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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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

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

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

W5DXP wrote:
Looks like Born and Wolf are wrong.

Not bloody likely.

ac6xg

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.

..

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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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

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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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

Peter O. Brackett wrote:
In a nutshell, photons do not cancel. :-)

Is this quote from _Optics_, by Hecht incorrect? "Unlike ordinary
objects, photons cannot be seen directly; what is known of them
comes from observing the results of their being either created or
annihilated."

Waves are an illusion, they are not the way Mother Nature works!

Note that waves can pass through two slits at the same time. :-)

A footnote from _Optics_: "... although the light-is-corpuscular
model has wide acceptance ... and is a crucial part of the
contemporary discourse, like almost everything else in physics,
it is not yet established beyond all doubt. See, for example,
the summary article by R. Kidd, J. Ardini, and A. Anton,
"Evolution of the modern photon", Am. J. Phys. 57 (1), 27 (1989)."

Deal with it!

Accept on faith the existence of God and virtual photons? :-)
To me, God and virtual photons seem to both be band aids for
things not yet understood by man. To me, the following two
statements don't seem all that different.

1. Although completely undetectable, virtual photons have to be the
vehicle by which electrons communicate their presence to each other.

2. Although completely undetectable, God has to be the vehicle by
which everything was created.

It just occurred to me that virtual photons might explain telepathy.
What do you think?
--
73, Cecil http://www.qsl.net/w5dxp



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2. Although completely undetectable, God has to be the vehicle by
which everything was created.

73, Cecil http://www.qsl.net/w5dxp

Last time when I tried to explain how the transistor radio works to my dog, she
wiggled her tail: "It can't be and it don't taste no good" :-)

Yuri da BUm

Good comments.

One of my professors made an astute observation that sums up your
conclusion. He said that science isn't set back so much by getting the
wrong results by using the right methods as it is by getting the right
results using the wrong methods. Amen to that.

Roy Lewallen, W7EL

David Robbins wrote:

yes, i still hold that they are different. they are from completely
different realms of electromagnetics. in the transmission line reflection
coefficient you are working with a distributed system that is modeled with
wave equations. there are delays, waves travel and reflections return after
a finite delay. in your example you have coerced the voltage and current
waves to be the same in the transmission line, but that is not the same as
modeling the line itself.

in the generator feeding a load you are looking at a lumped model. these
are normally not solved with wave equations, there are no delays or
reflections. the problem of calculating maximum power transfer from a
generator to a load of a given impedance is a simple problem solved with no
waves and only one pde used to find the max of the power to the load wrt the
generator impedance. it doesn't care what the load is, as long as it is a
linear time invariant impedance. now, you could create that impedance by
putting a load on the end of a transmission line and transforming it back
through the normal transmission line equations, but once you do that you
throw the line out and solve the simple equations in the lumped models.

these discussions always seem to end up in this same quagmire, one group
trying to solve everything with wave equations, sinusoidal steady states,
transmission line transformations, and the other holding on to the lumped
models and trying to make them fit the wrong problem domain. one thing i
did learn in ee classes was to use the right model for the job, choosing the
wrong one will give you wrong answers most of the time.... the real problem
is that sometimes you can get the right answer from the wrong model just
because the numbers work out... and when you have to show your work you lose
credit for using the wrong method but still getting the right numerical
answer. that is what i see here, someone has started with a simplified case
that just happens to work out to the same answer and has tried to generalize
it... -10 points for the wrong method!

Dave:

[snip]
these discussions always seem to end up in this same quagmire, one group
trying to solve everything with wave equations, sinusoidal steady states,
transmission line transformations, and the other holding on to the lumped
models and trying to make them fit the wrong problem domain. one thing i
did learn in ee classes was to use the right model for the job, choosing
the
wrong one will give you wrong answers most of the time.... the real
problem
is that sometimes you can get the right answer from the wrong model just
because the numbers work out... and when you have to show your work you
lose
credit for using the wrong method but still getting the right numerical
answer. that is what i see here, someone has started with a simplified
case
that just happens to work out to the same answer and has tried to
generalize
it... -10 points for the wrong method!
[snip]

Dave:

I disagree, I feel that for measurements and calculations involving the
impedance matching
dynamics at any reference plane in such systems the differences between
lumped and
distributed systems is trivial and unimportant.

The impedance matching dynamics at the reference plane/junction point are
the same for
distributed and lumped systems, they obey all the same equations, their
electrodyanamics
is the same, one simply cannot tell the difference at the driving point.

If you don't like lumped models, then make the internal resistance of the
Thevenin
generator out of a distributed line. It will work just the same way.

The disagreements that occur when there are discussions/arguments occur
because folks
often assume they are different situations, but they are not.

Every line has a driving point impedance of Zo and at that point the voltage
and current
are related by Ohms Law v = Zo*i just as for a lumped system of impedance
Zo.
It simply doesn't matter if Zo is lumped or distributed when making
observations at the
driving point. And this is what one is doing when making impedance matching
calculations.

