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

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

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

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

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.

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.

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

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

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

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.