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  #121   Report Post  
Old August 27th 03, 02:03 AM
W5DXP
 
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Peter O. Brackett wrote:
I am afraid that Cecil has been seduced by the "age" of some of the optical
publications into believing that the opticians are ahead of the elecricians,
but that is simply not so... we electricians know lots more about wave
dynamics than opticians.


So, Peter, what mechanism of physics causes the reversal in the momentum
of the reflected wave when a Z0-match exists?
--
73, Cecil http://www.qsl.net/w5dxp



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  #122   Report Post  
Old August 27th 03, 02:14 AM
Richard Clark
 
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On Wed, 27 Aug 2003 00:27:27 GMT, "Peter O. Brackett"
wrote:

Richard:

[snip]
I know you eschew academic references in favor of "first principles,"
but others may want more material than the simple puzzle aspect.
They can consult "Transmission Lines & Networks," Walter C. Johnson,
Chapter 13, "Insertion Loss and Reflection Factors." But lest those
who go there for the answer, I will state it is from another
reference, Johnson simply is offered as yet another reference to
balance the commonly unstated inference of SWR mechanics being
conducted with a Z matched source. "Transmission Lines," Robert
Chipman is another (and not the source of the puzzle either).

73's
Richard Clark, KB7QHC

[snip]

I hear you...

I don't eschew academic references, but when it comes to systems
Engineering, I do try to follow what our great President Regan once said,
"Trust, but Verify!"


Hi Peter,

I presume by this you have neither a solution nor an analysis by first
principles. Just as well, this stumper has baffled many. Two
resistors and a hank of line....

73's
Richard Clark, KB7QHC
  #123   Report Post  
Old August 27th 03, 04:45 AM
Tdonaly
 
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And why "Black-faced"?


It's a classical allusion - a test. You failed. Now I know
more about you.


And many of the other "adults" in this NG act infantile IN THEIR
OWN WAY.
Witness how Roy will compare someone to his adolescent son, when they
don't agreed with him. And then look at Cecil. Same sh**.


Classic case of projection.


And you know, you could be a high-school drop out for all i care.
The great thing about the NGs is that people don't judge you by your
age or looks, only by what you write.


In that case, you're in trouble.

If you wrote logically, like
Richard, then i would respect you...but you don't. I don't care where
Richard went to school, he knows his sh**.


Because you think he agreed with you.

Anyway, you've satisfied my curiosity. Thank you. If the other
members of the newsgroup want to read and respond to your posts,
fine. Most of us, though, went through middle school once and don't
want to repeat the experience. Good luck with your job search.
73,
Tom Donaly, KA6RUH



  #124   Report Post  
Old August 27th 03, 07:16 AM
Roy Lewallen
 
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I read the analysis earlier, and didn't and don't see how it constitutes
the proof I proposed.

Permit me to repeat the entirety of what I said, and not just the last
sentence. I said:

I'll restate something I mentioned before (first incorrectly, then
corrected). Connecting a load to a transmission line which is the
complex conjugate of the transmission line Z0 does *not* guarantee
maximum power delivery from the source, or to the load. The load
impedance which provides maximum load power is the complex conjugate of
the impedance looking back from the load toward the source. That
impedance is the source impedance transformed through the transmission
line between source and load, and it's not generally the same as the
line's Z0, or its complex conjugate. When this condition of maximum load
power is met, there will almost certainly be voltage and current wave
reflections on the line -- there would be none only if the optimum load
impedance coincidentally happened to be equal to the line Z0. So the
argument that there can be no reflection of the voltage wave under the
condition of maximum power transfer is wrong.

You didn't show differently in your analysis, and no one has stepped
forward with a contrary proof, derivation from known principles, or
numerical example that shows otherwise.


Now let's see if this constitutes such a proof.

Peter O. Brackett wrote:
Roy:

[snip]

You didn't show differently in your analysis, and no one has stepped
forward with a contrary proof, derivation from known principles, or
numerical example that shows otherwise.

