The dB Cheat Sheet

Wouldn't it be great if you could work all decibel problems in your head and never be far from the correct answer! I mean isn't the whole point of having the dB to make problems easier? So why not cheat a little? OK - so we lose a little accuracy - maybe you'd rather carry a TI Calculator around in your belt strap. Here's the plan... First, we note that in any dB problem only one of two things change - sound power (P) or the sound power reference (Pref). One of these two things gets multiplied by some factor and we estimate the resulting change in dB,

, or, .

This is what we did in class when we converted from one reference value to another, and when we computed dB changes due to the inverse-square law. OK - according to the first law of logarithms (earlier handout) we can simplify this expression,

Here the sign means we add if factor appeared in the numerator and we subtract if it appeared in the denominator. Hey - look at the first term on the right of the equals sign. This is just the dB value we started out with before any changes. In other words,

Now here's a formula we can all live with. It says that to get the answer we want, all we have to do is compute10log(factor) and add to (or subtract from) the original dB value. Simple right?! Well, this concludes the deeper appreciation portion of this handout. Now that you have a deeper appreciation you can use the table below and the expression above to do all the dB problems you will ever encounter in this class!


                         Some Factors and Their dB Values
                                      
                        2               1/2               10              1/10       
factor                                                                               
                      + 3 dB           - 3 dB          + 10 dB          - 10 dB      
10log(factor)*

^*Note: All dB entries are doubled if the reference is a pressure value (e.g. as in dB SPL).