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.