Cecil and others, even authors of books, have said -

- - - - |rho|^2 cannot be greater than 1.0 - - - -

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

Would you change your minds if I describe a
reflection-coefficient bridge, which anybody can
construct, which accurately measures values of | rho |
up to its greatest possible value in transmission lines
of 2.414 There's no catch!

For some reason Dr Slick has remained silent to my
acceptance of his challenge to find such an instrument.
Perhaps he's gone away to think about it.
---
Reg, G4FGQ

Yet in my example, |rho|^2 *is* greater than one.

Also, in the past, you and others have defined the "forward power" to be
the power calculated from the forward voltage and current waves, namely
Re(fE * fIconj) or |fE| * |fI| * cos(phiE - phiI). This is what you've
consistently been calling the "power of the forward wave" or some such.
Likewise for "reverse power". This is the definition I used for the
substitution for fP and rP in the equation for total average power.
And the result is that the total power *isn't* equal to fP - rP.
What you're doing now is lumping the extra power into fP or rP, now
making those terms mean something else. The additional power term has
two components, one arising from the product of forward current and
reverse voltage, and the other from the product of forward voltage and
reverse current. (I combined the two cosine functions with a trig
identitity into a product of two sine functions, but you should go back
a step or two in the analysis to get a clear idea of their derivation.)
I believe you've chosen to assign each of these, or the sine product, to
either "forward power" or "reverse power", depending on its sign, even
though they're a function of both forward and reverse voltage and
current waves. I can't imagine the justification for doing this, but
then there's quite a lot that people have been doing which I don't
understand. As part of the process, you might consider the consequence
of the sine or cosine function returning a negative value, which either
of course can.

Again, I welcome an alternate solution that accounts for all the
voltages, currents, and powers, including one that does it with rho 1.

Roy Lewallen, W7EL

Cecil Moore wrote:

This seems to me to be somewhat akin to the fact that s11 and
rho can have different values at an impedance discontinuity
where a 'third power' is commonplace. Roy's 'third power' at
the load appears to be analogous to a re-reflection of some
sort as the inductive load tries and fails to dump energy
back into the Z0=68-j39 transmission line. A re-reflection
is another component of forward power.

The ratio of reflected Poynting vector to forward Poynting
vector is |rho|^2. In Roy's example, the total average
Poynting vector points toward the load indicating that
(Pz+ - Pz-) 0. That means |rho|^2 cannot be greater
than 1.0.

Well, Cecil, you've redefined Pref and Pfwd. Pref used to be solely a
function of the forward voltage and current waves, and Pref a function
of the reverse voltage and current waves. But now you've chosen to add
an extra term to one or the other of those, or both -- a term which
contains components of both forward and reverse waves. You might recall
from the analysis that I originally had two cosine terms, one arising
from the product of forward voltage and reverse current, and the other
arising from the reverse voltage and forward current. Which of these do
you assign to the "forward power" and which to "reverse power"? If the
choice is based solely on the sign, does the choice automatically change
when the cosine function returns a negative value? (To be truthful, I
haven't checked to see if that is, in fact, possible for possible values
of the argument.) When combined into a product of two sine functions as
I did in the analysis, do you assign this combined function to Pref or
Pfwd? The combined sine functions can, I know, return either positive or
negative values, so what do you do when it returns a negative value? If
I use another trig identitity to convert it to some trig functions
having a different sign, does it then switch from being part of Pref to
part of Pfwd, or vice-versa?

So now when you say Pref and Pfwd, what do you mean?

If you were to stick with the definition you've always used in the past,
i.e., powers calculated from solely forward or reverse voltage and
current waves, the answer is yes. For evidence I offer my derivations.
Roy Lewallen, W7EL

Cecil Moore wrote:
William E. Sabin wrote:

4) The determination that rho magnitude in a transmission line can be
greater than 1.0 is correct. In a passively loaded line fed by an
oscillator, where there is no positive feedback from load to
oscillator, there is no problem about a rho magnitude greater than 1.0.


But can |rho|=Sqrt(Pref/Pfwd) ever be greater than 1.0 for a
passive load?

