in news:Xns9D9A202A37637nonenowhere@
61.9.191.5:

Owen Duffy wrote in
:

...
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.

Thinking about this some more, the last paragraph should say:

If the reflected wave was always, and entirely dissipated in Rs in that
type of source, then Prs=a+Vr^2/Zo=a+Vr^2/Rs (where a is some constant
value)... but it isn't, see the above expression.

To deal properly with the fact that Vr has magnitude and phase, the above
should we written as:

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=a+|Vr|^2/Zo=a+|Vr|^2/Rs (where a is some
constant value)... but it isn't, see the above expression.

Owen

Where does it go? (mismatched power)
 

On Jun 16, 12:09*pm, Owen Duffy wrote:
Clearly, dissipation in Rs is related to Vr, but the contribution due to Vr
is not simply Vr^2/Rs as believed by many who lack the basics of linear
circuit theory to come to a correct understanding.

Come on, Owen, that's just a straw man. Exactly who said that Vref^2/
Rs is dissipated in Rs?
--
73, Cecil, w5dxp.com

Where does it go? (mismatched power)
 

On Jun 16, 12:09*pm, Owen Duffy wrote:
If the reflected wave was always, and entirely dissipated in Rs in that
type of source, then Prs=a+Vr^2/Zo=a+Vr^2/Rs (where a is some constant
value)... but it isn't, see the above expression.

Again, a straw man. To the best of my knowledge, nobody has said that
reflected power is "entirely dissipated in Rs" (except for the special
case when interference doesn't exist.)
--
73, Cecil, w5dxp.com

Where does it go? (mismatched power)
 

Owen Duffy wrote:
. . .

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

Sorry, I don't think so. The very first equation, Vf=Vs/2, would be true
only if the load = Ro or for the time between system start and when the
first reflection returns. For the specified steady state and arbitrary
load, Vf would be a function of the impedance seen by the source which
is in turn a function of the line length and load impedance.

I'm not saying an equation relating "reflected power" and source
dissipation can't be written, but it does involve load impedance and
line length and Z0. (Equations written with Vf and Vr as variables tend
to conceal the inevitable dependence of these on line length and Z0 and
load impedance, but the dependence is still there.) And the equation
will show that there isn't any one-to-one correspondence between the two
quantities. For a given source voltage and resistance, the source
resistance dissipation depends only on the impedance seen by the source.
This in turn depends on the length and Z0 of the line and the load
impedance, and can be created by an infinite combination of loads and
lines having different "reflected powers". As I illustrated in an
earlier posting, a constant load with various lines having very
different "reflected powers" can present the same impedance to the
source and therefore result in the same source resistor dissipation.
Likewise, different lines having the same "reflected powers" can result
in different impedances seen by the source and therefore different
dissipations. So the two aren't really related, even though you could
write an equation which contains both terms.

Roy Lewallen, W7EL

Where does it go? (mismatched power)
 

Roy Lewallen wrote in
:

....

Sorry, I don't think so. The very first equation, Vf=Vs/2, would be true
only if the load = Ro or for the time between system start and when the
first reflection returns. For the specified steady state and arbitrary
load, Vf would be a function of the impedance seen by the source which
is in turn a function of the line length and load impedance.

Vf means the 'forward wave' voltage equivalent voltage at the source
terminals.

I provide the mathematical development of the case that Vf=Vs/2 where
Rs=Ro=Zo at http://vk1od.net/blog/?p=1028 .

Can you fault that development?

Owen

Where does it go? (mismatched power)
 

On Jun 16, 2:49*pm, Cecil Moore wrote:
On Jun 16, 1:29*am, 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, Owen's equation, Prs=(Vs/2-Vr)^2/Rs, and the following power
density equation are equivalent and yield the same answers for Prs.
However, the power density equation indicates the magnitude of
interference present at Rs and the sign of the interference indicates
whether the interference is destructive or constructive. This general
equation is how optical physicists track the power density in their
light and laser beams.

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

where A is the phase angle between the electric fields of wave1 and
wave2. In our case, it will be the phase angle between the forward
wave and the reflected wave at the source resistor.

For our purposes, based on your specifications, we can rewrite the
equation:

Prs = Pfor + Pref + 2*SQRT(Pfor*Pref)*cos(A)

Pfor is fixed at 50 watts. For a 50 ohm load, Pref=0.
For a 25 ohm or 100 ohm load, Pref=5.556 watts.
Angle A is either 0, 90, or 180 degrees depending upon which example
is being examined.

For all three of the RL=50 ohm examples, Pref=0, the line is matched
and Prs = Prl = 50 watts. There is no question of where the reflected
power goes because Pref=0.

For all three of the 0.125WL examples, since A=90 deg, the
interference term in the power density equation is zero so for the 25
ohm and 100 ohm loads:

Prs = Pfor + Pref = 55.556 watts

When there is no interference, all the reflected power is obviously
dissipated in the source resistor. This matches my "zero interference"
article.

For all four of the other examples, the SWR is 2:1.
Pfor = 50w, Pref = 5.556w, Pload = 44.444w

If the reflected wave arrives with the electric field in phase with
the forward wave,
angle A = 0 degrees so cos(A) = +1.0
The positive sign tells us that the interference is constructive.

