Where does it go? (mismatched power)
 

On Jun 12, 4:24*am, lu6etj wrote:
On 11 jun, 23:26, Cecil Moore wrote: On Jun 11, 5:03*pm, lu6etj wrote:

From my perspective your main differences are reducible

The basic argument revolves around what math shortcuts can be used to
solve a particular problem vs what is actually happening in reality
according to the accepted laws of physics. I agree one doesn't
necessarily need to understand the laws of physics to solve a problem
but one should probably know enough physics to recognize when those
laws of physics are being violated by one's argument.
--
73, Cecil, w5dxp.com
...........
of course, but that is no fun!

I agree :D :D
.....
As a courtesy to me, a foreigner tourist ham, would you mind stop for
a brief moment your more general differences and tell me if you agree
on the behavior of a Thevenin generator with a series resistance of 50
ohms in relation to changes in impedance of a lossless TL predicted by
the Telegrapher's equations solutions in terms of the power dissipated
on the load resistance and series resistence of Thevenin source?
I am pretty serious about this: until today I could not know if you
agree in that!! :)

sure, if you properly apply the telegrapher's equations and the
thevenin equivalent methods. The real problem is that if you try to
do that for most amateur radio transmitters the source impedance is
not linear, and even worse may be time varying, which renders the
thevenin equivalent source substitution invalid.

Note though that in real world cases you need to use the full set of
equations, usually called by engineers the 'general transmission line
equations'. beware, some places may over simplify the telegrapher's
equations which may make them invalid in some cases. The
Telegrapher's equations (http://en.wikipedia.org/wiki/Telegrapher
%27s_equations), are often considered a subset of the 'General
transmission line equations (http://en.wikipedia.org/wiki/
Transmission_line) that are taught in distributed circuits and fields
and waves courses in engineering schools.

Where does it go? (mismatched power)
 

On Jun 12, 2:26*am, Cecil Moore wrote:
On Jun 11, 5:03*pm, lu6etj wrote:

From my perspective your main differences are reducible

The basic argument revolves around what math shortcuts can be used to
solve a particular problem vs what is actually happening in reality
according to the accepted laws of physics. I agree one doesn't
necessarily need to understand the laws of physics to solve a problem
but one should probably know enough physics to recognize when those
laws of physics are being violated by one's argument.
--
73, Cecil, w5dxp.com

rather than worrying about basic physics, the real problem here is
that analysis of a non-linear circuit is being attempted using
techniques that are only valid for linear circuits.

Where does it go? (mismatched power)
 

On Jun 11, 11:24*pm, lu6etj wrote:
As a courtesy to me, a foreigner tourist ham, would you mind stop for
a brief moment your more general differences and tell me if you agree
on the behavior of a Thevenin generator with a series resistance of 50
ohms in relation to changes in impedance of a lossless TL predicted by
the Telegrapher's equations solutions in terms of the power dissipated
on the load resistance and series resistence of Thevenin source?
I am pretty serious about this: until today I could not know if you
agree in that!! :)

Miguel, I don't think there is much disagreement about things that are
easily measured, like voltage and current. One solution to the
telegrapher's equations involves forward and reflected waves of
voltage and current. The conventional way of handling power (energy/
unit-time) is to use the voltages and currents to calculate the power
at certain points of interest. The telegrapher's equations do not tell
us *why* the power is what it is and the energy is where it is. To
obtain the why, one must study the behavior of electromagnetic waves.
How does the energy in electromagnetic waves behave? The telegrapher's
equations and Thevenin source do not answer that question.

For instance: Most readers here seem to think that the only phenomenon
that can cause a reversal of direction of energy flow in a
transmission line is a simple EM wave reflection based on the
reflection model. When they cannot explain what is happening using
that model, they throw up their hands and utter crap like, "Reflected
wave energy doesn't slosh back and forth between the load and the
source". But not only does it "slosh back and forth", it sloshes back
and forth at the speed of light in the medium because nothing else is
possible.

These are the people who have allowed their math models to become
their religion. They will not change their minds even when accepted
technical facts are presented. One response was, "Gobblydegook". (sic)

There is another phenomenon, besides a simple reflection, that causes
reflected energy to be redistributed back toward the load and that is
wave cancellation involving two wavefronts. If the two wavefronts are
equal in magnitude and opposite in phase, total wave cancellation is
the result which, in a transmission line, redistributes the wave
energy in the only other direction possible which is, surprise, the
opposite direction. This is a well known, well understood,
mathematically predictable phenomenon that happens all the time in the
field of optics, e.g. at the surface of non-reflective glass. It also
happens all the time in RF transmission lines when a Z0-match is
achieved.

Using the s-parameter equations (phasor math) at a Z0-match point in a
transmission line:

b1 = s11*a1 + s12*a2 = 0 = reflected voltage toward the source
Square this equation to get the reflected power toward the source.

These are the two wavefronts that undergo total wave cancellation,
i.e. total destructive interference.

b2 = s21*a1 + s22*a2 = forward voltage toward the load
s22*a2 is the re-reflection. Square this equation to get the forward
power toward the load.

If one squares both of those equations, one can observe the
interference terms which indicate why and where the energy goes. All
of the energy in s11*a1 and s12*a2 reverses direction at the Z0-match
and flows back toward the load. All the things that Roy is confused
about in his food-for-thought article on forward and reflected power
are easily explained by the power density equation (or by squaring the
s-parameter equations).

Ptot = P1 + P2 + 2*SQRT(P1*P2)*cos(A)
--
73, Cecil, w5dxp.com

Where does it go? (mismatched power)
 

On Jun 12, 5:51*am, K1TTT wrote:
rather than worrying about basic physics, the real problem here is
that analysis of a non-linear circuit is being attempted using
techniques that are only valid for linear circuits.

Well, that's certainly one problem but I was talking about linear
circuit analysis. When will the effects of linear wave cancellation be
understood by the ham radio community?

Walt seems to have sidestepped the non-linear source problem by
defining the "source" as the first point in the system at which the
source signal becomes linear after filtering. He seems to draw a
system box boundary through that point and labels that signal, "the
source". The ratio of voltage to current at that point is, by
definition, the source impedance. That approach sure does simplify
things.

The output of a class-C amp is weighted with odd harmonics. The system
is obviously mismatched at the odd harmonic frequencies.
--
73, Cecil, w5dxp.com

Where does it go? (mismatched power)
 

On Jun 12, 1:23*pm, Cecil Moore wrote:
On Jun 12, 5:51*am, K1TTT wrote:

rather than worrying about basic physics, the real problem here is
that analysis of a non-linear circuit is being attempted using
techniques that are only valid for linear circuits.

Well, that's certainly one problem but I was talking about linear
circuit analysis. When will the effects of linear wave cancellation be
understood by the ham radio community?

they won't because waves don't 'cancel' any more than they 'interact'
in linear systems. they do superimpose which gives the illusion of
cancellation and intensification. the worst thing causing amateurs
problems is using power and trying to think about energy... neither
one is a linear phenomenon, but they are often treated as such because
of some simple cases where you can add and subtract powers... when
someone who doesn't understand the restrictions fully tries to use
powers on more general cases they get in trouble. that is why it is
always better to use current or voltage where you can (in many more
cases) add them linearly and be correct. of course then there are
those who try to push even those restrictions back into non-linear
realms, like amplifiers, and get into even more trouble.

Where does it go? (mismatched power)
 

On Jun 12, 9:00*am, K1TTT wrote:
they won't because waves don't 'cancel' any more than they 'interact'
in linear systems. *they do superimpose which gives the illusion of
cancellation and intensification.

