(David or Jo Anne Ryeburn) wrote in message .. .


You mean i can't use Zo=50 + j200 with

Rho = (Zload-Zo*)/(Zload+Zo), for complex Zo?

Only up to Zo=50 + j50?

If Z_0 is supposed to be the surge impedance of a transmission line, then
yes, you *can't* use Z_0 = 50 + 200j, because God doesn't make
transmission lines of such an impedance. If Z_0 is supposed to mean
something else, then I'm not talking about whatever your Z_0 is, except to
say that you shouldn't talk about the voltage standing wave ratio and
expect others to understand your English, if there isn't any transmission
line.


Ok, well use Zo=50 + j50, and Zload = 50 - j50 and
the conjugate formula is correct again.



Ok, well, the conjugate formula still makes more sense to
me.

And never mind that it gives measurably *wrong* answers when used to
determine the voltage standing wave ratio on transmission lines?



Where is it wrong? It's the "normal" equation that is wrong when
Zo is complex! Nobody has explained how a rho1 will NOT lead to a
power RC that is also greater than one, thus violating conservation of
energy.




It says exactly what the formula says. Translating mathematics into some
other language is dangerous.

Is
Besser and Kurokawa and the ARRL incorrect?

It depends upon what they expect to be able to do with their formula for
rho. If they expect to be able to use the conjugate formula for rho to
determine the voltage standing wave ratio, by using VSWR = (1 + |rho|)/|1
- |rho||, then yes, they're incorrect -- demonstrably, *measurably*,
incorrect.



Where and when and how did you do a measurement? When you get a rho
greater than 1, the VSWR = (1 + |rho|)/(1 - |rho|) gives ridiculous
NEGATIVE SWRs. And rho WILL give you the SWR, assuming your tranmission
isn't extremely lossy. This is why people say that the SWR meter has to
be at the antenna, instead of at the end of 100' of RG-58.




The conjugate formula simply gives wrong answers (wrong in the
sense of disagreeing with measurements) while the non-conjugate formula
gives right answers. It also conflicts, mathematically, with the usual
formula for which you have seen a proof (based upon the two Kirkhoff laws
and Ohm's law). So there are theoretical as well as experimental reasons
for rejecting the conjugate formula. It's hard to argue against success or
for failure.



I saw the derivation, but i don't totally agree with it, yet.

Again, what actual bench measurement did you make?

What test setup, circuit did you use?

Did you actually get more reflected power that incident?
I'd really like to see THAT!




If you're not too sure
and you don't wanna say, i wouldn't blame you.

I'm more sure of this than I am of many things. If I had any doubts, then,
logically, I would also have to have doubts about at least two of the
following: (a) the Kirkhoff voltage law; (b) the Kirkhoff current law; (c)
Ohm's law; (d) many empirical measurments that have confirmed the
predictions that can be made using the non-conjugate version of the
formula for rho. It would be logically inconsistent to have confidence in
all four of these four things and, at the same time, doubt that the model
in question describes reality accurately.

David, ex-W8EZE


Please! No Straw man! I never debated a-c.

For (d), certainly, assuming Zo is purely real.



Slick

"David Robbins" wrote in message ...
"Dr. Slick" wrote in message
m...
(David or Jo Anne Ryeburn) wrote in message
.. .


ok, i quit... i have shown several times that the calculation of vswr from
rho only applies in ideal lines, you can not use it in lossy lines because
of the simplifications that went into deriving it. if you insist on harping
on this point how can you ever open your eyes to see the rest of the errors
in your calculations.



25' of RG-8 isn't THAT lossy.

Once again, where and when and how did you do a measurement? When
you get a rho greater than 1, the VSWR = (1 + |rho|)/(1 - |rho|) gives
ridiculous
NEGATIVE SWRs. And rho WILL give you the SWR, assuming your
tranmission
isn't extremely lossy. This is why people say that the SWR meter has
to
be at the antenna, instead of at the end of 100' of RG-58.

You are quitting because you can't really answer this question.


Slick

"David Robbins" wrote in message ...

I did read some of Kurokawas paper, and it IS a bit confusing.
Have you figured it out David?

Please tell us how the conjugate equation was derived.

Please explain where the fallacy of my logic lies for you.


