Sunday, September 23, 2012

How loud is that whale, anyway?

When we're thinking about how loud a noise needs to be to damage a whale's hearing, it's important to consider how loud the sounds are that the whale makes itself. If a singing whale makes a sound itself that is at a certain sound level, it is pretty unlikely that it will be deafened by being exposed to a sound of the same level.  This is something like how you are unlikely to be deafened by the sound of your own voice. This isn't to say that continual exposure to sounds at or quieter than their own won't affect whales. It will - just like constant low-noise sounds such as traffic can damage human health.

Calibrating hydrophone sensitivity
curves is very pleasurable.
The first step in figuring out how loud your whale is to drop a hydrophone in the water. Hydrophones convert pressure into voltage (if you're interested in how, I explain here). Hydrophones also have a property called sensitivity, which is a measurement of how quiet of a sound that the hydrophone can hear. The difference in sound sensitivity between a very sensitive hydrophone and a very insensitive hydrophone is like the difference in touch sensitivity between your tounge and your heel. A light touch on the heel might not make any neurons fire, but...

As an aside, in the book "Last Chance to See," Douglas Adams puts a microphone in a condom and uses it to try and listen to Yangtzee river dolphins in China. (Insert nerdy bio-acoustition joke here).

Hydrophone sensitivity is usually listed as a single number, something you will notice if you are a buyer of hydrophones or microphones. This number is negative, and the less negative, the more sensitive the hydrophone. Listing a single number actually oversimplifies things. For example, let's look at the sensitivity of the Reson TC 4024 hydrophone at a range of frequencies between 0 and about 90 kHz. If you remember, frequency determines how shrill a sound is. Barry White, for example, has a low frequency voice, and the Chipmunks squeak their songs at high frequencies. OK, so the "typical receiving sensitivity of this hydrophone" is listed as -173 dB. But, as you can see, the sensitivity actually varies quite a lot with frequency:



Because of this, most people only use their hydrophones to record sounds in the "flat frequency response" region, which, for this hydrophone, is between about 4 Hz and 45 kHz. This flat frequency region is pretty much about equal to -173 dB (and if you want to be really specific about things you can always go back and correct for the variation later - as long as your hydrophone is calibrated!)

You'll probably also need to know your gain. Gain means amplification. Generally, the signals coming out of the hydrophone are still too quiet to use, so we amplify them again.

Finally, we can calculate how loud the whale is. So here we go:

Sound Pressure Level = Sensitivity - Gain + 20 * log (Voltage from the Hydrophone)

It's actually pretty simple once you know what all the numbers mean...

... except for the fact that, by convention, the Sound Pressure Level is measured at 1 m from the sound source. So, if we are working with wild animals, we will either have to get incredibly lucky, or we have to figure out a way to deal with attenuation of the sound in water. Attenuation is a fancy word that means "makes quieter." For example, sound-proof walls attenuate the sound of your neighbor's Eminem music.  I'll deal with that one a little bit later.

You almost never get this lucky. Photo by Flip Nicklin.

Tuesday, September 18, 2012

Why Can't We Directly Compare Sounds in Water and Air?

One of the biggest, and most common, mistakes that people make when writing about underwater acoustics is making direct comparisons between the loudness of things in water and in air.  Here's an example of something that you might read about underwater sound:

"The bottlenose dolphins were being exposed to sounds up to 160 dB.  That's  louder than a 747!"


Well, actually, it's not.   

The problem here is related to the units with which we measure sound. First of all, we measure sound using the decibel.  Decibels are logarithmic units. Basically, logarithmic scales allow us to make big numbers smaller so that they are easier to visualize. For example, in the chart below, you can see that when we take the log of value for x (in green), the log values (in blue) are decreased more for large values of x than for small values of x. The opposite is true of taking an exponential of x, such as x^2 (in red).


Why make things so complicated when we talk about sound? Well, human beings actually perceive sound pressure in a log scale.  For example, you wouldn't really be able to tell the difference in loudness between 6 trumpet players and 7 trumpet players, because your brain doesn't process sound lineally. In fact, every time you want to change the loudness level by just three measly decibels, you have to double the power.  So, if you want to increase the loudness of your trumpet players by 3 dB, you'll need 6 more trumpets.  