Prs = Pfor + Pref + 2*SQRT(Pfor*Pref)*cos(A)

Prs = 50 + 5.556 + 2*SQRT(50*5.556)

Prs = 55.556 + 33.333 = 88.89 watts

In addition to the sum of the forward power and reflected power being
dissipated in Rs, the source is forced to supply the additional power
contained in the 33.333 watt constructive interference term.

So Psource = 100w + 33.333w = 133.33w

Because of the constructive interference, the source not only must
supply the 100 watts that it supplies during matched line conditions,
but it must also supply the constructive interference power of 33.333
watts.

When the reflected wave arrives at Rs 180 degrees out of phase with
the forward wave,
cos(A) = cos(180) = -1.0. The negative sign indicates that the
interference is destructive.

Prs = 50 + 5.556 - 2*SQRT(50*5.556)

Prs = 55.556 - 33.333 = 22.223 watts

The source is forced to throttle back on its power output by an amount
equal to the destructive interference power of 33.33 watts. It is
supplying the 22.223w dissipated in Rs plus the difference in the
forward power and the reflected power.

Psource = Prs + Pfor - Pref = 66.667w

Note that the source is no longer supplying the reflected power. The
destructive interference has apparently redistributed the incident
reflected energy back toward the load as part of the forward wave.

In attempted chart form, here are the Prs values:

line length, 0.5WL, 0.25WL, 0.125WL

25 ohm load, 88.889w, 22.223w, 55.556w

50 ohm load, 50w, 50w, 50w

100 ohm load, 22.223w, 88.889w, 55.556w
--
73, Cecil, w5dxp.com

lets see one case off the top of my head... 100 ohms resistive at 1/4
wave, transformed back to the source results in 25 ohms, pure
resistive. 50 ohm source in series with 25 ohm transformed load gives
100v/75ohms = 1.333A from at the source terminals and 100v*25ohm/(25ohm
+50ohm) = 33.33V. power from the source has to equal power into the
load since the line is lossless so PL = 33.33v*1.333a = 44.44w.
power dissipated in the source resistor =1.333a*(100v-33.33v)=88.88w

no sines, no cosines, no square roots, no Pfor/Pref... so much simpler
using voltage or current and simple transformations.

Where does it go? (mismatched power)
 

On Jun 16, 4:44*pm, Roy Lewallen wrote:
Sorry, I don't think so. The very first equation, Vf=Vs/2, would be true
only if the load = Ro or for the time between system start and when the
first reflection returns.

I would really like to see you prove that assertion.
If Rs = 50 ohms and Z0 = 50 ohms, why would the reflected wave change
the forward power since the reflected wave sees a matched 50 ohm
"load"? Hint: The forward power won't change unless there is a re-
reflection at the source. You have deliberately designed your food-for-
thought example to avoid re-reflections at the source. Lumped-circuit
models seem to have atrophied a lot of brains.

Those are the kind of strange concepts that result from over-
simplification of the math models. When one denies the ExH energy
content of an EM wave (even a reflection) one chooses to follow
oneself down a primrose path. Optical physicists can track their light
wave energy down to the last photon. Why are some RF engineers so
ignorant in the tracking of energy that they assert there is no energy
in a reflected wave?
--
73, Cecil, w5dxp.com

Where does it go? (mismatched power)
 

On Jun 16, 2:49*pm, Cecil Moore wrote:
Ptot = P1 + P2 + 2*SQRT(P1*P2)*cos(A)

where A is the phase angle between the electric fields of wave1 and
wave2. In our case, it will be the phase angle between the forward
wave and the reflected wave at the source resistor.

For our purposes, based on your specifications, we can rewrite the
equation:

Prs = Pfor + Pref + 2*SQRT(Pfor*Pref)*cos(A)

Pfor is fixed at 50 watts. For a 50 ohm load, Pref=0.
For a 25 ohm or 100 ohm load, Pref=5.556 watts.
Angle A is either 0, 90, or 180 degrees depending upon which example
is being examined.

would you care to provide us with the general equation for A given a
complex load impedance and a line length other than a multiple of 90
degrees?

Where does it go? (mismatched power)
 

On Jun 16, 5:58*pm, K1TTT wrote:
no sines, no cosines, no square roots, no Pfor/Pref... so much simpler
using voltage or current and simple transformations.

Yes, but determining where the reflected energy goes is the title of
this thread. My method shows where the reflected energy goes and yours
does not! That's the entire point.
--
73, Cecil, w5dxp.com

Where does it go? (mismatched power)
 

On Jun 16, 6:04*pm, K1TTT wrote:
would you care to provide us with the general equation for A given a
complex load impedance and a line length other than a multiple of 90
degrees?

It's simply the relative phase angle between the forward voltage and
reflected voltage at the source resistor. I trust that you can manage
that without my help.
--
73, Cecil, w5dxp.com

Where does it go? (mismatched power)
 

On Jun 16, 11:10*pm, Cecil Moore wrote:
On Jun 16, 5:58*pm, K1TTT wrote:

no sines, no cosines, no square roots, no Pfor/Pref... so much simpler
using voltage or current and simple transformations.

Yes, but determining where the reflected energy goes is the title of
this thread. My method shows where the reflected energy goes and yours
does not! That's the entire point.
--
73, Cecil, w5dxp.com

ah, but it does... it gives the exact same results but without all the
power manipulations. the point is, in sinusoidal steady state you can
calculate the impedance seen by the source and get all those same
results much simpler than trying to track forward and reflected
powers. and if you really want to know the source and load powers
they are easy enough to get as i showed.

