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# 1- Acoustics fundamentals

## Sound waves

### Production of a sound wave

When a body (for us, the membrane of a loudspeaker) vibrates in a point of a medium (for us, the air), this vibration is transmitted to the other points, from near to near, by contact. It results in a propagation of the vibration in the medium, called wave.

On the left side of the figure, the loudspeaker is animated by a regular back and forth movement, of period $T$ and of frequency $f=1/T$ (number of periods per second). When it moves to the right, its membrane pushes the air creating an overpressure (it "squeezes" the air molecules). When it then moves to the left, it creates a vacuum (the vacuum left by the membrane "sucks" in the surrounding molecules).

Each point of the medium reproduces the vibration generated by the loudspeaker membrane as the wave passes, with a delay $\tau$ . This one depends on the distance $AB$ between the points and the speed of propagation $c$ of the wave in the medium:

$\tau = \frac{AB}{c}$

This results in the propagation of a sound wave that will set in motion another membrane, further away: the eardrum of the listener. The human ear is able to detect frequencies ranging from 20 Hz to 20 kHz. The speed at which the vibration propagates depends on the properties of the air (temperature, humidity, density), but it is about $c \approx 340 \, \text{m/s}$.

The curve represented above the figure gives the shape of the air pressure along the path of the wave (recall here that the international unit of pressure is the Pascal $\text{Pa}$. In this example, it corresponds to the simplest and most important case of wave physics: it is a sinusoidal wave.

### Wavelength

The pressure along the axis of propagation of a sine wave is periodic (it is described by a repeating pattern), the length of this pattern is called wavelength and labeled $\lambda$ :

The wavelength is the distance between two peaks on the curve. It is clear that it corresponds to the distance traveled by the disturbance during a period of the wave. Thus we have:

$\lambda = c \cdot T = \frac{c}{f}$

We can now determine the extreme wavelengths detectable in the air by the human ear:

• For the lowest sounds, $f=20 \hspace{0.1cm} \text{Hz}$ and: $\lambda_\text{grave} = \frac{340}{20}=17\,\text{m}$
• For the most acute sounds, $f=20 \, \text{kHz}$ and: $\lambda_\text{aigu} = \frac{340}{20\cdot 10^3}=17\cdot 10^{-3}\hspace{0.1cm}\text{m}=17\,\text{mm}$

### Wave equation

We will write $\rho_0$ and $P_0$ the density and pressure of the air at rest. In the usual conditions, we have $\rho_0 = 1,18 \,\text{kg}\cdot\text{m}^{-3}$ and $P_0 = 1,01 \times 10^5 \,\text{Pa}$.

We will stick on that course in the approximation of faibles signaux. We will thus consider that the variations of density and pressure are weak compared to these values:

$\left\{ \begin{array}{ll} \rho = \rho_0 + \rho\prime \\ P = P_0 + p \end{array} \right. \hspace{2cm} \text{avec} \hspace{2cm} \left\{ \begin{array}{ll} \rho\prime \ll \rho_0 \\ p \ll P_0 \end{array} \right.$

This simplification will be justified later, when we will see that the maximum overpressures we will have to deal with in acoustics are several thousand times lower than the atmospheric pressure.

#### Mass conservation

Let us consider an infinitesimal volume of air. The mass being a conserved quantity, the temporal variation of its density can only be explained by the mass flow crossing its surface. Keeping only the first order terms, we can write :

$$\displaystyle{ \boxed{\frac{\partial \rho\prime}{\partial t} + \rho_0 \vec{\nabla} \cdot \vec{u} = 0}}$$

$\vec{u}$ is the air velocity and $\vec{\nabla}\cdot \vec{u}$ its divergence. In Cartesian coordinates: $\vec{\nabla}\cdot \vec{u} = \frac{\partial u_x}{\partial x} + \frac{\partial u_y}{\partial y} + \frac{\partial u_z}{\partial z}$

#### Euler's equation

Let's consider a flat surface $A$, in red on the figure, vibrating in air of density $\rho$ autour de sa position d’origine d’abscisse nulle. Dans les conditions habituelles, la masse volumique de l’air est $\rho_0 = 1,18 \,\text{kg}\cdot \text{m}^{-3}$. We are interested in the vibrations of a slice included, at rest, between the abscissa $x$ and $x+\text{d}x$. Its mass is $\rho_0 \cdot A \cdot \text dx$.

