tuner - feedline - antenna question ?
 

Richard Clark wrote in
:

On Thu, 01 Mar 2007 19:21:43 GMT, Owen Duffy wrote:

If you are asserting that VSWR on a real or even theoretical line
varies sinudoidally with displacement, it is time to go back to
basics. You need some time with a reputable text book.

Hi Owen,

Under the right circumstances (and they have been presumed in some
discussion here), then the power terms (as expressed by a power meter
inserted into the line) will vary sinusoidally with displacement, even
if the SWR does not.

Richard,

I am not sure of what you mean by the "right circumstances".

Firstly, except in the very special case of an almost purely inductive
load on a lossy cable at low frequencies, the VSWR calculated from Vf and
Vr at a point on a line, decreases smoothly from the load end to the
source end. (Vf and Vr each decay exponentially in magnitude from source
to load, and the forward and reflected power values calculated from those
samples will vary as the square of the exponentially decaying Vf, Vr from
load to source.)

If you use an instrument that is calibrated for an impedance other than
the line under test, your measurement does not indicate VSWR on the line
under test, and the instrument readings will be different than I outlined
in the previous paragraph. Fig 3 in my article at
http://www.vk1od.net/VSWR/VSWRMeter.htm shows a line labelled "VSWR(50)"
that indicates the values that would be indicated / calculated using a 50
ohm instrument in a 75 ohm cable with a 1.5:1 VSWR.

Owen

tuner - feedline - antenna question ?
 

On Thu, 1 Mar 2007 20:13:32 +0000, Ian White GM3SEK
wrote:

I ordered the book from the other side of the world because I wanted to
be very sure of my answers next time around.

Hi Ian,

It will contain much of interest. For instance, it relates to Owen's
moribund thread "the power explanation." Page 205, third paragraph
from the bottom conforms to one of my recent posts the
"Although the power delivered by the source to the line is thus
shown to be reduced by the amount of the reflected power returning
to the input terminals ... the implication of the latter reasoning
that the reflected wave power is entirely absorbed in the source
impedance without affecting the total output of the signal source
generator, is incorrect."

Contrary to that teaching, is discussion on page 203, last paragraph.
It relates to figure 9-26, clearly illustrating a mismatched line fed
by a source with a source resistance. This may be upsetting to many:
"At the signal source end of the line ... none of the power
reflected by the terminal load impedance is re-reflected on
returning to the input end of the line."
The ellipsis reveals that the source Z matches the line Z.

To begin at the beginning of multiple reflection coverage, go to the
same named section (8.8) on page 174. It is not his complete say on
the topic, but it starts here formally. To add insult to someone's
injury, his math includes source Z. However, by the same token
Chipman explicitly states:
"... the shape of the standing wave pattern ... is in no way
affected by the quantities Vs, Zs and Rho-s at the source."

I would also note the irony in that Chipman expresses reflections in
lines in terms of power. To subdue that irony, I would also admit he
is quick to shift to energy when the usage of power is to lead to
problematic solutions (so, using power as an expression in this
context is allowable by precedent as being informal).

Of course, Chipman must be accepted as an authority for any of these
issues to be considered valid.

73's
Richard Clark, KB7QHC

tuner - feedline - antenna question ?
 

On Thu, 01 Mar 2007 21:04:36 GMT, Owen Duffy wrote:

I am not sure of what you mean by the "right circumstances".

Hi Owen,

It should be under "all circumstances." However, to reveal it
requires the "right circumstances." This again returns us to the
discussion of source resistance/impedance. A matching source driving
a mismatched load through a line of indeterminate length can exhibit
this variation, but it will require considerable skill to see it. If
you force the problem by mismatching the source as well (this then
means that the line is mismatched at both ends, much like your
halfwave model); then you can observe a variation in power readings
along its length that vary sinusoidally.

If you use an instrument that is calibrated for an impedance other than
the line under test,

That is not the case, although there are occasions where power has to
be determined in a heavily mismatched situation - this is done with
considerable error if the line lengths are unknown. If they are, then
corrections can be made.

your measurement does not indicate VSWR on the line

I have restricted myself to the cyclic display of powers.

under test, and the instrument readings will be different than I outlined
in the previous paragraph. Fig 3 in my article at
http://www.vk1od.net/VSWR/VSWRMeter.htm shows a line labelled "VSWR(50)"
that indicates the values that would be indicated / calculated using a 50
ohm instrument in a 75 ohm cable with a 1.5:1 VSWR.

Well, at a quick glance and noting no tabular form of data, what is
presented wouldn't reveal cyclic variation anyway for two reasons:
1. It lacks resolution (not enough places);
2. It lacks sufficient mismatch.
This is not a complaint, merely an observation because the evidence
for your example is hidden deep in the decimal places.

The solution is simple conceptually and mathematically. Reference any
discussion of Two Beam Interference as treated in Optics. For the
special case the math devolves to:
I = 2 · I1 · (1 + cos(theta2 - theta1 - delta))

Conceptually, it is only the combination of phase and amplitude from
two sources (each reflecting interface on the ends of the line).

