On 11/05/2011 12:21, Richard Fry wrote:
On May 10, 9:35 pm, wrote:9.542
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

Or expressed another way a 2:1 vswr equates to a return loss of 9.542dB
and so a mismatch loss of -0.512dB which equates to 0.888 or 88.8% .

Jeff