I agree that if you move off down the line in distance/space that you will
find
differences, but you certainly can't observe them at the driving points.

A slotted line device for measuring VSWR is a physical example of a
measurement
system that works on a distributed system that won't work for a lumped
system.
However you can exactly correlate the measurements taken with the slotted
line
to measurements taken at the driving point.

And so, although I agree with you that some can't seem to be able to see
that these
situations are the same, I don't agree that makes them different. They are
the same
at the driving point.

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

David Robbins wrote:
"Peter O. Brackett" wrote in message
link.net...



yes, i still hold that they are different. they are from completely
different realms of electromagnetics. in the transmission line reflection
coefficient you are working with a distributed system that is modeled with
wave equations. there are delays, waves travel and reflections return after
a finite delay. in your example you have coerced the voltage and current
waves to be the same in the transmission line, but that is not the same as
modeling the line itself.

In simulation programs, transmission lines are
solved for their two-port
parameters, and are then treated as lumped
circuits in the actual
simulation, just like any lumped-element circuit.
Which is a good way
to do it.

I notice that in the ARRL Antenna Book, 19th
edition , on page 24-7, it is stated with definite
finality that the reflection coefficient formula
uses the complex conjugate of Zo in the numerator.
I also understand that this has been established
by a "well-trusted authority".

I have used Mathcad to calculate rho and VSWR for
Reg's example, for many values of X0 (imaginary
part of Z0) from -0 to -250 ohms.

The data follows:

Note: |rho1*| is conjugated rho1, SWR1 is for
|rho1*|, |rho2| is not conjugated and SWR2 applies
to |rho2|

X0.......|rho1*|..SWR1.....|rho2|..SWR2
-250..... 0.935...30.0.....1.865...-3.30
-200..... 0.937...30.8.....1.705...-3.80
-150..... 0.942...33.3.....1.517...-4.87
-100..... 0.948...37.5.....1.320...-7.25
-050..... 0.955...43.3.....1.131...-16.3
-020..... 0.959...47.6.....1.030...-76.5
-015..... 0.960...48.4.....1.010...-204
-012..... 0.960...48.9.....0.997....+/- infinity
-010..... 0.960...49.2.....0.990....+305
-004..... 0.961...76.3.....0.974....+76.3
0000..... 0.961...50.9.....0.961....+50.9

The numbers for not-conjugate rho are all over the
place and lead to ridiculous numbers for SWR. It
is also obvious that for a low-loss line it
doesn't matter much. But values of rho greater
than 1.0, on a Smith chart correspond to negative
values of resistance (see the data).

Something is wrong here that we are overlooking.

The use of conjugate rho is so much better behaved
that I have some real doubts about some of our
conclusions on this matter.

What about it folks? How can we get to the bottom
of this?

Bill W0IYH

From your more recent posting, I misinterpreted what you mean by
"driving point impedance". It's simply the impedance of the transmission
line, not the impedance from the load looking back toward the source,
nor the source impedance. Based on that interpretation, I'll try again.

Peter O. Brackett wrote:
. . .
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.

That is the impedance of the line. Ok.

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.

Oops. No it's not. Consider circuit "A", a voltage source in series with
a lumped impedance equal to r + jx. And circuit "B", a voltage source in
series with a transmission line of length, say, one wavelength, with
characteristic impedance Z0 = r + jx. The two are not at all equivalent.
To show that they're not, connect a 50 ohm resistor to the output of
each. The resistor current in circuit "A" is Vi/(50 + r + jx). The
current in circuit "B" is Vi/50.

David Robbins recently made several postings explaining that these two
circuits are not equivalent. Hopefully the above example illustrates that.

Can't you do your "proof" with, say, a one wavelength transmission line
of characteristic impedance Z0?

Roy Lewallen, W7EL

William E. Sabin wrote:

David Robbins wrote:

"Peter O. Brackett" wrote in message
link.net...



yes, i still hold that they are different. they are from completely
different realms of electromagnetics. in the transmission line
reflection
coefficient you are working with a distributed system that is modeled
with
wave equations. there are delays, waves travel and reflections return
after
a finite delay. in your example you have coerced the voltage and current
waves to be the same in the transmission line, but that is not the
same as
modeling the line itself.


In simulation programs, transmission lines are solved for their two-port
parameters, and are then treated as lumped circuits in the actual
simulation, just like any lumped-element circuit. Which is a good way
to do it.

I notice that in the ARRL Antenna Book, 19th edition , on page 24-7, it
is stated with definite finality that the reflection coefficient formula
uses the complex conjugate of Zo in the numerator.
I also understand that this has been established by a "well-trusted
authority".

I have used Mathcad to calculate rho and VSWR for Reg's example, for
many values of X0 (imaginary part of Z0) from -0 to -250 ohms.

