Jim Kelley wrote:
"W5DXP" wrote in message
Exactly like the other scenario, yeah, like the extra .033 Kw makes no
difference. That's a good one, Cecil :-)

In the first scenario, the 0.033 Kw was loaded into the system before
steady-state conditions were established. The s-parameter equations
consider the power reflected from a mismatched load as another source
of power in the system. The two scenarios are very similar.

But forget the first scenario. What happens to Pref1 in this scenario?
It takes a 25 joules/sec wave to cancel a 25 joules/sec wave.

Vfwd1(rho) = 35.36V at zero degrees, Ifwd1(rho) = 0.707A at 180 degrees.

Vref2(tau) = 35.36V at 180 degrees, Iref2(tau) = 0.707A at zero degrees.

Each of these rearward-traveling waves contains |35.36V*0.707A| = 25 joules/sec

They superpose to zero. What happens to that 50 joules/sec of rearward-
traveling energy? We already know it winds up in a forward-traveling wave
toward the load. You have already admitted that wave cancellation is
responsible for Pref1 being zero. Waves simply cannot exist without energy.
When waves cease to exist, they are forced to give up their intrinsic energy.
We know that Vref2(tau) is traveling rearward. That's all we need to know.
Vref2(tau) is the same voltage term as s12*a2 in the s-parameter equation.

When b1=0=s11*a1+s12*a2, what happens to |s11*a1|^2 and |s12*a2|^2? We
know that s11*a1 and s12*a2 are rearward-traveling voltages.
--
73, Cecil http://www.qsl.net/w5dxp



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