Here's proof otherwise:

b1 = s11*a1 + s12*a2 = 0 = reflected voltage toward the source

The permanent result is that reflected voltage and power equal zero?
How can that possibly be an illusion?

(b1)^2 = (s11*a1 + s12*a2)^2 = 0 = reflected power toward the source.
s11*a1 is not zero, s12*a2 is not zero, they are equal in magnitude
and opposite in phase.

What happens to the non-zero (s11*a1 + s12*a2)^2 power? The answer to
that question is what a lot of people are missing.

Those two reflected wavefronts have interacted destructively and
canceled. That destructive interference energy is redistributed back
toward the load as constructive interference energy in the other s-
parameter equation.

b2 = s21*a1 + s22*a2 = forward voltage toward the load

(b2)^2 = (s21*a1 + s22*a2)^2 = forward power toward the load

Doesn't the fact that reflected energy cannot be traced using only the
wave reflection model indicate that there is something missing from
that procedure?
--
73, Cecil, w6dxp.com

Where does it go? (mismatched power)
 

On Jun 12, 2:37*pm, Cecil Moore wrote:
On Jun 12, 9:00*am, K1TTT wrote:

they won't because waves don't 'cancel' any more than they 'interact'
in linear systems. *they do superimpose which gives the illusion of
cancellation and intensification.

Here's proof otherwise:

b1 = s11*a1 + s12*a2 = 0 = reflected voltage toward the source

The permanent result is that reflected voltage and power equal zero?
How can that possibly be an illusion?

(b1)^2 = (s11*a1 + s12*a2)^2 = 0 = reflected power toward the source.
s11*a1 is not zero, s12*a2 is not zero, they are equal in magnitude
and opposite in phase.

What happens to the non-zero (s11*a1 + s12*a2)^2 power? The answer to
that question is what a lot of people are missing.

Those two reflected wavefronts have interacted destructively and
canceled. That destructive interference energy is redistributed back
toward the load as constructive interference energy in the other s-
parameter equation.

b2 = s21*a1 + s22*a2 = forward voltage toward the load

(b2)^2 = (s21*a1 + s22*a2)^2 = forward power toward the load

Doesn't the fact that reflected energy cannot be traced using only the
wave reflection model indicate that there is something missing from
that procedure?
--
73, Cecil, w6dxp.com

i don't do s stuff so i have no idea what you just proved... give me
the impedances and voltages/currents.

Where does it go? (mismatched power)
 

I see that I am going to have to re-assert your own standard:

On Fri, 11 Jun 2010 20:31:42 GMT, Owen Duffy wrote:

there is a proposition that a transmitter "designed/adjusted
for, and expecting a 50 + j 0 ohm load" can be well represented by a
Thevenin equivalent circuit and naturally has Zeq=50+j0.

On Sat, 12 Jun 2010 00:19:34 GMT, Owen Duffy wrote:

For the data I asked, you supplied:
I have performed many tests on many radios. One documented example is at
http://vk1od.net/blog/?p=1045 .
and offered:
I invite you and others to perform the same test. You will realise that
one, or even 100 supporting tests do not prove the proposition, but one
valid test to the contrary is damaging.

The test of proving the "proposition" invalid is, in part:
Adjusting the load impedance ... on this load with VSWR(50)=1.5
... is proof that the equivalent source impedance of the transmitter is not 50+j0Ohm.

I observed how you violated the adjusted-for part of
"designed/adjusted for, and expecting a 50 + j 0 ohm load."

As no claim has been made by anyone about a source being constant in Z
nor in Power across all loads and all frequencies, your response does
not conform to your own reprise of the "proposition."

*******

What you have performed is a load pull which constructs a curve of
complex source impedances around the point at which the transmitter
was adjusted for a 50 Ohm load. All well and good. However,
Thevenin's theorem says nothing of this. The correct test, to the
letter of the theorem is a test no one performs: the measured open
circuit voltage divided by the measured short circuit current.

*******

If I were to return to another statement from your link offered in my
quote above:
the transmitter is 50+j0Ohm
is in all likelihood incorrect. I am speaking strictly to what is
reported and to the implied accuracy of 50 ±0.5 Ohms. I seriously
doubt that you have the means to achieve the absolute accuracy of 1%.

To many, this is a trivial point - simply because they, too, are
wholly incapable of achieving even ten times this error. That is,
measure an RF power in the HF to within ±10%. So I often get shrugged
off with a dismissal of "so what?"

What is this scrabbling over decimal points that I am making? IF:
the equivalent source impedance of the transmitter is not 50+j0Ohms
then it could easily be satisfied by it being 51+j0Ohms, as you do not
report what the Z was. I am sure some would condemn 51 Ohms with
sneering contempt. Reports of what the Source Z "is not" is not
informative in the least. What the significance 1 Ohm has when the
likelihood of being able make a measurement with much less range of
error is approaching nil and thus does not stand as a very rigorous
proof AGAINST the "proposition."

Similarly, IF:
the equivalent source impedance of the transmitter is 50+j0Ohms
then even this may not qualify if the source impedance MUST be
50+j0Ohms as, again, the likelihood of being able make a measurement
within ±1% range of error is approaching nil and thus does not stand
as a very rigorous proof FOR the "proposition."

********

Last, I see the lack of rigor in reporting that is so evidenced by
Walt's work. I am familiar with EVERY one of his lab instruments
(enumerated elsewhere) as I have calibrated ALL like them in three
different RF Standards laboratories over my career as a Metrologist. I
am wholly informed as to their capabilities and capacities to perform
the tests he reports to the precision and accuracies he documents. I
am also trained in the methods of performing this work and I have read
from Walt's descriptions that he has taken care to observe the proper
methods. All of these considerations are scrupulous to his first two
steps of his description which amply demonstrate:
there is a proposition that a transmitter "designed/adjusted
for, and expecting a 50 + j 0 ohm load" can be well represented by a
Thevenin equivalent circuit and naturally has Zeq=50+j0.

73's
Richard Clark, KB7QHC

Where does it go? (mismatched power)
 

On Jun 12, 2:37*pm, Cecil Moore wrote:
On Jun 12, 9:00*am, K1TTT wrote:

they won't because waves don't 'cancel' any more than they 'interact'
in linear systems. *they do superimpose which gives the illusion of
cancellation and intensification.

Here's proof otherwise:

b1 = s11*a1 + s12*a2 = 0 = reflected voltage toward the source

The permanent result is that reflected voltage and power equal zero?
How can that possibly be an illusion?

(b1)^2 = (s11*a1 + s12*a2)^2 = 0 = reflected power toward the source.
s11*a1 is not zero, s12*a2 is not zero, they are equal in magnitude
and opposite in phase.

What happens to the non-zero (s11*a1 + s12*a2)^2 power? The answer to
that question is what a lot of people are missing.

Those two reflected wavefronts have interacted destructively and
canceled. That destructive interference energy is redistributed back
toward the load as constructive interference energy in the other s-
parameter equation.

b2 = s21*a1 + s22*a2 = forward voltage toward the load

(b2)^2 = (s21*a1 + s22*a2)^2 = forward power toward the load

Doesn't the fact that reflected energy cannot be traced using only the
wave reflection model indicate that there is something missing from
that procedure?
--
73, Cecil, w6dxp.com

i don't do s stuff so i have no idea what you just proved... give me
the impedances and voltages/currents.

