Hi all. I have a doubt about if this text from "Another look at
reflections" reflect the author ideas.

"Laboratory and experimental work often requires holding incident
voltage constant with variation in loading. A constant-voltage source
is usually obtained for this purpose by inserting a pad having 15 to
20 dB attenuation between the generator and the line to absorb the
reflected power, preventing it from reaching the generator where it
would alter the line coupling and cause the generator output voltage
to vary. Because of the absorption of the pad, the generator sees a
nearly perfect match for all load conditions and all reflected power
is lost - but these are laboratory control conditions required to
obtain valid test data."

For example this part = "by inserting a pad having 15 to 20 dB
attenuation between the generator and the line TO ABSORB the reflected
power"

¿Absorb the reflected power or amortiguate the effects of variyng load
impedance?

As I understand it, even without a conjugated match the reflected
power neither returns to the source (Thevenin equivalent source).

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 admire Walter's work and this consultation should only be read as a
question, not as a criticism)

Thank you very much in advance.

Miguel Ghezzi LU6ETJ

lu6etj wrote in news:62702483-6d3c-43e0-aa8b-
:

Hi all. I have a doubt about if this text from "Another look at
reflections" reflect the author ideas.

"Laboratory and experimental work often requires holding incident
voltage constant with variation in loading. A constant-voltage source
is usually obtained for this purpose by inserting a pad having 15 to
20 dB attenuation between the generator and the line to absorb the
reflected power, preventing it from reaching the generator where it
would alter the line coupling and cause the generator output voltage
to vary. Because of the absorption of the pad, the generator sees a
nearly perfect match for all load conditions and all reflected power
is lost - but these are laboratory control conditions required to
obtain valid test data."

For example this part = "by inserting a pad having 15 to 20 dB
attenuation between the generator and the line TO ABSORB the reflected
power"

I would not refer to the cascading of a practical source with a 15 to
20dB attenautor at a 'constant-voltage source' as Walt did above.

A Thevenin source with Zs=50+j0, with varying load impedances, will
always have terminal v/i such that the Vfwd component of a traveling wave
decomposition of v is constant. I give the explanation / proof at
http://vk1od.net/blog/?p=1028 .

Even if a given source has Zs not equal to 50+j0, a sufficiently large
cascaded 50 ohm attenuator will result in a combination that is very
close to the ideal Zs=50+j0, albeit at lower level. This is a common
technique for 'standardising' the equivalent source impedance of a
source.

Owen

On 24 mayo, 17:30, Owen Duffy wrote:
lu6etj wrote in news:62702483-6d3c-43e0-aa8b-
:





Hi all. I have a doubt about if this text from "Another look at
reflections" reflect the author ideas.

"Laboratory and experimental work often requires holding incident
voltage constant with variation in loading. A constant-voltage source
is usually obtained for this purpose by inserting a pad having 15 to
20 dB attenuation between the generator and the line to absorb the
reflected power, preventing it from reaching the generator where it
would alter the line coupling and cause the generator output voltage
to vary. Because of the absorption of the pad, the generator sees a
nearly perfect match for all load conditions and all reflected power
is lost - but these are laboratory control conditions required to
obtain valid test data."

For example this part = "by inserting a pad having 15 to 20 dB
attenuation between the generator and the line TO ABSORB the reflected
power"

I would not refer to the cascading of a practical source with a 15 to
20dB attenautor at a 'constant-voltage source' as Walt did above.

A Thevenin source with Zs=50+j0, with varying load impedances, will
always have terminal v/i such that the Vfwd component of a traveling wave
decomposition of v is constant. I give the explanation / proof athttp://vk1od.net/blog/?p=1028.

Even if a given source has Zs not equal to 50+j0, a sufficiently large
cascaded 50 ohm attenuator will result in a combination that is very
close to the ideal Zs=50+j0, albeit at lower level. This is a common
technique for 'standardising' the equivalent source impedance of a
source.

Owen- Ocultar texto de la cita -

- Mostrar texto de la cita -

Thank you for your answer Owen (but I think that it doesn't answer my
specific question, possibly because I am not able to express it well
due to my translation to English).
......
I agree with what you say in your answer and several things you write
in your good article "Is Zs of to HF ham tx typically 50+j0? ".

For example that for an idealized Theveninj generator, the incident
power is independent of the load impedance (net power Pf-Pr not)
(If an impedances adapting device is inserted between the line and
generator and we insert directional wattmeter between the matching
device and line, of course Pf varies and Pnet also do it).

Regarding that output impedance of a rig be quite near to 50+j0 I
think that depends on the concrete design; such a condition is not
necessary and in my opinion neither optimal.