73's
Richard Clark, KB7QHC

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Richard Clark wrote:

Is power/energy separable from its source?

What other point is there to attaching an antenna to a transmitter?

73, Jim AC6XG

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On Thu, 01 Mar 2007 14:28:58 -0800, Jim Kelley
wrote:

Richard Clark wrote:

Is power/energy separable from its source?

What other point is there to attaching an antenna to a transmitter?

Hi Jim,

I will take that as an affirmative.

When it is generally accepted that our sources do not exhibit 50 Ohms
source resistance/impedance, what resistance/impedance do they
exhibit?

73's
Richard Clark, KB7QHC

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Richard Clark wrote:
When it is generally accepted that our sources do not exhibit 50 Ohms
source resistance/impedance, what resistance/impedance do they
exhibit?

It doesn't matter. The net power supplied by the source
is *always* the difference between the forward power
and the incident reflected power *by definition*.

The definition seems to assume that 100% of the
incident reflected power is re-reflected *as if*
a 100% re-reflection condition exists.
--
73, Cecil http://www.w5dxp.com

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Richard Clark wrote in
:

I have restricted myself to the cyclic display of powers.
....
Well, at a quick glance and noting no tabular form of data, what is
presented wouldn't reveal cyclic variation anyway for two reasons:
1. It lacks resolution (not enough places);
2. It lacks sufficient mismatch.
This is not a complaint, merely an observation because the evidence
for your example is hidden deep in the decimal places.

The graph at http://www.vk1od.net/lost/RG58sol.gif has not been labelled
for presentation, so you will need to make some allowance in reading it.

The case that is plotted is an extreme mismatch, ou you might argue
impractical, but it is extreme enought to show the effects clearly on a
graph. The x axis is displacement from the load (-ve towards the
generator).

The red line is the so called "forward power" (Real(Vf^2/Zo)) that would
be indicated by a correctly calibrated sampler like a Bird 43, but
correctly calibrated means for the actual Zo, not the nominal 50+j0.

The dashed purple line is the so called "reflected power" (Real(Vr^2/Zo))
under the same conditions.

The power at a point is shown by the cyan line.

The dashed olive line is Real(Vf^2/Zo))-Real(Vf^2/Zo) (so called forward
power - reflected power). It is not the same as as the power because it
ignores two terms of the power expansion. A Bird 43 callibrated for the
actual Zo would lead you to this line.

This is a very detailed RLGC model, and it reveals from the P(x) line
that attenuation per unit length is not constant, a result of the loss
being higher in the region of a current maximum. Nevertheless, a sampler
that responds to Vf or Vr will not expose the true power curve (due to
the two missing terms).

I have thought at times of writing an article that explains the effects
on a true practical transmission line, and what practical instruments
calibrated for 50+j0 would indicate. I doubt that it would have appeal,
people like the "reflected power is dissipated in the transmitter and may
overheat it" explanation... it is easier to swallow.

Owen

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Owen Duffy wrote in
:


Hmmm, replying to my own postings again!

This is a very detailed RLGC model, and it reveals from the P(x) line
that attenuation per unit length is not constant, a result of the loss
being higher in the region of a current maximum. Nevertheless, a
sampler that responds to Vf or Vr will not expose the true power curve
(due to the two missing terms).

The two "missing" terms are the second and third lines in the legend, the
crossproducts in the power equation expansion. In the case where Zo is
purely real, these terms are equal and opposite in phase and cancel out.

Owen

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Cecil Moore wrote:
Dave wrote:
Cecil, as an engineer you should stick with standard vocabulary.

Just trying to appease the physicists, Dave. They are
arguing that it is not power until work is done. They
say that since reflected energy is not doing any work,
it cannot be reflected power. Therefore, reflected
power doesn't exist. It's purely semantics. The very
essence of an EM wave is its energy content.

So the real question is: Since standing waves obviously
exist and just as obviously cannot exist without two
coherent waves traveling in opposite directions, does
reflected energy exist? (That question seems to cause their
skivvies to get all bunched up.)

I will just be happy when they admit that reflected
EM waves possess a certain amount of energy that cannot
stand still and according to the theory of relativity
must necessarily travel at the speed of light.


Cecil,

Physicists know when power is equal to work and when it represents
undissipated flow of energy. No need to keep beating that dead horse.

I must have missed class the day they talked about obviousness. Why is
it "obvious" that standing waves cannot exist without coherent traveling
waves? Do you believe that traveling waves are somehow more pure or more
fundamental than standing waves?

Have you ever tried working out the mathematical details of the wave
equation when loaded with a standing wave trial solution? Did it explode
or otherwise fail? (Hint, the answer should be "no".)

The question of standing waves or traveling waves is purely one of
mathematical convenience. The physical phenomena are identical
regardless of your choice. Indeed, this is the point that seems to
always trip you. There is no added information from manipulating the
form of the equations. That is the sort of thing, if done carelessly,
that leads to adding power waves and other nonsense.

73,
Gene
W4SZ

tuner - feedline - antenna question ?
 

Owen Duffy wrote:
people like the "reflected power is dissipated in the transmitter and may
overheat it" explanation... it is easier to swallow.

