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Convert Decibels (Db) to Signal to Noise (S/N)?

Discussion in 'Electronic Basics' started by KP, Nov 17, 2003.

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  1. KP

    KP Guest

    Hi All

    I am abit stuck on converting a number in decibels (Db) to signal/noise

    I know the equation for converting into DB: -

    SN (Db) = 10 Log [Signal Power / Noise Power]

    The way which I have managed to conver the Decibel to S/N is using the
    following formula: -

    Inverse Log [XXXX db / 10]

    So using the above formula, 30 Db will work out to be 1000 s/n which is

    The problem is, explaining how I got this formula. Any ideas or better ways
    of working this out would be appreciated.


  2. Steve

    Steve Guest

    um, rearrange the formula? It's easier than you think.

    S/N[dB] = 10 Log (S/N ratio) - divide both sides by ten
    S/N[dB] / 10 = log (S/N ratio) - take antilog of both sides
    alog (S/N[dB] / 10) = S/N ratio - et voila!
  3. KP

    KP Guest

    Thanks Steve.... I ended up with the same formula as you got, just I couldnt
    explain how I got there.

    Thanks again for the help... Its very much appreciated.

  4. David

    David Guest

    Hi Steve

    Thanks for helping my son, now I am stuck on another equation (like fathe,
    like son).

    If I have been asked to calculate the signal power at a point in the system
    where the s/n ratio is 23Db and the Noise Power is 3mW, what would be first
    step be?

    I have looked through loads of books, and my son has only just got into this
    kind of stuff, so not much help!

    Any ideas?

  5. Ben Weaver

    Ben Weaver Guest

    Quick and dirty answer.

    Signal is going to be 23dB bigger than the noise. You just need to find
    the ratio form of 23dB and multiply it by the noise power.

    Whereas 30dB was a factor of 1000, 23dB is a factor 200. I know that
    because 20dB is 100 (dB's drop 10 for every power of ten change) and 3dB
    is a factor of 2. Now for an addition in dB you just multiply the
    factors. 100 x 2 = 200.

    Now just multiply the noise power by that factor. 3mW x 200 = 600mW.

    Or... You could covert 3mW into dBm by doing 10xlog(3)=4.77dBm. Then you
    can just add the 23dB to it. The answer is 27.77dBm. And if you really
    want, you can convert that back into a "proper" number by dividing by
    ten and doing 10^(answer). You get the same result.

  6. Guest

    10 dB in terms of power is ten times greater.
    3 dB in terms of power is well nigh double.


    10 + 10 + 3 = 23dB is the same as:-
    10 * 10 * 2 = 200 times

    This is the *real* reason for using dB i.e.:-
    1) Reduce the range to "human" proportions e.g. not many things are >
    120 dB
    2) Change multiplication to addition.

    If you are playing with some power levels that change from say 1 to
    1000000 you could re-label them 0dB to 60dB. This then allows you to
    stop at some level say 45.7 dB and, from that, double it (45.7 + 3 =
    48.7 dB) or halve it (42.7dB) (by doing the sums in your head).

    If you make a piece of equipment and label everything in dB then it is
    easier for the operator to do make such adjustments. E.g.

    "set output to the onset of clipping then back off 3dB". The
    alternative would have been :-

    "set output to the onset of clipping then back off until it becomes
    half of what it was."

    This might sound trivial (it is!) but it makes the *ratios* of things
    truly appear the same (when they *are*) e.g. +3dB for a power
    amplifier might mean 100 watts whereas +3dB for a front end might be a
    pico watt and a pico watt is certainly different to watts which is

  7. David

    David Guest

    Thank you very much for both of your help, its very much appreciated.

    While im here...... No, im only joking, you have all helped me more than

    Thanks again

    David and Karl
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