Where does it go? (mismatched power)
 

On Jun 12, 5:36*pm, Richard Clark wrote:
I see that I am going to have to re-assert your own standard:

On Fri, 11 Jun 2010 20:31:42 GMT, Owen Duffy wrote:

there is a proposition that a transmitter "designed/adjusted
for, and expecting a 50 + j 0 ohm load" can be well represented by a
Thevenin equivalent circuit and naturally has Zeq=50+j0.
On Sat, 12 Jun 2010 00:19:34 GMT, Owen Duffy wrote:

For the data I asked, you supplied:

I have performed many tests on many radios. One documented example is at
http://vk1od.net/blog/?p=1045.
and offered:
I invite you and others to perform the same test. You will realise that
one, or even 100 supporting tests do not prove the proposition, but one
valid test to the contrary is damaging.

The test of proving the "proposition" invalid is, in part:

Adjusting the load impedance ... on this load with VSWR(50)=1.5
... is proof that the equivalent source impedance of the transmitter is not *50+j0Ohm.

I observed how you violated the adjusted-for part of
"designed/adjusted for, and expecting a 50 + j 0 ohm load."

As no claim has been made by anyone about a source being constant in Z
nor in Power across all loads and all frequencies, your response does
not conform to your own reprise of the "proposition."

*******

What you have performed is a load pull which constructs a curve of
complex source impedances around the point at which the transmitter
was adjusted for a 50 Ohm load. *All well and good. *However,
Thevenin's theorem says nothing of this. *The correct test, to the
letter of the theorem is a test no one performs: the measured open
circuit voltage divided by the measured short circuit current.

that assumes the source is linear... with a non-linear source it is
much more complicated to describe the full range of it's response.
you would have to do a series of output voltage vs current curves for
a range of impedances.

Where does it go? (mismatched power)
 

On Fri, 11 Jun 2010 17:35:34 -0700 (PDT), K1TTT
wrote:

you can qualify an impedance as non-dissipative

It's called reactance.

not always. there is a non-dissipative resistance. a lossless
transmission line has a pure real impedance, but no dissipation.

Hi David,

Ah! The appeal of the infinite, lossless transmission line. You
forgot radiation.

I think we can both agree that it can be measured, even by the
inference of a chain of measurements and those measurements would
consistently show the absence of heat.

I think you would enjoy the absurdity of its appeal to being the plate
"resistance" when the 1400 Ohms there would be embodied in 28
infinitely long lossless cables being tucked inside the tube.

For others,

If it were so (where those 28 infinitely long lines are tucked into
another convenient dimension that sprang into being to satisfy
argument), then what is the source of the heat of the plate that is in
excess of that of the DC operating point?

If some feel compelled to jump on the losslessness of radiation loss
(now, if that isn't a paradox) - all fine and well for Art's short
wound self-resonant radiators that would eventually devolve in size to
the plate tank. If plate resistance was due to radiation, what need
for antennas? Of course, we could add another rhetorical dimension
that absorbs radiation....

73's
Richard Clark, KB7QHC

Where does it go? (mismatched power)
 

On Jun 12, 5:51*pm, Richard Clark wrote:
On Fri, 11 Jun 2010 17:35:34 -0700 (PDT), K1TTT
wrote:

you can qualify an impedance as non-dissipative

It's called reactance.

not always. *there is a non-dissipative resistance. *a lossless
transmission line has a pure real impedance, but no dissipation.

Hi David,

Ah! *The appeal of the infinite, lossless transmission line. *You
forgot radiation.


a lossless line would not be lossless if it lost energy due to
radiation!

Where does it go? (mismatched power)
 

Richard Clark wrote in
:

....
I observed how you violated the adjusted-for part of
"designed/adjusted for, and expecting a 50 + j 0 ohm load."

The IC7000 has no relevant user adjustments, it is designed for and
adjusted for in its design, manufacture, and alignment.


As no claim has been made by anyone about a source being constant in Z
nor in Power across all loads and all frequencies, your response does
not conform to your own reprise of the "proposition."

To be usable, any pretence of linearity at least over a limited range of
loads, power, frequency, Zeq must be sufficiently constant within that
range.

What you have performed is a load pull which constructs a curve of
complex source impedances around the point at which the transmitter
was adjusted for a 50 Ohm load. All well and good. However,

No, I have performed a go/nogo test on whether the transmitter delivers
sufficiently constant forward power into a quite limited range of loads.
The meaning of "sufficiently" is proposed in the reference article
describing the test and providing the mathematical basis for the test.

Thevenin's theorem says nothing of this. The correct test, to the
letter of the theorem is a test no one performs: the measured open
circuit voltage divided by the measured short circuit current.

No, Thevenin's theorem does not speak at all of how to test a source. For
a linear source, your proposed test of o/c voltage and s/c current would
provide sufficient data, as would ANY two points on the (complex) v/i
relationship.

I expect that a typical HF ham transmitter output is not linear over the
entire v/i characteristic, but my test just focuses on whether it is
approximately linear over a limited range of loads.

My recollection of Walt's tests were that they tested at points other
than Zl=0 and Zl=infinity.


*******

If I were to return to another statement from your link offered in my
quote above:
the transmitter is 50+j0Ohm
is in all likelihood incorrect. I am speaking strictly to what is
reported and to the implied accuracy of 50 ±0.5 Ohms. I seriously
doubt that you have the means to achieve the absolute accuracy of 1%.

You dwell at some further length on the implicit accuracy of the stated
quantity, but ignore the explicit discussion about a reasonable tolerance
for the test.

Owen

Where does it go? (mismatched power)
 

....
As a courtesy to me, a foreigner tourist ham, would you mind stop for
a brief moment your more general differences and tell me if you agree
on the behavior of a Thevenin generator with a series resistance of 50
ohms in relation to changes in impedance of a lossless TL predicted by
the Telegrapher's equations solutions in terms of the power dissipated
on the load resistance and series resistence of Thevenin source?
I am pretty serious about this: until today I could not know if you
agree in that!! :)

sure, if you properly apply the telegrapher's equations and the
thevenin equivalent methods. *The real problem is that if you try to
do that for most amateur radio transmitters the source impedance is
not linear, and even worse may be time varying, which renders the
thevenin equivalent source substitution invalid.

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)

Where does it go? (mismatched power)
 

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.

It is the implementation of the source that provides the answer to your
question, and the word "never" is too strong for the general case.

In respect of typical HF ham transmitters, you may find my article
entitled "Does SWR damage HF ham transmitters?" at
http://vk1od.net/blog/?p=1081 of interest.

Owen

Where does it go? (mismatched power)
 

On Sat, 12 Jun 2010 10:57:08 -0700 (PDT), K1TTT
wrote:

a lossless line would not be lossless if it lost energy due to
radiation!

Hmmm,

You want to reconsider that?

1. How could you tell if a lossless line lost energy by radiation?
B. Radiation was the second in your incomplete list, not a line
characteristic.

73's
Richard Clark, KB7QHC

Where does it go? (mismatched power)
 

On Sat, 12 Jun 2010 10:43:37 -0700 (PDT), K1TTT
wrote:

that assumes the source is linear... with a non-linear source it is
much more complicated to describe the full range of it's response.
you would have to do a series of output voltage vs current curves for
a range of impedances.

Ah!

But the difference is that no one is going to go to the bench to see
if this holds true and roll with the punch if they are surprised that
it works out perfectly.

73's
Richard Clark, KB7QHC

Where does it go? (mismatched power)
 

On Jun 12, 12:39*pm, K1TTT wrote:
i don't do s stuff so i have no idea what you just proved... give me
the impedances and voltages/currents.

Too bad about the "s stuff". Here are the RF equations for a Z01 to
Z02 impedance discontinuity in a transmission line. The forward
voltage on the Z01 side is Vfor1 and the reflected voltage from the
impedance discontinuity (back toward the source) is Vref1. The forward
voltage on the Z02 side is Vfor2 and the reflected voltage (from the
load) is Vref2. Hopefully, the reflection and transmission
coefficients are self-explanatory. rho1 is the reflection coefficient
encountered by Vfor1, etc.

