How do you convert watts per meter squared to decibels?

  • Thread starter Ed Aboud
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  • #1
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Main Question or Discussion Point

Hi.
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.
 

Answers and Replies

  • #2
199
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Its ok I figured it out.
[decibels] = 10log + 120

where S = sound intensity in watts per metre squared
 
  • #3
20
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I may be comeplete wrong but aren't decibels 20Log(gain)? Also, the argument of the log must be unitless.
[tex]
=\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.
 
  • #4
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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).
 
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  • #5
555
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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.
 
  • #6
rbj
2,226
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I may be comeplete wrong but aren't decibels 20Log(gain)?
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).

Also, the argument of the log must be unitless.
that was just playing fast and loose with the scalers.

actually Ed wasn't play so fast and loose:

[decibels] = 10log + 120

where S = sound intensity in watts per metre squared


the actual meaning of that is

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

where S0 = 1 W/m2
 
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  • #7
20
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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.
 
  • #8
rbj
2,226
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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.
 
  • #9
555
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Sound intensity is proportional to [itex]r^{-2}[/itex] while sound pressure is propertional to [itex]r^{-1}[/itex].
 

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