Richard Clark wrote:
You cannot show that any two powers used to compute Rho are negative
to fulfill this shift of your logic.

It doesn't require either power to be negative. All it requires is a
short circuit. +1 simply has two square roots. rho = -1 for a short
circuit and rho = +1 for an open circuit. All (rho = -1) requires is a
short at the end of a transmission line as explained in _Transmission_
Lines_and_Networks_, by Walter C. Johnson when he was chairman of the
Princeton EE Dept. Here's how he calculated rho for a short:

rho = (Z1-Z0)/(Z1+Z0) = (0-Z0)/(0+Z0) = -Z0/Z0 = -1

So your argument is with Dr. Johnson whom I am merely quoting. The
(rho = -1) simply indicates a 180 degree phase shift in the reflected
voltage at the short.

From Johnson, section 1.6, page 16:
Is NOT a citation. Who is Johnson?

It doesn't surprise me a bit that you are ignorant of Johnson. In 1950,
his book was one of the series of McGraw-Hill Electrical and Electronic
Engineering Series with Terman as the consulting editor and containing
textbooks by Kraus, Skilling, Terman, and others.

... it is not due to
my lack of communication ability (as I am probably the only one here
credentialed to that matter).

Must be really difficult to communicate when you are so special as to
be the "only one here" who is "credentialed to that matter". Many of
the posters to this newsgroup have written books and articles which
I find to be communicated rather well.
--
73, Cecil http://www.qsl.net/w5dxp



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