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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My, it sure didn't take long to get the discussion diverted from the
voltages, currents, and powers in the analysis. I'm sorry to say I
expected that.

Cecil Moore wrote:
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?


I'll leave the philosophical question to you of when a transmission line
is an n-port network and when it isn't, and which s parameter I
inadvertently calculated. Was I unclear about what I did calculate? What
part of it don't you understand?

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.


I don't know, and don't really care, where you're trying to go with your
S parameter analysis. But when you're all done, please translate all
that wonderful stuff to voltages, currents, and powers, using a finite
length transmission line, and present your analysis.

Are you having difficulty understanding what I've done simply with
voltages, currents, and powers?

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.

So, you mean that the term containing the product of two sine functions
is part of Pfwd when the angles are such that the sine functions return
a positive value, and part of Pref when they return a negative value?

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.

No, I did not derive an s parameter analysis. I derived voltages,
currents, and powers. Interpretation of this in terms of s parameters is
strictly your own doing, and it provides wonderful opportunities to
obscure and misinterpret what's really happening. If you're unable to
understand voltages, currents, and powers and want to argue instead
about s parameters (which indeed do represent voltages and powers, but
not necessarily in a one-to-one correspondence to those in the circuit I
analyzed), how many ports the circuit has, and the meaning of the power
reflection coefficient, have at it. But I won't participate. I'll simply
wait until you're done with your philosophising, calculations,
translation back and forth, and post your analysis with V, I, and P as
the variables.

Roy Lewallen, W7EL

Roy Lewallen wrote:

My, it sure didn't take long to get the discussion diverted from the
voltages, currents, and powers in the analysis. I'm sorry to say I
expected that.

Please calm down, Roy. Disagreeing with you is not a diversion. You made
a simple error. When you introduced the 'x' term, the distance away from
the load, you introduced a 2-port analysis. It is a well known fact that
there are four power terms involved in a 2-port analysis as explained in
another posting.

Was I unclear about what I did calculate?

Yes, you were, but it was inadvertent.

I don't know, and don't really care, where you're trying to go with your
S parameter analysis. But when you're all done, please translate all
that wonderful stuff to voltages, currents, and powers, using a finite
length transmission line, and present your analysis.

I'm just showing you what small error you made when you assumed that only
one of the four power terms was the forward power. There are four power
terms. They divide up and add to obtain the forward power and reflected
power. You neglected to do that.

Are you having difficulty understanding what I've done simply with
voltages, currents, and powers?

Nope, I recognize the tiny error you made and am trying to explain it
to you. You didn't include all the forward voltages in your forward
voltage. There are four voltage terms, two forward and two reflected.
You left out half the terms and got the wrong forward or reflected
voltage or both. The mistake is in assuming that rho = s11. It doesn't
in this case.

So, you mean that the term containing the product of two sine functions
is part of Pfwd when the angles are such that the sine functions return
a positive value, and part of Pref when they return a negative value?

No, after further thought, I think you should NOT have combined those two
terms. Four terms is what exists in the analysis so just leave it at
four terms. All the terms with a plus sign combine and all the terms
with a minus sign combine. Please publish the four term power equation
before you used a trig identity to combine the terms. Two of those terms
are forward power and two of those terms are reflected power.

No, I did not derive an s parameter analysis. I derived voltages,
currents, and powers.

You obviously did a something-parameter analysis (maybe a z-parameter
analysis?). Whatever you did results in four power terms, not two plus
a third. When you introduced 'x' you introduced an analysis that produces
a reflected wave on each side of 'x' and a forward wave on each side of
'x'. That's four waves. You went too far when you combined two of those
waves into one especially since one is a forward wave and one is a
reflected wave.
--
73, Cecil http://www.qsl.net/w5dxp



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So do it right and show us how it really should be done.

Roy Lewallen, W7EL

Cecil Moore wrote:

You obviously did a something-parameter analysis (maybe a z-parameter
analysis?). Whatever you did results in four power terms, not two plus
a third. When you introduced 'x' you introduced an analysis that produces
a reflected wave on each side of 'x' and a forward wave on each side of
'x'. That's four waves. You went too far when you combined two of those
waves into one especially since one is a forward wave and one is a
reflected wave.

Roy Lewallen wrote:
So do it right and show us how it really should be done.

Sorry, that diversion won't work. Correct your error first
and effort on my part will be avoided.
--
73, Cecil http://www.qsl.net/w5dxp



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What all you experts have forgotten is that SWR on a
lossless line is the ratio of two voltages, max and
min, SPACED APART BY 1/4-WAVELENGTH. That is if the
line is long enough to contain both a max and a min.

