> Here is some differnt info.
That data (0.15 dB @ 10 GHz) implies that the K in Joe's equation should
be a bit less than 0.05, rather than 0.5
>> Joe Reisert, W1JR wrote:
>>> Hi Manius,
>>> The insertion loss of a good N connector is equal to the square root
>>> of F (in GHz) times a constant (K).
>>> The constant for a single N connector is 0.5. Therefore, at 23 CM
>>> (1.3 GHz) the loss is approximately .057 dB so you can add up the
>>> losses based on your situation.
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