Vref1 = Vfor1(rho1) + Vref2(tau2) = 0

That is wavefront cancellation in action. The external reflection
phasor, Vfor1(rho1), is equal in magnitude and 180 degrees out of
phase with the internal reflection phasor, Vref2(tau2), arriving from
the mismatched load.

Vfor2 = Vfor1(tau1) + Vref2(rho2)

If these RF equations are normalized to SQRT(Z0), they are the same as
the s-parameter equations.
--
73, Cecil, w5dxp.com

Where does it go? (mismatched power)
 

On Jun 12, 8:59*pm, Richard Clark wrote:
On Sat, 12 Jun 2010 10:57:08 -0700 (PDT), K1TTT
wrote:

a lossless line would not be lossless if it lost energy due to
radiation!

Hmmm,

You want to reconsider that?

1. *How could you tell if a lossless line lost energy by radiation?

there would be missing energy?

Where does it go? (mismatched power)
 

On Jun 12, 9:17*pm, Cecil Moore wrote:
On Jun 12, 12:39*pm, K1TTT wrote:

i don't do s stuff so i have no idea what you just proved... give me
the impedances and voltages/currents.

Too bad about the "s stuff". Here are the RF equations for a Z01 to
Z02 impedance discontinuity in a transmission line. The forward
voltage on the Z01 side is Vfor1 and the reflected voltage from the
impedance discontinuity (back toward the source) is Vref1. The forward
voltage on the Z02 side is Vfor2 and the reflected voltage (from the
load) is Vref2. Hopefully, the reflection and transmission
coefficients are self-explanatory. rho1 is the reflection coefficient
encountered by Vfor1, etc.

Vref1 = Vfor1(rho1) + Vref2(tau2) = 0

That is wavefront cancellation in action. The external reflection
phasor, Vfor1(rho1), is equal in magnitude and 180 degrees out of
phase with the internal reflection phasor, Vref2(tau2), arriving from
the mismatched load.

Vfor2 = Vfor1(tau1) + Vref2(rho2)

If these RF equations are normalized to SQRT(Z0), they are the same as
the s-parameter equations.
--
73, Cecil, w5dxp.com

ok, so you defined a case where the superposition of the reflected and
refracted waves at a discontinuity results in a zero sum. is that
supposed to prove something? did i ever say that you could not define
such a case??

Where does it go? (mismatched power)
 

On Sat, 12 Jun 2010 19:52:15 GMT, Owen Duffy wrote:

To be usable, any pretence of linearity at least over a limited range of
loads, power, frequency, Zeq must be sufficiently constant within that
range.

Hi Owen,

Well I see neither data of that Z over Loads over Power over
Frequency, nor do I read what you would call what is "usable."

The meaning of "sufficiently" is proposed in the reference article
describing the test and providing the mathematical basis for the test.

"Sufficiently" is a subjective term for which I see no objective
criteria.

My recollection of Walt's tests were that they tested at points other
than Zl=0 and Zl=infinity.

Steps 1 and 2 are quite explicit.

You dwell at some further length on the implicit accuracy of the stated
quantity, but ignore the explicit discussion about a reasonable tolerance
for the test.

I see no further discussion of my observation of the possibility of 51
Ohms being, in your terms, "usable, sufficient, or reasonable," so I
dwell on what is known, and not on vacant adjectives.

73's
Richard Clark, KB7QHC

Where does it go? (mismatched power)
 

Richard,

The article does propose a practical tolerance. I quote the relevant
paragraph below.

"Of course, nothing is perfect so an acceptable tolerance for depending on
the assumed Zs=50+j0? for Mismatch Loss calculations might be that ‘forward
power’ doesn’t vary by more than 10% of the ‘reflected power’ at any point.
The worst case ‘reflected power’ for the experiment as described is 4%, so
a variation of ‘forward power’ by more than 0.4% (ie 0.4W in 100W) at any
point would indicate significant error in any Mismatch calculations based
on an assumed Zs=50+j0?."

I explained in the articles how I am able to reliably detect such small
relative power variation. In one of the tests, the measured variation was
21%.

My intention in publishing the two articles wasn't to prove the propostion
one way or the other, but to show practical amateurs a test that they could
make with relatively simple equipment, one that is relevant to the
application of Zs if it was known, and an explanation of why it works so
that they can understand the test and make their own interpretations about
significance.

Owen

Where does it go? (mismatched power)
 

On 12 jun, 17:28, 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.

It is the implementation of the source that provides the answer to your
question, and the word "never" is too strong for the general case.

In respect of typical HF ham transmitters, you may find my article
entitled "Does SWR damage HF ham transmitters?" athttp://vk1od.net/blog/?p=1081of interest.

Owen

Hello Owen thank you for your answer: Sorry I do not quite understand
your answer. I choose a Thevenin model of circuit theory because it is
an idealization consisting of an idealized constant voltaje source in
series with an idealized resistance without any relation with
practical implementation of such imaginary electrical (and
mathematical) entity.

I first interested get from you such idealized model answer as a
reductionistic aproximation method to try arrive later at subsequent
interpretations of practical situations. I think we all used to
working with idealized models and we accept its limitations, but we
also know frequently they are very useful to clear the "field" (as in
football "field")
(I said "never" because Cecil seem say "sometimes").
For example: ideal conjugate mirror in Maxwell article in my
interpretation implicates "never". Reflected power do not return to
the source in that context.
If you prefer I would be equally satisfied knowing who agree with
"never", who with "sometimes" and who with "always". But I would not
be too annoying :)

73

Miguel Ghezzi LU6ETJ

Where does it go? (mismatched power)
 

On Jun 12, 4:34*pm, K1TTT wrote:
ok, so you defined a case where the superposition of the reflected and
refracted waves at a discontinuity results in a zero sum. *is that
supposed to prove something? *did i ever say that you could not define
such a case??

I would call two waves superposing to zero indefinitely, "wave
cancellation". If that is not wave cancellation, where did the
reflected and refracted wavefronts go along with their energy
components? The answer to that question will reveal exactly what
happens to the reflected energy.

Here's a brain teaser for you and others. Given a Z01 to Z02 impedance
discontinuity with a power reflection coefficient of 0.25 at the '+'
discontinuity:

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

Pfor1 in the Z01 section is 100 watts. Pref1 in the Z01 section is
zero watts.

What is Pfor2, Pref2, and the SWR in the Z02 section?
--
73, Cecil, w5dxp.com

Where does it go? (mismatched power)
 

lu6etj wrote in
:

On 12 jun, 17:28, 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.

It is the implementation of the source that provides the answer to
your question, and the word "never" is too strong for the general
case.

In respect of typical HF ham transmitters, you may find my article
entitled "Does SWR damage HF ham transmitters?"
athttp://vk1od.net/blog/?p=1081of interest.

Owen

Hello Owen thank you for your answer: Sorry I do not quite understand
your answer. I choose a Thevenin model of circuit theory because it is
an idealization consisting of an idealized constant voltaje source in
series with an idealized resistance without any relation with
practical implementation of such imaginary electrical (and
mathematical) entity.

I first interested get from you such idealized model answer as a
reductionistic aproximation method to try arrive later at subsequent
interpretations of practical situations. I think we all used to
working with idealized models and we accept its limitations, but we
also know frequently they are very useful to clear the "field" (as in
football "field")

Miguel,

From Wikipedia: "Thévenin's theorem for linear electrical networks states
that any combination of voltage sources, current sources and resistors
with two terminals is electrically equivalent to a single voltage source
V and a single series resistor R. For single frequency AC systems the
theorem can also be applied to general impedances, not just resistors."

