Hi, I should learn to sleep at night and email during the day. The tl;dr should have read: Sensitivity for a Modulation Rate at a given bandwidth (from the spec sheet) minus the clean noise floor for that bandwidth (see below) equals the signal needed over the actual noise floor for a given modulation.
A rough estimate for the clean noise floor would be (50MHz=-97dBm, 40=-98dBm, 30=-99dBm, 20=-101dBm, 10=-104dBm). Chuck On Tue, Jan 13, 2015 at 8:39 AM, Chuck McCown <[email protected]> wrote: > Noodling around with Chucks info: > > Noise floor = –80 > Sensitivity = –82 > -80 – (-82) = 2 dB > RX signal required –80 + 2 = –78 dBm > > Noise floor = –40 > Sensitivity = –82 > -40 – (-82) =42 dB > -40 + 42 = +2dBm (Not sure I want to stand in front of the transmitter) > > Some simple algebra: > > Required RX signal =Noise floor + (Noise Floor – Sensitivity) > > - simplifying - > > RX signal = (2 * Noise floor )-Sensitivity > > So, every dB of noise floor == 2dB of RX signal. > > Have I got this right Chuck? > > *From:* Chuck Macenski <[email protected]> > *Sent:* Monday, January 12, 2015 11:25 PM > *To:* [email protected] > *Subject:* Re: [AFMUG] Af5 > > > tl;dr...the natural noise floor for the band minus the sensitivity of > that band for a given modulation equals the signal needed over the noise > floor for a given modulation. > >
