In articles and ,
"pez" wrote:

1.
The well known relation

|rho| = 1 + Sqrt[2]

has been sharpened
to be exactly

|rho| = Sqrt[(1+Sin|to|)/(1-Sin|to|)]

where 'to' is the 'angle' of rho.


'to' is not the 'angle' of the
Reflection Coefficient rho
but that of
Z0 = Ro +j Xo.

It has rather to do with
the emotional charge of the moments...

Sincerely,

pez
SV7BAX

In article
the argument I gave showed that if you look at an ellipse with foci at
Z_0 and at - Z_0 then among the points Z_L on any such ellipse that are
in the fourth and first quadrants, the one for which the magnitude of
the reflection coefficient (non-conjugate definition, of course) is
maximized is the one on the imaginary axis near - Z_0 and far from Z_0,
i.e. having opposite reactance to that of Z_0. If you then use geometry,
or algebra, or even calculus if you feel compelled to use it, to see
which such point on the imaginary axis is best (i.e., which of the
ellipses of various eccentricities has the best "best" point) you will
find that the maximum occurs when the magnitude of Z_L is the same as
that of Z_0. In symbols, you will want to have
Z_L = +/- j sqrt ((R_0)^2 + (X_0)^2), where Z_0 = R_0 + j X_0 and the
sign to be chosen (+/-) is the opposite to that of X_0. When you use
this value of Z_L you will indeed get a reflection coefficient
magnitude of sqrt [(1 + sin |t_0|)/(1 - sin |t_o|)], thereby
confirming your formula
|rho| = sqrt [(1 + sin |t_o|)/(1 - sin |t_o|)].

This is a
_claim_
that no one else has verified it.

Now some one has. Details left for the reader -- you can think of what I
wrote above as a "Heathkit proof" ;-)! I used to enjoy giving "Heathkit
proofs" for my students to build; as they became more proficient, the
construction manuals offered became sketchier and sketchier. Eventually
the students got to the point where they could design and build their own
rigs (er, proofs) and the days of "Heathkit proofs" were over for them.

2.
Also,
the well known relation

|Xo|/Ro = 1

is sharpened too, to be exactly

|Xo|/Ro = min{ a/b, b/a }.

This is as well a
_claim_
no one else has verified it yet.

I'll see if I can sketch a nice demonstration of this second result,
preferably one that does *not* involve calculus ;-)!

David, ex-W8EZE, retired math professor, and not quite finished with either
of these things

--
David or Jo Anne Ryeburn

To send e-mail, remove the letter "z" from this address.