Slick

it wasn't derived, it was defined. formula (1) defines the power waves.
formula (11) defines the 'power wave reflection coefficient' in terms of the
two waves in (1). it is then just algebra to rearrange the terms to get the
formula in (12).


Please show us if he correctly defines formula (1) and why. And i'd also
like to see if someone can derive these:

ai= (Vi+Zi*Ii)/(2*sqrt(Re(Zi))
bi= (Vi-conj(Zi)*Ii)/(2*sqrt(Re(Zi))

For what he calls the incident and reflected power waves.

And he does say that this is also the voltage RC when Zi is real and
positive.

And then he does square the MAGNITUDE of this, to get the Power RC.




i have given up on convincing any one on here that this equation can not be
applied to voltage and current waves on lossy lines. unfortunately it will
give the right answers but for the wrong reason on ideal lines. so go read
the paper, understand what he is doing, and realize that it is a different
domain than the 'classical' voltage and current waves that have been used
for many years and work just fine.


You are giving up because you don't understand the paper either.


Slick

wrote in message ...

You can say that a short is always at zero
volts, but that doesn't mean that there isn't a forward and reflected
voltage moving through it.

True. But if the voltage is always 0, then the sum of the forward
and reflected voltage is always 0.



Well, there is no phasor notation in your overly simplistic
example. So we don't see the forward and reflected waves.



Where does circuit theory predict
Vr = 0.5 * 8.062/_ -97.125 = 4.031/_ -97.125?

This isn't circuit theory, it is from the definition of reflection
coefficient: Vr = Vi * rho.

Right, but if you are going to compare...how does circuit
theory give you Vr = 0.5 * 8.062/_ -97.125 = 4.031/_ -97.125?

Circuit theory gives the voltage across the capacitor as
.02 * (0-j200) = 4/_ -90 (current time impedance)
Reflection theory (can I call it that) using classic rho
gives the voltage across the capacitor as Vr + Vi
0.5 + 4.031/_ -97.125 = 4/_ -90
So both provide the same voltage across the capacitor.



Right, but if you are going to compare...how does circuit
theory give you Vr = 0.5 * 8.062/_ -97.125 = 4.031/_ -97.125?


This would convince me quite a bit, if you could derive this
with circuit theory.




Yes, errrr no! I think that power reflection coefficient as normally
defined has no utility with lossy lines since it can not be used
to make any useful predictions.

Not too sure, are we! And when i say we, i mean WE. "Yes, errrr
no!"
is very much like "maybe" and "sometimes", which a lot of people use
for
fear of making an absolute statement which they may have to (God
Forbid!) retract later, as everything you type is recorded FOREVER.

("FOREVER" used for scary effect for those who are terrified to
admit they were WRONG!)

I find, in practice, that the answer to many apparently simple
questions is 'yes and no'. This simply means the question was
incompletely specified.


Or not understood.


If you define power reflection coeffecient as voltage reflection
coefficient squared then it is indeed around 64.

If you define power reflection coefficient as the actual reflected
power divided by the actual incident power (assuming such real
powers exist), then, indeed, it can not be 64 since that would
violate some generally accepted rules about conservation of energy.

So we have quite a contrast between the meaning implied by the name
and the resulting value. So the value IS 64, but the meaning is
NOT 'power reflection coefficient', though the name may be.



Very nice dancing around the point, Keith!

You're more confused than me!



I disagree with you completely, because a Bird or Daiwa meter will
do exactly that, measure Pfwd and Prev, and then with two cross
needles,
you read off the VSWR! That stands for "VOLTAGE stand wave ratio".

Since directional wattmeters simply compute Vf and Vi and square
the scalar, they will show that factor of 64 mentioned earlier.
This strongly suggests they are not doing a good job of computing
real powers.



I'll agree that power meters recitify the signal, and actually get
a DC voltage from the line. In a certain sense, they are more like
RMS voltmeters than power meters.


I'd like to see ANY power RC over 1 for a passive network,

please show us the circuit to build on the bench!




In practice, none of this matters (for RF -- to keep Peter quiet),
since the line losses are sufficiently low that the line impedance
is sufficiently close to 50 ohms, that the Daiwa gives sufficiently
useful information to allow matching.

Despite this, the examples we are using here have far from real Z0
and are interesting to further our understanding.



Or lack of understanding...




