Decibels made ultra-simple: Part 3
But what use are they?
Engineers are lazy people and don't do anything the hard way. We hate to multiply and divide. Adding and subtracting we can just about put up with.
Look again at the conversion from power ratios to decibels that we did before. (Since the power ratios are exact powers of 10, all we have to do to convert to decibels, is to count the zeroes and multiply by 10.)
|Power ratio||Level difference|
Starting with any amount of power, suppose we increase the power, first by 10 times, then by a further 100 times. The total increase would be 10 x 100 = 1000 times. That's what we get by multiplying the two numbers in the left hand column.
Now add the two numbers in the right hand column: 10 + 20 = 30. If we put 1000 in the left hand column and 30 dB in the right hand column, does it give the right answer?
|Power ratio||Level difference|
This is not coincidence! Adding the decibels always corresponds to multiplying the power ratios. So we can add instead of multiplying. And of course it works just as well backwards so we can subtract instead of dividing.
At last! A practical use
If you have only one calculation to make, that doesn't save a lot of multiplication, but very often we have lots. Think about a ventilation system, in which a fan feeds air to a number of rooms through a system of ducts. We often need to know now loud the fan noise will be in each room.
To do this calculation, we start with the noise the fan is putting into the first duct. Every item the air passes through will have its effect on the sound. Some sound leaks out through the duct walls, or is converted to vibrations in them, so we lose some sound depending on the length of the duct. Where the duct splits to feed two destinations, the sound will divide. Where the duct has sharp bends, some of the sound will be reflected backwards, and so less will travel onwards. In fact, each element will reduce the sound in a certain proportion, or ratio.
To calculate how much sound comes out at the end of each branch of the system, we have to multiply all of those proportions together, and then multiply by the fan power. (That's the sound power not the electrical input power! Manufacturers of fans used in the sort of complex systems that we're discussing here will provide this information.)
If the amount of sound lost in each element in the system is expressed in decibels, instead of multiplying the ratios, we simply add the numbers to find the total loss. Manufacturers will provide data for some elements in the system, and others can be looked up in standard tables. For example, to account for the sound that is lost in the duct, we can look up a figure of so many decibels per metre, depending on the type of duct.
If we have twelve metres of duct, and 0.7 dB per metre loss, then the total loss will be 8.4 dB - twelve lots of 0.7 dB. Even if your tables are a bit shaky, this is hardly a challenging calculation.
But without decibels? If we were told that the sound was 85% less for each metre of duct, we would multiply by 0.85 twelve times in succession, once for each metre of duct. OK you can do it easily enough with a spreadsheet but it won't be so immediately obvious if you make a mistake. And that's just one bit of a much larger calculation! In a nutshell, that's one of the major reasons why we use decibels.
Decibels for sound
But there is another reason why decibels have become the universal measurement for sound in particular. If there is a sound of 0.1 watt and we increase it by ten times to 1 watt, there will be a certain increase in loudness*. Now if we increase it again to 10 watts, this would seem to increase the loudness by about the same amount again. But, in the first case the power has increased by 0.9 watt, and in the second case it has increased by 9 watts. In decibels, however, it is the same increase (10 dB) in both cases. So decibels give a much better subjective idea of sounds, than simple measurements do.
*Pedants please go away! I know our ears don't respond to power but pressure (see next page).
Finally we might need to convert the other way. What is the power ratio that corresponds to, say, 20 dB?
The first thing in this sort of calculation is always to convert to bels, so divide by 10. This will make two bels.
Then we can simply write a one with two zeroes. So the power ratio is 100:1.
If the number of decibels doesn't end in 0, after converting to bels you'll need the calculator, to do the inverse of taking the log. Unfortunately different calculators have different names for it. The key you need could be labelled "antilog" - obvious enough. More likely you will get to it by pressing a "shift", "inv" or "2nd function" key, then "log". Or the key you need could well be labelled 10x (read as "ten to the x"). In the pop-up calculator, press "shift" to see the 10x key.
Is that it?
That is the basics, and I hope it has dispelled some of the mystery of decibels.
But if you want to see how we use what might be called a fiddle, to make them even more useful in practice - that's on the final decibel page.