W9DMK (Robert Lay) wrote:
On Sat, 02 Jul 2005 08:21:07 -0500, Cecil Moore
wrote:


The first example was much too easy. How about this one?

---50 ohm feedline---+---300 ohm feedline---
Pfwd1=100w-- Pfwd2 not given--
--Pref1=0w --Pref2 not given

Given a Z0-match at point '+':
Solve for Vfwd1, Ifwd1, Vref1, Iref1, Pfwd2, Vfwd2, Ifwd2,
Pref2, Vref2, Iref2, including magnitudes and phase angles
for all voltages and currents. Source is unknown. Load is
unknown. Lengths of feedlines are unknown.


Without disclosing the answers or the exact procedure for solving the
"brain teaser", I would like to draw attention to some of the implicit
relationships that "ought" to help.
1) It is assumed that both feelines have purely resistive
characteristic impedances (imaginary component, Xo, is zero).
2) Regardless of the length of the 300 ohm line and its termination
impedance, the standing wave pattern and the voltages and currents,
both incident and reflected as a function of distance x along that
line are determined completely by the requirement/condition that there
is a Z0 match at point "+".
3) There are an infinite number of lengths of the 300 ohm line and a
corresponding infinite number of termination impedances for that line
that will produce a Z0 match at point "+". However, because of (2),
above, some of those combinations are well known combinations with
well understood results (e.g., odd multiple of quarter wavelength or
an integer number of half wavelengths).
4) Due to conditions (1) and (2) above, the phase relations between
all of the voltages and currents immediately adjacent to either side
of point "+" are trivial (i.e., any two quantities chosen will be
either exactly in phase or exactly 180 degrees out of phase with one
another).

Due to (3) and (4) above, it would seem that an arbitrary choice of
either a quarter wave line with an 1800 ohm termination or a half wave
line with a 50 ohm termination would provide a convenient example with
which to begin an analysis. However, that is not necessary and only
provides a crutch to get off dead center.

If all of the above elements are kept in mind, then it becomes a
matter of solving a simple algebraic relationship involving 4
equations with 4 unknowns (the incident and reflected voltages and
currents at the right hand side of point "+").

The actual numerical answer to such a problem is irrelevant. The
points to be learned from all this are really the implicit
relationships (2), (3) and (4) above. Without an understanding of
those points, it is virtually impossible to even know where to start.
I think that is the real point that Cecil is trying to make.

Bob, W9DMK, Dahlgren, VA
Replace "nobody" with my callsign for e-mail
http://www.qsl.net/w9dmk
http://zaffora/f2o.org/W9DMK/W9dmk.html


Cecil already defined the voltage and current at
the match point when he gave the characteristic
impedances of the two lines and the rate of
energy transfer through them. Knowing the voltage
and current, anyone can calculate
Pfwd2 and Prev2 using Pfwd2 = |(V+IZ0)/2sqrt(Z0)|^2 and
Prev = |(V-IZ0)/2sqrt(Z0)|^2, where Z0 is the characteristic
impedance of the second transmission line.
Cecil's ability to add powers together, which he did in
this instance, isn't anything unique, and doesn't
really teach anything about the general case.
In fact, for a quarter wave transformer, you can
do the following trick: compute the value of the
power as it just comes through the impedance discontinuity
for the first time and call it Pa. Call Rho^2 at the
load P. Then the power delivered to the load will be
Pa( 1 + P + P^2 + P^3 + P^4 ....) which looks the
same as if the power reflection coefficient looking
back toward the generator was 1 and the power at the
load was the result of the addition of an infinite
number of reflections. Such an interpretation, though,
can be shown to be absolutely wrong. Can anyone see why?
73,
Tom Donaly, KA6RUH