Where does it go? (mismatched power)
 

On Jun 16, 11:16*pm, Cecil Moore wrote:
On Jun 16, 6:04*pm, K1TTT wrote:

would you care to provide us with the general equation for A given a
complex load impedance and a line length other than a multiple of 90
degrees?

It's simply the relative phase angle between the forward voltage and
reflected voltage at the source resistor. I trust that you can manage
that without my help.
--
73, Cecil, w5dxp.com

ah, so you do have to calculate the voltages, which would requires two
transformations of the length of the coax plus calculating the
magnitude and angle of the complex reflection coefficient then adding
them all up. still sounds like more work than necessary.

Where does it go? (mismatched power)
 

On Jun 16, 6:47*pm, K1TTT wrote:
ah, but it does... it gives the exact same results but without all the
power manipulations.

Give us a break. If what you say is true, why didn't you tell us where
the reflected power goes a long time ago? :-)
--
73, Cecil, w5dxp.com

Where does it go? (mismatched power)
 

On 16 jun, 20:50, K1TTT wrote:
On Jun 16, 11:16*pm, Cecil Moore wrote:

On Jun 16, 6:04*pm, K1TTT wrote:

would you care to provide us with the general equation for A given a
complex load impedance and a line length other than a multiple of 90
degrees?

It's simply the relative phase angle between the forward voltage and
reflected voltage at the source resistor. I trust that you can manage
that without my help.
--
73, Cecil, w5dxp.com

ah, so you do have to calculate the voltages, which would requires two
transformations of the length of the coax plus calculating the
magnitude and angle of the complex reflection coefficient then adding
them all up. *still sounds like more work than necessary.

Hello all

If I do not make any mistake, my numbers agree with yours Cecil, it is
a pleasure (my procedure was the old plain and simple standard) :)
I know, 1000 examples probe nothing, only one popperianists man can
bring a black swan at any time and falsify the induction :). but now
seems more clear to me you have a predictive model that render
identical numbers of my classical one, at least in basic tests, it is
a step forward to better considerate your efforts and with goodwill
begin to establish basic points of agreement.
I recognize it is a more simple approach the classic method (as said
K1TTT) to me too, but also I think not always one good method or model
it is more convenient to understand another useful view of phenomena.
Phasorial solutions are good, practical and likely complete electric
solutions but in my opinion they do not married very well with other
more general electromagnetic and physics models (wave guides perhaps?
I have not experience); unification has its advantages too (However my
favourite answer was definitely the Roy's one in the fourth post of
the thread, hi hi...)

73 - Miguel - LU6ETJ

PS: Owen do not be upset with me :) most of my available newsgroup
time it is spent in translate to english without flaws that may induce
to misinterpretations (all of you are very demanding with precise
wording and exact definitions), one mistake and I will need three or
four painly translations more to clarify :D

Today I tested your interesting formula with a Half wave 50 ohms TL,
loaded with 100 and 25 ohms respectively. Vs=100 Vrms, Rs=50 ohm give
2:1 VSWR. In both cases then Pf 50 W, Pr=5.5 W, Pnet=44.4 W. Giving
Vf=50 V and Vr=16,6 V aprox. for both loads.
I am using (Pf=Pnet / (1-Rho^2) and Pr=Pnet / ((1/1/Rho^2) -1) formula
which not gives different sign to Vr, in such case applying to
PRs=(((Vs/2)-Vr)^2)/Rs = ((50/2)-16.66)^2/50. I get PRs=22,2 with
both loads because the sign of Vf (always +) from simple Pr and Pf
formulas, changing the sign of Pr render Rs=88,8 W for Prs (OK).

I have in my disk a very descriptive, advisable and friendly article
downloaded from: http://www.ittc.ku.edu/~jstiles/622/...es_package.pdf

In page number 88 there is a agreement with your formula in "Is zo of
aa HF ham tx typycally 50+j0?".

Where does it go? (mismatched power)
 

Owen Duffy wrote:
Roy Lewallen wrote in
:

...
Sorry, I don't think so. The very first equation, Vf=Vs/2, would be true
only if the load = Ro or for the time between system start and when the
first reflection returns. For the specified steady state and arbitrary
load, Vf would be a function of the impedance seen by the source which
is in turn a function of the line length and load impedance.

Vf means the 'forward wave' voltage equivalent voltage at the source
terminals.

I provide the mathematical development of the case that Vf=Vs/2 where
Rs=Ro=Zo at http://vk1od.net/blog/?p=1028 .

Can you fault that development?

Owen

I stand corrected. In the special case of Rs = Z0, the steady state
forward voltage and therefore "forward power" become independent of the
transmission line length and load impedance, just as you said and show
in your analysis.

But this isn't true in the general case where Rs isn't equal to Z0. In
the general case, the forward voltage is

Vf = Vfi * exp(-j*theta) / (1 - Gs*Gl*exp(-j*2*thetal))

where Gs = reflection coefficient at the source looking back from the
line = (Zs - Z0)/(Zs + Z0)
Gl = reflection coefficient at the load = (Zl - Z0)/(Zl + Z0)
theta = distance along the line from the source end
thetal = length of the line in radians
Vfi = initial forward voltage = Vs * Z0 / (Zs + Z0)

At the input end of the line, theta = 0 so

Vf1 = Vfi / (1 - Gs*Gl*exp(-j*2*thetal))

and the reverse voltage at the input is

Vr1 = Gl * Vf1 * exp(-j*2*thetal)

Note that although Vf is a function of Gs, reflected voltage Vr is also
a function of Gs and their ratio is not. So VSWR and Pf/Pr don't depend
on Gs. But the general equation for Vf does include line length and Z0
as well as both source and load Z.