The data follows:

Note: |rho1*| is conjugated rho1, SWR1 is for |rho1*|, |rho2| is not
conjugated and SWR2 applies to |rho2|

X0.......|rho1*|..SWR1.....|rho2|..SWR2
-250..... 0.935...30.0.....1.865...-3.30
-200..... 0.937...30.8.....1.705...-3.80
-150..... 0.942...33.3.....1.517...-4.87
-100..... 0.948...37.5.....1.320...-7.25
-050..... 0.955...43.3.....1.131...-16.3
-020..... 0.959...47.6.....1.030...-76.5
-015..... 0.960...48.4.....1.010...-204
-012..... 0.960...48.9.....0.997....+/- infinity
-010..... 0.960...49.2.....0.990....+305
-004..... 0.961...76.3.....0.974....+76.3
0000..... 0.961...50.9.....0.961....+50.9

The numbers for not-conjugate rho are all over the place and lead to
ridiculous numbers for SWR. It is also obvious that for a low-loss line
it doesn't matter much. But values of rho greater than 1.0, on a Smith
chart correspond to negative values of resistance (see the data).

Something is wrong here that we are overlooking.

The use of conjugate rho is so much better behaved that I have some real
doubts about some of our conclusions on this matter.

What about it folks? How can we get to the bottom of this?

Bill W0IYH


The equation in the ARRL Antenna Book is identical
to the equation for rho that is in the Power Wave
literature (see Gonzalez and also see Kurokawa).
Also, numerous literature sources describe how an
open-circuit generator with internal impedance Z0,
connected directly to load ZL, is actually a power
wave setup that leads to a rho formula that is
identical to the formula in the ARRL Antenna Book.
When calculating rho, it is not necessary to fool
around with the wave equations, because frequency
is constant and everything is steady-state.

Bill W0IYH

Roy:

[snip]
Oops. No it's not. Consider circuit "A", a voltage source in series with
a lumped impedance equal to r + jx. And circuit "B", a voltage source in
series with a transmission line of length, say, one wavelength, with
characteristic impedance Z0 = r + jx. The two are not at all equivalent.
To show that they're not, connect a 50 ohm resistor to the output of
each. The resistor current in circuit "A" is Vi/(50 + r + jx). The
current in circuit "B" is Vi/50.
[snip]

Y'all are missing my point.

First off... you, and Dave have just changed situation I proposed. I did
not say
that there was an unterminated line of one wavelength! You and Dave said
that. Forget it. That has nothing to do with the discussion/proof!

Read my "typing"! I said you must have *either* a semi-infinite line, whose
driving point impedance is known by all to be exactly Zo, *or*, but it is
simply beside the point, if you must insist on using a finite length
line then that is OK as well as long as it is terminated at it's far end
in an impedance equal to Zo.

In either case of the semi-infinite Zo line or the finite line terminated in
Zo the
driving point impedance is identical, i.e. it is Zo and you can't tell the
difference
between the two at the driving point.

[snip]
David Robbins recently made several postings explaining that these two
circuits are not equivalent. Hopefully the above example illustrates that.
[snip]

I disagree with Dave. A semi-infinite Zo line and a finite Zo line
terminated
in Zo are indistinguisable at the driving point and at all points along
the lines, until you come to the end of the finite line, but even there the
voltage and current across then terminating Zo is identical to the voltage
and current at the same point on the semi-infinite line.

All of this discussion about there being a difference is moot. There is no
difference. That's exactly why I set it up that way, simply because there
is no difference. You guys are missing the point.

Check out any book on transmission lines...

When you look into the end of a transmission line of semi-infinite extent
you
see a driving point impedance of Zo. That is the very definition of Zo!

If you cut the line off to a finite length and terminate it in it's
characteristic
impedance Zo the driving point remains the same, i.e. Zo. That again is
part of the defining charateristic of characteristic or surge impedance!

Take a chunk of 50 Ohm lossless coax and terminate it in 50 Ohms. What's
the impedance you see looking into the end? 50 Ohms. Now change the length
to something else and again terminate it in 50 Ohms, what 's the impedance?
50 Ohms, it doesn't matter how long the line is if it is terminated in it's
characteristic impedance it's driving point impedance is Zo! End of story.

You simply can't tell the difference between a line terminated in it's
characteristic
impedance and the characteristic impedance of an semi-infinite line as
found by
observiations at the driving point, they are one and the same. Zo!

[snip]
Can't you do your "proof" with, say, a one wavelength transmission line
of characteristic impedance Z0?

Roy Lewallen, W7EL
[snip]

Yep, sure can... sigh, why is this soooo hard.

Use a one wavelength line if you insist, but terminate it at the far end
with an
impedance equal to Zo!,

The length of the line has nothing to do with the proof.

As a matter of fact that is one of the neat things about the so-called
"proof" I cooked up.

The "proof" is just about whether there are reflections at a conjugate
match or not.

The proof, mathematical or experimental, just proves that there are no
reflections [as defined
by the classical rho] when the generator sees it's image Zo, but there are
reflections when the
generator sees it's conjugate! And the same in the reverse direction. i.e.
there are no reflections
when the line sees it's image Zo but there are reflections when it sees it's
conjugate.

