I have been searching and working to understanding this equation below in relation to FM signals minimum threshold.

Pr [dBm]= (Eb/N0)+10∗log10(Rb)+N0 −Gm

Eb is the energy per bit to noise power spectral density ratio
N0 is the signal to noise ratio per bit
Rb is the bit rate
Gm is miscellaneous gain

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?

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.

This is my first stack question and really would love some help.

Thanks in advance!

  • \$\begingroup\$ Hmm, Try searching for the term descriptions singly and in groups on Slideshare, there may be something relevant there. slideshare.net/MazumderAlam/… \$\endgroup\$
    – KalleMP
    May 9, 2016 at 5:40
  • \$\begingroup\$ There is no such thing as an "FM antenna". It will work equally well with AM or single sideband. Are you looking to derive the -154 dBm + 10 log(baud rate)dBm formula? \$\endgroup\$
    – Andy aka
    May 9, 2016 at 7:32
  • \$\begingroup\$ Your best bet for a simple rule of thumb is: Reciever sensitivity (dB) + antenna gain (dB) - 10log(baud rate) ~= your minimum discernable signal (threshold dB), that'd be a good way to at least get a ball park figure so that you can do a sanity check on your equations \$\endgroup\$
    – Sam
    May 9, 2016 at 8:11
  • \$\begingroup\$ Wow everyone thanks for all the help and educating me on the antenna portion I'll gladly update it above as well. \$\endgroup\$
    – Shay9
    May 9, 2016 at 10:46

1 Answer 1


Firstly there is no such thing as an FM antenna - an antenna doesn't understand the modulation method.

E\$_b\$/N\$_0\$ - what practical values are needed?

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.

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.

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).

Converting to dBm we get -154 dBm + 10log\$_{10}\$(bit rate) dBm.

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.

  • \$\begingroup\$ Thanks very much Andy.The detailed answer really helps me out a lot. I appreciate you taking the time to help. \$\endgroup\$
    – Shay9
    May 9, 2016 at 10:47

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