- #1

Ed Aboud

- 201

- 0

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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- Thread starter Ed Aboud
- Start date

- #1

Ed Aboud

- 201

- 0

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

Ed Aboud

- 201

- 0

[decibels] = 10log

where S = sound intensity in watts per metre squared

- #3

Jwink3101

- 20

- 0

[tex]

=\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

Lojzek

- 249

- 1

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).

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:

- #5

Nick89

- 555

- 0

[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

- 2,227

- 9

I may be comeplete wrong but aren't decibels 20Log(gain)?

when

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/S

where S

Last edited:

- #7

Jwink3101

- 20

- 0

- #8

rbj

- 2,227

- 9

- #9

Nick89

- 555

- 0

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