Walter Maxwell wrote:
I'll make it a little easier for you, Cecil. I' ll give you two ways to
determine Vfwd.

OK, here's the example:

100W XMTR---50 ohm line---x----1/2WL 150 ohm line---50 ohm load
PF1=100W-- PF2=133.33W--
--PR1=0W --PR2=33.33W

Way 1. Vfwd = sqrt(Power fwd x Zo)

So Vfwd = sqrt(133.33W x 150ohms) = 141.4 volts (that's what I said)

Way 2. Vfwd = Vsource x sqrt[1/(1- rho^2)]
Now plug rho = 0.5 into the expression for Way 2 and see what comes up.

OK, Vfwd = 70.7V x sqrt[1/0.75] = 81.6 volts (something wrong there)
We know Vfwd has to be 141.4V as calculated above using Ohm's law.

Walt, your two ways don't yield the same value. Perhaps that is the
source of your confusion about what Dr. Best said. Your Way 2 appears
to be a calculation of the voltage delivered to a mismatched load.
--
73, Cecil http://www.qsl.net/w5dxp




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