You've noted what I've noted before in this forum, if you say that
"(V)SWR" is the SWR that would result if the forward and reverse waves
at a particular point on the line (the point at which you've measured
rho) were allowed to develop a voltage maximum and a voltage minimum
(or a current maximum and minimum). I happen to believe that is more
in keeping with the original meaning of (V)SWR than the formula which
says VSWR=(1+rho)/(1-rho).

Of course, this all gets quickly into people trying to assign physical
significance to rho which is not really there; they get confused when
rho1.

If you are clear and consistent with your definitions, I don't see that
any problems result either way. I happen to believe that the formula
you came up with is a better one than the one you commonly see in
texts, but I also agree with Roy that you'd better be clear about it if
you use it, because it goes against the commonly accepted grain. On
the other hand, I've seen texts that derive it in the same way you
have, which come up with the "wrong" answer (without the abs) based on
their initial premises.

Cheers,
Tom

In article ,
Owen wrote:

Indeed. It occurs to me that if one was to try to measure VSWR (say
using a slotted line with probe) on a lossy line operating at high VSWR,
the best estimate would come from finding a minimum, measuring it and
the adjacent maxima and calculating VSWR=(Vmax1+Vmax2)/Vmin/2 so that
the measurement is biassed towards the notional VSWR at the point of the
minimum (the minimum being much more sensitive to line attenuation than
the maxima).

If you think about exponential decay along the line, the geometric mean
sqrt(Vmax1 * Vmax2) would be an even better replacement for Vmax, to use
in the numerator, than the arithmetic mean (Vmax1 * Vmax2)/2.

David, ex-W8EZE

--
David Ryeburn

To send e-mail, use "ca" instead of "caz".

Tam/WB2TT wrote:
There was a big discussion about this last year, and somebody posted that
the ARRL was going to eliminate the conjugate reference.

Another problem that needs to be fixed is the difference between
the "virtual" rho and the physical rho. What is rho looking
into point '+' from the XMTR side?

XMTR---50 ohm coax---+---1/4WL 75 ohm coax---112.5 ohm load

The physical rho is (75-50)/(75+50) = 0.2 which is the same
as 's11' in an S-parameter analysis.

The "virtual" rho is SQRT(Pref/Pfor) which, in a Z0-matched
system is zero. (The 50 ohm coax "sees" a V/I ratio of 50 ohms)

Rho, looking into the load, is (112.5-75)/(112.5+75) = 0.2.

The virtual rho, looking back at point '+' from the load side
is |1.0| but that same reflection coefficient, s22 for an
S-parameter analysis, is (50-75)/(50+75) = -0.2.
--
73, Cecil http://www.qsl.net/w5dxp


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On Fri, 17 Jun 2005 22:44:13 -0400, "Walter Maxwell"
wrote:

[snip]


Hi Owen,

From the general use I'm familiar with, rho alone refers to the abs value, while
the two vertical bars on each side of rho indicates the magnitude alone.
However, following Hewlett-Packard's usage in their AP notes, in Reflections I
use a bar over rho for the absolute, and rho alone for the magnitude. However, I
explain the term in the book to avoid confusion.

Confusion reigns.

Four years ago in another thread I posted thus:

Quote

On Mon, 12 Feb 2001 15:25:28 -0800, Roy Lewallen
wrote:

Just a point of clarification. Rho in these equations is the magnitude
of the reflection coefficient, not the reflection coefficient itself.
The reflection coefficient is actually a complex number. Rho is
unfortunately used to sometimes represent the (complex) reflection
coefficient and sometimes (like here) its magnitude, although some
people (me included) prefer to use uppercase gamma for the complex
reflection coefficient and lowercase rho for its magnitude.

Roy raises a good point. Tom Bruhns already took me to task for a
somewhat careless use of rho. Although I did define it below, as Roy
and Tom said, it is often used as a complex number.

I too prefer upper case Gamma for the complex number and rho for the
magnitude but unfortunately the literature is full of confusing usage.
Some of the literature was even published by Tom's employer, the
former H-P, now Agilent (how do you pronounce that again?)

My autographed copy of Steve Adam's, book "Microwave Theory and
Applications", published by H-P, shows on page 23:

" |Gamma| = rho "

Similarly, my handy dandy H-P "Reflectometer calculator" sliderule
says that SWR = (1 + rho) / (1- rho) which bears a striking
resemblence to what I wrote below.

But then in H-P's App Note 77-3, "Measurement of Complex Impedance
1-1000 MHZ", it says that rho is a vector quantity and it shows:

SWR = (1 +|rho| ) / (1 - |rho| )

Finally, the best reference I have is General Radio's "Handbook of
Microwave Measurements" (out of print but reissued by Gilbert
Engineering) and it says that Gamma is complex and rho isn't.

End quote.

