Click here to purchase the entire book in PDF format. Power and Bels Sound in the air is a change in pressure. The greater the change, the louder the sound. The softest sound you can hear according to the books is $20 * 10^{-6}$ (or 0.00002) Pascals (abbreviated ``Pa'') (it doesn't really matter how big a Pa is - you just need to know the number for now...)3.1. The loudest sound you can tolerate without screaming in pain is about $200 000 000*10^{-6}$ Pa (or 200 Pa). This ratio of the loudest sound to the softest sound is therefore a 10,000,000:1 ratio (the loudest sound is 10,000,000 times louder than the softest). This range is simply too big to put on the fader of a mixing console. So a group of people at Bell Labs decided to represent the same scale with smaller numbers. They arrived at a unit of measurement called the Bel (named after Alexander Graham Bell - hence the capital B.) The Bel is a measurement of power difference. It's really just the logarithm of the ratio of two powers (Power1:Power2 or Power1/Power2). So to find out the difference in two power measurements measured in Bels (B). We use the following equation. $$\Delta \textrm{Power (in Bels)} = \log_{10} \left ( \frac{\textrm{Power 1}}{\textrm{Power 2}} \right )$$ (3.14) Let's leave the subject for a minute and talk about measurements. Our basic unit of length is the metre (m). If I were to talk about the distance between the wall and me, I would measure that distance in metres. If I were to talk about the distance between Vancouver and me, I would not use metres, I would use kilometres. Why? Because if I were to measure the distance between Newfoundland and my house in Denmark in metres the number would be something like 4,176,120 m. This number is too big, so I say I'll measure it in kilometres. I know that 1 km = 1000 m therefore the distance between Newfoundland and me is 4,176,120 m / 1 000 m/km = 4,176 km. The same would apply if I were measuring the length of a pencil. I would not use metres because the number would be something like 0.15 m. It's easier to think in centimetres or millimetres for small distances - all we're really doing is making the number look nicer. The same applies to Bels. It turns out that if we use the above equation, we'll start getting small numbers. Too small for comfort; so instead of using Bels, we use decibels or dB. Now all we have to do is convert. There are 10 dB in a Bel, so if we know the number of Bels, the number of decibels is just 10 times that. So: $$1 \textrm{dB} = \frac{1 \textrm{ Bel}}{10}$$ (3.15) $$\Delta \textrm{Power in dB} = 10 \log_{10} \left ( \frac{\textrm{Power 1}}{\textrm{Power 2}} \right )$$ (3.16) So that's how you calculate dB when you have two different amounts of power and you want to find the difference between them. The point that I'm trying to overemphasize thus far is that we are dealing with power measurements. We know that power is measured in watts (or Joules per second if you're reading an older book) so we use the above equation only when the ratio is comparing two measurements in watts. What if we wanted to calculate the difference between two voltages (or electrical pressures)? Well, Watt's Law says that: $$\textrm{Power} = \frac{\textrm{Voltage}^{2}}{\textrm{Resistance}}$$ (3.17) or $$P = \frac{V^{2}}{R}$$ (3.18) Therefore, if we know our two voltages (V1 and V2) and we know the resistance stays the same: $$\Delta \textrm{Power (in dB)} = 10 \log \left ( \frac{\textrm{Power 1}}{\textrm{Power 2}} \right )$$ (3.19) $$= 10 \log \left ( \frac{\frac{V1^{2}}{R} }{ \frac{V2^{2}}{R}} \right )$$ (3.20) $$= 10 \log \left ( \frac{V1^{2}}{R} * \frac{R}{V2^{2}} \right )$$ (3.21) $$= 10 \log \left ( \frac{V1^{2} }{ V2^{2}} \right )$$ (3.22) $$= 10 \log \left ( \frac{V1 }{ V2} \right )^{2}$$ (3.23) $$= 2 * 10 \log \left ( \frac{V1 }{ V2} \right )$$ (3.24) $$\textrm{(because} \log A^{B} = B * \log A\textrm{)}$$ (3.25) $$= 20 \log \left ( \frac{V1 }{ V2} \right )$$ (3.26) That's it! (Finally!) So, the moral of the story is, if you want to compare two voltages and express the difference in dB, you have to go through that last equation. Remember, voltage is analogous to pressure. So if you want to compare two pressures (like $20 * 10^{-6}$ Pa and $200 000 000*10^{-6}$ Pa) you have to use the same equation, just substitute V1 and V2 with P1 and P2 like this: $$\Delta \textrm{Power (in dB)} = 2 * 10 \log \left ( \frac{\textrm{Pressure 1}}{\textrm{Pressure 2}} \right )$$ (3.27) This is all well and good if you have two measurements (of power, voltage or pressure) to compare with each other, but what about all those books that say something like ``a jet at takeoff is 140 dB loud.'' What does that mean? Well, what it really means is ``the sound a jet makes when it's taking off is 140 dB louder than...'' Doesn't make a great deal of sense... Louder than what? The first measurement was the sound pressure of the jet taking off, but what was the second measurement with which it's compared? This is where we get into variations on the dB. There are a number of different types of dB which have references (second measurements) already supplied for you. We'll do them one by one.