Minimum threshold fm equation - Electrical Engineering Stack Exchange most recent 30 from electronics.stackexchange.com 2019-08-20T01:06:21Z https://electronics.stackexchange.com/feeds/question/233113 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://electronics.stackexchange.com/q/233113 0 Minimum threshold fm equation Shay9 https://electronics.stackexchange.com/users/109475 2016-05-09T04:51:29Z 2016-05-09T09:33:16Z <p>I have been searching and working to understanding this equation below in relation to FM signals minimum threshold.</p> <p>Pr [dBm]= (Eb/N0)+10∗log10(Rb)+N0 −Gm</p> <p>Eb is the energy per bit to noise power spectral density ratio<br/> N0 is the signal to noise ratio per bit<br/> Rb is the bit rate<br/> Gm is miscellaneous gain<br/></p> <p>I cannot seem to figure out for an fm antenna how I could apply this formula to determine the dBm? For example how to know Eb and Rb from an FM antenna?</p> <p>Any help is appreciate, even academic sources I could reference (books, papers, journals) to give me more clues on how to apply this to a fm antenna. Even the name of this theorem would be helpful.</p> <p>This is my first stack question and really would love some help.</p> <p>Thanks in advance!</p> https://electronics.stackexchange.com/questions/233113/-/233148#233148 0 Answer by Andy aka for Minimum threshold fm equation Andy aka https://electronics.stackexchange.com/users/20218 2016-05-09T09:33:16Z 2016-05-09T09:33:16Z <p>Firstly there is no such thing as an FM antenna - an antenna doesn't understand the modulation method.</p> <p>E\\$_b\\$/N\\$_0\\$ - what practical values are needed? </p> <p>N\\$_0\\$ basically equals kT i.e. Boltzmann's constant x temperature of the object that the antenna is pointed at. It is usually accepted that T = 290 kelvin and hence kT = 4 x \\$10^{-21}\\$ joules.</p> <p>It is also generally accepted that for a decent bit error rate (BER), E\\$_b\\$ should be 100 times higher than N\\$_0\\$ i.e. 4 x \\$10^{-19}\\$ joules.</p> <p>To convert the energy needed per bit to power we multiply by the data rate hence, the received power needed is 4 x \\$10^{-19}\\$ x bit_rate (watts).</p> <p>Converting to dBm we get -154 dBm + 10log\\$_{10}\\$(bit rate) dBm.</p> <p>This the generally accepted rule of thumb formula when transmitting data. For digitized speech a lower energy per bit can be acceptable, maybe one-tenth of the above power.</p>