W9DMK (Robert Lay) wrote:
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.

Actually, it is one mm broader than that. In the above analysis,
an energy analysis works just as well as any other, contrary to
the Sacred Cow Lamentations I and II of some experts on this
newsgroup. There is so much redundancy built into the voltage,
current, and power relationships in a transmission line that there
are a number of valid ways to skin the cat. An energy analysis is
one of those valid ways. Two people have sent me emails with
correct solutions. Here's how to approach the solution from an
energy standpoint.

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

rho=250/350=0.7143, rho^2 = 0.51, (1-rho^2) = 0.49
rho^2 is the power reflection coefficient.
(1-rho^2) is the power transmission coefficient.

Pfwd1*rho^2 = 100*0.51 = 51w reflected back toward the source
at the match point. My article labels that quantity 'P3'

Pfwd1*(1-rho^2) = 100*0.49 = 49 watts transmitted through the
match point toward the load. My article labels that quantity
'P1' (as does Dr. Best's QEX article).

For a match to exist Pref2(1-rho^2) must equal 51w, the part
of Pref2 transmitted back through the match point, i.e. not
re-reflected. My article labels that quantity 'P4'

That makes Pref2 = 51w/0.49 = 104.1w, and makes
Pref2(rho^2) = 53.1w, the part initially re-reflected. My article
labels that quantity 'P2' as does Dr. Best's QEX article.

So Pfwd2 = P1 + P2 + P3 + P4 = 49w + 53.1w + 51w + 51w = 204.1w

Who said powers can never be added? Pfwd2 is indeed 204.1w.

Now that we know all the powers (without knowing a single voltage)
we can calculate the voltages and currents whose phase angles
are all either zero degrees or 180 degrees. As Bob sez, phase
angles are trivial at a Z0-match point.

Is there anybody out there who still believes that an energy
analysis is impossible and/or "gobbledegook"?

Incidentally, the two 51w component powers represent the amount
of destructive interference energy involved in wave cancellation
and the amount of constructive interference energy re-reflected
toward the load as a result of that wave cancellation. This is
something that Dr. Best completely missed in his QEX article.
He correctly identified P1 and P2 but completely ignored P3 and P4.
Thus he came up with the equation: Ptot = 75w + 8.33w = 133.33w.
Remember that argument on this newsgroup from spring of 2001?
--
73, Cecil http://www.qsl.net/w5dxp


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