Owen Duffy wrote:
Breaking out of the previous thread to explore the "power explanation" in
a steady state situation:

The scenario for discussion is a transmitter connected to a half wave of
600 ohm lossless transmission line connected to an antenna with a
feedpoint impedance of 70+j0.

The transmitter is rated for 100W output, 100W is developed in the 70 ohm
load, the VSWR on the transmission line is 8.6, the "forward power"
(meaning Vf^2/Zo) on the transmission line is 267W, the "reflected
power" (meaning Vr^2/Zo) on the transmission line is 167W, the DC input
power to the transmitter is 200W.

What part of the "reflected power" of 167W is dissipated in the
transmitter (and why)?

There is a conceptual exercise that I use on such problems.

Source---1WL 70 ohm line---+---1/2WL 600 ohm line---70 ohm load
Pfor1=100w-- Pfor2=267w-- 100w
--Pref1=0 --Pref2=167w

Now the picture becomes perfectly clear. The insertion of 1WL
70 ohm line doesn't affect the steady-state condition. A 70
ohm Z0-match is achieved at '+' so none of the 167w is dissipated
in the transmitter. What happens at point '+' toward the source
is total destructive interference, i.e. no reflected energy. What
happens at point '+' toward the load is total constructive
interference. Reference "Optics", by Hecht, 4th edition, page 388.

rho at '+' is 0.791. That makes rho^2 = 0.626.

The amount of Pfor1 that makes it through the impedance discontinuity
at '+' is Pfor1(1-rho^2) = 100w(0.374) = 37.4 watts - call that P1

The amount of Pref2 that is reflected from the impedance discontinuity
at '+' is Pref2(rho^2) = 167w(0.626) = 104.5 watts - call that P2

The power equation says that Pfor2 = P1 + P2 + 2*SQRT(P1*P2)cos(A)

Since it is a Z0-match point A=0 so cos(A)=1. Therefore,

Pfor2 = 37.4w + 104.5w + 2*SQRT(37.4w*104.5w) = 267 watts

What do you know? The power equation works perfectly. Pfor2 didn't
even need to be given.
--
73, Cecil http://www.w5dxp.com