"Dr. Slick" wrote in message
om...
"David Robbins" wrote in message
...


I believe this line (3) is only true if Zo is purely real.

If Zo is complex, i don't think you can apply
this.

I swore that I wouldn't get into this one, but enough's enough.

Equation (1) is an application of Kirchoff's voltage law.
Equation (2) is an application of Kirchoff's current law.
Equation (3) results from (2) if you apply Ohm's law three times, to
the
three terms in Equation (2).

Which of these three principles (Kirchoff's voltage law, Kirchoff's
current law, or Ohm's law) is the one you don't believe? Or do you
disbelieve more than one of the three?

now, now, take it easy on him... he didn't say he didn't believe kcl or
kvl
or ohm's law... he just doesn't understand that they still do apply to
phasor notation used in sinusoidal steady state analysis. an easy
misunderstanding.


Gee, thanks David. I was wrong! This was a little review for me!
Hehe... owww..

But it still doesn't answer my question.

I don't think Kurokawa and Besser and the ARRL just pulled it
out of thin air.

And how do you explain the rho 1 for a passive network?
Shouldn't be possible. And neither should a negative SWR.

I'm not sure what is wrong with your derivation, but there
must be something that they are missing to not have the conjugate
in the numerator. Or there is a particular step that you cannot do
with complex impedances.

Again, the normal equation is only for purely real Zo, or
when Zo*=Zo. If Zo is complex, you have to use the conjugate
equation.

Could you email a scan of some of the pages? Not that it
would absolutely help me too much, but perhaps you are missing
something.


Slick
sorry, no scanner here.

how do you get rho1? please give me the Zo and Zl to try out, i have been
playing for a while with the basic equations and haven't found a case where
either formulat gives rho1.

and of course if |rho|=1 then swr can never be negative.

Reg Edwards wrote:
Cecil and others, even authors of books, have said -
- - - - |rho|^2 cannot be greater than 1.0 - - - -

Would you change your minds if I describe a
reflection-coefficient bridge, which anybody can
construct, which accurately measures values of | rho |
up to its greatest possible value in transmission lines
of 2.414 There's no catch!

Note that I didn't say |rho| couldn't be greater than one.
I said |rho|^2, the power reflection coefficient, cannot
be greater than 1.0 for a passive load, i.e. you cannot
get more power out of a passive load than you put into it.
It follows that the conservation of energy principle will
not allow the square of rho to be the power reflection
coefficient if rho is greater than 1.0.

For some reason Dr Slick has remained silent to my
acceptance of his challenge to find such an instrument.
Perhaps he's gone away to think about it.

There is an answer here. I suspect you can answer it by
answering the following question about s-parameters.
Consider the following example:

Source--50 ohm feedline--+--1/2WL 150 ohm feedline--50 ohm load

s11 is 0.5 but rho, on the 50 ohm feedline, is zero.

|s11|^2 is defined in the HP AN 95-1 Ap note as the ratio of
the "Power reflected from the network input" to the "Power
incident on the network input" Assuming we have 100 watts of
power incident on the network input, the power reflected from
the network input would have to be 25 watts. But the actual
reflected power on the 50 ohm feedline measures to be zero
watts. Hint: |s12|^2 must also be taken into account.
--
73, Cecil http://www.qsl.net/w5dxp



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On Wed, 3 Sep 2003 18:19:52 +0000 (UTC), "Reg Edwards"
wrote:

Cecil and others, even authors of books, have said -

- - - - |rho|^2 cannot be greater than 1.0 - - - -

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

Would you change your minds if I describe a
reflection-coefficient bridge, which anybody can
construct, which accurately measures values of | rho |
up to its greatest possible value in transmission lines
of 2.414 There's no catch!

For some reason Dr Slick has remained silent to my
acceptance of his challenge to find such an instrument.
Perhaps he's gone away to think about it.
---
Reg, G4FGQ


Ah Reg,

(Too many, like this season's crop of presidential hopefuls, have
usurped the role of clown, sorry to demote you - but you know the
irony in that gesture, you at least gained it honestly. ;-)

No catch? You stand little chance of interest as that would imply an
end to it - what fun when the stream of debate circles endlessly
around simple issues of arithmetic gone bad?