The theorem does not state or imply that the Thevenin equivalent circuit
dissipates the same internal power as the real source, just that any
*linear* two terminal circuit containing sources and impedances can be
reduced to this two component equivalent (at a single frequency), and V
and I at the network terminals will be the same as the original network,
irrespective of the external load attached to the network terminals.

It is a simple exercise to develop two source networks with the same
Thevenin equivalent circuit, but that have quite different internal
efficiencies. It is easy to demonstrate that both networks deliver the
same power to any given load, but that the internal dissipation of those
source networks is different in both cases, and not explainable simply as
absorbing 'reflected power'.

This is basic linear circuit theory.

If there was a valid Thevenin equivalent circuit for a transmitter (and
that is questionable), then you can not use that equivalent circuit to
make any inference about the internal dissipation of the source (the
transmitter in this case), or its efficiency. Nevertheless, I see people
trying to do this one way or another in the various threads here.

(I said "never" because Cecil seem say "sometimes").
For example: ideal conjugate mirror in Maxwell article in my
interpretation implicates "never". Reflected power do not return to
the source in that context.
If you prefer I would be equally satisfied knowing who agree with
"never", who with "sometimes" and who with "always". But I would not
be too annoying :)

I know that in this age of instant gratification, people reading posts in
these fora tend to accept simple dogmatic statements as sure sign of
author credibility, and qualifications such as 'often', 'usually' etc as
a sign of uncertainty, of a lack of understanding, of weakness in the
author. The opposite is often, if not usually true.

In English, we have a saying "never say never".

What 'never'? 'Hardly ever'... to borrow some dialogue.

A man who is hardly ever wrong doesn't use words like 'always' and
'never' much, or imply as much in general statements.

Owen

Where does it go? (mismatched power)
 

On Sat, 12 Jun 2010 14:21:29 -0700 (PDT), K1TTT
wrote:

1. *How could you tell if a lossless line lost energy by radiation?

there would be missing energy?

A lossless line infinite in extent is the typical definition that
corresponds to its characteristic "non-dissipative" resistance. That,
or it is terminated in a lossy resistor (and the loss is suddenly in
everone's face, full and foursquare).

How long are you going to wait to finish the energy budget to measure
IF there is missing energy from an infinite line?

73's
Richard Clark, KB7QHC

Where does it go? (mismatched power)
 

On Sat, 12 Jun 2010 21:52:41 GMT, Owen Duffy wrote:

Richard,

The article does propose a practical tolerance. I quote the relevant
paragraph below.

"Of course, nothing is perfect so an acceptable tolerance for depending on
the assumed Zs=50+j0? for Mismatch Loss calculations might be that ‘forward
power’ doesn’t vary by more than 10% of the ‘reflected power’ at any point.
The worst case ‘reflected power’ for the experiment as described is 4%, so
a variation of ‘forward power’ by more than 0.4% (ie 0.4W in 100W) at any
point would indicate significant error in any Mismatch calculations based
on an assumed Zs=50+j0?."

Hi Owen,

Excuse me for incorrectly referencing your material as lacking in that
regard.

I explained in the articles how I am able to reliably detect such small
relative power variation. In one of the tests, the measured variation was
21%.

What is the residual VSWR of your directional wattmeter?

My intention in publishing the two articles wasn't to prove the propostion
one way or the other, but to show practical amateurs a test that they could
make with relatively simple equipment, one that is relevant to the
application of Zs if it was known, and an explanation of why it works so
that they can understand the test and make their own interpretations about
significance.

On Fri, 11 Jun 2010 20:31:42 GMT, Owen Duffy wrote:

However, that
proposition is easily proven wrong by valid experiments

But what lead us here was that you had proof, at least through some
third party (the experience of your having seen something that
diminished the "proposition"). I asked for that specifically. I have
not seen any reputable counter to Walt's data supporting the
"proposition."

The point remains that out of tolerance loads do not invalidate the
"proposition:"
On Fri, 11 Jun 2010 20:31:42 GMT, Owen Duffy wrote:

there is a proposition that a transmitter "designed/adjusted
for, and expecting a 50 + j 0 ohm load" can be well represented by a
Thevenin equivalent circuit and naturally has Zeq=50+j0.
In fact they announce themselves as violating the "proposition."

I hope this does not return us to the vacant adjectives, which as an
English major, I can full well argue to greater advantage. I don't
think that is productive, but choices remain to pursue them, or to
introduce new facts.

***********

Now, let's return to your paragraph, how do you know where the 50 Ohm
point was? You are relying on relative accuracy to make an absolute
measurement. This only works if you have a standard to make a
transfer from. Walt did this but it is arguable that his 1400 Ohm
resistor was, in fact, that value - however, he had the correct tools
to measure it.

Your two test results could be inverted (fail/pass for pass/fail), or
they could both be pass, or both fail.

I have calibrated many loads from DC to 12 GHz - and most of them
failed for many and sundry reasons and had to be "qualified." In the
vernacular of the standards laboratories that means a table of
correction values that are valid as long as certain characteristics
are met. I prepared many such lengthy reports as no one was going to
buy a new load each calibration cycle (typically 3 or 6 months).

I own one that met 1% accuracy out to 1GHz (@150W), but that was luck
of the draw and I wouldn't be so naive as to claim it retains that
accuracy - however, experience has revealed that with care it should.
Aging loads is like aging wine - there is no stasis. Here we get back
into subjectives, but I know the limits of care, and I know how to
practice it. If pressed to prove my inference of 1%, I also have the
skill to do so to that accuracy according to methods practiced by
Metrologists. It would be exceedingly tedious. Exceedingly tedious.
Or I could pay for it to be done.

73's
Richard Clark, KB7QHC

Where does it go? (mismatched power)
 

On 12 jun, 22:52, Owen Duffy wrote:
lu6etj wrote :





On 12 jun, 17:28, 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.

It is the implementation of the source that provides the answer to
your question, and the word "never" is too strong for the general
case.

In respect of typical HF ham transmitters, you may find my article
entitled "Does SWR damage HF ham transmitters?"
athttp://vk1od.net/blog/?p=1081ofinterest.

Owen

Hello Owen thank you for your answer: Sorry I do not quite understand
your answer. I choose a Thevenin model of circuit theory because it is
an idealization consisting of an idealized constant voltaje source in
series with an idealized resistance without any relation with
practical implementation of such imaginary electrical (and
mathematical) entity.

I first interested get from you such idealized model answer as a
reductionistic aproximation method to try arrive later at subsequent
interpretations of practical situations. I think we all used to
working with idealized models and we accept its limitations, but we
also know frequently they are very useful to clear the "field" (as in
football "field")

Miguel,

From Wikipedia: "Thévenin's theorem for linear electrical networks states
that any combination of voltage sources, current sources and resistors *
with two terminals is electrically equivalent to a single voltage source
V and a single series resistor R. For single frequency AC systems the
theorem can also be applied to general impedances, not just resistors."

The theorem does not state or imply that the Thevenin equivalent circuit
dissipates the same internal power as the real source, just that any
*linear* two terminal circuit containing sources and impedances can be
reduced to this two component equivalent (at a single frequency), and V
and I at the network terminals will be the same as the original network,
irrespective of the external load attached to the network terminals.

It is a simple exercise to develop two source networks with the same
Thevenin equivalent circuit, but that have quite different internal
efficiencies. It is easy to demonstrate that both networks deliver the
same power to any given load, but that the internal dissipation of those
source networks is different in both cases, and not explainable simply as
absorbing 'reflected power'.

