On May 27, 9:50*am, lu6etj wrote:
I am reading this newsgroup through Google groups web page and I just
realized that later replies to previous post are intercalated in the
thread, while I expected to see it always at the end of it, for that
reason I did not ACK before to it. (I hope yours be the only one, I
will review all thread tho chek for others).

I am also using Google since ATT dropped Usenet. I liked Thunderbird a
lot better than Google's usenet interface but I am adapting. The above
information is good to know. Thunderbird has a way to keep up with
unread vs read postings but Google doesn't seem to - at least I don't
know how to do it on Google.

In a early post I wrote = "of course if we insert a circulator to
separate both powers, generator now would see 1 ohm load, could
develope 1 W incident, 0 W reflected (Pn=1W) on circulator input, 0.36
W would be outputting on the other port to render 0.64 W (Pn) to the
load with 1 W Pf and 0,36 W Pr again"
Is this result OK for you?.

The SGCR source is usually designed for 50 ohms, i.e. the signal
generator always "sees" a 50 ohm load because it does not "see" any
reflected energy. The ideal circulator is usually designed with 50 ohm
line and a 50 ohm load resistor. If we could stick with that
particular configuration for the SGCR source, it would aid in my
understanding what is the actual system configuration, i.e. not your
fault but I am confused by your above posting.

I am interested in your optic analogy, I can imagine the load as a
partially reflecting surface, real part of it as absorbance
(transmittance if it was a radiator). line as a unidimensional medium
and reflection as the form of "redistribute energy" (is it OK?) and a
coherent light source for the voltage source, but I am still trying to
visualze the optical equivalent of source resistance and its job to be
a good analog, Also I am interested in check other values and
conditions in your other article (first part) with 45 degree line.

I don't think a laser source handles reflected energy like an RF amp
does. So, to start with, let's avoid reflected energy being incident
upon the laser source. Here is a good example to start with, a 1/4WL
non-reflective coating on glass.

Laser-----air-------|--1/4WL thin-film, r = 1.2222---|---Glass, r =
1.4938---...

The 1/4WL thin-film coating on the glass acts exactly like a 1/4WL
matching section of transmission line. Reflections at the air to thin-
film interface are eliminated by wave cancellation just as the FSU web
page says,

micro.magnet.fsu.edu/primer/java/scienceopticsu/interference/
waveinteractions/index.html

"... when two waves of equal amplitude and wavelength that are 180-
degrees ... out of phase with each other meet, they are not actually
annihilated, ... All of the photon energy present in these waves must
somehow be recovered or redistributed in a new direction, according to
the law of energy conservation ... Instead, upon meeting, the photons
are redistributed to regions that permit constructive interference, so
the effect should be considered as a redistribution of light waves and
photon energy rather than the spontaneous construction or destruction
of light."

Note that the reflection coefficient, r, is 1.0 for air. Thus the
SQRT[(1.0)(1.4938)] = 1.2222 ensures that reflections are eliminated
by the r = 1.2222 thin-film coating.

The same thing happens at the '+' Z0-match in the following RF system.

XMTR---50 ohm coax---+---1/4WL 300 ohm feedline---1800 ohm load

Note that SQRT[(50)(1800)] = 300 ensuring that reflections are
eliminated.
--
73, Cecil, w5dxp.com

On May 27, 9:50*am, lu6etj wrote:
Also I am interested in check other values and
conditions in your other article (first part) with 45 degree line.

Sorry, I forgot to comment on this. If the line length is fixed at 45
degrees, the reflected wave arrives back at the 50 ohm source resistor
90 degrees out of phase with the source's forward wave. When two waves
are 90 degrees out of phase, there is zero interference between them
because cos(90) = 0 and the interference term in the following
equation disappears.

