What the h*ll is a decibel?

The Transmission Unit and Telephone Transmission Reference Systems (pdf)

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Decibel — The Name for the Transmission Unit (pdf)

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The decibel is based on human hearing. It is a logarithmic quantity intended to accommodate very large ranges of levels. The decibel allows computing gain/loss by addition/subtraction. The decibel has no units (it’s a ratio) but it can be used as a unit of measure by defining the ratio relative to a reference value (see examples below). It is important to know which category a measurement belongs when using decibels. The decibel is a logarithmic ratio between two numbers (a measured value and a reference value).

These ratios are specified differently for power quantities and root-power quantities (but always in terms of power!). See [here] for a basic explanation and rules for working with logarithms.

(Root-)power quantities

Energy and power are closely related but are not the same physical quantity. Energy (Joule; $J$ ) is the ability to cause change; Power is how fast energy is used or transmitted. It is the amount of energy divided by the time it took to use the energy. Its unit is the watt, which is one joule per second of energy used ( $Watt =J/s$ ).

A power quantity (see also [Wikipedia]) is a power, or a quantity directly proportional to power. Energy is also labeled as a power quantity in this context. For example, sound power (a.k.a. acoustic power; Watt; scalar quantity) and sound intensity ( $Watt/m^2$ ; vector quantity with magnitude and direction). Both are power quantities. Sound intensity is defined as the power carried by sound waves per unit area in a direction perpendicular to that area.

A root-power quantity (a.k.a. amplitude quantities). Examples are temperature, voltage ( $V$ ), and sound pressure level (SPL; Pascal ( $Pa$ )). Its square is proportional to power. Root-power quantities are sometimes also referred to as field quantities. In physics a field is a quantity that depends on the position. However, not all root-power quantities (e.g., voltage) are fields and, therefore, the use of ‘field quantity’ can be confusing and should not be used. See Figure 1 for a visual explanation of sound power, sound intensity, and sound pressure.

Figure 1. A heater analogy: sound power (=acoustic power) , sound intensity, and sound pressure. Figure copied from Siemens

Power quantities

Examples are sound power $P$ (a.k.a. acoustic power; Watts), sound intensity ( $Watt/m^2$ ) and electrical power (Watts). The level ( $L$ ) for these quantities is specified as a ratio relative to a reference value:

$\boxed{ \!\begin{aligned} L_{pq} = \textbf{10} log_{10} \left( \frac{P}{P_{ref}} \right) \end{aligned} }$

Note that $log_{10}$ denotes the common logarithm with base 10, i.e., $log_{10}(a) = b \leftrightarrow a = 10^b$ .

If we talk about acoustic power then the normal reference level is $P_{ref}=10^{-12} W$ (i.e., 1 picowatt), which is the lowest sound persons of excellent hearing can discern. Note that some older books use a reference value of $P_{ref}=10^{-13} W$

If acoustic power (AP) is halved (i.e., $P = \frac{1}{2}P$ ) then its level drops by 3dB:

$\boxed{ \!\begin{aligned} L_{AP} &= 10 log_{10} \left( \frac{0.5*P}{P_{ref}} \right) \\ &= 10 log_{10} (0.5) + 10 log_{10} \left( \frac{P}{P_{ref}} \right) \\ &= \textbf{-3dB }+ 10 log_{10} \left( \frac{P}{P_{ref}} \right) \end{aligned} }$

See [here] for logarithmic rules used in this calculation.

Root-power quantities

Examples of root-power quantities are sound pressure ( $p$ ) level (SPL; Pa) and voltage ( $V$ ). Note that we use small letter $p$ for sound pressure while we use capital letter $P$ for sound power. For root-power-quantities, the level is specified as:

$\boxed{ \!\begin{aligned} & L_{rpq} = \textbf{20} log_{10} \left( \frac{p}{p_{ref}} \right) \\ \end{aligned} }$

If we consider voltage then for studio equipment one often refers to dBu; let’s take this as an example. dBu is an abbreviation for dB unloaded which denotes the level relative to $V_{ref}=0.774.5967 V$ . If we use dBu and if we half the voltage $V$ (i.e., $V = \frac{1}{2}V$ ) then its level (dBu) drops by 6dB:

$\boxed{ \!\begin{aligned} L _{dBu} &= 20 log_{10} \left( \frac{0.5*V}{V_{ref}} \right) \\ &= 20 log_{10} (0.5) + 20 log_{10} \left( \frac{V}{V_{ref}} \right) \\ &= \textbf{-6dB} + 20 log_{10} \left( \frac{V}{V_{ref}} \right) \end{aligned} }$

In the next section, I explain how this voltage ratio is still related to a power ratio. Remember, a decibel is always specified in terms of power.

Why use decibels?

The Human ear is sensitive to the sound of frequency from approximately 20 Hz to 20 kHz, that is, in the range of three decades in frequency, and of intensity (energy of a sound wave passing through a unit area ( $1 m^2$ ) per second) from $10^{−12}$ to $1 W/m^2$ , that is, twelve decades in the intensity! The response of the ear to the sound intensity is remarkable. We can hear both very quiet and very loud sounds. The intensity response of the ear is logarithmic. What does this mean?

On a linear scale, a change between two values is perceived on the basis of the difference between the values:
- e.g., a change from 1 to 2 would be perceived as the same increase as from 4 to 5.
On a logarithmic scale, a change between two values is perceived on the basis of the ratio of the two values:
- e.g., a change from 1 to 2 would be perceived as the same increase as a change from 4 to 8.

This is also shown in Figure 2, which shows the Fletcher Munson curves of equal loudness. We perceive a change from 20 to 40dB the same as a change from 100 to 120dB. In terms of sound intensity these changes correspond to $0.000000000099$ and $0.99 W/m^2$ respectively. Or….both correspond to a change in ratio of $100 W/m^2$ .

Thus, a decibel let’s us conveniently represent variables that have a huge range of values (like sound intensity and sound pressure; see also [here]) but it also more closely resembles the way in which we perceive changes in sound level.