Roy Lewallen, W7EL


[snip]

Yes I did. I guess that you missed that post.

I'll paste a little bit of that posting here below so that you can see it
again.

[begin paste]
We are discussing *very* fine points here, but...

[snip]
ratio of the reflected to incident voltage as rho = b/a would yeild the
usual formula:

rho = b/a = (Z - R)i/(Z + R)i = (Z - R)/(Z+ R).


Here I'll assume that you've defined a and b as equalling the forward
and reverse voltages in a transmission line, which is a different
definition than in your other posting. With that definition of a and b,
and if R is the transmission line characteristic impedance, then the
equation is valid for a transmission line, and ok so far. Otherwise,
you're talking about something other than a transmission line, so the
"proof" doesn't apply at all.

In which no conjugates appear!

Now if we take the internal/reference impedance R to be complex as R = r +
jx then for a "conjugate match" the unknown Z would be the conjugate of the
internal/reference impedance and so that would be:

Z = r - jx


So the load impedance is the complex conjugate of the transmission line
characteristic impedance. Ok.


Thus the total driving point impedance faced by the incident voltage a would
be 2r:

R + Z = r + jx + r - jx = 2r


Here you've lost me. a is the forward voltage in the transmission line.
What can be the meaning of its facing a driving point impedance? The
forward wave sees only the characteristic impedance of the line; at all
points the ratio of forward voltage to forward current is simply the Z0
of the line. So I don't believe that it sees 2r anywhere.

This is where I'm stuck. If you can show where along the line the
forward voltage wave "faces" 2r, that is, Vf/If = 2r, I can continue.

. . .


Roy Lewallen, W7EL

  #125   Report Post  
Old August 27th 03, 06:46 PM
Jim Kelley
 
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W5DXP wrote:
Looks like Born and Wolf are wrong.


Not bloody likely.

ac6xg


  #126   Report Post  
Old August 27th 03, 11:53 PM
Peter O. Brackett
 
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Roy:

[snip]
Here you've lost me. a is the forward voltage in the transmission line.
What can be the meaning of its facing a driving point impedance? The
forward wave sees only the characteristic impedance of the line; at all
points the ratio of forward voltage to forward current is simply the Z0
of the line. So I don't believe that it sees 2r anywhere.

This is where I'm stuck. If you can show where along the line the
forward voltage wave "faces" 2r, that is, Vf/If = 2r, I can continue.

Roy Lewallen, W7EL

[snip]

I appologize if I am "going to fast" for the limitations inherent in
NewsGroup postings.

Let me take it a little more slowly here...

To see that there is essentially no difference between my (apparently two)
different
definitions of the incident and reflected waves a and b, consider the
following scenario
of an actual transmission line having surge impedance [characteristic
impedance] Zo.
Let Zo be in general complex. i.e. Zo = sqrt[(R + jwL)/(G + jwC)] where R,
L, G,
and C are the primary parameters of R Ohms, L Henries, G Siemens, and C
Farads
per unit length.

Now at any particular frequency w = 2*p*f you will find that this general
complex
surge impedance Zo evaluates to a complex number, say Zo(jw) = r + jx.
Later
let's let r = 50 Ohms and x = 5 Ohms so that we can work out a numerical
example.

Consider either a semi-infinite length of this Zo line, or even a finite
length of the Zo line
terminated in an impedance equal to Z0. I am sure that you will agree that
both the
semi-infinite Zo line or the finite length Zo line terminated in Zo have the
same
driving point impedance namely Zo.

Now excite this semi-infinte Zo line by an ideal generator of open circuit
voltage
Vi = 2*a behind an impedance equal to the surge impedance Zo. In other
words
this is a Thevenin generator of ideal constant voltage 2a behind a complex
impedance
of Zo. And so... since an ideal voltage source has zero impedance, the
termination
at the source end of this semi-infinite Zo line is Zo, and so the line which
has a
driving point impedance of Zo is terminated in Zo. Thus the port or
reference plane
between the generator and the semi-infinite line is an "image match" point.