Doublemint gum only makes you 3 dB louder, just ask the dolphins.

OK, so now we understand decibels, why can't we compare dB in air with dB in water?

First of all, decibels must always be referenced to some standard. In air, decibels are referenced to 20 micropascals (pascals are a unit of pressure). In water, they are referenced to 1 micropascal.  This means that, from the start, 26 dB must be added to the sound level to make the two references levels the same.

Secondly, water is much denser than air, so the effects of pressure are also different. Although it is possible to compress water, it is much more difficult to do so.  In fact, when we add the difference in densities between air and water into the equation, we find that the sound pressure in water is 60 times greater (or 35.6 dB higher) than the same value in air.

Loud, but not as loud as a plane. Photo from the CRRU.
So, if white-beaked dolphins are producing whistles that are between 139 and 148 dB, that doesn't mean that they are making sounds louder than a 747. If we convert those numbers to values in air, we get somewhere between 103 and 113 dB.  Which is still pretty darn loud, but not quite as loud an airplane engine.  It's also good to point out that dolphin whistles are "narrowband" sounds. I explained narrowband in an earlier blog post, but basically what this means is that you can't imagine a whistling dolphin sounding like a jackhammer, because the sounds cover a very different range of frequencies and are produced by very different mechanisms.

So, next time you read something comparing underwater sound compared to sound in air, check to see if they did their conversions right!

Monday, September 17, 2012

A wave is a complicated thing to equate

I am only studying chapter 1.2.1 of my big scary bioacoustics book, and already I'm running into some scary maths (tm). However, my advisor has told me that I don't have to actually know how to derive the equations, just understand them conceptually (Apparently I am getting my PhD in Zoology and not Electrical Engineering - could have fooled me!).

I use sound to study whales in the ocean. Understanding sound is absolutely essential to my research, and to understand sound, you've got to understand the wave equation.

We're not going to actually derive the wave equation, but it's important to know what goes into it. (if you want to see the real math, look here). In order to understand waves, you need to understand four other fundamental laws. The great thing about these laws is that they pretty much are telling you that waves obey the laws of physics. By the combined powers of these laws, Captain Wave equation emerges!

Conservation of Mass: This means that even though the number of molecules in any part of the volume that  sound passes through may change, the total number of molecules in the volume stays the same.

In the sound wave, molecules get more
compressed, but don't appear out of thin air. 

Equation of Motion (Newton's Second Law): This means that we can calculate the force acting on particles by multiplying their density when a sound is passing through them by their acceleration.

Force = Acoustic Density * Acceleration

Equation of Force: With this, we can use the total density of the fluid (which is different from the acoustic density) and divide it by the x movement of the particles to find the force acting on the particles.

Force = Total Density differential * x differential
Equation of State: This tells us that acoustic pressure is related to how easily the medium through which a sound travels is compressed. Some people think that liquids and solids can't be compressed, but this isn't true. The reason we generally consider liquids to be incompressible is that it is REALLY REALLY hard to compress a liquid. I couldn't do it, even if I worked out a LOT.  However, liquids at the bottom of the ocean have the weight of tons of water sitting on top of them.  The weight of all this water presses down on the liquid, pushing the molecules closer together. In fact, if you have enough pressure, you can even compress rocks. Here's my version of this equation:
Pressure = (squishiness / acoustic pressure) * pressure

When you put these four equations together, you get the wave equation, which uses all four to describe the movement of a wave. This equation factors in pressure, time, the speed of sound, and the movement of particles.

Why is this important to a whale researcher?  Well, if I want to use sound to study animals underwater, I need to understand that sound travels differently in different types of water. In the ocean, things that might change how sound travels are depth, salinity, and temperature. If, for example, a beaked whale is clicking at 5000 m, and I am recording it at the surface, I can't assume that the sound is traveling in a straight line from the whale to me. As the depth and temperature changes, the way sound moves through it will change. I need to be mindful that these things can change how sound moves through the water.

I warned you this was going to be dense. And that was only chapter 1.2.1.

Also, if I made a math mistake, and I messed up on understanding any of the equations, please comment!