I'd like to know the answer to this question too!
If we are talking about only DISSIPATED power, do we have
to say P=Vrms**2/Re(Zo)? Taking only the real part of the Zo?
And if phase doesn't mean anything for power, how can we use a
complex Zo in the denominator?

And if the Pfrd originates in Zo, Prev is loaded by Zo,
then even if Zo is complex, can you still not say:

[rho]**2 = power RC? Because the Zo cancels out anyways
(a ratio)?

Yup, its a puzzler all right.

And the escape?



Ha! As if you knew! Should we say P=Vrms**2/Re(Zo)? Taking
only the real part of the Zo?



-- Stick to incident and reflected voltage (or current) waves for
analysis. They work.


You don't do this in your circuit theory example. How does
circuit
theory give you Vr = 0.5 * 8.062/_ -97.125 = 4.031/_ -97.125?


-- Don't bother trying to compute the incident and reflected powers.
-- They have no meaning in the general case.
-- Only instantaneous and time averaged power really exist and
they CAN be computed in all cases.


You can't compute average incident and reflected powers when
Zo is complex? Your proof or references, please?


-- Forget power reflection coefficient for it only misleads.
-- And never forget that all a 'directional watt' meter does is
compute and respond to Vf and Vr.

...Keith


I'll agree that power meters recitify the signal, and actually
get
a DC voltage from the line. In a certain sense, they are more like
RMS voltmeters than power meters.


I'd like to see ANY power RC over 1 for a passive network,

please show us the circuit to build on the bench!



Slick

"Dr. Slick" wrote in message
om...
"David Robbins" wrote in message
...
"Dr. Slick" wrote in message
m...
(David or Jo Anne Ryeburn) wrote in message
.. .


ok, i quit... i have shown several times that the calculation of vswr
from
rho only applies in ideal lines, you can not use it in lossy lines
because
of the simplifications that went into deriving it. if you insist on
harping
on this point how can you ever open your eyes to see the rest of the
errors
in your calculations.



25' of RG-8 isn't THAT lossy.

Once again, where and when and how did you do a measurement? When
you get a rho greater than 1, the VSWR = (1 + |rho|)/(1 - |rho|) gives
ridiculous
NEGATIVE SWRs. And rho WILL give you the SWR, assuming your
tranmission
isn't extremely lossy. This is why people say that the SWR meter has
to
be at the antenna, instead of at the end of 100' of RG-58.

You are quitting because you can't really answer this question.


Slick
you really are hopeless... that formula for VSWR is not valid when the line
has losses, and you can only get rho 1 when the line has losses.... see the
catch 22????

"David Robbins" wrote in message ...


Please show us if he correctly defines formula (1) and why. And i'd
also
like to see if someone can derive these:

ai= (Vi+Zi*Ii)/(2*sqrt(Re(Zi))
bi= (Vi-conj(Zi)*Ii)/(2*sqrt(Re(Zi))

For what he calls the incident and reflected power waves.

re-read my reply above until you understand it... (1) is GIVEN, it is the
starting point, he DEFINED ai and bi to be those values... the then goes on
for several paragraphs explaining why he thinks those are better waves than
other types of waves.


Well, I certainly didn't expect you to provide the derivation,
but maybe someone else can.

And there is the point that if Zi is real and positive, the
power wave is actually a voltage wave.


Slick

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


Please show us if he correctly defines formula (1) and why. And
i'd
also
like to see if someone can derive these:

ai= (Vi+Zi*Ii)/(2*sqrt(Re(Zi))
bi= (Vi-conj(Zi)*Ii)/(2*sqrt(Re(Zi))

For what he calls the incident and reflected power waves.

re-read my reply above until you understand it... (1) is GIVEN, it is
the
starting point, he DEFINED ai and bi to be those values... the then goes
on
for several paragraphs explaining why he thinks those are better waves
than
other types of waves.


Well, I certainly didn't expect you to provide the derivation,
but maybe someone else can.

no, no one can derive something that is defined... it is a given. it is the
author's choice to define waves the way he wants, and then to define
whatever he wants to all his reflection coefficient from those waves. that
does not mean it can be generalized to other waves. the waves defined that
result in the normal reflection coefficient happen to be a simple solution
to a second order partial differential equation that results when you
analyze the voltage or current waves in a transmission line.... a different
type of wave, requiring a different reflection coefficient... and never the
twain shall meet!

wrote in message ...

