Category Archives: Acoustics

Nonlinear Response and Harmonics

Because we are so often solving problems in quantum mechanics, it is sometimes easy to forget that certain effects also show up in classical physics and are not “mysterious quantum phenomena”. One of these is the problem of avoided crossings or level repulsion, which can be much more easily understood in the classical realm. I would argue that the Fano resonance also represents a case where a classical model is more helpful in grasping the main idea. Perhaps not too surprisingly, a variant of the classical harmonic oscillator problem is used to illustrate the respective concepts in both cases.

There is also another cute example that illustrates why overtones of the natural harmonic frequency components result when subject to slightly nonlinear oscillations. The solution to this problem therefore shows why harmonic distortions often affect speakers; sometimes speakers emit frequencies not present in the original electrical signal. Furthermore, it shows why second harmonic generation can result when intense light is incident on a material.

First, imagine a perfectly harmonic oscillator with a potential of the form V(x) = \frac{1}{2} kx^2. We know that such an oscillator, if displaced from its original position, will result in oscillations at the natural frequency of the oscillator \omega_o = \sqrt{k/m} with the position varying as x(t) = A \textrm{cos}(\omega_o t + \phi). The potential and the position of the oscillator as a function of time are shown below:

harmpotentialrepsonse

(Left) Harmonic potential as a function of position. (Right) Variation of the position of the oscillator with time

Now imagine that in addition to the harmonic part of the potential, we also have a small additional component such that V(x) = \frac{1}{2} kx^2 + \frac{1}{3}\epsilon x^3, so that the potential now looks like so:

nonlinearharm

The equation of motion is now nonlinear:

\ddot{x} = -c_1x - c_2x^2

where c_1 and c_2 are constants. It is easy to see that if the amplitude of oscillations is small enough, there will be very little difference between this case and the case of the perfectly harmonic potential.

However, if the amplitude of the oscillations gets a little larger, there will clearly be deviations from the pure sinusoid. So then what does the position of the oscillator look like as a function of time? Perhaps not too surprisingly, considering the title, is that not only are there oscillations at \omega_0, but there is also an introduction of a harmonic component with 2\omega_o.

While the differential equation can’t be solved exactly without resorting to numerical methods, that the harmonic component is introduced can be seen within the framework of perturbation theory. In this context, all we need to do is plug the solution to the simple harmonic oscillator, x(t) = A\textrm{cos}(\omega_0t +\phi) into the nonlinear equation above. If we do this, the last term becomes:

-c_2A^2\textrm{cos}^2(\omega_0t+\phi) = -c_2 \frac{A^2}{2}(1 + \textrm{cos}(2\omega_0t+2\phi)),

showing that we get oscillatory components at twice the natural frequency. Although this explanation is a little crude — one can already start to see why nonlinearity often leads to higher frequency harmonics.

With respect to optical second harmonic generation, there is also one important ingredient that should not be overlooked in this simplified model. This is the fact that frequency doubling is possible only when there is an x^3 component in the potential. This means that the potential needs to be inversion asymmetric. Indeed, second harmonic generation is only possible in inversion asymmetric materials (which is why ferroelectric materials are often used to produce second harmonic optical signals).

Because of its conceptual simplicity, it is often helpful to think about physical problems in terms of the classical harmonic oscillator. It would be interesting to count how many Nobel Prizes have been given out for problems that have been reduced to some variant of the harmonic oscillator!

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Loudspeaker Crossover

The goal of an audio speaker is to reproduce the sound of the input as best as possible. This may sound like a simple statement, but it is a notoriously difficult one to engineer. Furthermore, sound to peoples’ ears is subjective, which only serves to complicate this task.

Today, with currently available technology, it is not possible to reproduce a flat frequency response across the entire spectrum of sound within the human-audible regime (i.e. ~20Hz to ~20kHz) with a single driving cone. This is why a decent set of speakers usually comprise two or more components, for example, a sub-woofer and a tweeter. The sub-woofer plays low-frequency sound, while the tweeter plays the high-frequency sound, which you would probably guess just from the onomatopoeic value of their names. Here is an example of a sound pressure level vs. frequency graph for this sub-woofer:

eminence_subwoofer

Black Line: Sound Pressure Level in dB vs. Frequency

You can see that the sound level dies off dramatically at higher frequencies, which is why we need the tweeter. For those that are curious, the plots like the one above are usually obtained by hanging a microphone in front of the speaker cone in an anechoic chamber and sweeping through the frequencies with a frequency generator. Due to the directional nature of sound from a speaker, the microphone is hung directly in front of the cone at a distance of 1 meter.

Because sound has to be routed through two separate cones, the speaker manufacturer has to make an electronics decision about how to do this. The electronics components that are used to do this are usually referred to as the “crossover circuit”. This is really just a fancy name for a high- and low-pass filter (or band-pass filter if one is also using a mid-range speaker). Crossover circuits tend to come in two varieties: active and passive. Active crossovers usually use operational amplifiers (or op-amps) to make the filters (which require the use of external power), while passive crossovers use inductors and capacitors (and do not need to be plugged into the wall).

For the sake of simplicity, let’s suppose that we have both 8 Ohm tweeters and sub-woofers for our speaker system. Suppose we want our cross-over frequency to be 500Hz. We can make the world’s simplest high-pass filter like so for the tweeter:

High-PassFilter

High-Pass Filter

So what would the capacitance need to be? Well, simply enough, it’s just the cut-off frequency of an RC circuit. In this case, for the 500 Hz crossover, we can calculate it as so:

f_c = \frac{1}{2 \pi R*C} = \frac{1}{2 \pi *8*C}

Solving for C with f_c = 500 Hz gives 39.79\mu F.