For reference, a rough ASCII style schematic of this situation would be as
follows.

Generator = Vi = 2aVolts - generator impedance Z0 - impedance Zo of
semi-infinite line.

Clearly the the generator impedance of Zo and the line's driving point
impedance of Zo
constitute a voltage divider hanging off the generator with open circuit
volts Vi = 2a Volts,
and since Zo = Z0 they divide the generator voltage, which equals 2a volts
exactly by 2, and
so the voltage across the input of the semi-infinte line is exactly a volts.
This a is the incident
voltage wave of the Scattering Formalism and it is exactly the same as the
classical forward
voltage for transmission lines. Note that it is exactly half of the
generator's open circuit volts.
i.e. the voltage across the perfectly terminated [image matched] line is
exactly a Volts or 1/2
of Vi Volts. [Aside: That is why I said in an earlier post that your
forward wave and my
incident wave only differ by a factor of two! They are really no different
it's only a scale
factor. You can compensate as I have by calling the Zo generator voltage
2a, but for
simplicity I often just call it a to eliminate the factors of 2 that occur
all over the place. Sorry
if that is confusing, I'll try not to do that any more. :-).

Now everybody knows and everybody will agree that at the junction of the Zo
line
terminated in Zo [NOT a conjugate match] where the Thevenin generator is
connected
to the semi-infinite Zo line there is an image match AND there will be no
reflected voltage
waves. What is more since the Zo line is semi-infinite, or the alternative
of a finite length
and terminated in Zo at the far end, there will be no reflections at the
"far end" either. i.e. in this
situation there are no reflections and the incident wave "a" just propagates
into the line and
there is no reflected wave b to interfere with it. Note that this does not
mean that maximum
power is transmitted into the line. You and I both agree that when a
generator is image
matched [Cecil likes to call "image match" a Zo match] that in the general
case of complex
Zo maximum power is not transferred but there is NO reflected voltage when
measured
with a reflectometer which uses Zo as it's reference impedance.

This scenario at the Zo generator driving the Zo line is the same situation
found when you cut
into an infinite length of Zo line and insert a reflectometer with reference
impedance Zo,
the reflectometer will read out the reflected voltage b as:

b = rho*a = (Zo-Zo)/(Zo+ Zo) = 0 Zero, nada, nil... no reflections.

Now lets do a simple numerical example.

Consider that at some frequency the surge impedance evaluated to say Zo = r
+ jx = 50 + j5
Ohms and that we set the generator open circuit voltage to be Vi = 2a = 2
volts. Thus the "incident
voltage wave is a = 1 Volt.

The maximum power available from this Zo generator will occur when the load,
call it Z, on the
generator is a conjugate match to the generator impedance Zo. Thus for
maximum power
transfer Z must be the complex conjugate of Zo, i.e. we must let Z = 50 -
j5, to extract maximum
power not Z = 50 +j5.

Under these conjugate matched conditions, the total impedance faced by the
generator is the sum of
its' internal impedance Zo and the external Z, i.e. the total impedance is

Zo + conj(Zo) = 50 + j5 + 50 - j5 = 100 Ohms

And so the generator impresses it's 2 Volts of open circuit volts across the
resulting totalof 100 Ohms
supplying a current of I = 2/100 = 20mA to the real part of the load of 50
Ohms for a maximum
power of:

Pmax = I*I*R = 0.02*0.02*50 = 0.0004*50 = 0.02 = 20 mWatts.

Under this conjugate matched condition, where the maximum power of 20mW is
transferred, if we
use the classical definition of rho = (Z - Zo)/(Z + Zo) with Z = 50 - j5 and
Zo = 50 + j5
we get:

rho = b/a = ((50 - j5) - (50 + j5))/((50 - j5) + (50 + j5)) = (-j5 -
j5)/(100) = -j10/100 = -j/10

Since the incident wave is a = 1 Volt, then:

b = rho* a = -j/10 Volts.