Well, there is no phasor notation in your overly simplistic
example. So we don't see the forward and reflected waves.

If you follow back through the posts you will find that this
started with phasors.



You wrote:

Let us apply 1 volt to this circuit...
Total impedance
50+j200+0-j200 = 50 ohms
Total current (volts/impedance)
1/50 = .02 A
Voltage across resistor
.02 * 50 = 1 V
Voltage across inductor
.02 * (0+j200) = 4/_ 90 Volts
Voltage across capacitor (the load)
.02 * (0-j200) = 4/_ -90

Now for the check...
Vi = 0.5 V
With classic rho
Vr = 0.5 * 8.062/_ -97.125 = 4.031/_ -97.125
Vload = Vi+Vr = 0.5 + 4.031/_ -97.125 = 4/_ -90
The same as computed using circuit theory.

So your Vi is a DC voltage here, not a phasor.




Right, but if you are going to compare...how does circuit
theory give you Vr = 0.5 * 8.062/_ -97.125 = 4.031/_ -97.125?

This would convince me quite a bit, if you could derive this
with circuit theory.

I am unsure what you mean here. The equation Vr = Vi * rho is
used in both cases. In one, 'revised rho' is used. In the other,
'classic rho' is used. The results with 'revised rho' do not
agree with results from circuit theory. Does this not cast
some doubt on the validity of 'revised rho'?



Circuit theory may be wrong.

You never derived how circuit theory gives you
Vr = 0.5 * 8.062/_ -97.125 = 4.031/_ -97.125?





Or not understood.

The questioner possed a question. The answerer provided two answers
depending on how the question was interpreted. Wath's the problem?



The problem is you don't really know the answer either, that's for
certain, but you are responding as if you did.





Very nice dancing around the point, Keith!

You're more confused than me!

It does seem that I am having some difficulty conveying the concept.

Just because a thing has a name, does not mean the name accurately
describes the thing.



Just because you can use a few fancy RF terms, doesn't mean you
understand them fully.




I'll agree that power meters recitify the signal, and actually get
a DC voltage from the line. In a certain sense, they are more like
RMS voltmeters than power meters.

I'd like to see ANY power RC over 1 for a passive network,
please show us the circuit to build on the bench!

The passive network you provided in your first post fulfills this
requirement if you define 'power RC' as |rho|^2.


I disagree that the voltage RC will be greater than 1.

The tricky part is measuring this correctly, because you
would need an SWR meter that is calibrated for the same Z as
Zo.


Slick

"Dr. Slick" wrote:

wrote in message ...
The passive network you provided in your first post fulfills this
requirement if you define 'power RC' as |rho|^2.

I disagree that the voltage RC will be greater than 1.

The tricky part is measuring this correctly, because you
would need an SWR meter that is calibrated for the same Z as
Zo.

It is not nearly that tricky. 'Revised' rho, as you state,
predicts 0 Volts across the capacitor. This will be easy to
measure with any AC voltmeter that can handle your test frequency.

I predict, using circuit theory, that if you excite the
test circuit with a 1 Volt sinusoid at a frequency that
makes the impedances j200 and -j200, that you will measure
4 Volts across the capacitor, not 0. This aligns with the
result expected from 'classic' rho.

....Keith

PS Please consider dispensing with the insults. They do not
further the discussion and are quite unbecoming.

wrote in message ...

The tricky part is measuring this correctly, because you
would need an SWR meter that is calibrated for the same Z as
Zo.

It is not nearly that tricky. 'Revised' rho, as you state,
predicts 0 Volts across the capacitor. This will be easy to
measure with any AC voltmeter that can handle your test frequency.


Perhaps, but i'm interested in the forward and reflected
waves, which you can only get with directional couplers on a line of
the same Z as the Zo, i suspect. So even if you get 0 volts, there
are still fwd and rev waves.



I predict, using circuit theory, that if you excite the
test circuit with a 1 Volt sinusoid at a frequency that
makes the impedances j200 and -j200, that you will measure
4 Volts across the capacitor, not 0. This aligns with the
result expected from 'classic' rho.

...Keith

PS Please consider dispensing with the insults. They do not
further the discussion and are quite unbecoming.


Go ahead and bench test it, and let us know what you find.


Slick