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November 23rd 03, 04:27 PM

Cecil Moore

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(Tom Bruhns) wrote:
So what do YOU get? As I recall, it comes out to a voltage equivalent
to 112.5W for Pf2, and 12.5W for Pr1.

Yep, that's what I got with an s-parameter analysis. After all the negative
things said here about an s-parameter analysis, I was expecting someone to
post an answer different from the s-parameter answer.

In the s-parameter equation, b2 = s21*a1 + s22*a2, |b2|^2 is Pf2. If the
right side of the equation is squared, the (2*s12*a1*s22*a2) term is the
magnitude of the interference which is constructive in this case. On the
other side of the impedance discontinuity, b1 = s11*a1 + s12*a2. |b1|^2
equals Pr1. If the right side of that equation is squared, the
(2*s11*a1*s12*a2) is the magnitude of the interference which is destructive
in this case. Note that |(2*s11*a1*s12*a2)| = |(2*s21*a1*s22*a2)|, i.e.,
the constructive interference magnitude on one side of the impedance
discontinuity always equals the destructive interference magnitude on the
other side.
--
73, Cecil, W5DXP

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