For the sub-woofer’s low-pass filter, we would replace the capacitor with the inductor and the equation would be f_c = \frac{L}{2 \pi R}. I should mention that these simple circuits also give one a 6dB/octave rolloff. This point is explained well on Wikipedia, so I don’t think I need to repeat it here.

The stunning thing about this simple passive crossover is that manufacturers can make a set of pretty high-end speakers with just these circuits. Of course, one would have to use some reasonable capacitors and inductors — but it can be that simple.

Manufacturers have many other design considerations besides the electronics to make as well (such as the enclosure!), but I hope this helps you understand the basics of what’s inside your speaker box.

Sounding Out Krakatoa

I recently watched an interesting documentary on Krakatoa, which is what inspired this post.

The 1883 eruption of Krakatoa, a volcanic Indonesian island, was one of the largest in recorded history, killing between 30,000 – 100,000 people. Wikipedia gives a good overview of its remarkable destructive power. The sound that emanated from the eruption was perhaps the loudest in recorded history — reports suggest that sailors ruptured their eardrums, and subsequently went deaf, up to 80 miles away from the island. The eruption was heard across huge distances, from Sri Lanka to Australia (see pg 80-87). People on Rodrigues Island, close to Madagascar, were reported to have heard the eruption from across the Indian Ocean. Rodrigues Island is about 5,700 km or 3,800 miles from Krakatoa. Here is a Google map showing their separation (click to enlarge):

krakatoa_map

Furthermore, inaudible (to the human ear) acoustic signals were said to have circled the earth up to seven times, and were detected using infrasonic detection.

In this post, I intend to make a couple calculations to discuss the following:

  1. The approximate sound level at Rodrigues Island
  2. The maximum distance at which the volcano was probably heard
  3. The possibility of acoustic circumnavigation of the world

There are three facts that are needed to discuss the above points:

  1. Reports suggest that the sound level of the eruption was approximately a whopping 175 dB at 100 miles from the volcano.
  2. Sound intensity falls as 1/r^2.
  3. Acoustic damping in air is generally lower for lower frequencies. (A great little applet where one can calculate the sound absorption coefficient of air can be found here.)

In addition, we can make some decent estimates by using the sound level formula:

SL (dB) = 10*Log_{10}(I/I_0) - \alpha*r,

where I_0 is 10^{-12} W/m^2 is the threshold of human hearing, \alpha is the coefficient of sound absorption in dB/m and r is the distance in m.

Here is a plot for the sound level in dB as a function of distance from the eruption site with Rodrigues Island marked by the dots (click to enlarge):

KrakatoaLogLog

There are immediately a couple things to note:

  1. I have plotted three curves: the blue curve is a calculation that does not consider any damping from air at all, the yellow curve considers a low but audible frequency sound that includes damping, while the green curve considers an infrasonic sound wave that also includes damping. (The damping coefficient was obtained from this link assuming a temperature of 20C, pressure of 1atm and humidity of 75%).
  2. Close to the volcano, this calculation gives us an unrealistically large value for the sound level. It turns out that sound cannot exceed ~194dB because this is the sound level at which rarefaction of air corresponds to a vacuum. Values greater than this correspond to a shock wave.

Keeping these things in mind, the yellow curve probably is the best estimate of the sound level on Rodrigues Island, since it includes damping for an audible signal (humans can’t hear below about 20 Hz). Therefore, we can estimate that the eruption was heard with about 70 dB on Rodrigues Island! This is approximately the sound level of a noisy restaurant.

If we follow the yellow line to about the 40 dB mark, which is an approximate value where someone may still notice the sound, this would be at a distance of about 4,800 miles! This is approximately the distance from Cape Town, South Africa to Baghdad, Iraq.

The last point to address is the seven-time infrasonic global circumnavigation. It turns out that if one follows the green line on the plot out to where it reaches the 0 dB level, this would be approximately at a distance of about 11 million meters. The earth’s circumference is approximately 40 millions meters, however, and if we were to circumnavigate the world seven times, the required distance of travel would be 280 million meters. What went wrong in the calculation?

There is one major factor to consider. Very low frequency sound basically propagates with very little damping through air. For sub-Hz infrasonic sound, values for the absorption coefficient don’t seem to be very easy to find! (If you know of a database for these, please share and I’ll update this post). Let us then consider the case of no damping (the blue curve). The blue curve actually crosses the 0 dB mark at a distance of approximately 85 trillion meters. This way over-steps the mark (corresponding to circumnavigation 2.125 million times!). Even though this is a ridiculous estimate, at least it beat the 280 million meter mark, which suggests that with the right absorption coefficient, we may be in the right ballpark. A quick calculation shows that for realistic values of the absorption coefficient (about half the value of the 5Hz sound absorption coefficient), we would be very close to the 280 million meter mark (in fact, I get about 225 million meters for this absorption coefficient). This tells us that it is indeed possible for low frequency sound to circumnavigate the planet this way!

Interestingly, we can learn quite a bit concerning the sound propagation of the Krakatoa eruption using relatively simple physics.

Note: Throughout, we have neglected one very important effect — that of reflection. Anyone who has been inside an anechoic chamber will be acutely aware of the effects of sound reflection. (In an anechoic chamber, one can actually stand at different spots and hear the interference pattern when playing a sine wave from a pair of speakers). Even with this oversight, it seems like we have been able to capture the essential points, though reflection probably had a non-trivial effect on the acoustic propagation as well.