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 :)

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?

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.

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.

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

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

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?

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!

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

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