In the example I posted earlier (Transmitter power 100 watts, load Z =
50 + j0, line length 1/2 wavelength), no mention was made of the source
impedance, only the power being delivered by the transmitter, which was
the same for both cases since the impedance seen by the transmitter
didn't change. In the first case, where Z0 = Zl = 50 + j0, the forward
power on the line was 100 watts and reverse power was zero. In the
second case, the line Z0 was changed to 200 ohms. This caused an
increase in forward power to 156.25 watts (calculated from VSWR). From
these we can infer that the forward voltage increased by a factor of 2.5
from sqrt(100 * 50) = 70.71 Vrms to sqrt(156.25 * 200) = 176.8 Vrms.
Let's see if this agrees with the equations.

When the line is a half wavelength long, thetal = pi, so the equation
for Vf1 simplifies to:

Vf1 (half wavelength line) = Vfi / (1 - Gs*Gl)

Expanding Vfi, Gs, and Gl results in

Vf1 (half wavelength line) = Vs * (Z0 + Zl)/(2 * (Zs + Zl))

Various combinations of Vs and Zs can give us the 100 assumed watts, but
we don't need to specify any specific values. Leaving Vs and Zs as
unknown constants and Zl as fixed, we can find the ratio for two
different values of Z0:

Vf1 = Vs * (Z01 + Zl)/(2 * (Zs + Zl))
Vf2 = Vs * (Z02 + Zl)/(2 * (Zs + Zl))

Vf2/Vf1 = (Z02 + Zl)/(Z01 + Zl)

In the example, case 1 had Z0 = Z01 = 50; case 2 had Z0 = Z02 = 200, and
Zl was 50 for both. So Vf2/Vf1 = (200 + 50)/(50 + 50) = 2.5, which
agrees with the calculation based on VSWR.

Once again, the general equations for Vf and Vr and therefore Pf and Pr
include Zs, Zl, line length, and line Z0. While one or more of these
terms might disappear in special cases or ratios, they all have to
appear in any equation of general applicability.

Roy Lewallen, W7EL

Where does it go? (mismatched power)
 

lu6etj wrote in
:

....
PS: Owen do not be upset with me :) most of my available newsgroup
time it is spent in translate to english without flaws that may induce
to misinterpretations (all of you are very demanding with precise
wording and exact definitions), one mistake and I will need three or
four painly translations more to clarify :D

Today I tested your interesting formula with a Half wave 50 ohms TL,
loaded with 100 and 25 ohms respectively. Vs=100 Vrms, Rs=50 ohm give
2:1 VSWR. In both cases then Pf 50 W, Pr=5.5 W, Pnet=44.4 W. Giving
Vf=50 V and Vr=16,6 V aprox. for both loads.

I posted a correction to the formula to properly account for the fact
that Vr is a complex quantity.

The corrected expression is Prs=|(Vs/2-Vr)|^2/Rs . Apologies if you
missed it. Of course, the V quantities are RMS, it is a bit of a botch of
RMS with phase as we often do in thinking... but I see you using the same
shorthand in your calcs. Although we are using RMS values, don't overlook
that where they add (eg Vf+Vr) you must properly account for the phase.

For your cases:

For the 25+j0 load, Vr=-16.67+j0, so Prs=|50--16.67|^2/50=88.9W.

For the 100+j0 load, Vr=16.67+j0, so Prs=|50-16.67|^2/50=22.2W.

Of course, for a 50+j0 load, Vr=0, so Prs=|50-0|^2/50=50.0W.

As you can see, the first two cases have the same |Vr| (though different
phase), the same 'reflected power', and yet Prs is very different.

Consider an extreme case, Zl=1e6, VSWR is extreme, almost infinity, Vref=
50+j0, so Prs=|50-50|^2/50=0.0W. Here, your 'reflected power' is as large
as it gets, but the power dissipated in the source is zero.

The notion that reflected power is simply and always absorbed in the real
source resistance is quite wrong. Sure you can build special cases where
that might happen, but there is more to it. Thinking of the reflected
wave as 'reflected power' leads to some of the misconception.

Owen

Where does it go? (mismatched power)
 

Roy Lewallen wrote in
:

Owen Duffy wrote:
Roy Lewallen wrote in
:

...
Sorry, I don't think so. The very first equation, Vf=Vs/2, would be
true only if the load = Ro or for the time between system start and
when the first reflection returns. For the specified steady state
and arbitrary load, Vf would be a function of the impedance seen by
the source which is in turn a function of the line length and load
impedance.

Vf means the 'forward wave' voltage equivalent voltage at the source
terminals.

I provide the mathematical development of the case that Vf=Vs/2 where
Rs=Ro=Zo at http://vk1od.net/blog/?p=1028 .

Can you fault that development?

Owen

I stand corrected. In the special case of Rs = Z0, the steady state
forward voltage and therefore "forward power" become independent of
the transmission line length and load impedance, just as you said and
show in your analysis.

Thanks Roy.

There may have been some misunderstanding. I thought your challenge was
issued in the context of the ideal 50 ohm source that Miguel was using as
a test case.

