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September 9th 04, 04:08 AM
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Available power from an RF field
What is the available power, Pa in picowatts, that can be extracted from a free
space field of Ea microvolts per meter? A reference I have says it is: Pa = Ea squared multiplied by Ae (the antenna effective area) all divided by 2Zo, where Zo = 377 ohms, the impedance of free space. My question is why 2Zo rather than just Zo. Is it because only half of the power in the field can be extracted and delivered to a load? Ron, W4TQT |
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