This is basic linear circuit theory.

If there was a valid Thevenin equivalent circuit for a transmitter (and
that is questionable), then you can not use that equivalent circuit to
make any inference about the internal dissipation of the source (the
transmitter in this case), or its efficiency. Nevertheless, I see people
trying to do this one way or another in the various threads here.

(I said "never" because Cecil seem say "sometimes").
For example: ideal conjugate mirror in Maxwell article in my
interpretation implicates "never". Reflected power do not return to
the source in that context.
If you prefer I would be equally satisfied knowing who agree with
"never", who with "sometimes" and who with "always". But I would not
be too annoying :)

I know that in this age of instant gratification, people reading posts in
these fora tend to accept simple dogmatic statements as sure sign of
author credibility, and qualifications such as 'often', 'usually' etc as
a sign of uncertainty, of a lack of understanding, of weakness in the
author. The opposite is often, if not usually true.

In English, we have a saying "never say never".

What 'never'? 'Hardly ever'... to borrow some dialogue.

A man who is hardly ever wrong doesn't use words like 'always' and
'never' much, or imply as much in general statements.

Owen- Ocultar texto de la cita -

- Mostrar texto de la cita -

Hello Owen, good day in Australia I hope!

Sorry, with due respect, your answer throws back the ball out of the
soccer field :)
I like poetry also but would you mind search the web for another
scientific uses of "never" word?, such as in
http://www.upscale.utoronto.ca/PVB/H...y/Entropy.html.

Of course many thanks for your time and your kind reply.

Miguel LU6ETJ

Where does it go? (mismatched power)
 

lu6etj wrote in news:5856cdb7-211c-47d6-9b19-
:

Hello Owen, good day in Australia I hope!

Sorry, with due respect, your answer throws back the ball out of the
soccer field :)
I like poetry also but would you mind search the web for another
scientific uses of "never" word?, such as in
http://www.upscale.utoronto.ca/PVB/H...y/Entropy.html.

Of course many thanks for your time and your kind reply.


Hi Miguel,

Lastly first, your reference uses "never" in the statement of a law of
physics. That might be a safer place to use it than your typical ham
assertions like "a non-resonant antenna is never going to work as good as
a resonant one".

I am not sure what "throws the ball out of field". Is it the misuse of
Thevenin equivalent circuits?

I tempted you to an exercise. In case you didn't try it, try this one.

I have three ideal components, a 1000V voltage generator, a 999.5 ohm
resistor, and a 1 ohm resistor, all in series. They are in a black box,
and both ends of the 1 ohm resistor are bought out to two terminals on
the black box. It is a trivial exercise to show that load that will
develop the most power is approximately 1 ohms. At that point, the load
voltage is 0.5V, and load power is 0.25W. Efficiency of the system (ie
Pout/Pin) is 0.25/1000=0.025%. The o/c voltage is approximately 1V and
internal dissipation is approximately 1000W, the s/c current is
approximately 1A and internal dissipation is approximately 1000W.

I have another black box with a 1.0005V generator and a series resistance
of 0.999000 ohms bought out to its two terminals. It produces exactly
the same o/c current, but with internal dissipation of 0W, exactly the
same s/c current with internal dissipation of approximately 1W, maximum
power of approximately 0.25W in an approximately 1 ohm load for system
efficiency of 50%.

Both of these networks have the same Thevenin equivalent circuit (in fact
one is the equivalent circuit), they produce the same load current, load
power, load voltage on the same load impedance, but their internal
dissipation and efficiency is quite different.

The Thevenin equivalent circuit cannot be used to explain the internal
dissipation or efficiency of a source network, but it can be used the
explain the behaviour of the load network.

Owen

Where does it go? (mismatched power)
 

On 13 jun, 03:07, lu6etj wrote:
On 12 jun, 22:52, Owen Duffy wrote:





lu6etj wrote :

On 12 jun, 17:28, 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.

It is the implementation of the source that provides the answer to
your question, and the word "never" is too strong for the general
case.

In respect of typical HF ham transmitters, you may find my article
entitled "Does SWR damage HF ham transmitters?"
athttp://vk1od.net/blog/?p=1081ofinterest.

Owen

Hello Owen thank you for your answer: Sorry I do not quite understand
your answer. I choose a Thevenin model of circuit theory because it is
an idealization consisting of an idealized constant voltaje source in
series with an idealized resistance without any relation with
practical implementation of such imaginary electrical (and
mathematical) entity.

I first interested get from you such idealized model answer as a
reductionistic aproximation method to try arrive later at subsequent
interpretations of practical situations. I think we all used to
working with idealized models and we accept its limitations, but we
also know frequently they are very useful to clear the "field" (as in
football "field")

Miguel,

From Wikipedia: "Thévenin's theorem for linear electrical networks states
that any combination of voltage sources, current sources and resistors *
with two terminals is electrically equivalent to a single voltage source
V and a single series resistor R. For single frequency AC systems the
theorem can also be applied to general impedances, not just resistors."

The theorem does not state or imply that the Thevenin equivalent circuit
dissipates the same internal power as the real source, just that any
*linear* two terminal circuit containing sources and impedances can be
reduced to this two component equivalent (at a single frequency), and V
and I at the network terminals will be the same as the original network,
irrespective of the external load attached to the network terminals.

It is a simple exercise to develop two source networks with the same
Thevenin equivalent circuit, but that have quite different internal
efficiencies. It is easy to demonstrate that both networks deliver the
same power to any given load, but that the internal dissipation of those
source networks is different in both cases, and not explainable simply as
absorbing 'reflected power'.

This is basic linear circuit theory.

If there was a valid Thevenin equivalent circuit for a transmitter (and
that is questionable), then you can not use that equivalent circuit to
make any inference about the internal dissipation of the source (the
transmitter in this case), or its efficiency. Nevertheless, I see people
trying to do this one way or another in the various threads here.

(I said "never" because Cecil seem say "sometimes").
For example: ideal conjugate mirror in Maxwell article in my
interpretation implicates "never". Reflected power do not return to
the source in that context.
If you prefer I would be equally satisfied knowing who agree with
"never", who with "sometimes" and who with "always". But I would not
be too annoying :)

I know that in this age of instant gratification, people reading posts in
these fora tend to accept simple dogmatic statements as sure sign of
author credibility, and qualifications such as 'often', 'usually' etc as
a sign of uncertainty, of a lack of understanding, of weakness in the
author. The opposite is often, if not usually true.

In English, we have a saying "never say never".

What 'never'? 'Hardly ever'... to borrow some dialogue.

A man who is hardly ever wrong doesn't use words like 'always' and
'never' much, or imply as much in general statements.

Owen- Ocultar texto de la cita -

- Mostrar texto de la cita -

Hello Owen, good day in Australia I hope!

Sorry, with due respect, your answer throws back the ball out of the
soccer field :)
I like poetry also but would you mind search the web for another
scientific uses of "never" word?, such as inhttp://www.upscale.utoronto.ca/PVB/Harrison/Entropy/Entropy.html.

Of course many thanks for your time and your kind reply.

Miguel LU6ETJ- Ocultar texto de la cita -

- Mostrar texto de la cita -

Sorry I ommited one comment:
The final Thevenin circuit is an idealization built with an ideal
voltage source in series with an ideal resistor. This new idealized
circuit It is a new born entity whose properties are now fully
described for these only two idealized circuit elements. These are the
virtues and the defects of reductionist models :(

Where does it go? (mismatched power)
 

On 13 jun, 04:13, lu6etj wrote:
On 13 jun, 03:07, lu6etj wrote:





On 12 jun, 22:52, Owen Duffy wrote:

lu6etj wrote :

On 12 jun, 17:28, 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.