And if Zo is a real resistance say Zo = R, then both situations are
identical.

That's all, nothing complicated, simple straightforward stuff, when Zo is
complex, there *will* always
be reflections at a conjugate match.

This means that when there is a conjugate match on a complex line a
"classical" reflectometer
will not indiacate zero reflected voltage. And if the scale on the
reflectometer is actually calibrated
in "Watts" even though it measures Volts, it will indicate the presence of
[[a "false"] reflected power
at a conjugate match.

But then everybody already knew that!

QED!

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

Not QED at all. You claimed to have proved that maximum power is
delivered to a load when a transmission line is terminated such that the
reflected voltage on the line is zero. Here, you're agreeing that the
reflected voltage is zero when the line is terminated in its
characteristic impedance. So where's the proof that this condition leads
to maximum power to the load?

Roy Lewallen, W7EL

Peter O. Brackett wrote:
Roy:

[snip]

Oops. No it's not. Consider circuit "A", a voltage source in series with
a lumped impedance equal to r + jx. And circuit "B", a voltage source in
series with a transmission line of length, say, one wavelength, with
characteristic impedance Z0 = r + jx. The two are not at all equivalent.
To show that they're not, connect a 50 ohm resistor to the output of
each. The resistor current in circuit "A" is Vi/(50 + r + jx). The
current in circuit "B" is Vi/50.

[snip]

Y'all are missing my point.

First off... you, and Dave have just changed situation I proposed. I did
not say
that there was an unterminated line of one wavelength! You and Dave said
that. Forget it. That has nothing to do with the discussion/proof!

Read my "typing"! I said you must have *either* a semi-infinite line, whose
driving point impedance is known by all to be exactly Zo, *or*, but it is
simply beside the point, if you must insist on using a finite length
line then that is OK as well as long as it is terminated at it's far end
in an impedance equal to Zo.

In either case of the semi-infinite Zo line or the finite line terminated in
Zo the
driving point impedance is identical, i.e. it is Zo and you can't tell the
difference
between the two at the driving point.

[snip]

David Robbins recently made several postings explaining that these two
circuits are not equivalent. Hopefully the above example illustrates that.

[snip]

I disagree with Dave. A semi-infinite Zo line and a finite Zo line
terminated
in Zo are indistinguisable at the driving point and at all points along
the lines, until you come to the end of the finite line, but even there the
voltage and current across then terminating Zo is identical to the voltage
and current at the same point on the semi-infinite line.

All of this discussion about there being a difference is moot. There is no
difference. That's exactly why I set it up that way, simply because there
is no difference. You guys are missing the point.

Check out any book on transmission lines...

When you look into the end of a transmission line of semi-infinite extent
you
see a driving point impedance of Zo. That is the very definition of Zo!

If you cut the line off to a finite length and terminate it in it's
characteristic
impedance Zo the driving point remains the same, i.e. Zo. That again is
part of the defining charateristic of characteristic or surge impedance!

Take a chunk of 50 Ohm lossless coax and terminate it in 50 Ohms. What's
the impedance you see looking into the end? 50 Ohms. Now change the length
to something else and again terminate it in 50 Ohms, what 's the impedance?
50 Ohms, it doesn't matter how long the line is if it is terminated in it's
characteristic impedance it's driving point impedance is Zo! End of story.

You simply can't tell the difference between a line terminated in it's
characteristic
impedance and the characteristic impedance of an semi-infinite line as
found by
observiations at the driving point, they are one and the same. Zo!

[snip]

Can't you do your "proof" with, say, a one wavelength transmission line
of characteristic impedance Z0?

Roy Lewallen, W7EL

[snip]

Yep, sure can... sigh, why is this soooo hard.

Use a one wavelength line if you insist, but terminate it at the far end
with an
impedance equal to Zo!,

The length of the line has nothing to do with the proof.

As a matter of fact that is one of the neat things about the so-called
"proof" I cooked up.

The "proof" is just about whether there are reflections at a conjugate
match or not.

The proof, mathematical or experimental, just proves that there are no
reflections [as defined
by the classical rho] when the generator sees it's image Zo, but there are
reflections when the
generator sees it's conjugate! And the same in the reverse direction. i.e.
there are no reflections
when the line sees it's image Zo but there are reflections when it sees it's
conjugate.

And if Zo is a real resistance say Zo = R, then both situations are
identical.

That's all, nothing complicated, simple straightforward stuff, when Zo is
complex, there *will* always
be reflections at a conjugate match.

This means that when there is a conjugate match on a complex line a
"classical" reflectometer
will not indiacate zero reflected voltage. And if the scale on the
reflectometer is actually calibrated
in "Watts" even though it measures Volts, it will indicate the presence of
[[a "false"] reflected power
at a conjugate match.

But then everybody already knew that!

QED!

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

"Peter O. Brackett" wrote in message
hlink.net...
Roy:


This is a long winded explanation of a simple concept comprising a very
simple circuit
containing an ideal generator, a complex impedance and a terminated line,
along with
no more than a couple of equations that constitutes a clear and simple
mathematical
and/or experimental "proof" that the reflected voltage is non-zero for a
conjugate match
and is zero for an image match.

