Group:

Another very good and inexpensive reference on waves that might be of
interest to this group is:

William C. Elmore and Mark A Heald, "Physics of Waves", first published in
1969, but
most recently published in paperback by Dover Publications, New York, 1985
and
generally available in reprints even today for around US$17.00. A real
bargain.
ISBN: 0-486-64926-1 LCCN: 85-10419.

Be aware, since this is a Physics book it is loaded with gratuitous partial
differential
equations.

This book is interesting because it covers the *whole* field of waves, not
just
electromagnetic waves.

Electromagnetic waves are particularly simple when compared to general wave
phenomena,
since em waves are transverse only.

This book covers waves on strings and membranes, waves in and on fluids,
waves in compressible
media such as the earth [the seismic wavefield]etc... and so covers many of
the "analogies" that
posters to this news group like to draw upon, often drawing false
conclusions.

Are ocean waves at the beach *really* analogous to em waves?

This book will explain why not.

The earth supports both transverse and compressive-dillutive waves as well
as surface seismic waves. Are compressive-dillutive waves different either
qualitatively or quantitatively from electromagnetic transverse waves, this
book answers the question.

etc, etc...

Expand your wave horizons beyond mere em waves, if you are deeply interested
in waves, this book is more than worth the price at US$17.00. It's
available from Amazon.

--
Peter K1PO
Indialantic By-the-Sea, FL


"David or Jo Anne Ryeburn" wrote in message
...
I've enjoyed reading this and related threads. Some comments have been
made about using calculus. Though I spent a significant portion of my
pre-retirement life attempting to teach that subject to undergraduates, I
don't believe in using calculus whenever simple geometry and/or algebra
makes it unnecessary. A proof that avoids calculus can be meaningful for
those who don't know calculus, or who haven't used it for a while.

With that said, I have a few comments to make about some of the assertions
I have read here recently, some of which have appeared without explicit
proof.

(1) The surge impedance of a (lossy) transmission line cannot have an
angle more than 45 degrees away from the real axis.

This is true. Z_0 = sqrt((R + jwL)/(G + jwC)) (here I am using "w" instead
of omega). Both the numerator and denominator lie in the first quadrant,
so their quotient lies in the right half plane (angles subtract when one
divides), and the square root of that has an angle lying between - 45
degrees and + 45 degrees. (The branch of the square root with positive
real part has to be taken; if you can find coax whose surge impedance has
a negative real component, I'll pay you good money for it.)

[Since this angle lies between - 45 degrees and + 45 degrees, peculiar
consequences deduced from calculations involving surge impedances such as
50 - j200 can be ignored.]

(2) There is a nice geometrical interpretation for the reflection
coefficient, or rather for its magnitude. Since the coefficient is (Z_L -
Z_0)/(Z_L + Z_0), its magnitude expresses how much further Z_L is from the
surge impedance Z_0 than it is from the negative, - Z_0, of the surge
impedance. If Z_L is equidistant from Z_0 and - Z_0, then the magnitude of
the reflection coefficient is 1. If Z_L is closer in the complex plane to
Z_0 than it is to - Z_0, then the magnitude of the reflection coefficient
is less than 1. If Z_L is closer to - Z_0 than to Z_0, then the reflection
coefficient's magnitude exceeds 1. Now plot the points Z_0 and - Z_0 and
draw the perpendicular bisector of the segment joining them. If Z_L is on
that perpendicular bisector, the magnitude of the reflection coefficient
is 1; if it is on Z_0's side of the bisector, the magnitude is less than
1; if it is on - Z_0's side, the magnitude exceeds 1. Of course Z_L has to
stay in the right half plane; if it didn't have to do this, you could take
Z_L very close to - Z_0 and get enormous reflection coefficient
magnitudes.

(3) Consider an ellipse having Z_0 and - Z_0 as its foci. There are
infinitely many such ellipses, including a degenerate one (just the
segment between the "foci"). All these different ellipses fill up the
complex plane, and no point in the plane is on more than one of them. On
any one such ellipse, the sum of the distances from a point on the ellipse
to the two foci Z_0 and - Z_0 is constant (definition of an ellipse), the
value of that constant depending upon which ellipse it is but the constant
has to be at least as large as the interfocal distance. We should ignore
points on the ellipse that are in the left half plane. A portion of the
ellipse will be in the same quadrant as Z_0, and a portion will be in the
quadrant that contains the conjugate of Z_0. (Remember we are ignoring the
points in the left half plane.) All of the points on the ellipse that are
in Z_0's quadrant are closer to Z_0 than to - Z_0, so they'll give
reflection coefficients with magnitude less than 1. So will those of the
points on the ellipse in the other quadrant under consideration that are
between the real axis and the perpendicular bisector. But those that are
between the perpendicular bisector and the imaginary axis will be closer
to - Z_0 than to Z_0 and thus will yield reflection coefficients with
magnitudes greater than 1. It should be obvious that, along any given
ellipse, the one for which the magnitude of the reflection coefficient is
greatest is the one on the imaginary axis, since as we move along the
ellipse towards that point, the distance to - Z_0 decreases and the
distance from Z_0 increases (remember, their sum is constant along the
ellipse). So on any *one* ellipse, the largest reflection coefficient
magnitude occurs where the ellipse meets the imaginary axis, and thus Z_L
has real part 0 and imaginary part of opposite sign to that of Z_0.

(4) Start with Z_0 real, and slowly rotate Z_0 into either the first or
fourth quadrant, but not more than 45 degrees in either direction, keeping
the same magnitude while you rotate. The segments joining Z_0 and - Z_0,
their perpendicular bisectors, and the various ellipses will all
simultaneously rotate. It's now obvious that for ellipses of any fixed
size, the one producing the largest magnitude for the reflection
coefficient will occur when Z_0 is at + 45 degrees or - 45 degrees. So if
we want to maximize the reflection coefficient magnitude, we can restrict
attention to those two cases. The - 45 degree case (capacitive surge
impedance) is the more familiar one, but the math is the same either way.
The only question is, which one of the ellipses should we use, if we wish
to maximize the magnitude of the reflection coefficient?

(5) So now we're going to assume Z_0 = k(1 - j), and thus - Z_0 = k(-1 +
j), while Z_L = ktj. It's clear that the factor k is going to cancel out
when calculating the reflection coefficient, so I will henceforth ignore
it (i.e., normalize it to k = 1 by appropriate choice of units). If you
are of a geometrical turn of mind, you can produce a geometrical argument
showing that the best one can do is to make sure the ellipse meets the
imaginary axis at the same distance from the origin as the two foci, i.e.
at j*sqrt(2). If you are of an algebraic turn of mind, you can make an
algebraic argument involving completing the square to demonstrate the same
thing. If you insist on using calculus, it's now just one variable
calculus, not multivariable calculus, since the only independent variable
is t, which will turn out to be sqrt(2) at the maximum. (Hint: don't look
at the ratio of distances; look at the square of that ratio, so as to get
rid of all those square roots.)

(6) Once all that is done, it's just a bit of algebra to show that when t
= sqrt(2) then the magnitude of the reflection coefficient is 1 + sqrt(2).
That's the best (worst?) you can do. And if you can find some coax whose
surge impedance angle is - 45 degrees, you can indeed do it.

All the above was done from first principles. I am not fortunate enough to
own a copy of Chipman, though I wish I were, but if this is what he says,
then I am in full agreement with him.

David, ex-W8EZE, willing to part with some of my pension money for a copy
of Chipman if you know where one can be found (Powell's doesn't have any
copies)

--
David or Jo Anne Ryeburn

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