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

There is no re-reflection of the reflected wave from the 50 ohm source
resistor because it matches the coax Z0. There is no redistribution of
constructive/destructive interference energy because there is zero
interference. Therefore, for the special case where Vfor is 90 degrees
out of phase with Vref at the source resistor, all of the reflected
power will be dissipated in the source resistor.

If the interference at the source resistor is constructive, i.e. less
than 90 degrees difference between Vfor and Vref, the power dissipated
in the source resistor will be all of the reflected power plus some of
the source power.

If the interference at the source resistor between Vfor and Vref is
destructive, i.e. between 90 degrees and 180 degrees, some of the
reflected power will be redistributed (by wave cancellation) back
toward the load.

Note that the I and Q method of transferring two streams of
information on the same carrier relies on this same no interference
(at 90 degrees) concept.
--
73, Cecil, w5dxp.com

On May 27, 9:50*am, lu6etj wrote:
I am reading this newsgroup through Google groups web page and I just
realized that later replies to previous post are intercalated in the
thread, while I expected to see it always at the end of it, for that
reason I did not ACK before to it.

FYI, It appears that Google has a "Reader" available at the top of the
web page, that has the mark-as-read/unread ability with which one can
keep track of the read/unread postings.
--
73, Cecil, w5dxp.com

On 25/05/2010 07:06, lu6etj wrote:
....

I give the following example: If we load a generator directly with a
resistance of 10 ohms, without any transmission line, there are not
traveling waves interfering, therefore there are not stationary waves,
there is only a impedance mismatch affecting the source voltage ¿OK?.
If then I load it with a line of half wave finished in the same 10
ohms resitance, the generator "will see" a 10 ohms load, and it will
behave identically as it made it with the original "direct" load.
In both cases there is not reflected power that it can return to the
generator...

I used a similar problem in my discussion entitled "A simple VSWR
analysis without mirrors" at http://vk1od.net/blog/?p=1259. The article
then goes on to solve a real problem (ie a practical low loss line
rather than the lossless line in the first example, and in your example.

You should find them relevant.

Owen

On 24 mayo, 22:09, Owen wrote:
On 25/05/2010 07:06, lu6etj wrote:
...

I give the following example: If we load a generator directly with a
resistance of 10 ohms, without any transmission line, there are not
traveling waves interfering, therefore there are not stationary waves,
there is only a impedance mismatch affecting the source voltage ¿OK?.
If then I load it with a line of half wave finished in the same 10
ohms resitance, the generator "will see" a 10 ohms load, and it will
behave identically as it made it with the original "direct" load.
In both cases there is not reflected power that it can return to the
generator...

I used a similar problem in my discussion entitled "A simple VSWR
analysis without mirrors" athttp://vk1od.net/blog/?p=1259. The article
then goes on to solve a real problem (ie a practical low loss line
rather than the lossless line in the first example, and in your example.

You should find them relevant.

Owen

Well Owen...I have just read your article ¡and you have put me
definitively in a problem! :)
I was never happened to question the Walter's work, I made effort in
learning it being a young student. Now you have put me in a corner and
I will must meditate hard about the issue (with your help,
certainly).

In principle I take a risk to think that somehow it is necessary to
solve "what to do with the reflected traveling wave" because the model
seems to demand "come off of her" somehow that it doesn't return into
the generator, and an explanatory model (and IMHO appropriate) of
making it is to postulate its re-reflection to be consistent with it.

In conventional theory of circuits we simply apply the second law of
Kirchoff because we do not have traveling waves and we accept that the
energy can only flow from the generator toward the load (supposing a
circuit without reactances), there is not anything to come off (in
spanish = deshacerse, desprenderse) :)

I am happened to think that your article combine concepts of both
models, those of basic circuit theory (without traveling waves) and
those of the traveling waves -same as my simple example of course-
with the difference that in mine one I only wanted to point that the
reflected power is not "absorbed" by the pad, and yours is critic of
conjugate mirror model and other postulations.