If the forward current and reflected current are in phase
at the source, it may indeed become overheated due to an
over-current condition.

But just as likely (with random feedline lengths) is that
the forward voltage and reflected voltage may be in phase
at the source and blow out the finals due to over-voltage
conditions.

The two above events never occur at the same time. Over-
current conditions occur at low voltages. Over-voltage
conditions occur at low currents.
--
73, Cecil http://www.w5dxp.com

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Gene Fuller wrote:
Physicists know when power is equal to work and when it represents
undissipated flow of energy. No need to keep beating that dead horse.

Not all physicists, including one physics professor who
frequents this newsgroup, knows that, Gene. Maybe if you
sent him a private email, you could convince him that what
you say above is true ...

Why is it "obvious" that standing waves cannot exist without
coherent traveling waves?

Please draw us a picture of an example where standing
waves exist without a foundation of coherent traveling
waves. Here's a little help from Hecht of "Optics" fame.

(quote)
E(x,t)=2E0t*sin(kx)*cos(wt)

This is the equation for a *standing wave*, as opposed
to a traveling wave. Its profile does not move through
space; it is clearly not of the (traveling wave) form
f(x +/- vt) ...

Let the phasor E1 represent a (traveling) wave to the
left, and E2 a (traveling) wave to the right. ... (The
sum) doesn't rotate at all, and the resultant wave it
represents doesn't progress through space - it's a
standing wave.
(end quote)

Sure is hard to sweep the facts under the transmission
line rug when the EM waves are in empty space, huh?

Have you ever tried working out the mathematical details of the wave
equation when loaded with a standing wave trial solution? Did it explode
or otherwise fail? (Hint, the answer should be "no".)

Of course the answer is "no". Have you ever tried generating
a standing wave in the complete absence of traveling waves
in opposite directions? Exactly how did you do it?

The question of standing waves or traveling waves is purely one of
mathematical convenience.

Of course, that is a copout unrelated to reality. I concede
that you can perform miracles within your own mind.

The physical phenomena are identical regardless of your choice.

This from the man who asserted that standing wave phase is
meaningless (with which I agree). Point is that traveling
wave phase is NOT meaningless. So which is it?
--
73, Cecil http://www.w5dxp.com

tuner - feedline - antenna question ?
 

On 1 Mar, 17:22, Gene Fuller wrote:
Cecil Moore wrote:
Dave wrote:
Cecil, as an engineer you should stick with standard vocabulary.

Just trying to appease the physicists, Dave. They are
arguing that it is not power until work is done. They
say that since reflected energy is not doing any work,
it cannot be reflected power. Therefore, reflected
power doesn't exist. It's purely semantics. The very
essence of an EM wave is its energy content.

So the real question is: Since standing waves obviously
exist and just as obviously cannot exist without two
coherent waves traveling in opposite directions, does
reflected energy exist? (That question seems to cause their
skivvies to get all bunched up.)

I will just be happy when they admit that reflected
EM waves possess a certain amount of energy that cannot
stand still and according to the theory of relativity
must necessarily travel at the speed of light.

Cecil,

Physicists know when power is equal to work and when it represents
undissipated flow of energy. No need to keep beating that dead horse.

I must have missed class the day they talked about obviousness. Why is
it "obvious" that standing waves cannot exist without coherent traveling
waves? Do you believe that traveling waves are somehow more pure or more
fundamental than standing waves?

Have you ever tried working out the mathematical details of the wave
equation when loaded with a standing wave trial solution? Did it explode
or otherwise fail? (Hint, the answer should be "no".)

The question of standing waves or traveling waves is purely one of
mathematical convenience. The physical phenomena are identical
regardless of your choice. Indeed, this is the point that seems to
always trip you. There is no added information from manipulating the
form of the equations. That is the sort of thing, if done carelessly,
that leads to adding power waves and other nonsense.

73,
Gene
W4SZ- Hide quoted text -

- Show quoted text -

Gene,
there are 12 authors on this thread dealing with a very, very
complicated question. All are self perceived experts and after 130
posts they are still in deadlocked positions. The clock is ticking and
there has been no white smoke pour out of the chimney
as yet. The fighting is rough since only one can be right so he can
rise above the others and where the the others descend into the rable.
This question is one of the most serious questions to have faced
mankind for eons and it correct determination is so imperitive to the
universe and to ham radio. For that reason the debate is being
recorded for prosterity so that all who follow can judge what manner
of men they are indeed what manner of self perceived experts they are.
The winner will be the one that is more elequent more overbearing and
more obnoxious than any other in the group such that it may well be
not the best scientist to rise to the top but one who is most deviled
where others no longer can stay in their presence. This has happened
before where we have lost many an experienced antenna person to
venture off into other fields to the loss of all antenna enthusiasts.
This is a fight to maintain a position or to attain a position as an
expert in amateur radio So it is not to be taken lightly as past
winners have brought this newsgroup down to the level which we now
hold. There is not room for a lot of experts on this group so
descisions about who is most obnoxious have to be decided. So stand
clear of the garbage that is now flying in all directions and keep
your powder dry.Sooner or later one will be bragging that they
invented the WWW and then another will claim something else and it is
then that real savagery will enter the fray.
The clock is still ticking and still no white smoke comming out of the
chimney!
Art

tuner - feedline - antenna question ?
 

art wrote:
This question is one of the most serious questions to have faced
mankind for eons and it correct determination is so imperitive to the
universe and to ham radio. For that reason the debate is being
recorded for prosterity so that all who follow can judge what manner
of men they are indeed what manner of self perceived experts they are.