Tam/WB2TT wrote:
"Owen" wrote in message
...
.............................
I do note that my ARRL Antenna Handbook (18th edition) and ARRL Handbook
(2000) both use rho, however they reckon that rho=(Za-Zo*)/(Za+Zo) (where
Zo* means the conjugate of Zo). They do this without derivation, and seem
to be in conflict with the derivation in most texts. I suppose the
derivation is buried in some article in QST and in the members only
section of the ARRL website.


Owen,

There was a big discussion about this last year, and somebody posted that
the ARRL was going to eliminate the conjugate reference.


Les Besser agrees with the ARRL Handbook except he
uses gamma for the complex reflection coefficient:

gamma=(Za-Zo*)/(Za+Zo) (where Zo* means the conjugate of Zo)


But for most practical calculations, the Zo is assumed to
be purely real, so many texts give gamma=(Za-Zo)/(Za-Zo).

rho=[gamma]=absolute value of gamma=magnitude of gamma


Rho can never be greater than one going into a passive
network. Only when you have an active device, or gain, can
you move outside of the unity circle on the Smith Chart.


Slick




Tam/WB2TT

Back to notation, accepting that the preferred pronumeral for the voltage
reflection coefficient is rho, is there a pronumeral used for abs(rho)?

Owen

On 18 Jun 2005 14:20:34 -0700, wrote:

[snip]

Rho can never be greater than one going into a passive
network. Only when you have an active device, or gain, can
you move outside of the unity circle on the Smith Chart.


You better raise your deflection shield.

Go ahead Tom, you have the honor.

Makes things easy, gamma always equal to 1

wrote in message
oups.com...
-
But for most practical calculations, the Zo is assumed to
be purely real, so many texts give gamma=(Za-Zo)/(Za-Zo).

In article ,
David Ryeburn wrote:

In article ,
Owen wrote:

Indeed. It occurs to me that if one was to try to measure VSWR (say
using a slotted line with probe) on a lossy line operating at high VSWR,
the best estimate would come from finding a minimum, measuring it and
the adjacent maxima and calculating VSWR=(Vmax1+Vmax2)/Vmin/2 so that
the measurement is biassed towards the notional VSWR at the point of the
minimum (the minimum being much more sensitive to line attenuation than
the maxima).

If you think about exponential decay along the line, the geometric mean
sqrt(Vmax1 * Vmax2) would be an even better replacement for Vmax, to use
in the numerator, than the arithmetic mean (Vmax1 * Vmax2)/2.
?

I should have typed (Vmax1 + Vmax2)/2 and not (Vmax1 * Vmax2)/2.
Shouldn't post at three minutes before midnight.

Also, in article ,
Wes Stewart wrote:

On 18 Jun 2005 14:20:34 -0700, wrote:

[snip]

Rho can never be greater than one going into a passive
network. Only when you have an active device, or gain, can
you move outside of the unity circle on the Smith Chart.


You better raise your deflection shield.

Go ahead Tom, you have the honor.

I did my best with this same issue a year or two ago, and in the light
of what transpired then I suggest you not waste your time, Tom! The
easiest way to see that (under very special circumstances) rho can
exceed 1 in a passive network uses a simple geometrical argument*.
Unfortunately geometry has even less attention paid to it today than it
did when I was in high school.

*Tradition has it that words translatable as "Let no one ignorant of
geometry enter" were written on the door of Plato's Academy. This is
probably not true, though Plato certainly had a high opinion of
geometry. See

http://plato-dialogues.org/faq/faq009.htm

David, ex-W8EZE, retired SFU mathematician, and Google addict

--
David Ryeburn

To send e-mail, use "ca" instead of "caz".

Wes Stewart wrote:
On 18 Jun 2005 14:20:34 -0700, wrote:

[snip]

Rho can never be greater than one going into a passive
network. Only when you have an active device, or gain, can
you move outside of the unity circle on the Smith Chart.


You better raise your deflection shield.


No need to, i have the laws of physics
on my side!

Hint: Conservation of Energy!


Slick

Rho can never be greater than one going into a passive
network. Only when you have an active device, or gain, can
you move outside of the unity circle on the Smith Chart.

Slick

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

Yes, Rho CAN exceed unity when the termination is a passive network.

For example, when on a real line, Zo = Ro - jXo and the termination Zt
= Rt + jXt then Rho can exceed unity.

Rho has an absolute maximum value which approaches 1 + Sqrt(2) =
2.4142 which occurs when the angle of Zo approaches -45 degrees and Zt
is purely inductive.

It arises because of a weak resonant effect between -jXo and + jXt.

The angle of Zo of real lines always becomes more negative as
frequency decreases. Mr Smith's Chart does not recognise this. He did
it knowingly and deliberately. There are other departures from
reality. But at least it is fit for its few intended purposes.
----
Reg, G4FGQ