wrote:
You CANNOT superimpose POWERS, ...

Hi Tor, I appologize that the earlier energy flow chart
was not drawn as well as it might have been. I've expanded
that chart in the following and provided the power management
equations associated with the superposition of EM fields. The
'X' event is the initial reflection event at the first
discontinuity, point A. The 'Y' event is the initial reflection
event at the second discontinuity, point B. The 'Z' event is the
initial re-reflection event back at point A.

100W 1/4WL
XMTR---50 ohm---+--------100 ohm---------+---200 ohm---200 load
feedline A feedline B feedline

Just before the first reflection arrives back at A

Pfor(50) | |
100W----X |
X----+ |
11.11W----X \ Pfor(100) |
Pref(50) | +----88.89W----Y
| Y----79.01W
| +----9.876W----Y Pfor(200)
| / Pref(100) |
| --+ |
| |

Just after the first reflection arrives back at A

Pfor(50) | Pfor(100) |
100W----X 109.67W |
X----+ |
11.11W----X \ P1 |
P3 | +----88.89W----Y
| Y----79.01W
Pref(50) | +----9.876W----Y Pfor(200)
0.139W | / Pref(100) |
Z----+ |
8.779W----Z |
P4 Z----1.097W |
| P2 |

P1 = Pfor(50)*(1-rho^2), P2 = Pref(100)*rho^2

P3 = Pref(100)*(1-rho^2), P4 = Pfor(50)*rho^2

How much destructive interference between wave(P3) and wave(P4)?

Pref(50) = P3 + P4 - 2*sqrt(P3*P4) = 11.11W + 8.779W - 19.75W = 0.139W

19.75W of destructive interference, negative since cos(180)=-1

How much constructive interference between wave(P1) and wave(P2)?

Pfor(100) = P1 + P2 + 2*sqrt(P1*P2) = 88.89W + 1.097W + 19.75W = 109.67W

19.75W of constructive interference, positive since cos(0)=+1

Destructive interference energy = constructive interference energy
(toward the source) (toward the load)

It follows that |P1*P2| = |P3*P4| in order to satisfy the conservation
of energy principle.
--
73, Cecil, http://www.qsl.net/w5dxp


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