Let us note the subsequent displacements of the two surfaces delimiting the air slice $\psi (x, t)$ and $\psi (x + \text{d}x, t)$. The thickness of the air slice is $\text dx$ at rest and $\text dx + \psi (x + \text{d}x, t) - \psi (x, t)$ at time $t$.

The forces acting on the air slice are the pressure forces exerted on its two sides, so Newton's second law yields :

$$\displaystyle{\rho_0 \cdot A \cdot \text dx \cdot \frac{\partial u}{\partial t} = A\cdot p(x) - A \cdot p(x+\text dx)}$$

where $u = \frac{\partial \psi}{\partial t}$ is the velocity of air at abscissa $x$.

We then obtain the -linearized- Euler equation:

$$\displaystyle{\rho_0 \cdot \frac{\partial u}{\partial t} = - \frac{p(x+\text dx) - p(x)}{\text dx} }$$

$$\displaystyle{\iff \rho_0 \cdot \frac{\partial u}{\partial t} = - \frac{\partial p}{\partial x}}$$

The 3-dimensional genaralization is immediate:

$$\displaystyle{\boxed{\rho_0 \cdot \frac{\partial \vec{u}}{\partial t} = - \vec{\nabla}p }}$$

where $\vec{\nabla}p$ represents the gradient of $p$. In Cartesian coordinates: $\vec{\nabla}p = \frac{\partial p}{\partial x}\vec{e}_x + \frac{\partial p}{\partial y}\vec{e}_y + \frac{\partial p}{\partial z}\vec{e}_z$

#### Propagation equation

The velocity of the gas appears in the conservation of mass equation and in the Euler equation. To get rid of it, let's take the time derivative of the former and the divergence of the latter:

$$\displaystyle{\frac{\partial^2 \rho\prime}{\partial t^2} + \rho_0 \vec{\nabla} \cdot \frac{\partial\vec{u}}{\partial t} = 0}$$

$$\displaystyle{ \rho_0 \cdot \vec{\nabla} \cdot \frac{\partial \vec{u}}{\partial t} + \vec{\nabla}\cdot\vec{\nabla}p = 0 }$$

That is:

$$\displaystyle{\frac{\partial^2 \rho\prime}{\partial t^2} + \Delta p = 0}$$

where $\Delta p$ denotes the Laplacian of p. In Cartesian coordinates : $\Delta p = \frac{\partial^2 p}{\partial x^2} + \frac{\partial^2 p}{\partial y^2} + \frac{\partial^2 p}{\partial z^2}$

We still have to relate pressure and density to one another. In the linear approximation, a Taylor expansion to the first order yields:

$\displaystyle{ P(\rho) = P(\rho_0) + (\rho - \rho_0) \left.\frac{\partial P}{\partial \rho} \right| _{\rho_0} }$

$\displaystyle{ \iff p = \rho\prime \left.\frac{\partial P}{\partial \rho} \right| _{\rho_0} }$

The derivative appearing in the right-hand term has dimension $M^2 T^{-2}$ : it is the square of a speed. We then pose $c^2 \equiv \frac{\partial P}{\partial \rho}$, and get $p = \rho\prime \cdot c^2$ , the equation verified by the pressure is written :

$$\displaystyle{ \boxed{ \Delta p + \frac{1}{c^2} \frac{\partial^2 p}{\partial t^2} = 0 }}$$

This propagation equation is the fundamental equation of linear acoustics.