I think the mixture or combination of models -maybe- it would not be
"elegant" or consistent although it can arrive to the same numerical
results, but I don not dare to advance more than that in my
speculations :)

73

Miguel Ghezzi - LU6ETJ

On 25/05/2010 16:10, lu6etj wrote:
....
I think the mixture or combination of models -maybe- it would not be
"elegant" or consistent although it can arrive to the same numerical
results, but I don not dare to advance more than that in my
speculations :)

Miguel,

My explanation uses standard linear circuit theory, and the RLGC model
of a transmission line (captured in the Telegrapher's equation). It is a
frequency domain model, and it is complete and consistent.

One of the things that creates confusion in some peoples minds is that
they want one foot in the frequency domain (where you can talk about
concepts like reactance, complex impedance, VSWR) and simultaneously,
one in the time domain taking about re-re-reflected waves.

You can work in either domain, and you can transform between domains,
but trying to be in both at the same time creates problems.

BTW, if you think the problem is challenging to solve in the frequency
domain, don't even think about trying to solve it in the time domain.

So, do not worry about re-reflection, it is dealt with as you have
discovered by the steady state solution when you load the source with
the (steady state) impedance seen looking into the line. The resolution
of the wave component voltages and currents with KVL and KCL at the
circuit nodes gives the steady state solution. The Telegrapher's
equation gives you the amplitude and phase relationship of the wave
components for the transmission line, not just for fictitious lossless
lines, but for practical lines as the example demonstrates.

A steady state frequency domain analysis is quite adequate for most ham
problems. You don't see it spelled out as such, but that is how the ham
handbooks describe and solve problems.

As far as the myths about PAs destroyed by absorbing reflected power,
see "Does SWR damage HF ham transmitters?" at
http://vk1od.net/blog/?p=1081 .

Owen

On 25 mayo, 03:56, Owen wrote:
On 25/05/2010 16:10, lu6etj wrote:
...

I think the mixture or combination of models -maybe- it would not be
"elegant" or consistent although it can arrive to the same numerical
results, but I don not dare to advance more than that in my
speculations :)

Miguel,

My explanation uses standard linear circuit theory, and the RLGC model
of a transmission line (captured in the Telegrapher's equation). It is a
frequency domain model, and it is complete and consistent.

One of the things that creates confusion in some peoples minds is that
they want one foot in the frequency domain (where you can talk about
concepts like reactance, complex impedance, VSWR) and simultaneously,
one in the time domain taking about re-re-reflected waves.

You can work in either domain, and you can transform between domains,
but trying to be in both at the same time creates problems.

BTW, if you think the problem is challenging to solve in the frequency
domain, don't even think about trying to solve it in the time domain.

So, do not worry about re-reflection, it is dealt with as you have
discovered by the steady state solution when you load the source with
the (steady state) impedance seen looking into the line. The resolution
of the wave component voltages and currents with KVL and KCL at the
circuit nodes gives the steady state solution. The Telegrapher's
equation gives you the amplitude and phase relationship of the wave
components for the transmission line, not just for fictitious lossless
lines, but for practical lines as the example demonstrates.

A steady state frequency domain analysis is quite adequate for most ham
problems. You don't see it spelled out as such, but that is how the ham
handbooks describe and solve problems.

As far as the myths about PAs destroyed by absorbing reflected power,
see "Does SWR damage HF ham transmitters?" athttp://vk1od.net/blog/?p=1081.

Owen

Hi Richard and Owen

To Richard: What I mean is irrelevant :) relevant is what Walt
wanted to say in this sentence: "Because of the absorption of the
pad, the generator sees a nearly perfect match for all load conditions
and all reflected power is lost "
Pllease, tell me what in english means "all reflected power is lost"?
I understood (or translate or interpret) that reflected power is
dissipated in the pad: Is it a bad translation/interpretation?

To Owen: Sincerely thanks for your reasons. You can be sure I will
take note about your explanation an take some time tu analize it, but
I am not sure about to arrive at at the right conclusion because what
I read in this newsgroup is a long-standing discussion here.