Figure 2. Fletcher-Munson curves. The left y-axis (linear scale) shows the Sound Intensity Level (SIL) in decibels. The right y-axis (logarithmic scale) shows the sound intensity ( $Js^{-1}m^{-2}$ . A phon is a unit of perceived loudness. At 1kHz the loudness value (phon) corresponds to the sound intensity level (SIL). Thus, following the 20 phons contour, a 1 kHz tone with a SIL of 20dB is perceived as loud as a 50Hz tone with a SIL of 55dB.

From power decibels to voltage decibels.

A root-power quantity can be related to a power quantity. As mentioned above, the square of a root-power quantity is proportional to a power quantity. Or, vice versa, a root-power quantity is the root of power quantity (hence its name). For example, $P \propto V^2 \rightarrow V = \sqrt P$ (see below).

Let us consider the relation between voltage $V$ and acoustic power $P$ . The power level can be expressed as a voltage if we know the resistance $R$ (see [here]):

$\boxed{ \!\begin{aligned} P_{1} &= \frac{V_{1}^{2}}{R_{1}} \\ P_{2} & = \frac{V_{2}^{2}}{R_{2}} \end{aligned} }$

Now we can express the level for the acoustic power ( $L_{AP}$ } as a function of voltage $V$ :

$\boxed{ \!\begin{aligned} L_{AP} &= 10 log_{10} \left( \frac{P_{1}}{P_{2}} \right) dB \\ &= 10 log_{10} \left( \frac{V_{1}^{2}}{V_{2}^{2}} \cdot \frac{R_{2}}{R_{1}} \right) dB \end{aligned} }$

If we assume that $R_{1} = R_{2}$ (which in our context is the case) we can rewrite to get

$\boxed{ \!\begin{aligned} L_{AP} &= 20 log_{10} \left( \frac{V_{1}}{V_{2}} \right) - 10 log_{10} \left( \frac{R_{1}}{R_{2}} \right) dB \\ &= 20 log_{10} \left( \frac{V_{1}}{V_{2}} \right) dB \end{aligned} }$

Here we made use of the fact that $log_{10} \left( \frac{R_{1}}{R_{2}} \right) = log_{10} (1) = 0$ .

This derivation shows the change in acoustic power level $L_{AP}$ if we change a corresponding root-power quantity ( $V$ ). It also shows how $20log$ for voltage ratios gets introduced.

Putting it together: voltage and power

We already noted that halving of acoustic power $P$ results in a 3dB drop in the acoustic power level $L_{AP}$ . In the previous equation we now see that halving the voltage $V$ gives a 6dB decrease in the acoustic power level. Vice versa, doubling $P$ or $V$ gives a +3dB or +6dB increase in level respectively.

Let’s try to put these numbers together by doing another calculation. Suppose we have an +3dB increase in the acoustic power. Expressing acoustic power in terms of voltage $V$ , we have

$\boxed{ \!\begin{aligned} & L_{AP} = +3dB + 20 log_{10} \left( \frac{V_{1}}{V_{2}} \right) \\ \end{aligned} }$

Let us simply write $log$ instead of $log_{10}$ from now on.

We can now ask: what was the increase in voltage that resulted in this +3dB power level increase.

As we have seen earlier, the +3dB in this last equation originates from $20log(x)$ , where $x$ is the increase required to get a +3dB power increase.

$\boxed{ \!\begin{aligned} & L_{AP} = \stackbelow{20log(x)}{+3db} + 20 log \left( \frac{V_{1}}{V_{2}} \right) \\ \end{aligned} }$

We can rewrite to

$\boxed{ \!\begin{aligned} & L_{AP} = 20 log \left( \frac{x V_{1}}{V_{2}} \right) \\ \end{aligned} }$

If we want to calculate $x$ then we can simply solve $+3dB = 20 log(x)$ , which gives $x \approx 1.4 = \sqrt 2$ .

However, recall that +3dB is actually 3.010299957 dB, that is, $10log(2)$ . Thus, if we want to make an exact calculation, we can also write

$\boxed{ \!\begin{aligned} & L_{AP} = \stackbelow{10log(2)}{+3db} + 20 log \left( \frac{V_{1}}{V_{2}} \right) \\ \end{aligned} }$

Combining these insights lead to

$\boxed{ \!\begin{aligned} 20log(x) &= 10log(2) \\ log(x) &= \frac{1}{2}log(2) \\ &= log \left( 2^\frac{1}{2} \right) = log \left(\sqrt 2 \right) \end{aligned} }$

Plugging $log(x)$ back into the equation, we get

$\boxed{ \!\begin{aligned} L_{AP} &= 20 log \left( \sqrt 2 \right) + 20 log \left( \frac{V_{1}}{V_{2}} \right) \\ &= 20 log \left( \frac{\sqrt 2 \cdot V_{1}}{V_{2}} \right) \\ &= 10 log \left( \frac{\sqrt 2 \cdot V_{1}}{V_{2}} \right)^2 \\ &= 10 log \left( \frac{2 \cdot V_{1}^2}{V_{2}^2} \right) \end{aligned} }$

What does this tell us?

To get a +3dB level increase in acoustic power (i.e., doubling of acoustic power), the voltage ratio needs to be increased with $\sqrt 2 \approx 1.4$ .
To get a +6dB level increase in acoustic power (i.e., quadruple of acoustic power), the voltage ratio needs to be increased with a factor 2.

$\boxed{ \!\begin{aligned} L_{AP} &= 10 log \left( \frac{2 \cdot V_{1}}{V_{2}} \right)^2 \rightarrow 20log(2)=+6dB \\ \end{aligned} }$

Or equally,

$\boxed{ \!\begin{aligned} L_{AP} &= 10 log \left( \frac{4 \cdot V_{1}^2}{V_{2}^2} \right) \rightarrow 10log(4)=+6dB\\ \end{aligned} }$

Thus, the increase (or decrease) always reflects the increase (or decrease) of the ratio between the parentheses.

Finally, we can approach the same thing from a slightly different angle.

