Oh, my! I really should be changing the name on the thread, as you
suggest, to something about viewing r.r.a.a. as a wonderful source of
humor! Thank you very much for your contribution.

(Dr. Slick) wrote in message . com...

Contrary to popular belief, you will still have reflections going
from Zo=50-j5 to Zload=50-j5.

So, you can't even connect a line to another piece of the same
impedance line without getting reflections? :-) Wonderful!

This is what impedance matching is all about: getting rid of the
reactance in the line and load, and making sure the resistive
impedances left are equivalent.

Only when Zload=Zo*, will this happen. I.E.: Zo=50-j5 and
Zload=50+j5.

If you plug these into Reflection Coefficient =

(Zload-Zo*)/(Zload-Zo)

Where * indicates conjugate, you will see that the reactances cancel
to zero, and the numerator is zero.

Um, you should try that in your equation above for Zo=50-j5 and
Zload=50-j4. What magnitude rho do you get for that, hmmm?? Oh, this
is great fun!

This doesn't happen with the regular (Zload-Zo)/(Zload-Zo).

No, of course that one never gets larger than unity. Nor does it get
smaller. It doesn't evaluate well at Zload=Zo, but we might surmise
it stays unity there too.

It almost time for me to leave this to others to prove to you, as
the communication here is almost non-existant.

Seems to be, indeed, though you never know. The lurkers may well have
learned a thing or two. The one reference you did post now disavows
the form you posted. You've been invited to do some simple math that
would show you the truth and you apparently refuse. You are posting
any number of ideas contrary to what's easy to show from fundamentals
and what's in a large number of published papers and texts and what
has been posted here by many contributors recently and over the years.
It's been done with both symbolic math and specific examples.
Several inconsistencies demonstrate clearly that Vr/Vf does NOT equal
(Zload-Zo*)/(Zload+Zo), and of course most certainly a (Zload-Zo)
denominator is going to get you quickly into trouble. The
inconsistencies have been pointed out here by me and by others, but
apparently you've missed them. I'm sure it's apparent to most lurking
where the communications is breaking down. But the formulas you
posted above have given me a good laugh tonight, at least! Thanks!

Cheers,
Tom