Miguel

Another bit of reading that might help shed light on the matter is
http://eznec.com/misc/Food_for_thought.pdf, written eight years ago
during one of the many times the subject has come up before on this
newsgroup.

The chart and discussion in the "Forward and reverse power" section show
that the concept of "reflected power" being absorbed in or dissipated by
the source is incorrect. I find the concept of traveling waves of
average power to be misleading at best, and analyses using this concept
lead to impossible conclusions like the supposed absorption of power in
the source resistance.

Roy Lewallen, W7EL

On 25/05/2010 18:45, lu6etj wrote:
....
To Owen: Sincerely thanks for your reasons. You can be sure I will
take note about your explanation an take some time tu analize it, but
I am not sure about to arrive at at the right conclusion because what
I read in this newsgroup is a long-standing discussion here.


Yes, it is a recurrent discussion item. No doubt someone will be along
shortly to add some confusion to the pot.

A parting thought, do not confuse the process of establishment of steady
state with steady state. The only reason that a steady state solution is
valid, is that the system spends most of its time in steady state, or
substantially so. If the solution needs to focus mainly on establishment
of steady state (in other words, the system never substantially
settles), then you should be doing a time domain solution, and VSWR,
reactance, complex impedance are not a time domain concepts.

If you need to convince yourself, do a hand workup of five of ten
transits with a sinusoidal excitation. See that is does converge, and
quickly, and that at the load end, the reflected wave relative to the
forward wave is a fixed ratio (and hence VSWR) that depends ONLY on Zo
and Zl, there is NO influence by the source or any perceived source
reflection on the steady state. There are some animations of this on the
net, but they seem to confuse people more than enlighten them.

Owen

On Tue, 25 May 2010 00:45:36 -0700 (PDT), lu6etj
wrote:

To Richard: What I mean is irrelevant :) relevant is what Walt
wanted to say in this sentence: "Because of the absorption of the
pad, the generator sees a nearly perfect match for all load conditions
and all reflected power is lost "
Pllease, tell me what in english means "all reflected power is lost"?
I understood (or translate or interpret) that reflected power is
dissipated in the pad: Is it a bad translation/interpretation?

Hi Miguel,

Your translation is fine.

However, I have no idea what the pad design looks like, nor do I know
the component values. I have calibrated thousands of standard pads at
frequencies up to the 12 GHz. They came in either a Pi design, or a T
design. Their intended use is in system isolation. That is, they
isolate the source from the load OR isolate the load from the source
OR isolate both. For certain component values, you can replace the
"OR" with "AND."

You would isolate the source to keep its frequency and power constant.
You would isolate the load to keep line SWR flat. For this line
application, it is assumed you are calibrating either a load equal or
nearly equal to Zc, or you are measuring RF power. These are the
purpose of pads (they also serve the same function in audio circuits).
Measuring power in the presence of SWR other than 1:1 requires
sophisticated math that is rarely discussed here. Most discussion
usually accepts the presumptions of special cases (which are often
sufficient) and employ less rigorous formulas (which serve well within
the unstated presumptions).

In conventional Kirchoff analysis, the resistor that bridges the
transmission line opening becomes the source (that is Vs and Rs). Pad
design usually makes that one resistor for the Pi pad, or two
resistors for the T pad. If you are working in accurate and precise
measurement, you then account for the input (source) resistance in
parallel/series combinations. This second computation is the numeric
analysis of isolation. The higher the ratios of these pad resistors,
the higher the isolation.

It doesn't normally serve any use to have the pad apparent resistance
(what I called Rs above) different from Zc or from Zload, but as this
component is a sacrificial one, the designer may choose to put it to
use to achieve a desired goal. Pad performance suffers with heat due
to energy combinations that come from multiple/single sources.

73's
Richard Clark, KB7QHC