What is the relation between loudness and decibels? Is 80 dB twice as loud as 40 dB? How do you translate from decibels to loudness?
Sound level in dB is a physical quantity and may be measured objectively. Loudness is a perceived quantity and one can only obtain measurements of it by asking people questions about loudness or relative loudness. (Of course different people will give at least slightly different answers.) Relating the two is called psychophysics. Psychophysics experiments show that subjects report a doubling of loudness for each increase in sound level of approximately 10 dB, all else equal. So, roughly speaking, 50 dB is twice as loud as 40 dB, 60 dB is twice as loud as 50 dB, etc.
Does adding two identical sounds give an increase of 3 dB or 6dB? Why?
A subtle question: if you look at the page What is a decibel?, you'll see that doubling the sound pressure gives an increase of four in the intensity, so an increase in the sound level of 6 dB, whereas doubling the power increases the intensity by a factor of two, so an increase of 3 dB. There is no paradox here, because to double pressure and to double power you do two different things. Let's see how, starting with a simplified system.
Let's imagine that we have a completely linear amplifier and speaker. If we double the voltage at the input (like adding two identical voltage signals), the we double the voltage to our linear loudspeaker and thus double the sound pressure it puts out, which gives us four times as much energy, four times as much intensity everywhere in the sound field, and thus a uniform increase in sound level of 6 dB. (The amplifier is also delivering four times as much electrical power.)
Amplitude as a function of x and y for a Young's Experiment Now imagine that we have two identical, linear amplifiers and speakers, and we'll forget about reflections from walls, floor etc. We put the same signal into the inputs of both. Now each amplifier is producing the same amount of electrical power, so the output sound power is only doubled. On average, the intensity in the sound field is only doubled. Let's consider just a single sine wave (or, if you like, one frequency component of the sound). If your ear is equidistant from the two speakers, then the two pressure waves will add in phase (constructive interference), so the sound pressure at your ear will be twice as great, so you'll a level 6 dB higher when the second amplifier is turned on. However, if you are half a wavelength further from one speaker than from the other, then two two pressure waves will add one half cycle out of phase (destructive interference) and the sound pressure will be zero. So the sound intensity is four times as great in some places, zero in others, and intermediate values in most places. Integrate this over all space and the total power is only doubled.
What if the sounds are not identical, but have the same level? Then the phase difference will be random. If two signals have pressure amplitude pm and a phase difference θ, then their sum is 2pm(1+cosθ), which of course varies from 0 to 4pm. But the power is proportional to the amplitude squared, and so proportional to 4pm2(1+cosθ)2. The average of (1+cosθ)2 over all θ is ½ so the average of the whole term over all possible phase differences is 2pm2. So again, on average we double the power and so get an increase of 3 dB in sound level.