Best regards

Miguel Ghezzi LU6ETJ

On Mon, 24 May 2010 14:23:59 -0700 (PDT), lu6etj
wrote:

Regarding that output impedance of a rig be quite near to 50+j0 I
think that depends on the concrete design; such a condition is not
necessary and in my opinion neither optimal.

Hi Miguel,

Problems arise from qualitative statements or questions instead of
quantitative statements or questions.
output impedance of a rig be quite near to 50+j0
is a quantitative fact (depending upon the qualitative "near"). A
fact can be measured and compared by in dependant and objective means.
in my opinion neither optimal
is a qualitative judgment. A judgment is only as authoritative as is
the authority making it - and even then it comes at a discount. We
get many judgments here by folks with little authority (and such
judgments are usually adorned with anti-authority messages).

I have measured the characteristic Z of my two transistorized HF rigs
by different methods that agree in their results. I have also
measured the Z of active loads that employ the same circuitry found in
power sources. The values of Z for my HF rigs wander from 35 Ohms to
75 Ohms. The variation is a function of
1. frequency;
2. power;
3. which of the two is being measured.

What is notable, for those values that are furthest from 50 Ohms (the
design goal), my rigs experience issues (instabilities, poor
efficiency, distortion...).

The variation by frequency is found at the margins (at the ends of the
HF band).

The variation by power is found in inabilities to maintain a constant
power deliver in every band - which weakly correlates to Z offset from
50. Variation by power also defines Z. Low power operation is not
going to give you a 50 Ohm source (the manufacturer designs Z for
rated power).

The variation by set is due to different capabilities: one set is
capable of 150W, and the other 100W; thus the first is more robust. As
these variations of Z are not remarkable, trying to turn them into
quantitative efficiency results are speculative at best (common
designs are not optimal by any stretch when you add in the
complication of modes).

73's
Richard Clark, KB7QHC

On 24 mayo, 19:33, Richard Clark wrote:
On Mon, 24 May 2010 14:23:59 -0700 (PDT), lu6etj
wrote:

Regarding that output impedance of a rig be quite near to 50+j0 I
think that depends on the concrete design; such a condition is not
necessary and in my opinion neither optimal.

Hi Miguel,

Problems arise from qualitative statements or questions instead of
quantitative statements or questions.output impedance of a rig be quite near to 50+j0

is a quantitative fact (depending upon the qualitative "near"). *A
fact can be measured and compared by in dependant and objective means.in my opinion neither optimal

is a qualitative judgment. *A judgment is only as authoritative as is
the authority making it - and even then it comes at a discount. *We
get many judgments here by folks with little authority (and such
judgments are usually adorned with anti-authority messages).

I have measured the characteristic Z of my two transistorized HF rigs
by different methods that agree in their results. *I have also
measured the Z of active loads that employ the same circuitry found in
power sources. *The values of Z for my HF rigs wander from 35 Ohms to
75 Ohms. *The variation is a function of
1. frequency;
2. power;
3. which of the two is being measured.

What is notable, for those values that are furthest from 50 Ohms (the
design goal), my rigs experience issues (instabilities, poor
efficiency, distortion...).

The variation by frequency is found at the margins (at the ends of the
HF band). *

The variation by power is found in inabilities to maintain a constant
power deliver in every band - which weakly correlates to Z offset from
50. *Variation by power also defines Z. *Low power operation is not
going to give you a 50 Ohm source (the manufacturer designs Z for
rated power).

The variation by set is due to different capabilities: *one set is
capable of 150W, and the other 100W; thus the first is more robust. As
these variations of Z are not remarkable, trying to turn them into
quantitative efficiency results are speculative at best (common
designs are not optimal by any stretch when you add in the
complication of modes).

73's
Richard Clark, KB7QHC

Hi Richard:

As I understand = "output impedance of a rig be near to 50+j0" it is
not a cuantitative fact except when it refers to specific study
objects, obvously not all study objects (transmitters) will
necessarily share those numbers.
I think a good general cuantitative assertion would be for example,
"all the rig on the market gives near 50 +j0, Zout". I am not deny
that experimental fact...

When I say "in my opinion" I am just not giving a formal hipothesis or
theory and I quite understand technical and logical limitations of
such words (for that I use them)
Certainly I am definiteley not an authoritative guy on any
matter :(

(I'm so sorry Richard, although theory of knowledge is a topic of my
interest I am not in conditions of discussing about it in a (for me)
foreign language :( )

I saw some of this topics discussed time ago in this newsgroup and it
was not my intention to return to them because I know they motivated
good ponderings and respectfully points of view in its moment.

