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Having trouble understanding the dB formula

  1. Sep 27, 2012 #1

    Cee

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    I know that to work out dB the formula is: dB = 10*log(P1/P2
    I just don't get what the P1 and P2 are exactly but I know that dB is a ratio that expresses 2 values?

    And also, how do you compare two values of dB. Say, there's 80dB and 105dB, how do you work out how many times louder is the 105dB compared to the 80dB?

    Any help will be greatly appreciated, thank you.
     
  2. jcsd
  3. Sep 27, 2012 #2

    PeterO

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    I have always had dB = 10*log(I1/I2) where I is the intensity - Watts per square metre.
    P implies Power, which would account for the Watts - but seems to assume equal areas?

    As for difference in sound

    a 3dB change is both the smallest change the average human can notice, and also represents a doubling in intensity.

    Log(2) = 0.3, so 10*log(2) = 3

    Considering 80dB to 105dB

    80 = reference level
    83 = twice the intensity
    86 = twice that - or 4x the original
    89 = double again - 8x
    92 - 16x
    95 - 32x times

    Note that log(10) = 1 meaning 10*log(10) = 10 so 10x intensity = +10 dB

    95 --> 105 = +10dB so 10x intensity.

    so 95 --> 105 = 10x the already 32x

    SO 80 dB - 105 dB = 320x

    note: log(320) = 2.5, so 10*log(320) = 25

    so +25dB means 320x intensity.

    Or to go the other way

    80dB --> 105 dB = + 25dB which means +2.5 B

    [remember dB means deci-Bels or tenths of Bels so 25 dB = 2.5 B]

    102.5 = 320

    so + 25 dB = 320x intensity
     
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