How do you convert watts per meter squared to decibels?

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Discussion Overview

The discussion revolves around the conversion of sound intensity measured in watts per meter squared to decibels. Participants explore the mathematical relationships involved, the definitions of sound intensity and sound pressure levels, and the implications of these conversions in the context of acoustics.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant initially inquires about the conversion formula, expressing confusion about the logarithmic relationship.
  • Another participant provides a formula for converting sound intensity to decibels: [decibels] = 10log[S] + 120, where S is sound intensity in watts per meter squared.
  • Some participants question the formula, suggesting that decibels are typically calculated using 20Log for gain, particularly when dealing with pressure levels, and emphasize that the argument of the logarithm must be unitless.
  • It is noted that the power density must be divided by a reference power density (j0), commonly taken as 10^-12 W/m^2, to ensure a dimensionless argument for the logarithm.
  • Participants clarify that sound intensity level (SIL) and sound pressure level (SPL) have different formulas due to their relationship with pressure and intensity, with SPL involving a factor of 20 instead of 10.
  • One participant raises a question about how sound intensity falls off with distance, suggesting an inverse-square relationship, while another confirms that intensity decreases as the square of the distance from the source increases.
  • There is a discussion about the implications of energy conservation in relation to sound intensity and pressure, with some participants asserting that sound intensity is proportional to r^{-2} and sound pressure to r^{-1}.

Areas of Agreement / Disagreement

Participants express differing views on the correct formulas for decibels in relation to sound intensity and pressure, indicating that there is no consensus on the best approach to the conversion. The discussion remains unresolved regarding the nuances of these relationships.

Contextual Notes

Some participants highlight the importance of using a reference level for sound intensity and the distinction between sound intensity and sound pressure, which may not be fully resolved in the discussion.

Ed Aboud
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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.
 
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Its ok I figured it out.
[decibels] = 10log + 120

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

I would just look at your equation again.
 
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:
Sound intensity level is defined as:
L_J = 10 \log_{10}\left( \frac{J}{J_0} \right) \text{ dB}
where J is the sound intensity and J_0 is the reference intensity level.

In the case that J_0 = 10^{-12} \, \frac{ \text{W}}{\text{m}^2}, 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:
L_p = 10 \log_{10} \left( \frac{p^2}{p_0^2} \right) = 20 \log_{10} \left( \frac{p}{p_0} \right) \text{ dB}
where p is the sound pressure and p_0 is the reference sound pressure.

Again, in the case that p_0 = 2\times 10^{-5} \text{ Pa} (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.
 
Jwink3101 said:
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:

Ed Aboud said:
[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
 
Last edited:
So, when measuring sound intensity in decibels using the standard for J_0, by what power does sound fall off? My intuition says it should be r^{-2} because it spreads radially outwards but then there is air pressure and stuff like that in there.
 
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.
 
Sound intensity is proportional to r^{-2} while sound pressure is propertional to r^{-1}.
 

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