Since we can express the acoustic power level $L_{AP}$ as a function of power ( $P$ ) or as a function of voltage ( $V$ ), we can also directly related these two equations to determine the change in power if we change the voltage. Thus, we have two equations for the acoustic power level $L_{AP}$ :

$\boxed{ \!\begin{aligned} L_{AP}^1 &= 10 log \left( \frac{P_1}{P_0} \right) \\ L_{AP} ^2&= 20 log \left( \frac{V_1}{V_0} \right) \\ \end{aligned} }$

For a given change in level, these are equal. That is, $L_{AP}^1 = L_{AP}^2$ resulting in

$\boxed{ \!\begin{aligned} 10 log \left( \frac{P_1}{P_0} \right) &= 20 log \left( \frac{V_1}{V_0} \right) \\ log \left( \frac{P_1}{P_0} \right) &= 2 log \left( \frac{V_1}{V_0} \right) \\ \frac{P_1}{P_0} &= 10^{2 log \left( \frac{V_1}{V_0} \right) } \\ \end{aligned} }$

From this last equation we learn that if we double the voltage (i.e., $V_1 = 2V_0$ ), then the acoustic power gets four times as large:

$\boxed{ \!\begin{aligned} \frac{P_1}{P_0} &= 10^{2 log \left( \frac{V_1}{V_0} \right) } \\ &= 10^{2 log 2 } = 10^{0.6021} =4 \\ \end{aligned} }$

Thus, $P_1 = 4P_0$ .

In terms of decibels, doubling the voltage $=20log(2)=+6dB$ and quadrupling the power $=10log(4)=+6dB$ . Thus, both increased with the same level (decibels).

In other words: an $x$ dB increase/decrease in voltage gives gives the same $x$ dB increase/decrease in power. However, the absolute voltage ( $V$ ) and power ( $P$ ) do not change with the same amount (i.e., the ratio does not change equally).

Different power and voltage ratios are shown in Table 1.

Table 1. Conversion between decibels and linear values (copied from dB or not dB? See below)

Decibel Full Scale (dBFS)

dBFS is used in the digital domain. The unit of dBFS is defined in several standards (e.g., AES17-2020, see also [here]). In the AES17-2020 standard it is defined as the “RMS (root mean square) level expressed in decibels relative to full-scale (FS) level”. FS level is defined as a dc-free 997-Hz sine wave whose undithered positive peak value is positive digital full scale. A signal level ( $L$ ) is defined as the RMS signal level expressed as a fraction of the FS level. Consequently, dBFS is computed from the signal level $L$ using the relation $L_{dBFS} = 20 log (L)$ . In addition, sine waves whose peaks reaches digital full scale will read 0 dBFS. As a result dBFS is defined as:

$\boxed{ \!\begin{aligned} dBFS = 20 log \frac{\mbox{rms of signal}}{A_{max}} + 3dB \\ \end{aligned} }$

For a derivation of this relation see my post about sound concepts. For the current context, it is relevant to realize that doubling the digital RMS signal level corresponds to a +6dBFS increase (i.e., the constant +3dB is not relevant).

dBFS to Voltage Conversion

A fader on a mixing console or in your DAW is a ratio control device. The numbers next to the fader show the ratio output to input. 0dB is no change in level through the fader. Thus, -6dB is an output of half the incoming voltage or half the digital signal.

dBFS is not defined for analog levels, according to standard AES-6id-2006. No single standard converts between digital and analog levels, mostly due to the differing capabilities of different equipment. For example:

In Europe, the EBU recommend that −18 dBFS equates to the alignment level of +4 dBu or 0 VU
The American SMPTE standard defines −20 dBFS as the alignment level.

Nevertheless, if you double your digital signal level (+6dB) then also the output voltage (whatever that may be) will increase with +6dB, and consequently, the acoustic power increases with +6dB.

Sound Pressure Level (SPL)

Now let us consider sound pressure ( $p$ ; lowercase), which is a root-power quantity, like voltage. Sound pressure is what our ears are actually sensing. There is no way for us to sense acoustic power. We want to know how sound pressure level relates to changes in sound power ( $P$ ; uppercase) or voltage $V$ or dBFS.

As we have seen, the sound pressure level (SPL; Pa) is defined as:

$\boxed{ \!\begin{aligned} & L_{SPL} = \textbf{20} log \left( \frac{p}{p_{ref}} \right) \end{aligned} }$

Note: in literature, $L_{SPL}$ is often denoted as $L_p$ while $L_{AP}$ is often referred to as $L_w$ . By definition, a level of $L_{SPL}=0dB$ corresponds to 20 microPascals ( $p_{ref}=20 \mu Pa$ ).

Sound power level is a measure of the rate at which acoustic energy is radiated from a sound source. This is different from sound pressure level, which depends on additional factors, including the distance from the sound source and ‘directivity factor’ $Q$ .

In addition to sound power and sound pressure, we have sound intensity ( $I$ ; $Watss/m^2$ ), which a power quantity like acoustic power ( $P$ ) and, consequently, its level $L_I$ can be defined as

$\boxed{ \!\begin{aligned} L_I = \textbf{10} log \left( \frac{I}{I_0} \right) \end{aligned} }$

A level of $L_I=0dB$ corresponds to 1 picoWatt ( $I_{ref}=10^{-12}W/m^2$ ).

Note that sound intensity $I$ is defined as the acoustic power $P$ through a unit area ( $A$ ):

$\boxed{ \!\begin{aligned} I = \frac{P}{A} \end{aligned} }$

For a freely propagating plane wave, the average sound intensity is related to sound pressure by:

$\boxed{ \!\begin{aligned} I = \frac{p^2}{2 \rho c} \end{aligned} }$

where $\rho$ ( $kg/m^3$ ) is the density of the medium and $c$ ( $m/s$ ) is the speed of sound in the medium. For the derivation of the relation between sound intensity and sound pressure see my post [here]. Thus, these equations relate acoustic power and sound pressure. If we consider a single sound source radiating sound in all directions, and combining the previous two equations, and using $A=4 \pi \ r^2$ for the area of a sphere, we get

$\boxed{ \!\begin{aligned} \frac{p^2}{2 \rho c} = \frac{P}{4 \pi r^2} \end{aligned} }$

From here it takes some algebra (see [here]) to derive the relation between $L_{SPL}$ and $L_{AP}$ :

$\boxed{ \!\begin{aligned} L_{SPL} = L_{AP} + 10 log \left( \frac{Q}{4 \pi r^2} \right) \end{aligned} }$

Here, $Q$ denotes the so-called directivity factor to account for nearby reflective surfaces. However, assuming that $Q$ and $r$ are fixed we see that $L_{SPL}$ and $L_{AP}$ only differ by a constant value (i.e., $10log(Q/4 \pi r^2)$ ). Consequently, if the sound power level changes with 6dB then also the sound pressure level will change with 6dB.

