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Fiber Optics - Absolute and decibel Power Gains

  1. Jan 17, 2012 #1
    1. The problem statement, all variables and given/known data

    Problem is very simple but the text book used for this class is complete garbage and has confused me to no end.

    a) Convert the following absolute power gains (dBm) to decibel power gains (dB).
    10^-3 dBm, 2^n dBm

    b) Convert the following decibel power gains (dB) to absolute power gains (dBm).
    -30 dB, 10n dB

    2. Relevant equations

    Power Level in dBm = 10log(P (in mW)/ 1 mW)
    Power ratio in dB = 10log(P_2/P_1)


    3. The attempt at a solution

    a) 10^-3 -> 10log(10^-3) = -30 dB
    2^n -> 10log(2^n) = 10n dB

    b) Here is where I am having a issue with this problem. Part A is solvable according to my book by simply plugging the dBm value back into the same formula that is used to get it but if the same thing is attempted with the decibel power ratio of -30 well things don't go so well or make sense.

    I am more or less looking for help explaining dBms and dBs and how they are related more than I am help solving this particular problem. Any help would be greatly appreciated.
     
  2. jcsd
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