Sound Pressure Levels

Inverse Logarithms

Before we get back to dealing with sound, we need to understand how to "undo" a logarithm. Technically, this means taking the inverse logarithm of a number. We'll do this first with symbols and then with numbers as examples.

Putting the equations into words, the logarithm of a positive number (x) gives us the exponent of 10 (y); 10 raised to the power y gives us x. It might be that some examples will help:

If we're given y = 2 and we're asked to find the inverse logarithm of y; in other words what is the number x whose logarithm gives us y = 2? Here, the answer is simple:
x = 10^y = 10² = 100
If we're y = 5 and we're asked to find the inverse logarithm of y; in other words what is the number x whose logarithm gives us y = 5? Again, the answer is simple:
x = 10^y = 10⁵ = 100,000
If we're y = 3.699 and we're asked to find the inverse logarithm of y; in other words what is the number x whose logarithm gives us y = 3.699? This one's a little bit trickier:
x = 10^y = 10^3.699 = 5,000
Another way of writing this is
x = 10^{0.699 + 3} = 10^0.699 x 10³ = 5 x 10³
If you're like me, you'll often make mistakes in either entering numbers or hitting a wrong key on your calculator. So it pays to have simple expectations for your answer. The number y = 3.699 is between 3 and 4, so we expect the inverse logarithm of y to be between 10³ = 1,000 and 10⁴ =10,000.

Generally, your scientific calculator will be able to give you the "inverse logarithm" by first entering the number and then hitting the INV (or 2ND) key followed by the LOG key. For some calculators, you'll need to hit the INV (or 2ND) key followed by the LOG key, followed by the number and then the EQUAL key.

Sound Pressure Levels

Our perception of the loudness of sound operates on a logarithmic scale. If we listen to a series of sounds having, what we perceive, a constant step in loudness the corresponding sound waves in fact have a multiplicative factor in their amplitudes. Therefore, it is more convenient to use a logarithmic scale to measure loudness. It's not quite as simple as taking log(p), where p is the amplitude of the sound wave, beacuse p has units of N/m² (the same as Pascals) and we can't take the logarithm of a quantity that has units. So instead, we compare p to a reference pressure amplitude, p₀ = 2 x 10^-5 N/m². The ratio, p / p₀ is dimensionless, meaning it has no units, and tells us by what factor we have to multiply the reference pressure amplitude by to get the amplitude of a given sound wave. For the average person, p₀ = 2 x 10^-5 N/m² corresponds to where we can just start to hear a sound.

The sound pressure level is measured in decibels (dB) and is

L_p = 20 log(p / p₀) = 10 log(p² / p₀²)

The last step, where we replace the original factor (20) multiplying the logarithm and at the same time square p / p₀ uses a property of logarithms we talked about previously. There are a couple of other things to note

The "deci" prefix in decibel (the "bel" part is for Alexander Graham Bell, the inventor of the telephone, whose real area of research was in helping people with hearing problems) is a prefix meaning 1/10 and is accounted for by multiplying the logarithm by 10.
L_p is defined in terms of p² / p₀² so as to correspond to the sound intensity level which we'll talk about on Wednesday. Suffice it to say that the energy (and intensity) of a sound wave is proportional to the square of the amplitude of the wave oscillation (p²).
L_p gives us a logarithmic scale for loudness. Let's see what this really means. Consider two sound waves (a and b) that have amplitudes differing by a factor of 10: p_a = 10 p_b. Sound a is certainly louder than sound b but by how much? We find out by using the definition of sound pressure levels:
L_p(a) = 20 log(p_a / p₀) for sound a
L_p(b) = 20 log(p_b / p₀) for sound b
but, we can replace p_a in the definition of L_p(a) with 10 p_b
--> L_p(a) = 20 log(10p_b / p₀)
Now, we can can make use of some properties of logarithms,
log(AB) = log(A) + log(B), to write
L_p(a) = 20 [ log(10) + log(p_b / p₀)], or
L_p(a) = 20 dB + 20 log(p_b / p₀), or
L_p(a) = 20 dB + L_p(b)
This last equation tells us that a sound pressure level increase of 20 dB corresponds to a factor of 10 increase in the amplitude of the sound waves. You should consult Rossing's table 6.1 for some sound pressure levels.
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Last updated: 17 Oct 1999Comments: bland@indiana.edu