In principle, we could also have followed the same reasoning as we did for acoustic power $P$ and voltage $V$ . We can express acoustic power $P$ as a function of sound pressure level $p$ . If we have two different sound pressures, $p_1$ and $p_0$ at a same distance $r$ from the sound source, then we have

$\boxed{ \!\begin{aligned} P_1 = \frac{4 \pi r^2 p_1^2}{\rho c} \\ P_0 = \frac{4 \pi r^2 p_0^2}{\rho c} \end{aligned} }$

Then, the ratio between acoustic power $P_1$ and $P_0$ is

$\boxed{ \!\begin{aligned} \frac{P_1}{P_0} &= \frac{4 \pi r^2 p_1^2}{\rho c} \frac{\rho c}{4 \pi r^2 p_0^2}\\ &= \frac{p_1^2}{p_0^2} \end{aligned} }$

The acoustic power level $L_{AP}$ can be expressed in terms of acoustic power $P$ (power quantity)

$\boxed{ \!\begin{aligned} L_{AP} &= 10 log \frac{P_1}{P_0}\\ \end{aligned} }$

or in terms of sound pressure $p$ (root-power quantity)

$\boxed{ \!\begin{aligned} L_{AP} &= 10 log \left( \frac{p_1}{p_0} \right)\\ &=10 log \left( \frac{p_1^2}{p_0^2} \right)\\ &=10 log \left( \frac{p_1}{p_0} \right)^2\\ &= 20 log \left( \frac{p_1}{p_0} \right) \\ \end{aligned} }$

Thus, we see that the statements for acoustic power and voltage also apply to acoustic power and sound pressure. For example, if the acoustic power level $L_{AP}$ increases with 3dB then also the sound pressure level $L_{SPL}$ increases with 3dB.

Sound Pressure Level (SPL) and RMS

The sound pressure $p$ in $L_{SPL} = \textbf{20} log \left( \frac{p}{p_{ref}} \right)$ represents the root mean square (RMS) value of the sound pressure of the wave being measured. The RMS value of the sound pressure is derived from the instantaneous sound pressure values $p$ over a period of time $T$ . The RMS value provides a measure of the effective magnitude (average) of the varying sound pressure, which is particularly useful because sound pressure varies with time as the sound wave oscillates.

Instantaneous Sound Pressure $p(t)$ . This is the sound pressure at a specific time $t$ . It varies with time as the sound wave moves through the medium (e.g. air).
Square of the instantaneous pressure. To eliminate the effect of positive and negative values (since the sound wave oscillates around zero), the instantaneous sound pressure is squared (otherwise we would get an average pressure of zero). This gives $p(t)^2$ .
Average Over Time: The squared values are averaged over a period of time $T$ . This involves integrating the squared pressure over one period and then dividing by the period $T$ :

$\boxed{ \!\begin{aligned} \text{Average of } p(t)^2 = \frac{p(0)^2 + p(1)^2 + \cdots + p(T)^2}{T} \end{aligned} }$

which we can write as

$\boxed{ \!\begin{aligned} \text{Average of } p(t)^2 = \frac{1}{T} \int_{0}^{T} p(t)^2 dt \end{aligned} }$

Take the square root. Finally to obtain the RMS value, take the square root of the average of the squared pressures:

$\boxed{ \!\begin{aligned} \sqrt{\text{Average of } p(t)^2} = p_{rms} \longDefiningEquals \langle p \rangle\ = \sqrt{ \frac{1}{T} \int_{0}^{T} p(t)^2 dt } \end{aligned} }$

or maybe better:

$\boxed{ \!\begin{aligned} \langle p \rangle\ = \sqrt{ \lim_{T \to \inf} \frac{1}{T} \int_{0}^{T} p(t)^2 dt } \end{aligned} }$

The RMS value is a form of statistical measure that provides a single value representing the effective pressure level of the sound wave. This value is used in the sound pressure level (SPL) equation to relate the fluctuating sound pressure to a standard reference pressure.

Thus, to be precise, we should write

$\boxed{ \!\begin{aligned} & L_{SPL} = \textbf{20} log \left( \frac{\langle p \rangle }{p_{ref}} \right) \end{aligned} }$

Note that sound intensity $I$ itself is not an RMS value. However, $I \propto \langle p \rangle$ .

Root Mean Square (RMS)

Let’s dive a bit more into the RMS value, which is used for the calculation of the average sound pressure $p$ .

Compared to the arithmetic mean, the RMS represents a more natural way to represent the average of the sound pressures since, as we have seen, the acoustic power $P \propto p^2$ . Or, actually, $P \propto \langel p^2 \rangle$ . Thus, averaging $p(t)^2$ over time gives a measure of acoustic power while taking the square root converts this power measure back into the average sound pressure $\langle p \rangle$ .

To give some intuition, consider a harmonic pressure wave:

$\boxed{ \!\begin{aligned} p(t) = p(0) sin\omega t \end{aligned} }$

Here, the angular frequency $\omega = 2\pi/T$ . The angular frequency is measured in radians per second. The inverse of the wave period is the $f = 1/T$ . The frequency $f = 1/T = \omega/2\pi$ of the motion gives the number of complete oscillations per unit time.

