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

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