When the line is not lossless, ie., it has appreciable
attenuation in dB per 1/4-wavelength, then the ratio is
'distorted' and has a phase angle. So negative values
of indicated SWR can be expected at some values of |
Vmax | / | Vmin |

SWR is calculated from the square of | rho |. As I've
said before, immediately | rho | is squared, half the
information it contains is junked. Any
discussion/argument about power waves following
rho-squared on a lossy (a real ) line is meaningless
piffle.

Anybody who writes books about power waves, selling
them to make a living, is obtaining money under false
pretences.

On the other hand we should be kind to otherwise
unemployed Ph.D's. They too have wive's and kid's to
clothe, feed and provide a roof over their heads.
That's life!
---
Reg.

"Reg Edwards" wrote in message ...

When the line is not lossless, ie., it has appreciable
attenuation in dB per 1/4-wavelength, then the ratio is
'distorted' and has a phase angle. So negative values
of indicated SWR can be expected at some values of |
Vmax | / | Vmin |


What are you talking about? If it have losses, and they
are dissipative losses, the amplitude of the voltage will
decrease due to voltage drops. That would be moving AWAY
from having a greater reflected voltage than an incident one.

But, that's impossible anyways with a passive network.

The concept of Negative SWRs is rubbish.

SWR is calculated from the square of | rho |. As I've
said before, immediately | rho | is squared, half the
information it contains is junked. Any
discussion/argument about power waves following
rho-squared on a lossy (a real ) line is meaningless
piffle.

Anybody who writes books about power waves, selling
them to make a living, is obtaining money under false
pretences.

On the other hand we should be kind to otherwise
unemployed Ph.D's. They too have wive's and kid's to
clothe, feed and provide a roof over their heads.
That's life!
---
Reg.

Remind me not to be YOUR book when it comes out!



The ratio Pref/Pfwd is directly related to the ratio [rho].

Pref/Pfwd = [rho]**2 Absolute value brackets are a must!


Consider that after the absolute value brackets, the phase information
is gone. But since we are going to a ratio of average (RMS)
values OR peak values of power, it doesn't matter.

In other words, if you use V**2/R, the "V" can be either peak or
RMS, it doesn't matter, because it is a ratio. And of course, the "R"
doesn't matter either. And of course, the phase information is gone
with
the absolute value brackets.

If you agree that the Pref/Pfwd ratio cannot be greater than 1
for a passive network, then neither can the [Vref/Vfwd]= rho be
greater
than 1 either.

Some people wanna rewrite some books here.


Slick

If the loss per unit wavelength is large enough, and you produced a plot of
voltage vs. distance x. The voltage maximum would be at the source, and the
voltage minimum at the load. Try a thousand miles or so of RG58 at 60 Hz.
I suspect that to see anything that looks like a standing wave you would
have to look at dV/dx. Remember, I can always define a lossier line.

Tam/WB2TT

"Reg Edwards" wrote in message
...
What all you experts have forgotten is that SWR on a
lossless line is the ratio of two voltages, max and
min, SPACED APART BY 1/4-WAVELENGTH. That is if the
line is long enough to contain both a max and a min.

When the line is not lossless, ie., it has appreciable
attenuation in dB per 1/4-wavelength, then the ratio is
'distorted' and has a phase angle. So negative values
of indicated SWR can be expected at some values of |
Vmax | / | Vmin |

SWR is calculated from the square of | rho |. As

VSWR is defined as |Vmax|/|Vmin| and so can never be negative. in lossless
lines this expression can be reduced to a function of rho, but that method
is not valid in lossy lines. VSWR is not a constant in lossy lines and
probably doesn't really mean much of anything as each voltage maximum and
minimum is a different value, so which ones do you use???

VSWR is not a constant in lossy lines and
probably doesn't really mean much of anything as each
voltage maximum and
minimum is a different value, so which ones do you
use???

-------------------------------------------------------
---------

Dear David,

You have expressed my sentiments exactly. I have never
used either or any of them. What does anybody do with
value of SWR when they imagine they know it? I'm
pleased to make your acquaintance!

For some years I have mildly advertised the idea of
changing the name the name of the common-or-garden, so
called SWR meter / combined forward-and-reflected power
meter, to the TLI (Transmitter Loading Indicator) which
is all it does. Although I must admit, at the present
state of the art, it is a very useful instrument when
changing antennas.

Is the transmitter loaded with a resistance of 50 ohms
or is it not?

{ Actually, the meter on my top-band transmitter
indicates relative to 75 ohms }

And there HAS to be SOMETHING more than the weather to
talk about in QSO's and, of course, on this newsgroup.
;o)
----
Reg, G4FGQ