#### Velocity of sound waves

The law of perfect gases can be written $P = \rho r T$ with $T$ the temperature in Kelvin and $r = \frac{R}{M} = \frac{8,32}{0,029} = 287 \,\text{J }\,\text{kg}^{-1} \,\text{K}^{-1}$ is the specific constant of air (R is the constant of perfect gases and M the molar mass of air)

Moreover, the transformations undergone by air in linear acoustics are adiabatic (they are too fast for a volume of air to exchange heat with its neighbor). Such transformations are described by Laplace's law:

$$\displaystyle{ \frac{P}{\rho^\gamma} = \text{cte} }$$

where $\gamma = 1,4$ for the air, diatomic gas. Thus:

$$\displaystyle{ \frac{P}{(\rho_0 + \rho\prime)^\gamma} = \frac{P_0}{\rho_0^\gamma} =\frac{rT}{\rho_0^{\gamma -1}} }$$

At first order, we get:

$$\displaystyle{ P=rT\rho_0 \left( 1 + \gamma \frac{\rho\prime}{\rho_0} \right) }$$

Our definition of the speed then gives:

$c = \sqrt{\frac{\partial P}{\partial \rho}} = \sqrt{\gamma rT}$

The numerical application at 20°C gives 343 m/s, in good agreement with the experimental results.

### Sine solution: plane wave

The general solutions of the wave equation are: $p(r,t)=p_+(t-\frac{r}{c}) + p_-(t+\frac{r}{c})$

The first term represents a wave propagating in the direction of increasing $r$ and the second one a wave propagating in the direction of decreasing $r$  .

The simple solution of a plane wave propagating towards increasing $x$ reads:

$p(x,t) = p_0 \cdot \cos \left[\omega \left(t - \frac{x}{c} \right) + \phi \right] = p_0 \cdot \cos \left(\omega t - kx + \phi \right)$

Thus, all the points of the medium with the same abscissa (i.e. located in a plane perpendicular to the direction of propagation) are in the same vibratory state. It is easy to verify that this function is a solution of the wave equation.

• $p_0$ is the amplitude of the disturbance ;
• $\omega = 2\pi f$ is the pulsation, $f$ is the vibration frequency of the source ;
• $k = \omega/c = 2\pi / \lambda$ is the wave number, $\lambda$ the wavelength;
• $\phi$ is the phase at origin

The complex notation sine functions is easier to handle in calculations (deriving and integrating an exponential is trivial): $\alpha = A \cos{X} \rightarrow \underline \alpha = A \exp{(j X)} = A \cos X + j A \sin X$. The modulus $|\alpha| = A$ denotes the amplitude, and its argument $X$ the cosine's phase at origin.

Let's note $\underline\alpha^* = A \exp{(-jX)} = A \cos X - j A \sin X$ the conjugate of $\underline\alpha$The real part of $\underline\alpha$ is then: $\alpha = \Re (\underline{\alpha}) = \frac{1}{2} (\underline{\alpha} + \underline{\alpha}^*)$

The complex notation of the overpressure is:

$\underline{p}(x,t) = \underline{p}_0 \cdot \text{exp} \left[ j \left( \omega t - k x \right) \right]$

with $\underline{p}_0 = p_0 \cdot \exp{(j \phi)}$ the complex amplitude.

The Euler equation allows us to deduce the velocity of the fluid:

$$\displaystyle{ \rho_0 \cdot \frac{\partial \underline{u} }{\partial t} = - \frac{\partial \underline{p} }{\partial x}\iff j \omega \rho_0 \underline{u} = jk \underline{p}\iff \underline{u} = \frac{\underline{p}}{\rho_0c}}$$

Note here that the velocity is in phase with the overpressure.

