(Dr. Slick) wrote in message . com...
(Tom Bruhns) wrote in message om...

As others have noted, the magnitude of the reflection coefficient can
be greater than unity with a passive line and load. Don't try to read
too much physical significance into that, however.


That's impossible, unless you have a free source of energy, in
which case you should send me the schematic!

Do you believe everything people tell you? How did they set this
up? I don't think i will get an answer to this.

Excuse me? As a matter of fact, the way I discovered that you can
have a reflection coefficient greater than unity was to set up the
boundary conditions for a real transmission line feeding a moderately
high Q inductor and discover that |rho|1. Up to that point, nobody
had told me it was possible. At the time, it was a somewhat
surprising result to me. Try it yourself. On a line, Vf/If = -Vr/Ir
= Zo. At the end of the line connected to an load whose impedance is
Zl, the current is V/Zl, but V = Vf+Vr, and the current there must
also be If+Ir. You should be able to reduce those to the well-known
equation for reflection coefficient, and if you plug in numbers such
as Zo=50-j5 and Zl=1+j100, you'll see that |Vr/Vf|1. It only takes
three or four lines of simple linear algebra and perhaps plugging the
complex numbers into a calculator to find the result for that example.
Again, TRY it for yourself. DO NOT just believe what I tell you. If
you have trouble with the algebra, I'll be happy to help you out.

Cheers,
Tom

(Tom Bruhns) wrote in message om...

That's impossible, unless you have a free source of energy, in
which case you should send me the schematic!


Excuse me? As a matter of fact, the way I discovered that you can
have a reflection coefficient greater than unity was to set up the
boundary conditions for a real transmission line feeding a moderately
high Q inductor and discover that |rho|1. Up to that point, nobody
had told me it was possible. At the time, it was a somewhat
surprising result to me. Try it yourself. On a line, Vf/If = -Vr/Ir
= Zo. At the end of the line connected to an load whose impedance is
Zl, the current is V/Zl, but V = Vf+Vr, and the current there must
also be If+Ir. You should be able to reduce those to the well-known
equation for reflection coefficient, and if you plug in numbers such
as Zo=50-j5 and Zl=1+j100, you'll see that |Vr/Vf|1. It only takes
three or four lines of simple linear algebra and perhaps plugging the
complex numbers into a calculator to find the result for that example.
Again, TRY it for yourself. DO NOT just believe what I tell you. If
you have trouble with the algebra, I'll be happy to help you out.

Cheers,
Tom


Actually, my first posting was right all along, if Zo is always
real.

From Les Besser's Applied RF Techniques:

For passive circuits, 0=[rho]=1,

And strictly speaking: Reflection Coefficient =
(Zl-Zo*)/(Zl-Zo)

Where * indicates
conjugate.

But most of the literature assumes that Zo is real, therefore
Zo*=Zo.


And then i looked at the trusty ARRL handbook, 1993, page 16-2,
and lo and behold, the reflection coefficient equation doesn't have a
term for line reactance, so both this book and Pozar have indeed
assumed that the Zo will be purely real.

That doesn't mean Zl cannot have reactance.

Try your calculation again, and you will see that you can never
have a [rho] (magnitude of R.C.)greater than 1 for a passive network.

How could you get more power reflected than what you put in? If
you guys can tell us, we could fix our power problems in CA!

But thanks for checking my work, and this is a subtle detail that
is good to know.


Slick

|"Dr. Slick" wrote in message
| ...
| And strictly speaking:
| Reflection Coefficient = (Zl-Zo*)/(Zl-Zo)
| Where * indicates conjugate.
|...

Obviously by mistake.

pez
SV7BAX