The squared pressure is

$\boxed{ \!\begin{aligned} p(t)^2 = p(0)^2sin^2 \omega t \end{aligned} }$

The average of $p(t)^2 \longDefiningEquals p_{rms}^2$ is then given by

$\boxed{ \!\begin{aligned} p_{rms}^2 = \frac{1}{T} \int_{0}^{T} p(0)^2sin^2 \omega t dt \end{aligned} }$

where $T$ is one period of the wave and is given by $T=\frac{2\pi}{\omega}$ . This gives

$\boxed{ \!\begin{aligned} p_{rms}^2 &= \langle p^2 \rangle = \frac{1}{T} \int_{0}^{T} p(0)^2sin^2 \omega t dt \\ &= \frac{p(0)^2}{T} \int_{0}^{T} sin^2 \omega t dt\\ &= \frac{p(0)^2}{T} \int_{0}^{T} \frac{1-cos 2\omega t}{2} dt\\ &= \frac{p(0)^2}{2T} \int_{0}^{T} 1-cos 2\omega t dt\\ &= \frac{p(0)^2}{2T} \int_{0}^{T} 1 dt - \int_{0}^{T}cos 2\omega t dt\\ &= \frac{p(0)^2}{2T} (T - 0) \\ &= \frac{p(0)^2}{2} \end{aligned} }$

Thus,

$\boxed{ \!\begin{aligned} p_{rms}^2 &= \frac{p(0)^2}{2} & = \sqrt{ \frac{p(0)^2}{2} } \approx 0.707 p(0) \end{aligned} }$

Note, that $p_{rms}$ is $20log(0.707)=-3dB$ compared to the peak level $p(0)$ .

This is summarized in Figure 3.

Figure 3. Root mean square (RMS) pressure for harmonic waves (Figure copied from [here]). Top: $P_{RMS}$ represents the average pressure which equals to 0.707 of the pressure amplitude $p(o)$ . Bottom: $p_{RMS}^2 = \frac{p(0)^2}{2}$ , i.e., the sum of the $p(t)^2$ values above the dotted line equals the sum of the $p(t)^2$ values below the dotted line. Hence their average equals $0.5p^2$ . This situation is different for averaging the pressures $p$ in the top plot where the negative values below the solid line and the positive values above the solid line add to zero, i.e., would give an average pressure of zero (which is why we use the RMS value).

Note that is we have a sound composed of several ( $N$ ) harmonic waves:

$\boxed{ \!\begin{aligned} p = \sum_{i=1}^{N} p(0)^2sin^2 \omega_i t \end{aligned} }$

then it can be shown that the RMS pressure (for uncorrelated sound waves) of the resultant wave is:

$\boxed{ \!\begin{aligned} p_{rms} = \sqrt {p_{rms_1}^2 + p_{rms_2}^2 + \cdot + p_{rms_N}^2} \end{aligned} }$

Coherent and incoherent sound sources

One of the key concepts in acoustics is the difference between coherent and incoherent sound sources. Coherent sound sources are those that maintain a constant phase relationship with each other. This means that the waves emitted by these sources are synchronized in such a way that their peaks and troughs match up over time. Because of this synchronization, coherent sound waves can interfere with each other constructively and destructively. For example, if we play an sine tone of 500Hz from two speakers, then we have two coherent sound sources. Incoherent sound sources, on the other hand, do not maintain a constant phase relationship. The phase difference between their waves changes randomly over time. As a result, the interference effects are averaged out, and no clear interference pattern is produced. Incoherent sound sources tend to produce a more uniform sound field. For example, if we have two different signals (e.g., two sine waves of different frequencies, or two different instruments, or two different songs) then at certain moments the corresponding signals may cancel each other if they are completely out of phase (0 Pa), or their level is doubled if they are perfectly in phase (+4 Pa). Since sound waves vary over time the combined magnitude constantly varies between 0 and 4 Pascals over time (Figure 4).

Figure 4. Summation of two incoherent signals (the source signals do not maintain a constant phase relationship). Figure copied from Siemens.

Coherent signals

To define ‘coherent’ more precisely, consider two signals $S_1(t)$ and $S_2(t)$

$\boxed{ \!\begin{aligned} S_1(t) = A_1 \sin(\omega_1 t + \phi_1) \\ S_2(t) = A_2 \sin(\omega_2 t + \phi_2) \end{aligned} }$

where $A$ is the amplitude, $\omega$ is the angular frequency and $\phi$ represents the phase. For the sources to be coherent, the phase difference must remain constant over time:

$\boxed{ \!\begin{aligned} \Delta \phi = \phi_1 - \phi_2 = \text{constant} \end{aligned} }$

For incoherent sources $\Delta \phi$ would vary over time: $\Delta \phi (t) = \phi_1(t) - \phi_2(t)$ .

For coherent signals we also require that $\omega_1 = \omega_2$ since otherwise the phase difference between the two sources will change with time (as is shown in Figure 4). If $\omega_1 \neq \omega_2$

$\boxed{ \!\begin{aligned} \Delta \phi (t) &= (\omega_2t+\phi_2) - (\omega_1t+\phi_1)\\ &= (\omega_2 - \omega_1)t + (\phi_2 - \phi_1) \end{aligned} }$

Since $\omega_1 \neq \omega_2$ , the term $(\omega_2 - \omega_1)t$ causes the phase difference to vary linearly with time, thus preventing a constant phase relationship. This varying phase difference leads to an absence of consistent constructive or destructive interference, characteristic of incoherent sources.

This requirement does not exclude the possibility of coherent signals $S_1(t)$ and $S_2(t)$ that are superpositions of sine waves with different frequencies. Their general form can be expressed as:

$\boxed{ \!\begin{aligned} S_1(t) = \sum_{i} A_i \cos(\omega_i t + \phi_i)\\ S_2(t) = \sum_{i} B_i \cos(\omega_i t + \psi_i) \end{aligned} }$

where $\omega_i$ represents different angular frequencies, and $\phi_i$ and $\psi_i$ are the phases associated with each frequency component. For $S_1(t)$ and $S_2(t)$ to be coherent, each corresponding frequency component must maintain a constant phase difference. This means that for each frequency component $\omega_i$ the phase differences must be constant:

$\boxed{ \!\begin{aligned} \Delta \phi_i = \psi_i - \phi_i = \text{constant} \end{aligned} }$

If $S_1(t)$ and $S_2(t)$ are identical signals, then $S_1(t) = S_2(t)$ , which implies that:

$\boxed{ \!\begin{aligned} A_i = B_i \quad \text{and} \quad \phi_i = \psi_i \quad \forall i \end{aligned} }$

In this case, the phase difference for each frequency component is zero $\Delta \phi_i = 0$ , which is indeed constant. Therefore, identical signals are trivially coherent because their phase relationship does not change.

