Transmitter Output Impedance
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
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