It is the implementation of the source that provides the answer to
your question, and the word "never" is too strong for the general
case.

In respect of typical HF ham transmitters, you may find my article
entitled "Does SWR damage HF ham transmitters?"
athttp://vk1od.net/blog/?p=1081ofinterest.

Owen

Hello Owen thank you for your answer: Sorry I do not quite understand
your answer. I choose a Thevenin model of circuit theory because it is
an idealization consisting of an idealized constant voltaje source in
series with an idealized resistance without any relation with
practical implementation of such imaginary electrical (and
mathematical) entity.

I first interested get from you such idealized model answer as a
reductionistic aproximation method to try arrive later at subsequent
interpretations of practical situations. I think we all used to
working with idealized models and we accept its limitations, but we
also know frequently they are very useful to clear the "field" (as in
football "field")

Miguel,

From Wikipedia: "Thévenin's theorem for linear electrical networks states
that any combination of voltage sources, current sources and resistors *
with two terminals is electrically equivalent to a single voltage source
V and a single series resistor R. For single frequency AC systems the
theorem can also be applied to general impedances, not just resistors.."

The theorem does not state or imply that the Thevenin equivalent circuit
dissipates the same internal power as the real source, just that any
*linear* two terminal circuit containing sources and impedances can be
reduced to this two component equivalent (at a single frequency), and V
and I at the network terminals will be the same as the original network,
irrespective of the external load attached to the network terminals.

It is a simple exercise to develop two source networks with the same
Thevenin equivalent circuit, but that have quite different internal
efficiencies. It is easy to demonstrate that both networks deliver the
same power to any given load, but that the internal dissipation of those
source networks is different in both cases, and not explainable simply as
absorbing 'reflected power'.

This is basic linear circuit theory.

If there was a valid Thevenin equivalent circuit for a transmitter (and
that is questionable), then you can not use that equivalent circuit to
make any inference about the internal dissipation of the source (the
transmitter in this case), or its efficiency. Nevertheless, I see people
trying to do this one way or another in the various threads here.

(I said "never" because Cecil seem say "sometimes").
For example: ideal conjugate mirror in Maxwell article in my
interpretation implicates "never". Reflected power do not return to
the source in that context.
If you prefer I would be equally satisfied knowing who agree with
"never", who with "sometimes" and who with "always". But I would not
be too annoying :)

I know that in this age of instant gratification, people reading posts in
these fora tend to accept simple dogmatic statements as sure sign of
author credibility, and qualifications such as 'often', 'usually' etc as
a sign of uncertainty, of a lack of understanding, of weakness in the
author. The opposite is often, if not usually true.

In English, we have a saying "never say never".

What 'never'? 'Hardly ever'... to borrow some dialogue.

A man who is hardly ever wrong doesn't use words like 'always' and
'never' much, or imply as much in general statements.

Owen- Ocultar texto de la cita -

- Mostrar texto de la cita -

Hello Owen, good day in Australia I hope!

Sorry, with due respect, your answer throws back the ball out of the
soccer field :)
I like poetry also but would you mind search the web for another
scientific uses of "never" word?, such as inhttp://www.upscale.utoronto..ca/PVB/Harrison/Entropy/Entropy.html.

Of course many thanks for your time and your kind reply.

Miguel LU6ETJ- Ocultar texto de la cita -

- Mostrar texto de la cita -

Sorry I ommited one comment:
The final Thevenin circuit is an idealization built with an ideal
voltage source in series with an ideal resistor. This new idealized
circuit It is a new born entity whose properties are now fully
described for these only two idealized circuit elements. These are the
virtues and the defects of reductionist models :(- Ocultar texto de la cita -

- Mostrar texto de la cita -

R. R. Well... I forget to say the more important = For the sake of
the advance of the topic please do replace "Thevenin circuit" in my
original question for "an ideal constant voltage source in series with
an ideal resistance" equivalent only to itself :)

Where does it go? (mismatched power)
 

On Jun 13, 12:00*am, Cecil Moore wrote:
On Jun 12, 4:34*pm, K1TTT wrote:

ok, so you defined a case where the superposition of the reflected and
refracted waves at a discontinuity results in a zero sum. *is that
supposed to prove something? *did i ever say that you could not define
such a case??

I would call two waves superposing to zero indefinitely, "wave
cancellation". If that is not wave cancellation, where did the
reflected and refracted wavefronts go along with their energy
components? The answer to that question will reveal exactly what
happens to the reflected energy.

i don't care, i know that the superimposed voltage or current is
zero. from that i can calculate the power or energy anywhere i
want.

why does anyone care about 'energy' anyway, that is even worse to
think about in transmission lines than power. at least you can
measure, or at least calibrate your meters, in power units. have you
ever seen an amateur station that had an energy meter on their
transmitter? and isn't the term 'reflected energy' kind of an
oxymoron anyway? for energy to be reflected it has to be moving, so
isn't that just another word for power?



Here's a brain teaser for you and others. Given a Z01 to Z02 impedance
discontinuity with a power reflection coefficient of 0.25 at the '+'
discontinuity:

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

Pfor1 in the Z01 section is 100 watts. Pref1 in the Z01 section is
zero watts.

What is Pfor2, Pref2, and the SWR in the Z02 section?
--
73, Cecil, w5dxp.com

so? what does this special case prove that hundreds of others
doesn't?

Where does it go? (mismatched power)
 

On Jun 13, 3:22*am, Richard Clark wrote:
On Sat, 12 Jun 2010 14:21:29 -0700 (PDT), K1TTT
wrote:

1. *How could you tell if a lossless line lost energy by radiation?

there would be missing energy?

A lossless line infinite in extent is the typical definition that
corresponds to its characteristic "non-dissipative" resistance. *That,
or it is terminated in a lossy resistor (and the loss is suddenly in
everone's face, full and foursquare).

How long are you going to wait to finish the energy budget to measure
IF there is missing energy from an infinite line?

73's
Richard Clark, KB7QHC

infinitely long of course. i think it was you who added the infinite
part. i don't care as long as it is not losing cecil's precious
energy.

Where does it go? (mismatched power)
 

On Jun 13, 12:00*am, Cecil Moore wrote:
On Jun 12, 4:34*pm, K1TTT wrote:

ok, so you defined a case where the superposition of the reflected and
refracted waves at a discontinuity results in a zero sum. *is that
supposed to prove something? *did i ever say that you could not define
such a case??

I would call two waves superposing to zero indefinitely, "wave
cancellation". If that is not wave cancellation, where did the
reflected and refracted wavefronts go along with their energy
components? The answer to that question will reveal exactly what
happens to the reflected energy.

Here's a brain teaser for you and others. Given a Z01 to Z02 impedance
discontinuity with a power reflection coefficient of 0.25 at the '+'
discontinuity:

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

Pfor1 in the Z01 section is 100 watts. Pref1 in the Z01 section is
zero watts.

What is Pfor2, Pref2, and the SWR in the Z02 section?
--
73, Cecil, w5dxp.com

just for fun... explain why i can see that discontinuity between z01
and z02 when i hook up my tdr.

Where does it go? (mismatched power)
 

On Jun 12, 8:52*pm, Owen Duffy wrote:
If there was a valid Thevenin equivalent circuit for a transmitter (and
that is questionable), then you can not use that equivalent circuit to
make any inference about the internal dissipation of the source (the
transmitter in this case), or its efficiency. Nevertheless, I see people
trying to do this one way or another in the various threads here.

