Cecil Moore wrote in message ...
Dr. Slick wrote:
And how do you explain the rho 1 for a passive network?
Shouldn't be possible. And neither should a negative SWR.

This seems to me to be somewhat akin to the fact that s11 and
rho can have different values at an impedance discontinuity
where a 'third power' is commonplace. Roy's 'third power' at
the load appears to be analogous to a re-reflection of some
sort as the inductive load tries and fails to dump energy
back into the Z0=68-j39 transmission line. A re-reflection
is another component of forward power.

The ratio of reflected Poynting vector to forward Poynting
vector is |rho|^2. In Roy's example, the total average
Poynting vector points toward the load indicating that
(Pz+ - Pz-) 0. That means |rho|^2 cannot be greater
than 1.0.


Cecil,

The ratio Pref/Pfwd is directly related to the ratio [rho].
Consider that after the absolute value brackets, the phase information
is gone. But since we are going to a ratio of average (RMS)
values OR peak values of power, it doesn't matter.

In other words, if you use V**2/R, the "V" can be either peak or
RMS, it doesn't matter, because it is a ratio. And of course, the "R"
doesn't matter either. And of course, the phase information is gone
with
the absolute value brackets.

If you agree that the Pref/Pfwd ratio cannot be greater than 1
for a passive network, then neither can the [Vref/Vfwd]= rho be
greater
than 1 either.

Some people wanna rewrite some books here.


Slick

Dr. Slick wrote:
If you agree that the Pref/Pfwd ratio cannot be greater than 1
for a passive network, then neither can the [Vref/Vfwd]= rho be
greater than 1 either.

But apparently rho=(Z2-Z1)/(Z2+Z1) can be greater than unity. So
those are not the same reflection coefficients. A physical rho
and an image rho are quite often different values. s11 is a physical
rho. Vref/Vfwd is an image rho.
--
73, Cecil http://www.qsl.net/w5dxp



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