### RMS acoustic pressure

The RMS acoustic pressure $p_\text{eff}$ is calculated as the RMS voltage of an AC source, it is the square root of the mean value of the square of the overpressure:

$p_\text{eff}=\sqrt{\frac{1}{T} \int_{0}^{T} \Re(\underline{p})^2 \,\text{d}t}$

We thus have: $p_\text{eff}^2=\frac{1}{4T} \int_{0}^{T} (\underline{p} + \underline{p}^*)^2 \,\text{d}t = \frac{1}{4T} \int_{0}^{T} (\underline{p}^2 + \underline{p}^{*2} + 2 \underline{p}\underline{p}^*) \,\text{d}t$

The first two terms are sinusoids of pulsation $2\omega$ : their average value over a period is zero. The last term is independent of time, it leaves the integral and the remaining integral, trivial, is $T$. Finally:

$$\displaystyle{ \boxed{ p_\text{eff}=\sqrt{\frac{\underline{p}\underline{p}^*}{2}} }}$$

In the case of a plane wave, we get:

$P_\text{eff}=\frac{P_0}{\sqrt{2}}$

The human ear can perceive sound waves whose RMS pressure is greater than or equal to $p_\text{ref} = 20 \,\mu\text{Pa}$. Below this level, the auditory system is not sensitive enough to trigger a nerve signal. The sensation becomes painful when $p \geq 20 \,\text{Pa}$.

## Energy carried by the wave

### Sound intensity

Thesound intensity $I$ indicates the power carried by the wave passing through a surface of $1\hspace{1mm}\text{m}^2$ perpendicular to the direction of propagation. $I$ is in units of $\text{W}/\text{m}^2$.

The sound intensity is therefore equal to the ratio of the sound power $\mathcal{P}$ of the source to the area $A$ on which it spreads:

$$\displaystyle{I=\frac{\mathcal{P}}{A} }$$

Now the power is :

$$\displaystyle{\mathcal{P} = \vec{F} \cdot \vec{{u}} = A\cdot \Re \left(\underline{p}\right) \cdot \Re \left( \underline{u}\right) }$$

We have therefore, developing as in the previous paragraph :

$I= \Re \left(\underline{p}\right) \cdot \Re \left( \underline{u}\right) = \frac{1}{4} \left[ \left( \underline{p} \underline{u}^* + \underline{p}^* \underline{u} \right) + \left(\underline{p}\underline{u} + \underline{p}^*\underline{u}^* \right) \right] = \frac{1}{2} \left( \Re(\underline{p}\underline{u}^*) + \Re(\underline{p}\underline{u}) \right)$

The second term oscillates with a pulsation $2\omega$, its mean value is thus zero. The relationship becomes:

$\displaystyle{ \boxed{ I = \frac{1}{2} \Re(\underline{p}\underline{u}^*) }}$

In the case of a plane wave, we get:

$$\displaystyle{ I = \frac{1}{2} \Re(\underline{p}\frac{\underline{p}^*}{\rho_0 c})= \frac{p_\text{eff}^2}{\rho_0 c}}$$

Thus, the threshold of audition (minimum sound intensity perceptible by the human ear) is:

$I_0=\frac{p_\text{ref}^2}{\rho_0 \cdot c}=1 \times 10^{-12} \,\text{W}/\text{m}^2$

On the other hand, for painful sound intensities of the order of $1\, \text{W} \cdot \text{m}^{-2}$ (emergency siren), the effective sound pressure is :

$p_\text{eff}=\sqrt{I \cdot \rho_0 \cdot c} \approx 20 \, \text{Pa}$

This value, corresponding to 0.02% of the atmospheric pressure, justifies the use of weak signal acoustics for the situations we're dealing with.

## Ondes sphériques

If the plane wave is an interesting limit case, it is clear that it is insufficient to fully describe the radiation of a loudspeaker. The solution of spherical waves will be necessary and we give here some important results.

We use the spherical coordinates. If we assume that the sound source, placed at the origin of the reference frame, has an isotropic radiation, the acoustic quantities will only depend on their distance to the source $r$. All the points located at the same distance from the origin will therefore have the same vibratory state.