An example comes to mind. Two obscure Australian doctors
advanced the theory that most stomach ulcers were caused
by a bacterial infection. They were mercilessly ridiculed
for many years. Then they won the Nobel Prize.
--
73, Cecil http://www.w5dxp.com

tuner - feedline - antenna question ?
 

On 1 Mar, 18:58, Cecil Moore wrote:
art wrote:
This question is one of the most serious questions to have faced
mankind for eons and it correct determination is so imperitive to the
universe and to ham radio. For that reason the debate is being
recorded for prosterity so that all who follow can judge what manner
of men they are indeed what manner of self perceived experts they are.

An example comes to mind. Two obscure Australian doctors
advanced the theory that most stomach ulcers were caused
by a bacterial infection. They were mercilessly ridiculed
for many years. Then they won the Nobel Prize.
--
73, Cecil http://www.w5dxp.com

Cecil, this question that is posed is not to be ridiculed. Putting an
antenna in a attic is one of the most exciting thing to have faced
antenna experts for a few years. Its importance to the advance in
antenna is unrivalled and more than one believes he is worthy of the
nobel prize. Who will it be? I look for the smoke for guidance.
Art

tuner - feedline - antenna question ?
 

Cecil Moore wrote:
Gene Fuller wrote:

Why is it "obvious" that standing waves cannot exist without coherent
traveling waves?

Please draw us a picture of an example where standing
waves exist without a foundation of coherent traveling
waves. Here's a little help from Hecht of "Optics" fame.


[ It's all math, not physical truth. You still cling to the mistaken
notion that there is something more fundamental about traveling waves
than standing waves. That is simply nonsense. ]

(quote)
E(x,t)=2E0t*sin(kx)*cos(wt)

This is the equation for a *standing wave*, as opposed
to a traveling wave. Its profile does not move through
space; it is clearly not of the (traveling wave) form
f(x +/- vt) ...

Let the phasor E1 represent a (traveling) wave to the
left, and E2 a (traveling) wave to the right. ... (The
sum) doesn't rotate at all, and the resultant wave it
represents doesn't progress through space - it's a
standing wave.
(end quote)

Sure is hard to sweep the facts under the transmission
line rug when the EM waves are in empty space, huh?

Have you ever tried working out the mathematical details of the wave
equation when loaded with a standing wave trial solution? Did it
explode or otherwise fail? (Hint, the answer should be "no".)

Of course the answer is "no". Have you ever tried generating
a standing wave in the complete absence of traveling waves
in opposite directions? Exactly how did you do it?

The question of standing waves or traveling waves is purely one of
mathematical convenience.

Of course, that is a copout unrelated to reality. I concede
that you can perform miracles within your own mind.

The physical phenomena are identical regardless of your choice.

This from the man who asserted that standing wave phase is
meaningless (with which I agree). Point is that traveling
wave phase is NOT meaningless.

[ Oh, yes it IS meaningless, when the traveling waves are exactly the
components of a standing wave. ]

So which is it?


Cecil,

You still choose not to "get it", and I don't plan to engage in a
marathon thread. Nothing has changed for a long time. I will hold to my
previous post 100%. I don't know Prof. Hecht, but I am confident he
would agree with me completely.

73,
Gene
W4SZ

tuner - feedline - antenna question ?
 

On 1 Mar, 20:40, Gene Fuller wrote:
Cecil Moore wrote:
Gene Fuller wrote:

Why is it "obvious" that standing waves cannot exist without coherent
traveling waves?

Please draw us a picture of an example where standing
waves exist without a foundation of coherent traveling
waves. Here's a little help from Hecht of "Optics" fame.

[ It's all math, not physical truth. You still cling to the mistaken
notion that there is something more fundamental about traveling waves
than standing waves. That is simply nonsense. ]





(quote)
E(x,t)=2E0t*sin(kx)*cos(wt)

This is the equation for a *standing wave*, as opposed
to a traveling wave. Its profile does not move through
space; it is clearly not of the (traveling wave) form
f(x +/- vt) ...

Let the phasor E1 represent a (traveling) wave to the
left, and E2 a (traveling) wave to the right. ... (The
sum) doesn't rotate at all, and the resultant wave it
represents doesn't progress through space - it's a
standing wave.
(end quote)

Sure is hard to sweep the facts under the transmission
line rug when the EM waves are in empty space, huh?

Have you ever tried working out the mathematical details of the wave
equation when loaded with a standing wave trial solution? Did it
explode or otherwise fail? (Hint, the answer should be "no".)