If $S_1(t)$ and $S_2(t)$ are not identical but have components with fixed phase differences, the signals can still be coherent. For example:

$\boxed{ \!\begin{aligned} S_1(t) &= \sum_{i} A_i \cos(\omega_i t + \phi_i)\\ S_2(t) &= \sum_{i} A_i \cos(\omega_i t + \phi_i + \Delta \phi_i) \end{aligned} }$

Here, if each $\Delta \phi_i$ is constant, $S_1(t)$ and $S_2(t)$ are coherent because the phase difference between corresponding components remains fixed.

Figure 5. Summation of two complex incoherent signals $S1$ and $S2$ . Each signal has three underlying components.

Incoherent signals

The RMS (root mean square) is used for combining sound pressures from multiple sources that do not have a fixed phase relationship (incoherent signals), which implies that their their phases are randomly related, meaning they are uncorrelated.

Incoherent sources have random phase relationships, leading to constructive and destructive interference averaging out over time. Therefore, the sound pressures do not add linearly. The RMS value captures the average energy $E$ over time, making it the appropriate measure for combining incoherent sources. The RMS pressure directly relates to the energy, which is what ultimately matters for sound perception and measurement. Since sound energy is proportional to the square of the pressure $p$ , the RMS pressure ensures that we correctly account for the total energy contribution from all sources. Let’s consider two incoherent sound sources with sound pressures $p_1$ and $p_2$ .

For incoherent sources, the total energy (or power) is the sum of the individual energies. The energy of a sound wave is proportional to the square of its pressure:

$\boxed{ \!\begin{aligned} E \propto p^2 \end{aligned} }$

Therefore, the total energy $E_{\text{total}}$ from the two sources is:

$\boxed{ \!\begin{aligned} E_{\text{total}} = E_1 + E_2 = p_1^2 + p_2^2 \end{aligned} }$

The RMS pressure $p_{\text{rms}}$ of the combined sound sources is derived from the total energy:

$\boxed{ \!\begin{aligned} p_{\text{rms}}^2 \propto E_{\text{total}} \end{aligned} }$

Since energy is proportional to the square of the pressure:

$\boxed{ \!\begin{aligned} p_{\text{rms}}^2 = p_1^2 + p_2^2 \end{aligned} }$

Taking the square root to find the RMS pressure:

$\boxed{ \!\begin{aligned} p_{\text{rms}} = \sqrt{p_1^2 + p_2^2} \end{aligned} }$

For $N$ incoherent sources with pressures $p_1, p_2, \ldots, p_N$ , the RMS pressure is:

$\boxed{ \!\begin{aligned} p_{\text{rms}} = \sqrt{p_1^2 + p_2^2 + \cdots + p_N^2} \end{aligned} }$

This shows that the RMS pressure is the square root of the sum of the squares of the individual pressures.

Decibel levels with one speaker

So far, we implicitly assumed that we had a single sound source (e.g., one speaker). Let us consider the sound pressure level:

$\boxed{ \!\begin{aligned} L_{SPL} &= 20 log \frac{p_1}{p_0}\\ \end{aligned} }$

The reference value $p_0$ has been defined as $20 \muPa$ (i.e., 0.00002 Pascal) because that is near the absolute threshold for a sound frequency of 1 kHz. Thus, a sound pressure of 2 Pa corresponds to $L_{SPL}=20log(2/0.00002)=100dB$ .

If a signal is added to itself (mixed on two inputs at equal levels, e.g., a mixbus), then this will double the voltage giving a $20 log \frac{2p}{p} = 6dB$ increase (Figure 6). Note, that in this case the signals are, by definition, of equal amplitude and are in phase. Thus, these two signals are coherent (correlated).

Figure 6. Voltage, (acoustic) power, and sound pressure level (SPL). Figure modified from Don Davies, Eugene Patronis, Jr. (2006) Sound System Engineering. Third Edition. Elsevier Focal Press, Amsterdam. If we double the voltage ( $20log(2)=+6dB$ by feeding two identical signals to a mixbus (audio source) then also the sound power level increases with 6dB (hence, 4 times increase in power $P$ : $10log(4)$ ). Consequently, also the sound pressure level increases with 6dB corresponding to a doubling of the sound pressure $p$ .

Decibel levels with two speakers

Complexities

At the moment we consider two (or more) speakers, then it becomes much more complex to determine how much a level increases or decreases. In addition, there are many other factors that play a role. Some of these also apply to single speakers.

If the amplitudes of sound pressures, intensities, or powers are not equal then we get different numbers than the standard 3dB and 6dB.
Phase (see below)
Comb filtering if the listener is not exactly positioned between the speakers (point x in Figure 7)
Mutual coupling between speakers (see video below)
Surface coupling with speaker
Reflecting surfaces
Reverberant vs anechoic rooms
……

Summation of sound pressures (coherent signals)

Suppose you measure the sound pressure level $L_{SPL}$ in front of your speaker (at a certain fixed distance), and you get a reading of 100dB. Next you add a second speaker that also produces a sound pressure of 100dB (Figure 7). Also assume that the signals from both sources are coherent (correlated, identical).

Figure 7. At point x you experience 100dB SPL from each speaker. The total sound pressure level at x will be $L_{SPL}=106dB$ . Note: here we calculated the increase in SPL if the SPL doubles. This situation is not the same as doubling the speakers, which doubles the acoustic power, which is $+3dB$ , which also gives a $+3dB$ increase in SPL. Doubling speakers does not double SPL because acoustic power $P \propto p^2$ . If speakers are very close then the level (for lower frequencies) might increase more than $+3dB$ because of mutual coupling.

What will be the total sound pressure you will now measure in front of your speakers? You cannot simply add the sound pressures levels, which would result in 100dB+100dB=200dB, which is well beyond the threshold of pain. Instead, you must first convert back to sound pressure (Pa) values and sum these values together. Thus, in this example, you would

$\boxed{ \!\begin{aligned} L_{SPL}&=100dB = 20 log \frac{p}{p_{ref}} \\ &=20 log \frac{p}{20 \cdot 10^{-6}Pa} p &= 10^5 \cdot 20 \cdot 10^{-6} = 2Pa \end{aligned} }$

Thus, we get 2Pa +2 Pa = 4Pa, which corresponds to $L_{SPL}=20log(4/0.00002)=106dB$ . Thus, we only get a 6dB increase!

