Steve etal,
Sorry but the 0.5 in the first line on losses. That was obviously a
typo. It should have been 0.05. You can see that by seeing that I
gave an example and that value is correct.
Thanks for bringing the typo to my attention. I better get my glasses updated!
73,
Joe, W1JR
\At 06:50 AM 8/26/2008, 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.
>
>73,
>Steve G3TXQ
>
>
>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.
> >
> > 73,
> >
> > Joe, W1JR
> >
> >
> >
>
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