The Laplacian in the propagation equation is written with these simplifications: $\Delta = \frac{\partial^2}{\partial r^2} + \frac{2}{r} \frac{\partial}{\partial r}$

The pressure propagation equation then becomes :

$\frac{\partial^2 p}{\partial r^2} + \frac{2}{r} \frac{\partial p}{\partial r} - \frac{1}{c^2} \frac{\partial^2 p}{\partial t^2} = 0$

Developing $\frac{\partial^2(rp)}{\partial r^2}$we realize that we can rewrite:

$\frac{\partial^2 (rp)}{\partial r^2} - \frac{1}{c^2} \frac{\partial^2 (rp)}{\partial t^2} = 0$

And the sound pressure of a spherical wave is therefore :

$$\displaystyle{\boxed{ \underline{p} = \frac{\underline{A}}{r} \exp{ \left[ j(\omega t - kr) \right] } }}$$

where $\underline{A}$ is a constant to be determined according to the properties of the sound source. The pressure decreases as $1/r$.

The Euler equation in sinusoidal regime allows to deduce the radial component of the air speed:

$$\displaystyle{\rho_0 \cdot \frac{\partial \vec{\underline{u}}}{\partial t} = - \vec{\nabla}\underline{p} \Rightarrow \underline{u}_r = -\frac{1}{j \omega \rho_0} \frac{\partial \underline{p}}{\partial r} = \frac{1}{j \omega \rho_0} \left( \frac{1}{r}+jk\right) \underline{p} = \left( 1 - \frac{j}{kr}\right) \frac{\underline{p}}{\rho_0 c} }$$

We find the expression of the air speed of a plane wave at the limit $r \rightarrow +\infty \iff kr \gg 1$ but, closer to the source, the speed is no longer in phase with the overpressure.

Let us calculate the sound intensity in the case of a spherical wave :

$\displaystyle{ I = \frac{1}{2} \Re(\underline{p}\underline{u}^*) }$

$\displaystyle{ I = \frac{1}{2} \Re \left[ \underline{p}\left( 1 + \frac{j}{kr}\right) \frac{\underline{p}^*}{\rho_0 c} \right] = \frac{\underline{p}\underline{p}^*}{2\rho_0 c} \Re \left( 1 + \frac{j}{kr}\right) = \frac{p_\text{eff}^2}{\rho_0 c}}$

We thus find the same simple formula as in the case of plane waves.

The pressure varies as $1/r$the sound intensity therefore varies as $1/r^2$. When the sound wave propagates, the energy transported is diluted over a larger and larger surface: the sound intensity decreases. We speak ofgeometric attenuation :

If we multiply the distance $r$ to the source by a factor two, we multiply the area $A$ by a factor four: the sound intensity $I$ is therefore divided by four.

## Sound level

### Definition

The higher the sound intensity, the louder the sound is perceived? Yes, but ... The human ear will not perceive as twice louder a sound whose intensity is, precisely, twice as important! The sensitivity of the ear is not linear, but logarithmic.

To get closer to the physiological sensations related to sound, we define the sound level $L$ :

$L=10 \cdot \log{\frac{I}{I_0}}$

where $I_0=10^{-12} \,\text{W} \cdot \text{m}^{-2}$ is the hearing threshold. The unit of sound level is the decibel $\text{dB}$.

Thus, while the sound intensities perceptible by the human ear range over 12 orders of magnitude, the corresponding sound levels will take values between $0$ and $120\, \text{dB}$. Furthermore, an increase of $2\, \text{dB}$ of the sound level will cause the same sensation to the ear, regardless of the initial level.

### Sound level variations

We said that the sound intensity was divided by four when the distance to the source was multiplied by two: what happens to the sound level?

$L'=10 \cdot \log{\frac{I'}{I_0}} = 10 \cdot \log{\frac{I}{4\cdot I_0}} = 10 \cdot \log{\frac{I}{I_0}} - 10\cdot \log{4} = L -6$

Thus, the sound level decreases by 6dB when the distance to the source is multiplied by 2.

What happens now to the sound level if we double the power of the source (or if we put two speakers of the same power together)?

$L'=10 \cdot \log{\frac{I'}{I_0}} = 10 \cdot \log{\frac{2 \cdot I}{I_0}} = 10 \cdot \log{\frac{I}{I_0}} + 10\cdot \log{2} = L +3$

When we double the power, we increase the sound level by 3dB. In the same way, it decreases by 3dB when the sound intensity is divided by two.