In his food-for-thought article on forward and reflected power, Roy
(w7el) says: "So we can model a 100 watt forward, 50 ohm nominal
transmitter as a 141.4 volt (100 * sqrt(2)) RMS voltage source in
series with a 50 ohm resistance." He goes on to calculate power
dissipation in the source resistor.
--
73, Cecil, w5dxp.com

Where does it go? (mismatched power)
 

On Jun 13, 5:35*am, K1TTT wrote:
On Jun 13, 12:00*am, Cecil Moore wrote:
The answer to that question will reveal exactly what
happens to the reflected energy.

i don't care, i know that the superimposed voltage or current is
zero. *from that i can calculate the power or energy anywhere i
want. why does anyone care about 'energy' anyway, ...

You get exactly the same answers doing it my way but my way yields the
additional information of exactly what happens to the energy
components. When two wavefronts superpose to zero indefinitely, I
would take that as proof of interaction and wave cancellation.

This is what invariably happens to the discussion. After being told
that I am absolutely wrong about energy flow, I introduce the known
laws of EM physics from the field of optics and prove that they
provide exactly the same answers as a conventional RF analysis. After
some discussion, it is asserted that the person (not only you) doesn't
care and it doesn't matter anyway. W7EL says in his food-for-thought
article, "I personally don’t have a compulsion to understand where
this power 'goes'."

A 1/4WL series matching stub is essentially the same function in
concept as a 1/4WL thin-film coating on non-reflective glass. How the
non-reflective glass works is perfectly understood and a 1/4WL series
matching section works the same way. Why not glean some knowledge from
the field of optics if it helps hams to understand "where the power
goes"? Optical physicists were forced to track power density from the
very start of their science because they didn't have the luxury of
tracking the voltage and current.

Here's a brain teaser for you and others. Given a Z01 to Z02 impedance
discontinuity with a power reflection coefficient of 0.25 at the '+'
discontinuity:

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

Pfor1 in the Z01 section is 100 watts. Pref1 in the Z01 section is
zero watts.

What is Pfor2, Pref2, and the SWR in the Z02 section?

so? *what does this special case prove that hundreds of others
doesn't?

The magnitudes of the voltages and currents in the above example are
indeterminate. Can you (or anyone else) solve the problem without
resorting to voltage and current calculations? I am just trying to get
people to think outside of their rigid concrete voltage/current boxes.
--
73, Cecil, w5dxp.com

Where does it go? (mismatched power)
 

On Jun 13, 2:04*pm, Cecil Moore wrote:
On Jun 13, 5:35*am, K1TTT wrote:

On Jun 13, 12:00*am, Cecil Moore wrote:
The answer to that question will reveal exactly what
happens to the reflected energy.

i don't care, i know that the superimposed voltage or current is
zero. *from that i can calculate the power or energy anywhere i
want. why does anyone care about 'energy' anyway, ...

You get exactly the same answers doing it my way but my way yields the
additional information of exactly what happens to the energy
components. When two wavefronts superpose to zero indefinitely, I
would take that as proof of interaction and wave cancellation.

This is what invariably happens to the discussion. After being told
that I am absolutely wrong about energy flow, I introduce the known
laws of EM physics from the field of optics and prove that they
provide exactly the same answers as a conventional RF analysis. After
some discussion, it is asserted that the person (not only you) doesn't
care and it doesn't matter anyway. W7EL says in his food-for-thought
article, "I personally don’t have a compulsion to understand where
this power 'goes'."

A 1/4WL series matching stub is essentially the same function in
concept as a 1/4WL thin-film coating on non-reflective glass. How the
non-reflective glass works is perfectly understood and a 1/4WL series
matching section works the same way. Why not glean some knowledge from
the field of optics if it helps hams to understand "where the power
goes"? Optical physicists were forced to track power density from the
very start of their science because they didn't have the luxury of
tracking the voltage and current.

Here's a brain teaser for you and others. Given a Z01 to Z02 impedance
discontinuity with a power reflection coefficient of 0.25 at the '+'
discontinuity:

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

Pfor1 in the Z01 section is 100 watts. Pref1 in the Z01 section is
zero watts.

What is Pfor2, Pref2, and the SWR in the Z02 section?

so? *what does this special case prove that hundreds of others
doesn't?

The magnitudes of the voltages and currents in the above example are
indeterminate. Can you (or anyone else) solve the problem without
resorting to voltage and current calculations? I am just trying to get
people to think outside of their rigid concrete voltage/current boxes.
--
73, Cecil, w5dxp.com

why are they indeterminate? i can calculate them, why can't you?

Where does it go? (mismatched power)
 

On Jun 13, 2:04*pm, Cecil Moore wrote:
On Jun 13, 5:35*am, K1TTT wrote:

On Jun 13, 12:00*am, Cecil Moore wrote:
The answer to that question will reveal exactly what
happens to the reflected energy.

i don't care, i know that the superimposed voltage or current is
zero. *from that i can calculate the power or energy anywhere i
want. why does anyone care about 'energy' anyway, ...

You get exactly the same answers doing it my way but my way yields the
additional information of exactly what happens to the energy
components. When two wavefronts superpose to zero indefinitely, I
would take that as proof of interaction and wave cancellation.

This is what invariably happens to the discussion. After being told
that I am absolutely wrong about energy flow, I introduce the known
laws of EM physics from the field of optics and prove that they
provide exactly the same answers as a conventional RF analysis. After
some discussion, it is asserted that the person (not only you) doesn't
care and it doesn't matter anyway. W7EL says in his food-for-thought
article, "I personally don’t have a compulsion to understand where
this power 'goes'."

hams 'know' where the power goes, their swr meter tells them it goes
back to the transmitter!

why don't physicists working in optics calculate fields? the electric
and magnetic field models work just as well at those frequencies as
they do on hf. indeed, why don't optical physicists learn something
from rf designers and model their thin films with transmission lines!

Where does it go? (mismatched power)
 

On Jun 13, 6:16*am, K1TTT wrote:
just for fun... explain why i can see that discontinuity between z01
and z02 when i hook up my tdr.

That's easy, because it *physically exists in reality* with a voltage
reflection coefficient of (Z02-Z01)/(Z02+Z01) = 0.5. It is related to
the indexes of refraction in the field of optics from which the same
reflection coefficient can be calculated.

The difficult question is: Exactly why doesn't that same physical
reflection coefficient reflect half of the forward voltage when it is
Z0-matched during steady-state?

The answer is that it indeed does reflect 1/2 of the forward voltage
during steady-state but that wavefront interacts with 1/2 of the
reflected voltage returning from the mismatched load which is equal in
magnitude and 180 degrees out of phase. In this case, superposition of
the two waves results in wave cancellation (total destructive
interference). The energy components in those two waves are combined
and redistributed back toward the load as constructive interference in
phase with the forward wave from the source. That's where the
reflected energy goes.

That's why the s-parameter analysis theory could be important to hams.
By merely measuring the four reflection/transmission coefficients
(s11, s12, s21, s22) one learns the basics of superposition. S-
parameter analysis was covered in my 1950's college textbook, "Fields
and Waves in Modern Radio", by Ramo and Whinnery (c)1944. I don't know
how or why the younger generation missed it.
--
73, Cecil, w5dxp.com

Where does it go? (mismatched power)
 

On Jun 13, 9:34*am, K1TTT wrote:
why are they (voltages) indeterminate? *i can calculate them, why can't you?

Purposefully, the numerical values of Z01 and Z02 are not given and
unknown. The answer to the problem would be the same if Z01=50 and
Z02=150, or if Z01=100 and Z02=300, or an infinite number of other
combinations. Please tell me how you can calculate an absolute voltage
when Z0 is an unknown variable?
--
73, Cecil, w5dxp.com