| ??? 11/23/09 16:28 Read: times |
#171076 - Questions Responding to: ???'s previous message |
Okay, now I have a question.
According to Kai, 100% efficiency is achieved when an antenna is lambda/2 (lambda/4 + image according to Daniel). I interpret 100% efficiency to mean that 100% of the initial signal is transferred to free space and none is reflected back toward the source. According to Daniel, this lambda/2 (lambda/4 + image) always constitutes a purely resistive 50 ohm element. If both of these statements are true, is it then assumed that the source has a purely resistive 50 ohm output? Stated differently, if one were required to use a source with other than purely resistive 50 ohm output impedence, couldn't one achieve the same 100% efficiency by designing the antenna to match that impedance? Or, in better keeping with the OP's original question, supposing that one were forced to use an antenna of some other impedance. Could you once again achieve 100% efficiency by designing the output impedance of the source to match the impedance of the antenna? Finally, could it be that this is effectively what is being done by creating reflections to cancel out reflected signal? Or are these two separate issues? Joe |
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