On May 13, 7:07*pm, K7ITM wrote:
On May 11, 4:21*am, Richard Fry wrote:









On May 10, 9:35 pm, walt wrote:

If you are so confident that the number is 88.8889%,
please derive the conditions that yield that number.

à = R(load) - R(source) / R(load) + R(source)

SWR = (1 + |Ã|) / (1 - |Ã|)

Power Accepted by the Load = Incident Power * (1 - Ã^2)

For a 50 ohm source connected to a 100 ohm load:

à = (100 - 50) / (100 + 50) = 50 / 150 = 0.333333...

SWR = (1 + 0.333333) / (1 - 0.333333) = 1.333333 / 0.666666 = 2:1

Load Power = Incident Power * (1 - 0.333333^2) = Incident Power *
0.888889 (or 88.8889%)

The answers are the same for a 50 ohm source with a 25 ohm load.

RF

Richard F is, of course, correct. *You can just as well do it for DC:
2V O.C., 1 ohm source will deliver 1A, 1V, 1 watt to a 1 ohm load.
For a 2 ohm load, the current is 2/3 amp; the power is 8/9 watt. *For
an 0.5 ohm load, the current is 4/3 amp; the power is 8/9 watt. *It's
not terribly difficult to show, for the AC case, that the same is true
for any 2:1 load, regardless of phase angle--trivial, in fact, if you
accept that the SWR along a lossless TEM line is constant.

Cheers,
Tom


My apology, Wim, the equation I quoted in my last post is incorrect--I
mean't to say R = E/I, not E/R. Sorry.

Walt