So, in their stead and knowing that anything practical is anathema,
and that anything observed as being cut-and-paste without context is
shunned as a cheap smear, give us the works. [Here's hoping that it
adds to the bottom line of my bountiful discredit.]

73's
Richard Clark, KB7QHC

Roy Lewallen wrote:

Yet in my example, |rho|^2 *is* greater than one.

If so, |rho|^2 is NOT the power reflection coefficient. The conservation
of energy principle will not allow the power reflection coefficient to
be greater than 1.0.

If you calculate a forward Poynting vector and a reflected Poynting vector
at a passive load, you will find that the forward Poynting vector always has
a larger magnitude than the reflected Poynting vector. Thus,
if Pz-/Pz+ = |rho|^2, as asserted in Ramo & Whinnery,
|rho| cannot be greater than 1.0. I suspect you have stumbled upon a
single-port case where rho and s11 are not equal.

You have apparently calculated an s11 reflection coefficient and called
it "rho" under conditions where s11 doesn't have to equal rho.

Also, in the past, you and others have defined the "forward power" to be
the power calculated from the forward voltage and current waves, namely
Re(fE * fIconj) or |fE| * |fI| * cos(phiE - phiI). This is what you've
consistently been calling the "power of the forward wave" or some such.

ONLY for lossless lines. I never said or implied that it would work for
lossy lines. I have carefully avoided making any assertions about lossy
lines. The only assertion that I will make about lossy lines is that
they obey the conservation of energy principle.

Likewise for "reverse power". This is the definition I used for the
substitution for fP and rP in the equation for total average power.

Well, that's apparently a boo-boo for lossy lines. Apparently, Vfwd*Ifwd*
cos(theta) equals forward power only for lossless lines.

And the result is that the total power *isn't* equal to fP - rP.
What you're doing now is lumping the extra power into fP or rP, now
making those terms mean something else.

Yes, for lossy lines, they apparently do mean something else. It reminds
me of the s-parameter equations for power. Like your calculations, there
are four powers, not just two. They are |s11|^2, |s22|^2, |s21|^2, |s12|^2.
It looks as if you have set fP = |s22|^2 and rP = |s11|^2 and your other
two power components are |s12|^2 and |s21|^2. But in real life, these last
two powers are forced to join either the forward wave or the reflected wave.

The additional power term has
two components, one arising from the product of forward current and
reverse voltage, and the other from the product of forward voltage and
reverse current. (I combined the two cosine functions with a trig
identitity into a product of two sine functions, but you should go back
a step or two in the analysis to get a clear idea of their derivation.)
I believe you've chosen to assign each of these, or the sine product, to
either "forward power" or "reverse power", depending on its sign, even
though they're a function of both forward and reverse voltage and
current waves. I can't imagine the justification for doing this, ...

The justification is two, and only two directions, in a transmission line.
All coherent components are forced to superpose into Total Forward Power
or Total Reflected Power depending on the direction (sign).
--
73, Cecil http://www.qsl.net/w5dxp



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Roy Lewallen wrote:

Well, Cecil, you've redefined Pref and Pfwd.

Nope, I haven't, Roy. You have somehow arrived at the equations for
a four-port network while dealing with what appears to be a two-port
network. Inadvertently, you seem to have calculated |s11|^2, |s12|^2,
|s21|^2, and |s22|^2 for what appears to be a two-port network. Is a
two-port lossy line network with inductive load really a four-port
network in disguise? Does the delay in the inductor returning energy
to the system constitute an 'a2' term in the s-parameter analysis?

Pref used to be solely a
function of the forward voltage and current waves, and Pref a function
of the reverse voltage and current waves. But now you've chosen to add
an extra term to one or the other of those, or both -- a term which
contains components of both forward and reverse waves.

Roy, that is built right into the s-paramater analysis. For instance,
for a Z0 (image) matched system:

Forward Power = |s11|^2 + |s12|^2 + |s21|^2 + |s22|^2

For a matched system, Forward Power contains four power terms.
In fact, Forward Power can contain from one to four terms depending
on system configuration.