The maths for the ideal 50 ohm source test case is so simple, that it is
evident the |Vr|^2/Ro is not simply summed into Rs's dissipation.

Of course, in the general case, the maths is uglier, but since the
special case of Rs=Ro=Zo proves the 'reflected power' is not simply
dissipated in Rs, it (the general case) must give the same outcome.

Owen

Where does it go? (mismatched power)
 

On Jun 17, 12:38*am, Owen Duffy wrote:
The notion that reflected power is simply and always absorbed in the real
source resistance is quite wrong. Sure you can build special cases where
that might happen, but there is more to it. Thinking of the reflected
wave as 'reflected power' leads to some of the misconception.

Again, exactly who proposed that notion? The "power" in a reflected
wave is actually reflected energy measured at some point on the
transmission line. Reflected EM waves cannot exist without ExH energy
per unit time per unit area.

The special case where 100% of the reflected power incident from a
Z0=50 ohm transmission line is dissipated in the 50 ohm source
resistor happens when there is zero interference between the forward
wave and reflected wave. Since it is not true when interference
exists, it is logical to conclude that interference has something to
do with the reflected power results deviating from the zero
interference case - and so it does. The interference term in the power
density equation indicates what kind of interference exists and what
happens to the interference energy.

Roy's experiment was designed to eliminate re-reflections from the
source and it does exactly that. What that means is that all of the
variations in power distribution are associated with interference, the
other mechanism by which RF wave energy can be redistributed. Contrary
to w7el's assumption, EM wave reflection is NOT the only mechanism for
redistributing RF energy in a transmission line. That glaring fact of
physics is what most of the RF experts are missing.
--
73, Cecil, w5dxp.com

Where does it go? (mismatched power)
 

On Jun 16, 11:47*pm, lu6etj wrote:
If I do not make any mistake, my numbers agree with yours Cecil, it is
a pleasure (my procedure was the old plain and simple standard) *:)

If, instead of a voltage analysis, one does an energy analysis, what
happens to the reflected energy becomes obvious. I posted an earlier
example that nobody solved. So let me repeat it here.

------Z01------+------Z02------

The power reflection coefficient at point '+' is
rho^2 = 0.5. The power transmission coefficient
is (1-rho^2) = 0.5
Pfor1 on the Z01 line is 100w
Pref1 on the Z01 line is 0w
What are Pfor2 and Pref2 on the Z02 line?
What is the SWR on the Z02 line?

The power reflected back toward the source at point '+' is

Pfor1(rho^2) = 100(0.5) = 50w

We know that Pref1 is zero, so

Pref2(1-rho^2) = 50w

Since 1-rho^2 is 0.5, Pref2 must be 100w and

Pref2(rho^2) = 50w

We also know that Pfor1(1-rho^2) = 50w

So we have two 50w waves trying to flow toward the source and two 50w
waves trying to flow toward the load. Do we have 100 watts flowing in
both directions? No, that would be superposition of power which is a
no-no. Using the power density equation indicates what is happening.

Pref1 = 0 = Pfor1(rho^2) + Pref2(1-rho^2) minus total destructive
interference at the Z0-match

Pref1 = 50 + 50 - 2*SQRT(50*50) = 0

Pfor2 = Pfor1(1-rho^2) + Pref2(rho^2) plus total constructive
interference

Pfor2 = 50 + 50 + 2*SQRT(50*50) = 200w

Since Pref2/Pfor2 = 0.5, the SWR on the Z02 line is 5.83:1

Note that we have solved the problem without knowing the value of
Vfor1, Vfor2, and Vref2. But let's assume Z01 is 50 ohms which makes
Z02 equal to 291.5 ohms. Now we can calculate the voltages assuming
Vfor1 at '+' is our phase reference:

Vfor1 = 70.7 volts at zero degrees at '+'
Vref1 = 0
Vfor2 = 241.4 volts at zero degrees at '+'
Vref2 = 170.7 volts at 180 deg at '+'

Vfor1(rho1) = 50v at zero deg
Vref2(tau2) = 50v at 180 deg

Vfor1(tau1) = 120.7v at zero deg
Vref2(rho2) = 120.7v at zero deg

There's the interference, voltage style. Two voltages superpose to
zero toward the source while engaged in total destructive
interference. What happens to the energy components that are
supporting those two voltages? There's only one thing that can happen.
They are obviously not flowing in their original directions so they
must necessarily flow in the only other direction possible. Total
destructive interference toward the source results in total
constructive interference toward the load.

Someone earlier remarked about the fact that when two 50w waves
combine, the result is one 200w wave. That is total constructive
interference in action and that extra energy had to come from
somewhere.
--
73, Cecil, w5dxp.com

Where does it go? (mismatched power)
 

On 17 jun, 11:30, Cecil Moore wrote:
On Jun 16, 11:47*pm, lu6etj wrote:

If I do not make any mistake, my numbers agree with yours Cecil, it is
a pleasure (my procedure was the old plain and simple standard) *:)

If, instead of a voltage analysis, one does an energy analysis, what
happens to the reflected energy becomes obvious. I posted an earlier
example that nobody solved. So let me repeat it here.

------Z01------+------Z02------

The power reflection coefficient at point '+' is
rho^2 = 0.5. The power transmission coefficient
is (1-rho^2) = 0.5
Pfor1 on the Z01 line is 100w
Pref1 on the Z01 line is 0w
What are Pfor2 and Pref2 on the Z02 line?
What is the SWR on the Z02 line?