Of course the answer is "no". Have you ever tried generating
a standing wave in the complete absence of traveling waves
in opposite directions? Exactly how did you do it?

The question of standing waves or traveling waves is purely one of
mathematical convenience.

Of course, that is a copout unrelated to reality. I concede
that you can perform miracles within your own mind.

The physical phenomena are identical regardless of your choice.

This from the man who asserted that standing wave phase is
meaningless (with which I agree). Point is that traveling
wave phase is NOT meaningless.

[ Oh, yes it IS meaningless, when the traveling waves are exactly the
components of a standing wave. ]

So which is it?

Cecil,

You still choose not to "get it", and I don't plan to engage in a
marathon thread. Nothing has changed for a long time. I will hold to my
previous post 100%. I don't know Prof. Hecht, but I am confident he
would agree with me completely.

73,
Gene
W4SZ- Hide quoted text -

- Show quoted text -





Darn it !
All those degrees in his possesion and all we get is a belching black
smoke column. It's going to be another long night Cecil!


Art

tuner - feedline - antenna question ?
 

On Fri, 02 Mar 2007 00:18:02 GMT, Owen Duffy wrote:

The case that is plotted is an extreme mismatch, ou you might argue
impractical, but it is extreme enought to show the effects clearly on a
graph. The x axis is displacement from the load (-ve towards the
generator).

Hi Owen,

Chipman shows much the same work in Chapter 8. If you got a copy you
might find it an useful resource as his math and discussion goes well
beyond the usual coverage.

73's
Richard Clark, KB7QHC

tuner - feedline - antenna question ?
 

Richard Clark wrote in
:

On Fri, 02 Mar 2007 00:18:02 GMT, Owen Duffy wrote:

The case that is plotted is an extreme mismatch, ou you might argue
impractical, but it is extreme enought to show the effects clearly on a
graph. The x axis is displacement from the load (-ve towards the
generator).

Hi Owen,

Chipman shows much the same work in Chapter 8. If you got a copy you
might find it an useful resource as his math and discussion goes well
beyond the usual coverage.

Re Chipman, no I don't have Chipman on the shelf. It sounds like I am
poorer for that, but there you go. Allmost all of what I have stated here
is based on just two things:
- that V/I=Zo for a travelling wave in a transmission line, and that
(Vf+Vr)/(If-Ir) at the load end of the line must equal Zload;
- that the voltage or current decays as e^(gamma*x).
Both are explained in probably any transmission line text, but the graphs
I created show a picture that, IMHO, is worth the proverbial thousand
words. Exploring the shape of the lines is revealing.

For example, you will remember Dr Ace (IIRC) asserting that rho cannot be
greater than 1, and supporting that with the challenge to demonstrate
rho1 with a Bird 43. Of course the Bird 43 cannot demonstrate rho1, it
is calibrated for Zo=50+j0 and rho1 is only possible if Zo has
sufficient -ve reactance to create an observable rho1 on a suitably
inductive load. So, at the risk of exciting another debate, rho can be
greater than one, but the maths that supports that proposition also
explains why you wont observe it on a Bird 43.

Owen

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Cecil Moore wrote:

Dave wrote:

Cecil, as an engineer you should stick with standard vocabulary.


Just trying to appease the physicists, Dave. They are
arguing that it is not power until work is done.

A Poynting vector is watts/square angle [watts/degree^2]. It is not being
dissipated in free space. It is Diverging [vector relationship]. How do the
physics type adjust their definition to include the Poynting Vector?

I'll sit back and read the follow up posts for the next few weeks :-)

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Gene Fuller wrote:
[ It's all math, not physical truth. You still cling to the mistaken
notion that there is something more fundamental about traveling waves
than standing waves. That is simply nonsense. ]

You apparently still cling to the mistaken notion that
standing waves can exist without their forward and
reflected wave building blocks. It is the standing
wave equation that is the math construct that only
exists in the human mind. EM waves cannot stand still.
They must necessarily move at the speed of light.
Anything else violates the principles of relativity.

I don't know Prof. Hecht, but I am confident he
would agree with me completely.

If I see Prof. Hecht, I will be sure to tell him that
you are engaging in mind-fornication with him. :-)

All you have to do to prove that you are right and I
am wrong is to provide an example of an independent
standing wave existing without the underlying forward
and reverse traveling wave components. If you can do
that, as you imply, you will also prove Einstein's
relativity to be wrong. That might be worth a Nobel
Prize.
--
73, Cecil http://www.w5dxp.com

tuner - feedline - antenna question ?
 

art wrote:
All those degrees in his possesion and all we get is a belching black
smoke column. It's going to be another long night Cecil!

Art, I'm still waiting for someone to prove that
an EM wave can stand still and continue to exist
in violation of Einstein's theory that light (and
RF waves) always travel at the speed of light.
--
73, Cecil http://www.w5dxp.com

tuner - feedline - antenna question ?
 

Owen Duffy wrote:
- that V/I=Zo for a travelling wave in a transmission line, and that
(Vf+Vr)/(If-Ir) at the load end of the line must equal Zload;

It seems that we can draw some conclusions from these
assumptions.