Summation of sound intensities (coherent signals)

Now suppose one speaker results in a intensity level of $L_{I}=100dB$ in front of your speaker at a certain fixed distance. Next you add a second speaker that also results in an intensity of 100dB. What will be the total sound intensity level produced by both speakers? We first calculate the sound intensity making use of the fact that the reference level is 1 picoWatt ( $I_{ref}=10^{-12}W/m^2$ ). Thus, $100dB=10log(I/10^{-12}) \rightarrow I=0.01 W/m^2$ .

The thing to keep in mind here is that we cannot directly sum the intensities to get a total intensity. Thus, adding a second speaker will not result in $I_{\text{total}}=I_1+I_2$ . Instead, we first need to convert the intensities to pressure $p$ , then sum the pressures, and finally convert back to intensity.

Recall that for a freely propagating plane wave, the sound intensity is related to sound pressure by:

$\boxed{ \!\begin{aligned} I &= \frac{p^2}{2 \rho c}\\ p &= \sqrt{2 \rho c I} \end{aligned} }$

where $\rho \quad (kg/m^3)$ is the density of the medium and $c (m/s)$ is the speed of sound in the medium.

For an intensity of $I=0.01 W/m^2$ we obtain a sound pressure of $p=\sqrt{2 \rho c \cdot 0.01} + \sqrt{2 \rho c \cdot 0.01} = 2 \sqrt{2 \rho c \cdot 0.01}$ . Converting back to the total intensity, we get

$\boxed{ \!\begin{aligned} I &= \frac{p^2}{2 \rho c}\\ I_{\text{total}} &= \frac{\left( 2 \sqrt{2 \rho c \cdot 0.01} \right)^2}{2 \rho c} =\frac{4 \cdot 0.01 \cdot 2\rho c}{2 \rho c} = 0.04 \end{aligned} }$

This corresponds to a level of $L_I_{\text{total}}=10log(0.04/10^{-12})=106dB$ .

We observe that the total intensity is 0.04 which is not equal to the sum of the individual intensities 0.01+0.01=0.02. In fact, we have summed the pressures, which as we know gives a +6dB increase in sound pressure level. Correspondingly, also the intensity shows a +6dB increase (compared to a single speaker). Implicitly, we assumed that both signals are coherent because we added pressures.

Let’s change the intensity levels a little bit: suppose you one speaker results in a intensity level of $L_{I}=100dB$ in front of your speaker at a certain fixed distance. Next you add a second speaker that results in an intensity of $L_{I}=90dB$ . If we calculate the intensity level for this situation then you see that we cannot simply add intensities:

100dB corresponds to an intensity of $I=0.01 W/m^2$ which corresponds to a sound pressure of $p=\sqrt{2 \cdot 0.01 \rho c}$ .

90dB corresponds to an intensity of $I=0.001 W/m^2$ which corresponds to a sound pressure of $p=\sqrt{2 \cdot 0.001 \rho c}$ .

Thus, the total pressure is $p=\sqrt{2 \cdot 0.01 \rho c} + \sqrt{2 \cdot 0.001 \rho c} = \sqrt{2 \rho c} (\sqrt{0.01} + \sqrt{0.001})$ .

Converting back to total intensity

$\boxed{ \!\begin{aligned} I &= \frac{p^2}{2 \rho c}\\ I_{\text{total}} &= \frac{\left( \sqrt{2 \rho c} (\sqrt{0.01} + \sqrt{0.001}) \right)^2}{2 \rho c} &= \left( \sqrt{0.01} + \sqrt{0.001} \right)^2 = 0.017325 \end{aligned} }$

We see that the individual intensities (0.01 and 0.001) do not add up to 0.017325. The total intensity level now becomes $L_I_{\text{total}}=10log(0.131623/10^{-12})=102dB$ . If we would have simply summed the intensities, we would have gotten: $L_I_{\text{total}}=10log(0.011/10^{-12})=100.41dB$ .

In the next section you see the same principle again.

Summation of sound powers 1 (coherent signals)

We can also consider the summation of powers $P$ . Suppose that we have one speaker of 10W and a second speaker of 20W which are 2m apart. We want to know the total intensity at point x at a distance of 2m. Point x is at equal distance ( $\sqrt 5 m$ ) from both speakers (Figure 8).

Figure 8. Summation of powers. Figure according to [here].

To calculate the total intensity $L_{\text{total}}$ , we make use of the following two relationships:

$\boxed{ \!\begin{aligned} I &= \frac{p^2}{2 \rho c} \rightarrow p=\sqrt{2 \rho \cdot c \cdot I} \\ I &= \frac{P}{A}=\frac{P}{4 \pi r^2} \end{aligned} }$

We first calculate the intensities for speaker 1 and speaker 2 at point x:

$\boxed{ \!\begin{aligned} I_1 &= \frac{10}{4 \pi \left( \sqrt 5 \right)^2} = \frac{1}{2 \pi} = 0.1592 W/m^2 \rightarrow 112.02dB\\ I_2 &= \frac{20}{4 \pi \left( \sqrt 5 \right)^2} = \frac{1}{\pi} = 0.3183 W/m^2 \rightarrow 115dB \end{aligned} }$

Next, as before, we convert the intensities to pressures and sum them together:

$\boxed{ \!\begin{aligned} p_{\text{total}} &= \sqrt \frac{ 2 \rho c}{2 \pi} + \sqrt \frac{2 \rho c}{\pi}\\ &= \sqrt{2 \rho v} \left[ \sqrt \frac{1}{2 \pi} + \sqrt \frac{1}{\pi} \right] \end{aligned} }$

Now, we convert the pressure back to a total intensity:

$\boxed{ \!\begin{aligned} I_{\text{total}} &= \frac{p^2}{2 \rho c} \\ &= \frac{ (\sqrt{2 \rho c})^2 \left[ \sqrt \frac{1}{2 \pi} + \sqrt \frac{1}{\pi} \right]^2}{2 \rho c} \\ &= \left[ \sqrt \frac{1}{2 \pi} + \sqrt \frac{1}{\pi} \right]^2 \\ &=0.9276 W/m^2 \rightarrow 119.67dB \end{aligned} }$

Thus, we set that the total intensity is 119.67 – 112.02 = 7.65dB larger compared to speaker 1.

