# How sound intensity (dB) and sound pressure level (dB) are related?

Both sound intesity and pressure level are measured in dB. Given a specific sound, are these two dB values the same?

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The first thing to understand is that dB is a logarithm of a ratio. More specifically, a logarithm of a ratio of power/power (which is why dB has no units). A Bel is the base-10 logarithm of a power ratio. A decibel slices each Bel into ten parts.

Power is power. It does not matter whether it is sound or electricity or whatever. And dB always refers to a ratio of power.

In most sound references, "dB" means "dB in reference to (as a ratio over) 'X'." Where 'X' is a reference level. I think the reference used is the sound pressure which is the threshold of human hearing. So a sound with 10 times the POWER of that reference would be 10 dB. A sound with POWER 100 times the reference would be 20 dB.

The answer to your question is: It can be. Sound intensity (if expressed as a pressure) and Sound Pressure Level (SPL) are the same. But neither is power. Sound power is proportional to sound pressure squared. Thus, if you increase sound PRESSURE by a factor of 10, you have increased POWER by a factor of 10^2 = 100. These two actions are synonomous, and result in a 20dB increase. Note that I did not explicitly state what increased by 20dB in that last sentence... because it is ALWAYS power.

So, you need to watch your units, and make sure the other guy is watching his. If he doubles his POWER level, that is a 3 dB increase. If he doubles his sound PRESSURE, that is a 6dB increase. If he doubles his sound INTENSITY, you need to nail down whether that is in units of pressure or power.

I hope that helped.

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I think so, yes. They are both measures of the loudness of what we hear. If $L_p$ is the sound pressure level and $L_i$ is the sound intensity, then: $$L_p =20 \log_{10}\left(\frac{p_{\mathrm{rms}}}{p_{\mathrm{ref}}}\right)\mbox{ dB} = 10 \log_{10}\left(\frac{{p_{\mathrm{{rms}}}}^2}{{p_{\mathrm{ref}}}^2}\right)\mbox{ dB}$$ The intensity of the sound wave, however, is proportional to the square of the sound pressure, or $I \propto p_{\mathrm{rms}}^2$. This gives: $$L_p = 10 \log_{10}\left(\frac{{p_{\mathrm{{rms}}}}^2}{{p_{\mathrm{ref}}}^2}\right)\mbox{ dB} = 10 \log_{10}\left(\frac{{I_{\mathrm{{rms}}}}}{{I_{\mathrm{ref}}}}\right)\mbox{ dB} = L_i$$

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