Vf*If*cos(0) = forward joules/sec

Vr*Ir*cos(0) = reflected joules/sec

The forward and reflected energy waves actually exist
and are the building blocks of the standing wave.

Vf and Vr are phasors and therefore subject to
superposition and interference.
--
73, Cecil http://www.w5dxp.com

tuner - feedline - antenna question ?
 

Dave wrote:
How do the physics type adjust their definition to include the Poynting
Vector?

The same way they handle the Power Flow Vector and
the "IEEE Dictionary", like a hot potato. :-)
--
73, Cecil http://www.w5dxp.com

tuner - feedline - antenna question ?
 

Cecil Moore wrote:
art wrote:
All those degrees in his possesion and all we get is a belching black
smoke column. It's going to be another long night Cecil!

Art, I'm still waiting for someone to prove that
an EM wave can stand still and continue to exist
in violation of Einstein's theory that light (and
RF waves) always travel at the speed of light.

Not everything about light (or EM waves) travels at the speed of light.

Standing waves are simply the RF equivalent of stationary optical
interference fringes.


--

73 from Ian GM3SEK 'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.co.uk/g3sek

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Cecil Moore wrote:


You apparently still cling to the mistaken notion that
standing waves can exist without their forward and
reflected wave building blocks. It is the standing
wave equation that is the math construct that only
exists in the human mind. EM waves cannot stand still.
They must necessarily move at the speed of light.
Anything else violates the principles of relativity.

Cecil,

You have fallen into the same trap as those who don't understand the
difference between phase velocity and group velocity. Relativity is a
physical concept, not a mere result of mathematical manipulation.

73,
Gene
W4SZ

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Ian White GM3SEK wrote:
Not everything about light (or EM waves) travels at the speed of light.

All EM waves (photons) travel at the speed of light.
Anything else would violate the principles of relativity.
Any detection of light results in the dissipation
(absorption) of the associated photons.

Standing waves are simply the RF equivalent of stationary optical
interference fringes.

That they appear to stand still is just an illusion.
Photons are continuously traveling at the speed of
light to get to your eyes where they are dissipated.
When you see an interference ring that appears to
stand still, you are actually receiving speed-of-
light photons into your eyes. (If the interference rings
were truly stationary, you wouldn't be able to see them.)

If the interference fringes are being absorbed, photons
are being dissipated. If the interference fringes are
being scattered, the photons are moving at the speed
of light. The photons either disappear or they travel
at the speed of light.

The photons in a transmission line cannot stand still.
Standing waves are caused by superposition and
interference between separate and distinct forward
and reflected energy waves. And the photons in the
standing wave are NOT standing still.

Consider two coherent nearly co-linear beams of light.
There are bright points of constructive interference
and dark points of destructive interference but the
interference has no effect on the two waves which
emerge from the superposition unaffected and continue
on their merry way. Those waves never slow down from
their speed-of-light velocity. The "stationary" bright
and dark points are an illusion.

At every point, photons are either being dissipated
or are traveling at the speed of light (including VF
of course). Nothing else is possible.
--
73, Cecil, http://www.qsl.net/w5dxp

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Dave wrote:
Cecil Moore wrote:

Dave wrote:

Cecil, as an engineer you should stick with standard vocabulary.


Just trying to appease the physicists, Dave. They are
arguing that it is not power until work is done.

A Poynting vector is watts/square angle [watts/degree^2]. It is not
being dissipated in free space. It is Diverging [vector relationship].
How do the physics type adjust their definition to include the Poynting
Vector?

I'll sit back and read the follow up posts for the next few weeks :-)



Dave,

Physicists are basically reasonable, but clever, folks. ;-) They define
power as the time rate at which work is done, OR, the time rate at which
energy is transferred. With some imagination applied to the context, the
definitions converge. In the case of the Poynting vector, convergence
(of the definitions) would require dealing with the ultimate destination
of the flowing energy. Might be troublesome in infinite space, but on
earth, all of the energy ultimately produces power.

The problem seems to be the careless use of "power flow" which, if not
meaningless, I find impossible to conceptualize. The flow of the flow of
energy? Sort of like common usage of "current flow", which is also the
flow of a flow.

73,

Chuck


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Gene Fuller wrote:
Relativity is a
physical concept, not a mere result of mathematical manipulation.

My point exactly! All your mathematical manipulation
is not going to get a photon to stand still.
--
73, Cecil, http://www.qsl.net/w5dxp

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Dave wrote:
Cecil Moore wrote:

Dave wrote:

Cecil, as an engineer you should stick with standard vocabulary.


Just trying to appease the physicists, Dave. They are
arguing that it is not power until work is done.

A Poynting vector is watts/square angle [watts/degree^2]. It is not
being dissipated in free space. It is Diverging [vector relationship].
How do the physics type adjust their definition to include the Poynting
Vector?

I'll sit back and read the follow up posts for the next few weeks :-)



And now one for the engineers!

How do you interpret a non-zero Poynting vector determined by static E-
and H- fields?

73,

Chuck

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On 2 Mar, 08:45, chuck wrote:
Dave wrote:
Cecil Moore wrote:

Dave wrote:

Cecil, as an engineer you should stick with standard vocabulary.

