Adding dB can be a challenge because dB, particularly when dealing with the world of acoustics, is based on the formula dB = 10 log_{10} (R), where (R) is a ratio between two powers or two pressures.

It is important to note that dB IL (intensity level) and dB SPL (sound pressure level) are equivalent, i.e., 20 dB IL ~ 20 dB SPL. It is only the reference values and units that differ. The reference for dB SPL is: .00002 Pascals (20 µPa) or .00002 Newtons/M^{2 } (20µ Nt/m^{2}); whereas the reference for sound intensity or power is 1 x 10^{-12} Watts/m^{2} or 1 x 10^{-16} Watts/cm^{2}.

The following two formulae help us solve for dB:

dB SPL = 20 log_{10 }(Po/Pr) where Po is the pressure being measured and Pr is the reference pressure.

dB IL is 10 log_{10} (Io/Ir) where Io is the power or intensity measured and Ir is the power/intensity reference.

Let’s say that I have two sources; one is 10 dB and the other is 15 dB and I want to know how many dB will result if the two sources are combined. Because the dB is not linear we know that adding of these two values will not result in a total of 25 dB, but rather in some lesser value.

The easy approach is to use a table or chart. Why go the easy route when we can figure it out for ourselves? It is important to understand the reasoning and the mechanism (procedure) behind the resultant dB value.

We have to break down the two sound sources into their basic components. By this I mean that we need to work backwards from what we know.

We know that 15 dB = 10 log_{10} (an unknown ratio we will call R); so how do we find out what (R) is? Well, I’m glad you asked J

We have to follow this formula:

dB = 10 log_{10} (R) can also be written (R) = antilog_{10 }* (dB/10)

Note that we are dividing both sides of the equation by 10 in order to solve for (R). Before we go any further, let’s define *antilog*.

According to Wikipedia (http://en.wikipedia.org/wiki/Antilog#Antilogarithms ) “the **antilogarithm** function antilog_{b}(*y*) is the inverse function of the logarithm function log_{b}(*x*).”

The formula is: Antilog of *x* is = 10^{x}

Example:

We know that the log_{10} of 100 is 2 which can also be written 10^{2} or 10 x 10 all of which equal 100. The exponent of 100 is 2; therefore a log and an exponent are the same thing; the antilog of 2 (also known as *x *in the formula above) is 10^{2 }or 100.

Now we are getting somewhere. So, R is some unknown ratio of two powers or pressures. We need to know the R for 10 dB and the R for 15 dB. Using our handy dandy formula we are going to solve for both ratios.

(R) = antilog_{10 } (dB/10)

(R) = antilog_{10} (1.5)

(R) = 31.62

To prove this let’s plug in our familiar formula for calculating dB

dB = 10 log_{10} (31.62)

dB = 10 * 1.49 (rounded off is 1.5)

dB = 15

So all we need to do now is figure out the ratio that results in 10 dB.

(R) = antilog (10dB/10)

(R) = antilog of 1

(R) = 10 (you already knew this because the log of 10 is 1!)

All we have left to do is add the two ratios to come up with a new ratio and plug it into our trusty dB formula and voila you have zee new value which is zee sum of zee original two sources.

dB = 10 log_{10} (R)

dB = 10 log_{10} (10+31.62)

dB = 10 log_{10 }(41.62)

dB = 10 x 1.62

dB = 16.2

Conclusion: When you add 15 dB and 10 dB together you get 16.2 dB.

Wait a minute here—the formula for pressure is dB SPL = 20 log_{10} (R), but I said earlier that 20 dB IL ~ 20 dB SPL. So how can this approach work?

Let’s plug in some numbers and see what happens.

dB SPL = 20 log_{10} (R) and can also be written (R) = antilog_{10} (15 dB/20), therefore:

(R) = antilog_{10} 0.75

(R) = 5.62

dB SPL = 20 log_{10} (5.62)

dB SPL = 20 * 0.749

dB SPL = 14.98 (rounded to the nearest decibel is 15!)

So whether you are using dB IL or dB SPL you still come up with the same answer. Remember we are dealing with ratios and that intensity is proportional to pressure squared and pressured squared is equal to the square root of intensity.

(I ∝ P^{2}) or (P^{2}∝√I)