The power reflected back toward the source at point '+' is

Pfor1(rho^2) = 100(0.5) = 50w

We know that Pref1 is zero, so

Pref2(1-rho^2) = 50w

Since 1-rho^2 is 0.5, Pref2 must be 100w and

Pref2(rho^2) = 50w

We also know that Pfor1(1-rho^2) = 50w

So we have two 50w waves trying to flow toward the source and two 50w
waves trying to flow toward the load. Do we have 100 watts flowing in
both directions? No, that would be superposition of power which is a
no-no. Using the power density equation indicates what is happening.

Pref1 = 0 = Pfor1(rho^2) + Pref2(1-rho^2) minus total destructive
interference at the Z0-match

Pref1 = 50 + 50 - 2*SQRT(50*50) = 0

Pfor2 = Pfor1(1-rho^2) + Pref2(rho^2) plus total constructive
interference

Pfor2 = 50 + 50 + 2*SQRT(50*50) = 200w

Since Pref2/Pfor2 = 0.5, the SWR on the Z02 line is 5.83:1

Note that we have solved the problem without knowing the value of
Vfor1, Vfor2, and Vref2. But let's assume Z01 is 50 ohms which makes
Z02 equal to 291.5 ohms. Now we can calculate the voltages assuming
Vfor1 at '+' is our phase reference:

Vfor1 = 70.7 volts at zero degrees at '+'
Vref1 = 0
Vfor2 = 241.4 volts at zero degrees at '+'
Vref2 = 170.7 volts at 180 deg at '+'

Vfor1(rho1) = 50v at zero deg
Vref2(tau2) = 50v at 180 deg

Vfor1(tau1) = 120.7v at zero deg
Vref2(rho2) = 120.7v at zero deg

There's the interference, voltage style. Two voltages superpose to
zero toward the source while engaged in total destructive
interference. What happens to the energy components that are
supporting those two voltages? There's only one thing that can happen.
They are obviously not flowing in their original directions so they
must necessarily flow in the only other direction possible. Total
destructive interference toward the source results in total
constructive interference toward the load.

Someone earlier remarked about the fact that when two 50w waves
combine, the result is one 200w wave. That is total constructive
interference in action and that extra energy had to come from
somewhere.
--
73, Cecil, w5dxp.com

Good day:

wave as 'reflected power' leads to some of the misconception.

Ugh! The slippery word again...! Please Owen, remember me what power
definition are you using here and expand the sentence idea.

EM waves cannot exist without ExH energy per unit time per
unit area.

This is what teachers taught me, but when the EM TL waves reachs a
"target" seems the issue arise in the newsgroup.
Puzzle to me why we can (want? :) ) not reconcile those physics
concepts with usual electricity concepts if today they arise from the
same place ultimately?
Certainly in electromagnetism we deal with vectors E,D,B,H,S and in
electricity with scalars V(t), I(t) or phasors, this is for
convenience and simplicity, but I accept we should be able to
understand them in both forms without contradiction.

73 - Miguel Ghezzi - LU6ETJ

Where does it go? (mismatched power)
 

On Jun 17, 1:36*pm, Cecil Moore wrote:
On Jun 17, 12:38*am, Owen Duffy wrote:

The notion that reflected power is simply and always absorbed in the real
source resistance is quite wrong. Sure you can build special cases where
that might happen, but there is more to it. Thinking of the reflected
wave as 'reflected power' leads to some of the misconception.

Again, exactly who proposed that notion? The "power" in a reflected
wave is actually reflected energy measured at some point on the
transmission line. Reflected EM waves cannot exist without ExH energy
per unit time per unit area.

The special case where 100% of the reflected power incident from a
Z0=50 ohm transmission line is dissipated in the 50 ohm source
resistor happens when there is zero interference between the forward
wave and reflected wave. Since it is not true when interference
exists, it is logical to conclude that interference has something to
do with the reflected power results deviating from the zero
interference case - and so it does. The interference term in the power
density equation indicates what kind of interference exists and what
happens to the interference energy.

Roy's experiment was designed to eliminate re-reflections from the
source and it does exactly that. What that means is that all of the
variations in power distribution are associated with interference, the
other mechanism by which RF wave energy can be redistributed. Contrary
to w7el's assumption, EM wave reflection is NOT the only mechanism for
redistributing RF energy in a transmission line. That glaring fact of
physics is what most of the RF experts are missing.
--
73, Cecil, w5dxp.com

but you have already admitted that it is the only mechanism... though
you will now undoubtedly argue against yourself. em wave reflection
IS the only mechanism for redistributing rf energy, you admitted that
when you agreed that superposition of the waves is the mechanism that
causes the interference in the first place. since interference is
caused by superposition then wave cancelation and this so called
redistribution of energy is also caused by the superposition of the
reflected wave with the incident wave. RF experts are not missing
anything, they just don't talk in your fancy energy and power terms
all the time or figure how to calculate power superpositions... read
the 500 pages BEFORE the s-parameter discussion in your version of the
fields and waves book and learn the simple way to handle the single
wave components that makes all that stuff you preach look over
complicated.

Where does it go? (mismatched power)
 

On Jun 17, 9:28*pm, lu6etj wrote:
On 17 jun, 11:30, Cecil Moore wrote:



On Jun 16, 11:47*pm, lu6etj wrote:

If I do not make any mistake, my numbers agree with yours Cecil, it is
a pleasure (my procedure was the old plain and simple standard) *:)

If, instead of a voltage analysis, one does an energy analysis, what
happens to the reflected energy becomes obvious. I posted an earlier
example that nobody solved. So let me repeat it here.

