> Since there are two inputs, we double that to be 40mW of power used up.
> 
>   10*log10( P / 1mW )
>   10*log10( 40mW / 1 mW )
>   16.02 dBm

Ah, and at this point I encourage Aadil to ignore my response, as I was
making a different (not very bright) assumption about why my numbers had
come out a factor of two off (that the 2V was supposed to have been
peak, not peak-to-peak).  Sorry about that, and thanks Brian for getting
the right answer out first.

--Scott



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