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How do you convert watts per meter squared to decibels?

  1. May 24, 2008 #1
    How do you convert watts per metre squared to decibels?
    I know you use logs but can't seems to work it out.
    Thanks for any help.
  2. jcsd
  3. May 24, 2008 #2
    Its ok I figured it out.
    [decibels] = 10log + 120

    where S = sound intensity in watts per metre squared
  4. May 25, 2008 #3
    I may be comeplete wrong but aren't decibels 20Log(gain)? Also, the argument of the log must be unitless.
    =\frac{W}{m^2}=\frac{J}{m^2\cdot s}=\frac{N}{m\cdot s}
    =\frac{kg\cdot m}{s^2}\frac{1}{m\cdot s}=\frac{kg}{s^3}\neq 1[/tex]

    I would just look at your equation again.
  5. May 25, 2008 #4
    The power density must be divided by a certain constant power density j0 to obtain dimensionless argument for log. I think they chose the minimum power density that can be heard by humans for j0 (this is supposed to be 10^-12 W/m^2). So 10*log(S)+120 is more consistently written as
    10*log(S/(1W/m^2))+120=10*log(S/(10^-12 W/m^2))

    20*log() is used for sound pressure amplitudes. The power density is proportional to the square of the pressure amplitude. Since log(x^2)=2*log(x), the multiplication constant must be twice smaller in the first case (so that we the same increase of decibels for the same power/pressure change).
    Last edited: May 25, 2008
  6. May 25, 2008 #5
    Sound intensity level is defined as:
    [tex]L_J = 10 \log_{10}\left( \frac{J}{J_0} \right) \text{ dB}[/tex]
    where [itex]J[/itex] is the sound intensity and [itex]J_0[/itex] is the reference intensity level.

    In the case that [itex]J_0 = 10^{-12} \, \frac{ \text{W}}{\text{m}^2}[/itex], the standard sound intensity reference level, the unit dB is sometimes changed to dB (SIL) where SIL stands for Sound Intensity Level.

    Sound pressure level on the other hand as a slightly different formula, since sound pressure is related to sound intensity squared:
    [tex]L_p = 10 \log_{10} \left( \frac{p^2}{p_0^2} \right) = 20 \log_{10} \left( \frac{p}{p_0} \right) \text{ dB}[/tex]
    where [itex]p[/itex] is the sound pressure and [itex]p_0[/itex] is the reference sound pressure.

    Again, in the case that [itex]p_0 = 2\times 10^{-5} \text{ Pa}[/itex] (root-mean-square pressure), the standard reference sound pressure, the unit dB is sometimes changed to dB (SPL) where SPL stands for Sound Pressure Level.

    These reference soundpressure/intensity are I believe related to the human threshold of hearing at 1000 Hz. I believe it is about the soundlevel of a mosquito flying at 3m distance.

    Note that the unit dB is not actually a unit, it is dimensionless.

    Also note that it's perfectly possible to have a negative sound level (-5 dB for example is not uncommon). It simply means that the soundlevel is below the average threshold of human hearing.
  7. May 25, 2008 #6


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    when gain is a voltage ratio or a current ratio (or in the case of acoustics, the ratio of pressure deviation levels) it's 20log10(gain). but if it's the ratio of power or energy (which are proportional to the square of voltage or current or acoustic pressure), then it's 10log10(gain).

    that was just playing fast and loose with the scalers.

    actually Ed wasn't play so fast and loose:

    the actual meaning of that is

    [decibels] = (10dB)log(S/S0) + 120dB

    where S0 = 1 W/m2
    Last edited: May 25, 2008
  8. May 25, 2008 #7
    So, when measuring sound intensity in decibels using the standard for [tex]J_0[/tex], by what power does sound fall off? My intuition says it should be [tex] r^{-2}[/tex] because it spreads radially outwards but then there is air pressure and stuff like that in there.
  9. May 25, 2008 #8


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    J is a symbol for intensity. if energy is conserved, spherically radiating energy has to have intensity (power per unit area at a right angle to the direction of propagation) that is inverse-square. distance doubles, surface area of sphere centered at the point source increases by a factor of 4, equal power is distributed over 4 times the area, intensity is reduced by a factor of 1/4, and, since this is about power not voltage, that's -6.02 dB.
  10. May 26, 2008 #9
    Sound intensity is proportional to [itex]r^{-2}[/itex] while sound pressure is propertional to [itex]r^{-1}[/itex].
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