With a conjugate match, and using the classical definition of rho, as
measured by a reflectometer using
Zo as it's reference impedance, there will be a reflected voltage of
magnitude 0.1 Volts, or one tenth of a volt at a phase lag of ninety
degrees.

It is interesting to see what value of reflected voltage would be indicated
by a Bird Model 43 in
this circumstance of a perfect conjugate match.

A Bird Model 43 uses an internal reference impedance of R = 50 Ohms and
it implements the "classical" definition of rho. i.e. The Bird Model 43
calculates the reflected
voltage b as b(Bird):

b(BIRD) = rho * a = (Z - 50)/(Z + 50) * 1 = (50 - j5 - 50)/(50 - j5 + 50)
= -j5/(100 - j5)

Actually the Bird cannot indicate phase angles, rather it just computes the
approximate magnitude
of the reflected voltage which in this case would be

|b(Bird)| = 5/sqrt(100*100 + 5*5) = 5/sqrt(1025) = 0.156 Volts.

Compare this to the magnitude of the "true" reflected voltage which is 0.1
Volts as computed by a true reflectometer which used Zo as its' internal
reference impedance.

And so... contrary to popular opinion, the Bird does not indicate zero
reflected power when it is inserted into a perfect conjugate matched
[maximum power transfer] situation when the Zo of the system is not a pure
50 Ohms!

Now if we choose to use a different definition of rho, say the one proposed
by Slick why then we will get different results for the reflected voltage,
in fact with his somewhat erroneous formula for rho he will read rho as zero
in the conjugate matched situation. However. even though one can use any
definition as long as one consistently uses it in all theoretical
developments and measurement, as in one of my other posts to this thread, I
do not believe Slick's or anybody elses re-definition of rho to be approved
by "Mother Nature".

Mother Nature uses rho = (Z - Zo)/(Z + Z0) simply because that is the
reflection coefficient that exists at any point along any infinitely long
transmission line of constant surge impedance Zo where looking to the right
and to the left at any point in the line one sees the same driving point
impedance Zo in both directions and there are no discontinuities in Zo to
cause reflectons, and so the reflection coefficient must be zero and (Zo -
Zo)/(Zo + Zo) is the only formulation of a reflection factor that supports
that condition.

I hope this "treatise" helps you to understand my thoughts on this.

--
Peter K1PO
Indialantic By-the-Sea, FL.

..


  #127   Report Post  
Old August 28th 03, 02:03 AM
W5DXP
 
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Peter O. Brackett wrote:
There is no reflection at a Zo-match!


The wave reflection model works just fine at a Z0-match.

There are no "waves", only particles, photons and electrons to be precise!


Please don't forget the necessary virtual photons and your inability to
even prove that they exist.

Check out Feynman's Lectures for an "undergraduate" introduction to QED, and
find out why "waves" are just plain WRONG!


Does he prove that virtual photons exist? Quoting Hecht: "We've learned from
Quantum Mechanics that at base interference is one of the most fundamental
mysteries in physics."

There is an interesting book out in the bookstores recently, written for the
layman, which details the strange life of quantum mechanics, the book is a
great read:


Thanks, I'll look for it. Right now, I'm about half way through _Quarks,_
Leptons,_and_the_Big_Bang_.
--
73, Cecil http://www.qsl.net/w5dxp



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  #128   Report Post  
Old August 28th 03, 03:52 AM
Richard Clark
 
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On Thu, 28 Aug 2003 00:33:08 -0000, "David Robbins"
wrote:

you have also hit on the basic problem in this discussion. there are two
totally different topics in play here and they are getting mixed up back and
forth all over this thread.

case 1: transmission line with a load..

case 2: generator feeding a load...

keep your conditions straight or these discussions will go on for ever with
both sides knowing they are right and that the other one is wrong.... but
only because they are discussing two different problems.