Just trying to appease the physicists, Dave. They are
arguing that it is not power until work is done.

A Poynting vector is watts/square angle [watts/degree^2]. It is not
being dissipated in free space. It is Diverging [vector relationship].
How do the physics type adjust their definition to include the Poynting
Vector?

I'll sit back and read the follow up posts for the next few weeks :-)

And now one for the engineers!

How do you interpret a non-zero Poynting vector determined by static E-
and H- fields?

73,

Chuck

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- Show quoted text -

That is a simple question for an engineer! You invest in a antenna
computor program that gives you the answer that you expect
and move on to the next problem. How would you solve it Chuck?
Art

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Cecil Moore wrote in news:AmWFh.2314$tv6.372
@newssvr19.news.prodigy.net:

Owen Duffy wrote:
- that V/I=Zo for a travelling wave in a transmission line, and that
(Vf+Vr)/(If-Ir) at the load end of the line must equal Zload;

It seems that we can draw some conclusions from these
assumptions.

Vf*If*cos(0) = forward joules/sec

What is this "we" business? That is your conclusion

What is the cos(0) term for Cecil? Are you implying that the phase angle of
Zo is 0? That isn't true in the general case, it is true ONLY for lossless
or distortionless lines.

Owen

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chuck wrote:
Dave wrote:

Cecil Moore wrote:

Dave wrote:

Cecil, as an engineer you should stick with standard vocabulary.



Just trying to appease the physicists, Dave. They are
arguing that it is not power until work is done.


A Poynting vector is watts/square angle [watts/degree^2]. It is not
being dissipated in free space. It is Diverging [vector relationship].
How do the physics type adjust their definition to include the
Poynting Vector?

I'll sit back and read the follow up posts for the next few weeks :-)



And now one for the engineers!

How do you interpret a non-zero Poynting vector determined by static E-
and H- fields?

73,

Chuck


Static fields, by definition, do not have a time varying divergence. No time
variation, no Poynting Vector. Nes Pas?

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On 2 Mar, 11:35, Dave wrote:
chuck wrote:
Dave wrote:

Cecil Moore wrote:

Dave wrote:

Cecil, as an engineer you should stick with standard vocabulary.

Just trying to appease the physicists, Dave. They are
arguing that it is not power until work is done.

A Poynting vector is watts/square angle [watts/degree^2]. It is not
being dissipated in free space. It is Diverging [vector relationship].
How do the physics type adjust their definition to include the
Poynting Vector?

I'll sit back and read the follow up posts for the next few weeks :-)

And now one for the engineers!

How do you interpret a non-zero Poynting vector determined by static E-
and H- fields?

73,

Chuck

Static fields, by definition, do not have a time varying divergence. No time
variation, no Poynting Vector. Nes Pas?- Hide quoted text -

- Show quoted text -

Then what you do is add a time varying field to a static field and add
something like " during a period of time " to Gausses law.
What on earth is your problem? Show a bit of inititive !
Art

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And now one for the engineers!

How do you interpret a non-zero Poynting vector determined by static E-
and H- fields?

What the Poynting vector actually tells us is that the flux into any closed
volume is equal to the rate of storage of energy within the volume. Since a
static field has no flux into any closed volume the poynting vector tells us
that no storage or dissipation of energy is taking place, ie it tells us
nothing and is therefore of no use in this situation.

In non-static fields it does mean something, but even here it may not
completely describe the energy flow. The Poynting Vector is just one of an
infinite number of way of describing energy flow. It is a convenient and
mathematically simple one, but not a complete solution. For instance an
alternative one is the Slepian Vector which is mathematically more
cumbersome but does include the static case.

73
Jeff

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On Fri, 2 Mar 2007 19:56:27 -0000, "Jeff" wrote:

Slepian Vector which is mathematically more
cumbersome but does include the static case.

Sorry Jeff, but I can just hear hope re-awakening beneath the frost of
disappointment, and the birth of a new theory emerging. I look
forward to a new thread
"Why haven't I received my Nobel Award from Ed McMahon yet?"

73's
Richard Clark, KB7QHC

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On Thu, 01 Mar 2007 17:36:12 -0600, Cecil Moore
wrote:

Basically nothing.

Richard Clark wrote:
When it is generally accepted that our sources do not exhibit 50 Ohms
source resistance/impedance, what resistance/impedance do they
exhibit?

To which I got no answers in stereo.

Some may sense frustration to which I once asked how much relief I
would need. Having already long anticipated a non-response (score
100% for accuracy in modeling); others may want to fumble rhetorically
with equally uninformed responses. In anticipation of that I provide
a multiple choice. Richard needs to take:
1. An aspirin;
2. A syringe of morphine;
3. Both of the above;
4. 50 Ohms of dissipation.
5. All of the above, except 3.
6. ____________

Now take that number, and divide it by the Sarcasm co-efficient.