QED

sorry, not always... another clear generalization from oversimplifying and
applying the wrong model. there are cases where there is zero reflected
voltage for a conjugate match.

"Peter O. Brackett" wrote in message
link.net...
Dave:


The impedance matching dynamics at the reference plane/junction point are
the same for
distributed and lumped systems, they obey all the same equations, their
electrodyanamics
is the same, one simply cannot tell the difference at the driving point.

If you don't like lumped models, then make the internal resistance of the
Thevenin
generator out of a distributed line. It will work just the same way.


i'm sorry, but that just isn't true. make me an equivalent, with a
distributed line, of a 1v step voltage source in series with a 1ohm
resistor.

"W5DXP" wrote in message
...
David Robbins wrote:
these discussions always seem to end up in this same quagmire, one group
trying to solve everything with wave equations, sinusoidal steady
states,
transmission line transformations, and the other holding on to the
lumped
models and trying to make them fit the wrong problem domain.

Plus a third group of particle physicists saying waves don't exist. :-)


what is really fun is watching those guys try to explain reflections with
photons getting absorbed and reemitted.

David Robbins wrote:

"W5DXP" wrote:
Plus a third group of particle physicists saying waves don't exist. :-)

what is really fun is watching those guys try to explain reflections with
photons getting absorbed and reemitted.

Yep, I notice I got no answer (so far) to my question about reflected
photons. And, apparently, when one electron gets too close to another,
they each emit virtual photons as a warning not to come any closer.
--
73, Cecil, W5DXP

David:

[snip]
sorry, not always... another clear generalization from oversimplifying and
applying the wrong model. there are cases where there is zero reflected
voltage for a conjugate match.
[snip]

David, my friend as I stated several times, there is only one such case NOT
"cases"
and occurs only when Zo is purely resistive.

In that case the conjugate is non-existent! What's your point?

Wanna proof:

The reflection coefficient is identically zero,

rho = (Z - Zo)/(Z + Zo) = 0

If and only if the numerator of rho is identically zero.

(Z - Zo) = 0

Solving for the unknown Z, this occurs whenever.

Z = Zo

For complex Zo = ro + jxo, conjugate match occurs whenever
Z = ro - jxo and so rho can only be zero when:

ro - jxo = ro + jxo

which only occurs when

-xo = xo

This can only happen if xo = 0, i.e. the reactive part of both Z and Zo
are identically zero.

i.e. -xo = xo if and only if xo = 0

And this only occurs when Zo is real, not when it is complex!

What exactly is your point?

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

Roy:

[snip]
"Roy Lewallen" wrote in message
...
Not QED at all. You claimed to have proved that maximum power is
delivered to a load when a transmission line is terminated such that the
reflected voltage on the line is zero.
[snip]

No I did not! Never said that, never have. Where did you get that idea?

I said that for a general complex Zo the reflected voltage is generally
NOT zero at maximum power transfer.

To make myself perfectly clear, let me repeat that...

I said that for a general complex Zo the reflected voltage is generally
NOT zero at maximum power transfer.

I said that for a general complex Zo the reflected voltage is generally
NOT zero at maximum power transfer.

I said that for a general complex Zo the reflected voltage is generally
NOT zero at maximum power transfer.

[snip]
Here, you're agreeing that the
reflected voltage is zero when the line is terminated in its
characteristic impedance. So where's the proof that this condition leads
to maximum power to the load?

Roy Lewallen, W7EL
[snip]

What I was trying to prove was that the reflected voltage is NOT zero
at conjugate match for the case of complex Zo!

I proved that by setting up a transmission line with perfect image [not
conjugate]
matching on both ends [Zo is seen looking in both directions from any point
in
the system] and driven by a generator set up to create the incident wave.

That system has no impedance discontinuities anywhere. The impedance
is Zo all along the line and into the generator looking in either direction.
No impedance discontinuities no reflections, period!

I then calculated the classical reflection coefficient and showed it to be
zero
confirming that rho = 0 when there are no impedance discontinuities and the
classical formula for rho is used, rho = (Z - Zo)/(Z + Zo).

As the last step I changed the termination from Z0 to conj(Zo) i.e. a
conjugate
match, NOT an image match and showed that rho is NOT zero in this case.

QED!

Summarizing...

Image Match: A line of surge impedance Zo terminated in Zo has no impedance
discontinuities and no reflections.

Conjugate Match: A line of surge impedance terminated in conj(Zo) has an
impedance
discontinuity and hence has reflections. [Unless in the one unique case
that Zo is purely real]

BTW...

Aside: From the postings of Dave and yourself along this thread, I get the
impression that
ya'll beleive that lumped systems obey different laws and should should be
modeled
differently than distributed systems. I am surprised by that claim. Surely
you don't mean that!