You might recall
from the analysis that I originally had two cosine terms, one arising
from the product of forward voltage and reverse current, and the other
arising from the reverse voltage and forward current. Which of these do
you assign to the "forward power" and which to "reverse power"?

You are talking about |s12|^2 and |s21|^2. The sign and phase of their
power flow vectors will indicate whether they are forward power or
reverse power.

When combined into a product of two sine functions as
I did in the analysis, do you assign this combined function to Pref or
Pfwd?

If the sign is positive, it is flowing toward the load, i.e. it will
superpose with the forward wave. If the sign is negative, it is flowing
toward the source, i.e. it will superpose with the reverse wave. The
conservation of energy principle will not allow the power in the reverse
wave to exceed the power in the forward wave for passive loads, no matter
what the value of rho.

So now when you say Pref and Pfwd, what do you mean?

What I have always meant. Pfwd is the total of all the coherent forward
components. Pref is the total of all the coherent reverse components.

If you were to stick with the definition you've always used in the past,
i.e., powers calculated from solely forward or reverse voltage and
current waves, the answer is yes. For evidence I offer my derivations.

All you have derived is the s-parameter analysis which is known to include
four power parameters. It is known that s11 doesn't always equal rho for
a four-ternimal network. You seem to have proven that to be true for what
appears to be a two-port network.
--
73, Cecil http://www.qsl.net/w5dxp



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Roy Lewallen wrote:
Again, I welcome an alternate solution that accounts for all the
voltages, currents, and powers, including one that does it with rho 1.

It dawned on me, just now in the shower, what is happening here. When
you introduced the 'x' parameter, the distance from the load, you
introduced a 2-port network analysis, be it an s-, h-, y-, z-, or
whatever-parameter analysis. And of course there are four power
terms in a 2-port analysis. There a

1. The power reflected from the network input back toward the source. |s11|^2

2. The power transmitted through the network port toward the load. |s21|^2

3. The power re-reflected from the network output back toward the load. |s22|^2

4. The power transmitted through the network port toward the source. |s12|^2

These are the four powers you calculated and you consider only |s12|^2 to
be forward power. That is an error. |s22|^2 is also forward power. These
two forward power flow vectors have to be added to obtain the total
forward Poynting vector. I do believe that clears up the confusion.
--
73, Cecil http://www.qsl.net/w5dxp



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Roy Lewallen wrote:
I didn't, and don't, claim to have derived a "power reflection
coefficient". What I calculated was the ratio of reflected voltage to
forward voltage at the load, and called its magnitude rho. If there's
any step in the analysis that's unclear, I'll be happy to explain it in
more detail.

What you apparently calculated is s11 which is not always equal to rho.

What I have calculated is the ratio of reflected voltage to forward
voltage at the load, no more and no less.

No, you have calculated the ratio of one of the reflected voltages to
one of the forward voltages. I believe you have calculated the ratio
of s21*a1 to s12*a2 when you should be calculating the ratio of
(s11*a1+s12*a2) to (s21*a1+s22*a2). You simply omitted half the terms.

Yet lossy lines are just what we're talking about now, isn't it?

Yes, and I am in the process of trying to understand them.

So what are the "forward power" and "reverse power" for lossy lines? Any
explanation for why they vary (other than with the expected attenuation)
with position along the line?

I don't know, yet.

I'm sure that with enough s parameter and optics references, the facts
of the matter can be satisfactorily obscured.

It is you who is using an s-, h-, y-, z-, or other-parameter analysis
and are inadvertently obscuring the facts. You left out half the voltage
terms that should be included in the forward voltage and reflected
voltage. Add all the reflected voltages together. Add all the forward
voltages together. Divide the total reflected voltage by the total
forward voltage.

Your view of how average powers add and travel do force that
restriction. I'm looking forward to your alternative analysis, which
shows the voltages, currents, and powers at both ends of the line while
simultaneously satisfying your notion of how average powers interact.

I think all that is built into your analysis. When you include all the
necessary terms, I will be surprised if everything doesn't fall out
consistently.
--
73, Cecil http://www.qsl.net/w5dxp



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