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 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.
