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.