------Z01------+------Z02------

The power reflection coefficient at point '+' is
rho^2 = 0.5. The power transmission coefficient
is (1-rho^2) = 0.5
Pfor1 on the Z01 line is 100w
Pref1 on the Z01 line is 0w
What are Pfor2 and Pref2 on the Z02 line?
What is the SWR on the Z02 line?

The power reflected back toward the source at point '+' is

Pfor1(rho^2) = 100(0.5) = 50w

We know that Pref1 is zero, so

Pref2(1-rho^2) = 50w

Since 1-rho^2 is 0.5, Pref2 must be 100w and

Pref2(rho^2) = 50w

We also know that Pfor1(1-rho^2) = 50w

So we have two 50w waves trying to flow toward the source and two 50w
waves trying to flow toward the load. Do we have 100 watts flowing in
both directions? No, that would be superposition of power which is a
no-no. Using the power density equation indicates what is happening.

Pref1 = 0 = Pfor1(rho^2) + Pref2(1-rho^2) minus total destructive
interference at the Z0-match

Pref1 = 50 + 50 - 2*SQRT(50*50) = 0

Pfor2 = Pfor1(1-rho^2) + Pref2(rho^2) plus total constructive
interference

Pfor2 = 50 + 50 + 2*SQRT(50*50) = 200w

Since Pref2/Pfor2 = 0.5, the SWR on the Z02 line is 5.83:1

Note that we have solved the problem without knowing the value of
Vfor1, Vfor2, and Vref2. But let's assume Z01 is 50 ohms which makes
Z02 equal to 291.5 ohms. Now we can calculate the voltages assuming
Vfor1 at '+' is our phase reference:

Vfor1 = 70.7 volts at zero degrees at '+'
Vref1 = 0
Vfor2 = 241.4 volts at zero degrees at '+'
Vref2 = 170.7 volts at 180 deg at '+'

Vfor1(rho1) = 50v at zero deg
Vref2(tau2) = 50v at 180 deg

Vfor1(tau1) = 120.7v at zero deg
Vref2(rho2) = 120.7v at zero deg

There's the interference, voltage style. Two voltages superpose to
zero toward the source while engaged in total destructive
interference. What happens to the energy components that are
supporting those two voltages? There's only one thing that can happen.
They are obviously not flowing in their original directions so they
must necessarily flow in the only other direction possible. Total
destructive interference toward the source results in total
constructive interference toward the load.

Someone earlier remarked about the fact that when two 50w waves
combine, the result is one 200w wave. That is total constructive
interference in action and that extra energy had to come from
somewhere.
--
73, Cecil, w5dxp.com

Good day:

wave as 'reflected power' leads to some of the misconception.

Ugh! *The slippery word again...! Please Owen, remember me what power
definition are you using here and expand the sentence idea.

EM waves cannot exist without ExH energy per unit time per
unit area.

This is what teachers taught me, but when the EM TL waves reachs a
"target" seems the issue arise in the newsgroup.
Puzzle to me why we can (want? :) ) not reconcile those physics
concepts with usual electricity concepts if today they arise from the
same place ultimately?
Certainly in electromagnetism we deal with vectors E,D,B,H,S and in
electricity with scalars V(t), I(t) or phasors, this is for
convenience and simplicity, but I accept we should be able to
understand them in both forms without contradiction.

73 - Miguel Ghezzi - LU6ETJ

except for the one guy in the other thread that likes his dc 'waves' i
think most everyone else gets the same answers but won't acknowledge
that the other peoples methods also work. even my simple transformed
impedance method shows the same results, you just have to look at what
it really means. the whole key is that it doesn't really matter, we
know how to use voltage/current, E/H, powers, s-parameters, etc to
design circuits, and as long as we know their limitations and stay
within them everything should work the same way no matter how its
designed on paper.

Where does it go? (mismatched power)
 

On Jun 17, 5:08*pm, K1TTT wrote:
but you have already admitted that it is the only mechanism... though
you will now undoubtedly argue against yourself. * em wave reflection
IS the only mechanism for redistributing rf energy, you admitted that
when you agreed that superposition of the waves is the mechanism that
causes the interference in the first place.

Superposition and reflection are NOT the same mechanism. You are
totally confused about what I have said. Wave REFLECTION (of one wave)
is not caused by SUPERPOSITION (of two waves) and vice versa. Wave
reflection happens to a single wave when it encounters an impedance
discontinuity. Superposition requires two or more waves. They are
clearly two completely different mechanisms. So I will repeat what I
said befo

There are two mechanisms for redistributing the reflected energy back
toward the load.

1. The re-reflection of the SINGLE reflected wave from the load at an
impedance discontinuity associated with the power reflection
coefficient, e.g. 0.5 in my earlier example. Thus in that earlier
example, 1/2 of the reflected energy from the load is re-reflected
back toward the load and joins the forward wave toward the load. That
leaves 1/2 of the reflected energy that is transmitted through the
impedance discontinuity toward the source without being reflected.

The second energy redistribution mechanism occurs associated with
superposition of MULTIPLE WAVES.