PSE, with the due respect and consideration toward you an the
distinguished colleagues and friends, Would you mind return to the
original question? (sorry if it is not this the most polite form to
ask it)

73

Miguel Ghezzi LU6ETJ

On May 24, 3:06*pm, lu6etj wrote:
For example this part = "by inserting a pad having 15 to 20 dB
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,

Yes, standing waves are hard to visualize, but there is indeed same-
cycle interference involving forward waves and reflected waves. There
is a certain delay from the source signal to the load and back that
can be calculated if one chooses. The wave reflection model is closer
to Maxwell's equations than is the lumped-circuit model where EM waves
propagate instantaneously. Maybe the concepts presented in the
following paper would help. There is no transmission line for a Tesla
coil but reflection effects still exist. The lumped circuit model, to
which you allude, incorrectly assumes that signals can travel at
faster than the speed of light, an obvious impossibility.

http://www.ttr.com/corum/index.htm

Incidentally, this is the reason that W8JI measured a 3 ns delay
through a foot-long 75m loading coil. There were same-cycle
reflections existing in a near-infinite SWR situation. In such a
configuration, the phase of the current doesn't change at all yet W8JI
assumed the measured phase change was proportional to the delay
through the coil. Nothing could be farther from the technical truth.
--
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, 21:30, Cecil Moore wrote:
On May 24, 3:06*pm, lu6etj wrote:

For example this part = "by inserting a pad having 15 to 20 dB
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,

Yes, standing waves are hard to visualize, but there is indeed same-
cycle interference involving forward waves and reflected waves. There
is a certain delay from the source signal to the load and back that
can be calculated if one chooses. The wave reflection model is closer
to Maxwell's equations than is the lumped-circuit model where EM waves
propagate instantaneously. Maybe the concepts presented in the
following paper would help. There is no transmission line for a Tesla
coil but reflection effects still exist. The lumped circuit model, to
which you allude, incorrectly assumes that signals can travel at
faster than the speed of light, an obvious impossibility.

http://www.ttr.com/corum/index.htm

Incidentally, this is the reason that W8JI measured a 3 ns delay
through a foot-long 75m loading coil. There were same-cycle
reflections existing in a near-infinite SWR situation. In such a
configuration, the phase of the current doesn't change at all yet W8JI
assumed the measured phase change was proportional to the delay
through the coil. Nothing could be farther from the technical truth.
--
73, Cecil, w5dxp.com


Hi Cecil. thank you very much for your answer. (I am a "fan" of your
rationale and very practical multiband open wire antenna feeding and I
do not know why it is no so popular as G5RV in my country).
.....
I know a model is necessary false by definition because is only a
model (an approximation to the hipothetical "real thing"), but I do
not imagine what is the tranmission line in the source-load
combination of my idealized model circuit example. I am not capable to
judge validity of idealized circuit theory in general to this example,
neither. Please, could you tell me which would be the transmission
line Zo in my resistive divider example?

Anyway, my question is about validity of the assertion that reflected
wave -in that example- IS ABSORBED by the pad. According to my simple
calculations this hipothesis, as I see it, it does not coincide with
my early learnings.

For example, with a 2 V generator with Zs=1 ohm, Zc=4 ohms and Zo=1
ohm, Pc is 0.64 W. Then Pf =1 W (as Owen says in his article), Pr=0.36
W. Rendering Ef=1 V, and Vf=0.6V. Summing both in phase gives 1,6 V on
the line-in point (same as on load point).
These are the same power and voltages values that simple resistive
divider predicts. The system fulfills the Kirchoff law and all power
(as we learn in circuits theory) it flows from source to load.
In the example I suggest a half wave transmission line loaded with 4
ohms it is at practical effects indistinguishable from a 4 ohms
resistive load directly connected to generator. What would be the
reflected power that would be dissipating in Zs (or the pad)?

This is not opposed to the conjugate mirror principle explained,
neither other propositions given in the cited article. IMHO reflected
power never "returns into" generators when they are active (and in
steady state); reflected power it is re-reflected in conjugated match,
and vectorially composed to render a load impedance to the generator
when this is directly connected to the line (when there is not any
matching devices inserted).
I do not suggesting that reflected power is "virtual" or any similar
thing, 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.

If I am in error please give me your explanations.

73

Miguel ghezzi LU6ETJ

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 Mon, 24 May 2010 17:17:41 -0700 (PDT), lu6etj
wrote:

PSE, with the due respect and consideration toward you an the
distinguished colleagues and friends, Would you mind return to the
original question? (sorry if it is not this the most polite form to
ask it)

Hi Miguel,

I presume by "original question" you mean:
On Mon, 24 May 2010 13:06:19 -0700 (PDT), lu6etj wrote:

¿Absorb the reflected power or amortiguate the effects of variyng load
impedance?

The answer is YES.

Now, if you mean by absorb that all absorbtion results in heat, then
the answer is NO.

If you mean by absorb that all energy is combined in a load, then the
answer is YES.

The difference between YES and NO is the PHASE differences of the two
energies that are combined.

73's
Richard Clark, KB7QHC