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*Subject*: Re: Path loss and signal calc...*From*: "daveb@xxxxxxxxxxx" <daveb@xxxxxxxxxxx>*Date*: Sun, 08 Feb 1998 22:08:37 -0500

Lisa R. McCarty wrote: > > > > How can one relate this mesurment with signal strengths > > > that are expressed in dbm or decibles as they relate to microvolts? > > > That is, how does a signal that is said to be -140dbm relate to > > > the measument of .2uv at (10db s+n/n) > > > > The .2 uv is into a 50 ohm load. Solve for power and then 10 times > > the log of the power is the dbm. The answer is -151 dbm. (I think). > > Remember this is for a certain bandwidth, narrow the bandwidth and > > the figure gets better. > > Oops, more like -120 dbm, -150 would be decibels below one watt or > dbw. > > Mike, KB8YHV/KH2 Or convert the .2 uV to dbuV ie: 20 log .2 = -13.98 Then simply add it to -107 dbuV (dbm relation) Hence: -107 + (-13.98) = 120.97 dbm --db-- (AA2GF)

**References**:**Re: Path loss and signal calc...***From:*Lisa R. McCarty

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