2. The percentage of the reflected energy from the load that is
transmitted through the impedance discontinuity toward the source
superposes with the reflection of the source forward wave from the
impedance discontinuity. In my earlier example, the following two
wavefronts superpose in the direction of the source.

Pfor1(rho^2) = 50w and Pref2(1-rho^2) = 50w

Pref1 = 50w + 50w - 2*SQRT(50w*50w) = 0

This is the second mechanism (wave cancellation) that redistributes
the energy in the canceled wavefronts back toward the load. This step
2 is technically NOT a reflection since it involves two waves. This is
the step that the RF gurus are missing.

I really wish you understood the s-parameter equations which are
easier to discuss than the above RF equations.

In the s-parameter equation:

b2 = s21*a1 + s22*a2

the term, s22*a2, is the SINGLE WAVE re-reflection term, i.e. a2 is
the reflected voltage from the load.

In the other s-parameter equation:

b1 = s11*a1 + s12*a2 = 0

s11*a1 and s12*a2 are the TWO WAVEFRONTS that superpose to zero, i.e.
engage in wave cancellation.
--
73, Cecil, w5dxp.com

Where does it go? (mismatched power)
 

On Jun 17, 5:12*pm, K1TTT wrote:
except for the one guy in the other thread that likes his dc 'waves' i
think most everyone else gets the same answers but won't acknowledge
that the other peoples methods also work. *even my simple transformed
impedance method shows the same results, you just have to look at what
it really means. *the whole key is that it doesn't really matter, ...

The subject of this thread is: "Where does it go? (mismatched power)".
So it did matter to the original poster. I have answered the question
by presenting the facts of EM wave physics known from the field of
optics. So now you say, "it doesn't really matter". That seems to be a
final fallback position.
--
73, Cecil, w5dxp.com

Where does it go? (mismatched power)
 

On Jun 17, 10:55*pm, Cecil Moore wrote:
On Jun 17, 5:08*pm, K1TTT wrote:

but you have already admitted that it is the only mechanism... though
you will now undoubtedly argue against yourself. * em wave reflection
IS the only mechanism for redistributing rf energy, you admitted that
when you agreed that superposition of the waves is the mechanism that
causes the interference in the first place.

Superposition and reflection are NOT the same mechanism. You are

i didn't say they were


totally confused about what I have said. Wave REFLECTION (of one wave)
is not caused by SUPERPOSITION (of two waves) and vice versa. Wave
reflection happens to a single wave when it encounters an impedance
discontinuity. Superposition requires two or more waves. They are
clearly two completely different mechanisms. So I will repeat what I
said befo

There are two mechanisms for redistributing the reflected energy back
toward the load.

1. The re-reflection of the SINGLE reflected wave from the load at an
impedance discontinuity associated with the power reflection

its really the voltage and current reflection coefficient.

coefficient, e.g. 0.5 in my earlier example. Thus in that earlier
example, 1/2 of the reflected energy from the load is re-reflected
back toward the load and joins the forward wave toward the load. That
leaves 1/2 of the reflected energy that is transmitted through the
impedance discontinuity toward the source without being reflected.

The second energy redistribution mechanism occurs associated with
superposition of MULTIPLE WAVES.

that is what i said, and what you agreed to earlier.


2. The percentage of the reflected energy from the load that is
transmitted through the impedance discontinuity toward the source
superposes with the reflection of the source forward wave from the
impedance discontinuity. In my earlier example, the following two
wavefronts superpose in the direction of the source.

Pfor1(rho^2) = 50w and Pref2(1-rho^2) = 50w

Pref1 = 50w + 50w - 2*SQRT(50w*50w) = 0

This is the second mechanism (wave cancellation) that redistributes

but that is NOT a "second' mechanism. it IS superposition, which is
what you agreed to earlier, maybe in a different thread. the
mechanism is superposition, the observation is that the waves
interfere either constructively or destructively.

the energy in the canceled wavefronts back toward the load. This step
2 is technically NOT a reflection since it involves two waves. This is
the step that the RF gurus are missing.

no we are not, we understand superposition which is the underlying
basic mechanism of your optical interference terms.



I really wish you understood the s-parameter equations which are
easier to discuss than the above RF equations.

In the s-parameter equation:

b2 = s21*a1 + s22*a2

the term, s22*a2, is the SINGLE WAVE re-reflection term, i.e. a2 is
the reflected voltage from the load.

In the other s-parameter equation:

b1 = s11*a1 + s12*a2 = 0

s11*a1 and s12*a2 are the TWO WAVEFRONTS that superpose to zero, i.e.

stop there and you are perfectly correct.

engage in wave cancellation.
--
73, Cecil, w5dxp.com

Where does it go? (mismatched power)
 

On Jun 17, 11:06*pm, Cecil Moore wrote:
On Jun 17, 5:12*pm, K1TTT wrote:

except for the one guy in the other thread that likes his dc 'waves' i
think most everyone else gets the same answers but won't acknowledge
that the other peoples methods also work. *even my simple transformed
impedance method shows the same results, you just have to look at what
it really means. *the whole key is that it doesn't really matter, ...

The subject of this thread is: "Where does it go? (mismatched power)".
So it did matter to the original poster. I have answered the question
by presenting the facts of EM wave physics known from the field of
optics. So now you say, "it doesn't really matter". That seems to be a
final fallback position.
--
73, Cecil, w5dxp.com

you edited too much out of context... it doesn't matter which method
you use to calculate where it goes as long as you know the limitations
of your method.