Steve Hunt wrote:
> Joe,
>
> Sorry to "nit pick", but if your formula is correct and the constant is
> 0.5, shouldn't the loss be 0.57dB? It makes a big difference if you have
> several connectors in the chain.
>
>
> 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.
>>
I suspect that 0.5 is an approximation (i.e. it's probably not 0.50000),
and expression in the first place is based on an assumption that the
loss is entirely due to ohmic losses (probably reasonable for an N
connector, since the dielectric is air).
If you're worried about 0.001 dB or even 0.01 dB in a chain of 10
connectors, I wouldn't go using a simple approximation. This reminds me
of folks designing spacecraft telecom links obsessing about 0.01 dB
better performance from coding or modem implementation, then putting in
6 ft of tiny coax between the radio and the antenna (at 7GHz).
Jim, W6RMK
_______________________________________________
_______________________________________________
TowerTalk mailing list
TowerTalk@contesting.com
http://lists.contesting.com/mailman/listinfo/towertalk
