wrote:
It is just that the
computation of Pfwd and Pref does not really produce real powers
(though, again unfortunately, the dimension of the quantity produced
is power).

Funny how they can heat resistors both at the load end and at the source
end when the source is equipped with a circulator+load.

Incidentally, I've come up with a proof that there are no reflections
at a voltage null in a homogeneous transmission line. Consider a 50
ohm lossless feedline. At a voltage null -

rho = (50-50)/(50+50) = 0 i.e. no reflections in either direction
--
73, Cecil http://www.qsl.net/w5dxp



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"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???

Cecil Moore wrote:

wrote:
It is just that the
computation of Pfwd and Pref does not really produce real powers
(though, again unfortunately, the dimension of the quantity produced
is power).

Funny how they can heat resistors both at the load end and at the source
end when the source is equipped with a circulator+load.

Not surprising. The energy comes from the source.

Incidentally, I've come up with a proof that there are no reflections
at a voltage null in a homogeneous transmission line. Consider a 50
ohm lossless feedline. At a voltage null -

rho = (50-50)/(50+50) = 0 i.e. no reflections in either direction

That's what you get when you use the surge impedance and compute
surge rho. Try steady state impedance for steady state rho.

rho = (0-50)/(0+50) = -1 as expected

....Keith

David Robbins wrote:
btw, for whom ever has it... i am still waiting to see the derivation of the
conjugate rho formula. i published one on here for the 'classical' version,
where is the other one???

It exists in the Kurokawa paper, "Power Waves and the Scattering Matrix".
He defines a new kind of wave, different from traveling waves, and calls
them "Power Waves". That conjugate term is apparently the result of this
new definition of waves. He says, "... when the main interest is in the
relation between various circuits in which the sources are uncorrelated,
the traveling waves are not considered as the best independent variables
to use for the analysis." Seems he is not talking about a system where
all the waves are coherent and has defined a new concept of a "Power Wave"
which includes an alternate definition of a reflection coefficient which
includes a conjugate term.
--
73, Cecil, W5DXP

"W5DXP" wrote in message
...
David Robbins wrote:
btw, for whom ever has it... i am still waiting to see the derivation of
the
conjugate rho formula. i published one on here for the 'classical'
version,
where is the other one???

It exists in the Kurokawa paper, "Power Waves and the Scattering Matrix".
He defines a new kind of wave, different from traveling waves, and calls
them "Power Waves". That conjugate term is apparently the result of this
new definition of waves. He says, "... when the main interest is in the
relation between various circuits in which the sources are uncorrelated,
the traveling waves are not considered as the best independent variables
to use for the analysis." Seems he is not talking about a system where
all the waves are coherent and has defined a new concept of a "Power Wave"
which includes an alternate definition of a reflection coefficient which
includes a conjugate term.
--
73, Cecil, W5DXP

is that paper on the web somewhere?? i figured it had to be something with
computing powers that was getting mixed in here some how, i think that is
the only place you can end up with conjugates in transmission lines. so i
assume its not a simple 1 page derivation from basic root principles, it
must take a whole new language to express it.

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

wrote in message ...

And yes, |rho| can be greater than unity for a passive load.

...Keith

Absolute Rubbish.. Could you produce a passive circuit that
will reflect a greater voltage than what you feed it? I'd
LOVE to see that...

Several examples have been presented, but rather than accepting
them, you changed the definition of rho. Perhaps you could build
one of these circuits to determine if modifying the definition
of rho was appropriate.



The definition of Rho has been set for "God-knows-how-long!"

Way before you or I were born, i would imagine!

Rho is the magnitude of Gamma, which is the complex voltage
reflection coefficient.

BUT, some people write Gamma is Rho, so just in case, it's wise
to say Power RC = [rho]**2 , with absolute value brackets.

Ok, show us a circuit, I'm waiting to see.



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

Which I haven't since Pref and Pfwd are just computed numbers and
the result for some circuits is that Pref/Pfwd is greater than 1.
Of course, Pnet is not equal to Pfwd-Pref in these circumstances
so there is no violation of basic physics. It is just that the
computation of Pfwd and Pref does not really produce real powers
(though, again unfortunately, the dimension of the quantity produced
is power).


ok, Keith, i look forward with great interest on your
imaginary passive circuit which can reflect more power than
what you feed it (incident power).

I can't wait to hook it up to see more reflected power than
incident on my DAIWA meter, that would be very interesting.


Slick

ok, Keith, i look forward with great interest on your
imaginary passive circuit which can reflect more power than
what you feed it (incident power).

I can't wait to hook it up to see more reflected power than
incident on my DAIWA meter, that would be very interesting.


Slick

"Reg Edwards" wrote in message ...
Dear Cec,

Your arithmetic is abominable. ;o) Dr Slick's
vanishing-act was a better tactic.

Your only avenue of escape is to prove the | rho |
meter gives incorrect meter readings.



ok, Reg, i look forward with great interest on your
imaginary passive circuit which can reflect more power than
what you feed it (incident power).

I can't wait to hook it up to see more reflected power than
incident on my DAIWA meter, that would be very interesting.


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