### Sound pressure level (dB SPL)

If we inject the formula for the sound intensity $I=p_\text{eff}^2 / (\, \rho \cdot c )\,$ in the definition of the sound level, we get:

$L=10 \cdot \log{\frac{p_\text{eff}^2}{p_\text{ref}^2}}=20 \cdot \log{\frac{p_\text{eff}}{p_\text{ref}}}$

where $p_\text{ref}=20 \,\mu\text{Pa}$ is the minimum perceptible effective pressure. When the effective sound pressure is divided by two, the sound level decreases by 6dB.

### Sensitivity of the human ear

#### Fletcher's curve

The thresholds defined above are determined for the reference frequency $f=1000 \,\text{Hz}$. But the human ear is not equally sensitive to all frequencies, as the Fletcher diagram shows:

The curves represented in blue are contours of the same perceived sound level for different frequencies ( isosonic curves).

For example, a note played at 100 Hz generating a sound pressure of 30 dB SPL will be perceived at only 10 dB, thus much weaker than the same sound level at 1000 Hz.

#### Decibel A - dB-A

The corresponding sound level scale, corrected for the sensitivity of the ear, is noted "dB-A". The corrections to be made to the sound pressure level are summarized in the following figure:

The fact that at 100 Hz the sound pressure level must be reduced by 20 dB can be seen more quickly on this curve: $L_\text{dB-A, 100 Hz} = L_\text{dB-SPL, 100 Hz}-20$.

## Superposition of frequencies, spectra

### Pure and complex sounds

A tuning fork emits a sinusoidal sound wave: we speak of a pure sound. Conversely, the same note played by a piano is not sinusoidal: it is a complex sound.

### Sound spectrum

A very important mathematical result in signal processing is that any signal can be decomposed into a sum of sinusoids. We already know this if we remember the dispersion of light by water droplets in a rainbow: the complex light wave of white sunlight is decomposed into a sum of sinusoidal (monochromatic) light waves.

We see this decomposition on the spectrum of the sound wave. On the abscissa is the frequency of the sinusoids, on the ordinate their amplitude. Not surprisingly, the spectrum of the note played by the tuning fork has only one peak, since only one sinusoid is needed to obtain the recording:

On the other hand, the spectrum of A played on the piano contains a large number of peaks:

Note the following important points:

• The frequency of the lowest frequency peak corresponds to the frequency $f_1$ of the played note (here $f_1=440 \,\text{Hz}$). This peak is called the fundamental.
• The other peaks have a frequency multiple of that of the fundamental ($2f_1=880 \,\text{Hz}\,;\, 3f_1=1320 \,\text{Hz}\,;\, ... \,;\,nf_1$). They are called harmonics of rank 2, 3, …, n.

Finally, in a real recording, one will observe at each moment a superposition of notes played by different instruments as well as "noises" (percussions). The spectrum of the sound will then be more complex to interpret.

### Timbre of an instrument

Identical notes played by different instruments do not cause the same sensation to the ear: we say that their timbre is different. It is the relative distribution of harmonics in the spectrum of the note that determines the timbre, the sound "signature" of the instrument.

## Conclusion

In this article, we have defined sound waves and studied their main characteristics as well as the relevant quantities to describe them and measure the energy they carry. The foundations of linear acoustics are now laid.

For a Hi-Fi application, we will seek to obtain a maximum sound level of approximately $L \approx 90\, \text{dB}$corresponding to a sound intensity $I \approx 10^{-3} \,\text{W} \cdot \text{m}^{-2}$ and an effective acoustic pressure $P_\text{eff} \approx 0,65 \,\text{Pa}$. We must also be able to produce sound over a frequency range as close as possible to the 20 Hz - 20 kHz range detectable by the ear, and with a response curve as flat as possible so as not to favor certain frequencies of the recording to be broadcast.

In the next article, before tackling the speaker, we will study the acoustic radiation of some simple shapes.