Walter Maxwell wrote:
However, I have another request for you. Can you find any forward voltages that
are appropriate to plug into Steve's Eq 9 and obtain 133.33 watts PFtotal ?

Here is a quote from an earlier posting:
----------------------
Continuing and calculating interference energy at point 'y':

V1y = VF2(1-rho) = 96.58(0.732) = 70.7V (100W)

V2y will be VR3(rho) = 40.82(0.268) = 10.94V (2.39W)

VF3 = 70.7V + 10.94V = 81.65V. (133.33W)

--------------- So ---

PFtotal = |VFtotal|^2/Z0 = 81.65V^2/50ohms = 133.33W

PFtotal = |V1+V2|^2/Z0 = (70.7V+10.94V)^2/Z0 = 133.33W

How's that?
--
73, Cecil http://www.qsl.net/w5dxp



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