Hi David,

If you don't respond with an answer, I will mark it up to your having
lost track of the thread through your computer swapping. I quoted
only as much of your post to offer that I have no interest in the
conjugation issue whatever logic it is cloaked in. This is a simple,
practical example expressed in heat lost in the transmission line.
Very simple conditions, two resistors and a hank of transmission line.

The scenario begins:

"A 50-Ohm line is terminated with a load of 200+j0 ohms.
The normal attenuation of the line is 2.00 decibels.
What is the loss of the line?"

Having stated no more, the implication is that the source is matched
to the line (source Z = 50+j0 Ohms). This is a half step towards the
full blown implementation such that those who are comfortable to this
point (and is in fact common experience) will observe their answer and
this answer a

"A = 1.27 + 2.00 = 3.27dB"

"This is the dissipation or heat loss...."

we then proceed:

"...the generator impedance is 100+0j ohms, and the line is 5.35
wavelengths long."

73's
Richard Clark, KB7QHC
  #129   Report Post  
Old August 28th 03, 04:44 AM
W5DXP
 
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Peter O. Brackett wrote:
There are no "waves", only particles, photons and electrons to be precise!


Given a coherent light source, like a laser, and exactly the right indices
of refraction, can you describe in a nutshell, how non-glare thin-film
coating works using only particles? Do out of phase photons cancel each
other when they are traveling in the same direction?
--
73, Cecil http://www.qsl.net/w5dxp



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  #130   Report Post  
Old August 28th 03, 10:11 AM
Roy Lewallen
 
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Thanks for hanging in there. Let's see where this one goes. . .

Peter O. Brackett wrote:
Roy:

[snip]

Here you've lost me. a is the forward voltage in the transmission line.
What can be the meaning of its facing a driving point impedance? The
forward wave sees only the characteristic impedance of the line; at all
points the ratio of forward voltage to forward current is simply the Z0
of the line. So I don't believe that it sees 2r anywhere.

This is where I'm stuck. If you can show where along the line the
forward voltage wave "faces" 2r, that is, Vf/If = 2r, I can continue.

Roy Lewallen, W7EL


[snip]

I appologize if I am "going to fast" for the limitations inherent in
NewsGroup postings.

Let me take it a little more slowly here...

To see that there is essentially no difference between my (apparently two)
different
definitions of the incident and reflected waves a and b, consider the
following scenario
of an actual transmission line having surge impedance [characteristic
impedance] Zo.
Let Zo be in general complex. i.e. Zo = sqrt[(R + jwL)/(G + jwC)] where R,
L, G,
and C are the primary parameters of R Ohms, L Henries, G Siemens, and C
Farads
per unit length.

Now at any particular frequency w = 2*p*f you will find that this general
complex
surge impedance Zo evaluates to a complex number, say Zo(jw) = r + jx.
Later
let's let r = 50 Ohms and x = 5 Ohms so that we can work out a numerical
example.


The numbers aren't necessary, but that's fine.


Consider either a semi-infinite length of this Zo line, or even a finite
length of the Zo line
terminated in an impedance equal to Z0. I am sure that you will agree that
both the
semi-infinite Zo line or the finite length Zo line terminated in Zo have the
same
driving point impedance namely Zo.


The terminated line I'm comfortable with. "Semi-infinite" isn't in my
lexicon, but I'll see where it goes.

Now excite this semi-infinte Zo line by an ideal generator of open circuit
voltage
Vi = 2*a behind an impedance equal to the surge impedance Zo. In other
words
this is a Thevenin generator of ideal constant voltage 2a behind a complex
impedance
of Zo. And so... since an ideal voltage source has zero impedance, the
termination
at the source end of this semi-infinite Zo line is Zo,
. . .


Maybe the trouble is with the "semi-infinite" aspect, but here we part
company again. The termination at the source end of the line is, by
definition, the source impedance, which is zero, not Z0.

Can't you do this analysis with a plain, ordinary, transmission line of
finite length? Lossless is ok, lossy is ok. Or is your proof true only
if the line is "semi-infinite" in length (whatever that is)?

Roy Lewallen, W7EL



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