Surely all electrical systems, lumped or distributed, must obey the laws of
electrodynamics as set
out by Maxwell-Heaviside. Do you know of any cases where they don't?

Regards,

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

David:

[snip]
i'm sorry, but that just isn't true. make me an equivalent, with a
distributed line, of a 1v step voltage source in series with a 1ohm
resistor.
[snip]

A 1volt step voltage in series with a zero length transmission line
terminated in a 1 Ohm resistor!

Dave, all electrical systems, lumped and distributed alike, obey
the Maxwell-Heaviside equations. They are all the same, not
different as you claim!

Prior to Maxwell [mid 1800's] folks believed that lumped and
distributed systems might obey different laws, but ever since
Maxwell wrote down his celebrated 22
equations, using quaternions, and Oliver Heaviside reduced
them to 4 neat little vector differential equations back in
the mid 1800's, most everyone, with the apparent exception
of yourself, has accepted that lumped and distributed systems
obey the same laws!

What exactly is your point?

Do you believe that lumped systems are described by different
laws than distributed systems? Are you trying to convince me
of that 150 year old discredited idea?

All macro - electrical systems confirm to the same laws of
electrodynamics, namely the Maxwell-Heaviside equations.

The only systems where Maxwell-Heaviside fails to predict
physical reality is when dealing with the "very" small, i.e.
quantum mechanics when one has to do Engineering and
make predictions and design one-photon-at-time. In this
case you have to use quantum electrodynamics or QED
but still all systems, lumped or distributed must obey QED
and Maxwell-Heaviside is just a special case or approximation
to QED laws in the aggregate when there are lots of photons.

Dave you will have a lot of arguing to do to convince modern
electro-technologists that lumped systems obey different laws
from distributed systems. Or that the equations of lumped
impedance matching are any different from transmission
line impedance matching. They are the same! The models
are the same, the mathematics are the same, the experiments
confirm that they are the same.

Exactly what is your point?

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

Peter O. Brackett wrote:
Roy:

[snip]
"Roy Lewallen" wrote in message
...

Not QED at all. You claimed to have proved that maximum power is
delivered to a load when a transmission line is terminated such that the
reflected voltage on the line is zero.

[snip]

No I did not! Never said that, never have. Where did you get that idea?

This is from your posting of August 25:

-----------

Roy:

[snip]

No one from "Camp B" has given any justification for the assumption that
the condition for minimum reflection is the condition for maximum power
transfer. We're lacking either a proof, a derivation from known
principles, or even a numerical example. I maintain that this assumption
is false.

[snip]

I did just that in a separate posting on this thread a couple of days ago.

-----------

Then on August 26, I posted:

-----------

. . .

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.

-----------

To which you replied, also on August 26:

-----------

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

-----------

I haven't been able to find this proof in your postings.


I said that for a general complex Zo the reflected voltage is generally
NOT zero at maximum power transfer.

Well, shoot, I agree with that, as I always have.

. . .

BTW...

Aside: From the postings of Dave and yourself along this thread, I get the
impression that
ya'll beleive that lumped systems obey different laws and should should be
modeled
differently than distributed systems. I am surprised by that claim. Surely
you don't mean that!

I believe you can build a bad or inappropriate model with lumped or
distributed components, and draw invalid conclusions from them. Perhaps
you missed my posting a day or two ago where I pointed out that your
model using lumped components was clearly not the same as one using a
transmission line, by means of the very simple test of observing the
current in a load resistor.

And yes, lumped systems should generally be modeled differently than
distributed ones.


Surely all electrical systems, lumped or distributed, must obey the laws of
electrodynamics as set
out by Maxwell-Heaviside. Do you know of any cases where they don't?

This argument of "you don't agree with my view of how things work, or my
inappropriate models, therefore you don't believe in the Laws of
Physics" is as tiresome as it is pompous.

Roy Lewallen, W7EL

"Peter O. Brackett" wrote in message
k.net...
David:

[snip]
i'm sorry, but that just isn't true. make me an equivalent, with a
distributed line, of a 1v step voltage source in series with a 1ohm
resistor.
[snip]

A 1volt step voltage in series with a zero length transmission line
terminated in a 1 Ohm resistor!


sri, you cut off one important thing... your last sentence before my
request...


If you don't like lumped models, then make the internal resistance of the
Thevenin
generator out of a distributed line. It will work just the same way.


i am asking you to show me a distributed model for the internal resistance
of a thevenin generator that as a lumped model is a 1v step voltage source
in series with a 1 ohm resistor. a zero length transmission line doesn't
exist so the model is still lumped.

Roy:

[snip]
This argument of "you don't agree with my view of how things work, or my
inappropriate models, therefore you don't believe in the Laws of
Physics" is as tiresome as it is pompous.

Roy Lewallen, W7EL
[snip]

Sorry if I offended. I didn't mean to.

And all along here I thought you were the one being tiersome and pompus!

Such is the medium of news group postings.

In summary, I believe that we agree completely, and that we were typing at
"cross purposes".