Plot this on a graph and draw an asymptote. You now have a diagram
for a Gaussian Bundle. Find the Pointing Vector for this (it points
to the same answer you picked in the multiple choice) to verify the
authenticity of your newsreader's software design. ;-)

73's
Richard Clark, KB7QHC

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Dave wrote:
chuck wrote:
Dave wrote:

Cecil Moore wrote:

Dave wrote:

Cecil, as an engineer you should stick with standard vocabulary.



Just trying to appease the physicists, Dave. They are
arguing that it is not power until work is done.


A Poynting vector is watts/square angle [watts/degree^2]. It is not
being dissipated in free space. It is Diverging [vector
relationship]. How do the physics type adjust their definition to
include the Poynting Vector?

I'll sit back and read the follow up posts for the next few weeks :-)



And now one for the engineers!

How do you interpret a non-zero Poynting vector determined by static
E- and H- fields?

73,

Chuck


Static fields, by definition, do not have a time varying divergence. No
time variation, no Poynting Vector. Nes Pas?


I think the Poynting vector can be calculated even when the E and H
fields are static. Of course, doing so would violate Poynting's
assumptions and thus be meaningless. But if one didn't know in advance
that an arbitrary closed surface contained static sources, and he found
the Poynting vector, S, for some small area, he could well get a
non-zero answer. Of course, the integral of S over the entire surface
would always be zero in the case of static sources.

To be applicable, the Poynting theorem requires that the E and H fields
arise from a single source, satisfying Maxwell's first two equations.
But that information may not be known in advance.

73,

Chuck








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Jeff wrote:
And now one for the engineers!

How do you interpret a non-zero Poynting vector determined by static E-
and H- fields?

What the Poynting vector actually tells us is that the flux into any closed
volume is equal to the rate of storage of energy within the volume. Since a
static field has no flux into any closed volume the poynting vector tells us
that no storage or dissipation of energy is taking place, ie it tells us
nothing and is therefore of no use in this situation.

My understanding is somewhat different (and quite possibly in error).
You can calculate a non-zero Poynting vector and it will not always tell
you that no storage or dissipation of energy is taking place. Assuming
you do not already know the source of fields inside a closed surface,
you can be misled by the Poynting vector. You need an additional step in
that case: i.e., integrating S over the entire surface. That will
produce a value of zero in the case of static fields.



In non-static fields it does mean something, but even here it may not
completely describe the energy flow. The Poynting Vector is just one of an
infinite number of way of describing energy flow. It is a convenient and
mathematically simple one, but not a complete solution. For instance an
alternative one is the Slepian Vector which is mathematically more
cumbersome but does include the static case.


There seems to be a divergence of views on the Slepian vector, but I am
not familiar with it.

73,

Chuck

73
Jeff



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On 2 Mar, 13:04, chuck wrote:
Dave wrote:
chuck wrote:
Dave wrote:

Cecil Moore wrote:

Dave wrote:

Cecil, as an engineer you should stick with standard vocabulary.

Just trying to appease the physicists, Dave. They are
arguing that it is not power until work is done.

A Poynting vector is watts/square angle [watts/degree^2]. It is not
being dissipated in free space. It is Diverging [vector
relationship]. How do the physics type adjust their definition to
include the Poynting Vector?

I'll sit back and read the follow up posts for the next few weeks :-)

And now one for the engineers!

How do you interpret a non-zero Poynting vector determined by static
E- and H- fields?

73,

Chuck

Static fields, by definition, do not have a time varying divergence. No
time variation, no Poynting Vector. Nes Pas?

I think the Poynting vector can be calculated even when the E and H
fields are static. Of course, doing so would violate Poynting's
assumptions and thus be meaningless. But if one didn't know in advance
that an arbitrary closed surface contained static sources, and he found
the Poynting vector, S, for some small area, he could well get a
non-zero answer. Of course, the integral of S over the entire surface
would always be zero in the case of static sources.

To be applicable, the Poynting theorem requires that the E and H fields
arise from a single source, satisfying Maxwell's first two equations.
But that information may not be known in advance.

73,

Chuck

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- Show quoted text -

Chuck,
I believe you would be much better off choosing Gausses law for
statics and adding a time element to the written law itself thus
providing all three cartesian coodinates reguired for the law.
I suppose one could say it is not Gaussian law anymore but
mathematically it fits quite well. This is the basis for Gaussian
antenna arrays for which I have submitted for consideration from the
patent office. As yet I have not heard any comment that invalidates
this concept other than it can't be done from psuedo experts and
frankly I feel that the addition speaks for itself.
Regards
Art

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Owen Duffy wrote:
What is this "we" business? That is your conclusion

Hams have a habit of saying "we", Owen. Are you a ham?
We have an IC-756PRO. We are a Texas Aggie. We drive a
GMC pickup. We are not married.

What is the cos(0) term for Cecil? Are you implying that the phase angle of
Zo is 0? That isn't true in the general case, it is true ONLY for lossless
or distortionless lines.

We sometimes assume lossless lines for the sake of discussion
which is often a close enough approximation to real-world
lines. We don't have our books with us but seems we remember
that the Z0 phase angle for common Z0 lines is about 1-2
degrees.
--
73, Cecil http://www.w5dxp.com