Your general accusation that no one from Camp B, what ever camp that was,
seemed to show
your "pique" and so I responded in kind.

A waste of time, or... a lesson learned.

Maxwell rules, lumped or distributive, there is no discrimination.

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

David:

[snip]
i am asking you to show me a distributed model for the internal resistance
of a thevenin generator that as a lumped model is a 1v step voltage source
in series with a 1 ohm resistor. a zero length transmission line doesn't
exist so the model is still lumped.
[snip]

Dave you are being picayune. No one wins a ****ing contest like this,
everyone just gets **** on their hands.

I could respond with... OK then... how about a transmission line of length
somewhat less than exp(-exp(-exp-1000))) meters in series with a 1 Ohm
resistor. Will that do? Or do I could use a "recursive" definition.

It simply doesn't address or affect the point at hand, which is that a
complex Zo line terminated in its' conjugate will exhibit a non-zero
reflected voltage. Do you agree? If not, what 's your point?

"When I use a word", Humpty Dumpty said, in a rather scornful tone,
"it means just what I choose it to mean, neither more nor less."

"The question is", said Alice, "whether you can make words mean so many
different things."

"The question is", said Humpty Dumpty, "Who is to be master: - that's all."

-- Lewis Carol, "Alice in Wonderland - The Turtle soup"

Dave, I am willing to help you understand my unimportant proof, I didn't
realize it was such a big
deal, but hey... I recall you asked or commented about my posting, but if
all you wanted is to fuss
with me over side issues such as if lumped systems obey different laws than
distributed systems, then
I presume that we will have to agree that we cannot have a productive
discussion..

If I have offended you in some way, I did not mean to, please accept my
appologies.

Best Regards,

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

Peter O. Brackett wrote:
In summary, I believe that we agree completely, and that we were typing
at "cross purposes".


If this newsgroup had its own logo, it would surely be two crossed
porpoises.


--
73 from Ian G3SEK 'In Practice' columnist for RadCom (RSGB)
Editor, 'The VHF/UHF DX Book'
http://www.ifwtech.co.uk/g3sek

It's often noted in texts that SWR is really a meaningless measure when
applied to lossy lines. So I wouldn't unduly worry about strange SWR
numbers for very lossy lines. Take a look at the analysis I just posted
on another thread, which gives voltages, currents, impedances, and
powers for an example case, and see if you can find anything wrong with
it. The calculation used for reflection coefficient is based on its
definition, namely reflected voltage divided by forward voltage. That
agrees with all the transmission line and electromagnetics texts I have,
which is getting to be quite a number now.

Roy Lewallen, W7EL

William E. Sabin wrote:

In simulation programs, transmission lines are solved for their two-port
parameters, and are then treated as lumped circuits in the actual
simulation, just like any lumped-element circuit. Which is a good way
to do it.

I notice that in the ARRL Antenna Book, 19th edition , on page 24-7, it
is stated with definite finality that the reflection coefficient formula
uses the complex conjugate of Zo in the numerator.
I also understand that this has been established by a "well-trusted
authority".

I have used Mathcad to calculate rho and VSWR for Reg's example, for
many values of X0 (imaginary part of Z0) from -0 to -250 ohms.

The data follows:

Note: |rho1*| is conjugated rho1, SWR1 is for |rho1*|, |rho2| is not
conjugated and SWR2 applies to |rho2|

X0.......|rho1*|..SWR1.....|rho2|..SWR2
-250..... 0.935...30.0.....1.865...-3.30
-200..... 0.937...30.8.....1.705...-3.80
-150..... 0.942...33.3.....1.517...-4.87
-100..... 0.948...37.5.....1.320...-7.25
-050..... 0.955...43.3.....1.131...-16.3
-020..... 0.959...47.6.....1.030...-76.5
-015..... 0.960...48.4.....1.010...-204
-012..... 0.960...48.9.....0.997....+/- infinity
-010..... 0.960...49.2.....0.990....+305
-004..... 0.961...76.3.....0.974....+76.3
0000..... 0.961...50.9.....0.961....+50.9

The numbers for not-conjugate rho are all over the place and lead to
ridiculous numbers for SWR. It is also obvious that for a low-loss line
it doesn't matter much. But values of rho greater than 1.0, on a Smith
chart correspond to negative values of resistance (see the data).

Something is wrong here that we are overlooking.

The use of conjugate rho is so much better behaved that I have some real
doubts about some of our conclusions on this matter.

What about it folks? How can we get to the bottom of this?

Bill W0IYH

"Roy Lewallen" wrote

It's often noted in texts that SWR is really a
meaningless measure when
applied to lossy lines.
=============================

In amateur SWR meter applications even the line is just
a figment of the imagination.

But that problem is easily solved - change the name
of the meter!
----
Reg

The equation in the ARRL Antenna Book is identical
to the equation for rho that is in the Power Wave
literature (see Gonzalez and also see Kurokawa).
Also, numerous literature sources describe how an
open-circuit generator with internal impedance Z0,
connected directly to load ZL, is actually a power
wave setup that leads to a rho formula that is
identical to the formula in the ARRL Antenna Book.
When calculating rho, it is not necessary to fool
around with the wave equations, because frequency
is constant and everything is steady-state.

Bill W0IYH



Also the same conjugate formula in Les Besser's
RF Fundamentals I, and Kurokawa, and the 1992 ARRL
general Handbook, which has NO term for Zo reactance, so
it assumes a purely real Zo.

This page also has the correct general formula:

http://www.zzmatch.com/lcn.html


Slick

A few days ago I posted a derviation of the (non-conjugate) formula for
voltage reflection coefficient on a transmission line. It required only
a few assumptions:

1. That the voltage reflection coefficient is the ratio of reverse to
forward voltage.
2. That the voltage at any point along the line, including the ends, is
the sum of the forward and reverse voltages, and that the current is the
sum of forward and reverse currents.
3. That the ratio of forward voltage to forward current, and the ratio
of reverse voltage to reverse current, equal the characteristic
impedance of the transmission line.

Given these assumptions, the derivation is a matter of straightforward
algebra.

For those promoting some other formula for voltage reflection
coefficient: Which of the above assumptions is false? What substitute
assumption is true? And what's *your* dervivation? Remember, we're
talking about transmission lines here, not a one- or two-port analysis
with a "reference impedance" instead of a transmission line, and where
there's no restriction that the total voltage and current are simply the
sum of the forward and reverse components.

Roy Lewallen, W7EL

Roy Lewallen wrote:
1. That the voltage reflection coefficient is the ratio of reverse to
forward voltage.
For those promoting some other formula for voltage reflection
coefficient: Which of the above assumptions is false?

Number 1 is not always true for s11, the s-parameter reflection
coefficient.

What substitute assumption is true?

For an s-parameter analysis, it's that s11 = b1/a1 when a2=0
Your definition above says that rho = b1/a1 no matter what
the value of a2. Some configurations have rho = s11 and some
don't.

There are differences between your transmission line analysis,
an s-parameter analysis, an h-parameter analysis, a y-parameter
analysis, or a z-parameter analysis. If they were all alike,
there would be no need for their separate existences. FYI:

s11=[(h11-1)(h22+1)-h12*h21]/[(h11+1)(h22+1)-h12*h21]

s11=[(1-y11)(1+y22)+y12*y21]/[(1+y11)(1+y22)-y12*y21]

s11=[(z11-1)(z22+1)-z12*z21]/[(z11+1)(z22+1)-z12*z21]

s11=Vref1/Vfwd1 when Vref2=0, i.e. Pref2=0
--
73, Cecil http://www.qsl.net/w5dxp



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Roy Lewallen wrote:
I tried really hard to say clearly that I'm speaking of transmission
lines and not one- or two-port analysis. How could I have phrased that
in a way it would be understood?

You can speak of anything you want. But a transmission line analysis
including a tuner and reflected waves *IS* a two-port analysis at the
tuner. It *IS* a one-port analysis at the load. There's just no getting
around it. Your transmission line analysis resembles a Z-parameter
analysis of one-port and two-port systems.
--
73, Cecil http://www.qsl.net/w5dxp



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"Roy Lewallen" wrote
I tried really hard to say clearly that I'm speaking
of transmission
lines and not one- or two-port analysis.

=======================

There's no difference between the two. If you produce
the symetrical S-matrix for a length of transmission
line, its 4 parameters contain complex hyperbolic
functions like (Ro+jXo)*Cosh(A+jB) . . . . . where A
is nepers and B is radians.

Whichever way you go, matrix-algebra or Heaviside, you
perform exactly the same set of calculations and, not
surpisingly, you end up with the same answers -
provided Zo in one place only is not followed with an
askerisk just to obtain another answer which is thought
to be more preferable.
---
Reg, G4FGQ

Roy:

[snip]
"Roy Lewallen" wrote in message
...
I tried really hard to say clearly that I'm speaking of transmission
lines and not one- or two-port analysis. How could I have phrased that
in a way it would be understood?
[snip]

!!!

Your phrasing is clear, but it is you that does not understand.

Since when has a transmission line NOT been a two port network?

A transmission line has two ports, period, end of story!

Why are you being so picayune about this issue? Something else must be
bothering
you here... I just can't figure out where you are coming from.

What is your purpose in differentiating between networks and transmission
lines?

As far as I understand an electrical network may consist of nothing more
than a simple transmission line, if so, what's your point?

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

Roy Lewallen wrote:
Someone, I don't even recall who now, noticed that the formula used for
calculating transmission line reflection coefficient allows a magnitude
greater than one when Z0 is complex. From there, the claim was made that
a reflection coefficient greater than one is impossible for a passive
network, since (they said) it implies the creation of energy.

I suspect the problem probably has something to do with squaring
the absolute magnitude of a complex voltage reflection coefficient.
Maybe only the real part should be squared to obtain the power
reflection coefficient?
--
73, Cecil http://www.qsl.net/w5dxp



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