Looking at the $L_{SPL}$ increase if comparing both speakers to speaker 1, we have

$\boxed{ \!\begin{aligned} L_{\text{SPL}} &= 20log \left( \frac{\sqrt{2 \rho c} \left[ \sqrt \frac{1}{2 \pi} + \sqrt \frac{1}{\pi} \right]}{\frac{\sqrt{2 \rho c}}{\sqrt{2 \pi}}} \right) \\ &= 20log \left( \frac{\left[ \sqrt \frac{1}{2 \pi} + \sqrt \frac{1}{\pi} \right]}{\frac{1}{\sqrt{2 \pi}}} \right) = 7.65dB \\ \end{aligned} }$

We can also calculate the effective sound power:

$\boxed{ \!\begin{aligned} P & = I \cdot A = 0.9276 \cdot 4 \pi \cdot \left( \sqrt 5 \right)^2 = 58 W \end{aligned} }$

Summation of sound powers 2 (coherent signals)

Now suppose we had two speakers with a acoustic power of $P=10W$ . As we have seen, for a single speaker, the intensity at point x is $I=\frac{1}{2 \pi} = 0.1592 W/m^2 \rightarrow 112.02dB$ . Considering both speakers, this gives a total pressure of

$\boxed{ \!\begin{aligned} p_{\text{total}} &= \sqrt \frac{ 2 \rho c}{2 \pi} + \sqrt \frac{2 \rho c}{2 \pi}\\ &= 2 \sqrt{2 \rho c} \left[ \sqrt \frac{1}{2 \pi} \right] \end{aligned} }$

Now, we convert the pressure back to the total intensity:

$\boxed{ \!\begin{aligned} I_{\text{total}} &= \frac{p^2}{2 \rho c} \\ &= \frac{ \left(2 \sqrt{2 \rho c} \right)^2 \left[ \sqrt \frac{1}{2 \pi} \right]^2 }{2 \rho c}\\ &= \frac{4}{2 \pi} \\ &=0.6366 W/m^2 \rightarrow 118.04dB \end{aligned} }$

Thus, we set that the total intensity is 118.04 – 112.02 = 6dB larger compared to a single speaker 1.

Looking at the $L_{SPL}$ increase if comparing both speakers to speaker 1, we have

$\boxed{ \!\begin{aligned} L_{\text{SPL}} &= 20log \left( \frac{\sqrt{2 \rho c} \left[ \sqrt \frac{1}{2 \pi} + \sqrt \frac{1}{2 \pi} \right]}{\frac{\sqrt{2 \rho c}}{\sqrt{2 \pi}}} \right) \\ &= 20log \left( \frac{ \left[ \sqrt \frac{1}{2 \pi} + \sqrt \frac{1}{2 \pi} \right]}{\frac{1}{\sqrt{2 \pi}}} \right) = 6dB \\ \end{aligned} }$

We can also calculate the effective sound power:

$\boxed{ \!\begin{aligned} P & = I \cdot A = 0.6366 \cdot 4 \pi \cdot \left( \sqrt 5 \right)^2 = 40 W \end{aligned} }$

Summation of sound pressures (incoherent signals)

We have seen that for incoherent signals we need to use the RMS pressure instead of simply summing two pressures. Thus we have:

$\boxed{ \!\begin{aligned} p_{\text{rms}} = \sqrt{p_1^2 + p_2^2} \end{aligned} }$

If we have two speakers with the same producing 100dB each at point x (Figure 9) but with incoherent signals, then these levels add to 103dB instead of 106dB. We can easily derive this:

$\boxed{ \!\begin{aligned} L_{SPL} &= 20 log \frac{p_1}{p_{ref}}\\ 100dB &= 20 log \frac{p_1}{20 \cdot 10^{-6}}\\ p_1 &= 2 Pa \end{aligned} }$

Thus $p_2$ is also 2Pa. Since $p_1 = p_2$ we have a total RMS pressure at point x of $p_{\text{rms}} = \sqrt{p_1^2 + p_2^2} = \sqrt{2p_1^2}$ . This corresponds to an increase of 3dB compared to speaker 1:

$\boxed{ \!\begin{aligned} L_{SPL} &= 20 log \frac{\sqrt{2p_1^2}}{p_1}\\ &= 20 log \frac{\sqrt{2} p_1}{p_1}\\ &= 20 log \sqrt2 = 3dB \end{aligned} }$

Figure 9. Summation of powers for two incoherent signals.

Perceived loudness (psychoacoustics)

The sound pressure $p$ determines how loud we perceive a sound. A doubling of sound pressure corresponds to a level increase of $L_{SPL} = 6dB$ . This also corresponds to a 6dB increase in the acoustic power level $L_{AP}$ and hence a 4 times increase in acoustic power $P$ .

However, sound pressure level is not a measure of perceived loudness. Loudness is defined as the attribute of auditory sensation that allows sounds to be ordered on a scale extending from quiet to loud. Two sounds with the same sound pressure level may not have the same (perceived) loudness. A difference of +6dB SPL between two sounds does also not equal a 2x increase in loudness.

In acoustics, loudness is a subjective measure of the sound pressure $p$ . From psychoacoustics we know that not +6dB SPL but +10dB SPL is perceived as a doubling in loudness. This corresponds to increasing the sound pressure by a factor 3.16 (i.e., $L_{SPL}=10=20log(p) \rightarrow p=3.16$ ). Note. it is more complex then this, i.e., it also depends on frequency (Fletcher–Munson curves).

A 10dB increase in sound pressure level corresponds to a 10 dB in acoustic power level. Consequently, to perceive a doubling of ‘volume’, the power $P$ should be increased 10 times ( $10=10log(P) \rightarrow P=10$ ).

Some other stuff

Mutual coupling

See also the AES paper in the references below

Serial and parallel connections of speakers to an amplifier

See also the comments below the YouTube video to see the confusion.

References

What is a decibel?
Martin, W.H. (1923) The Transmission Unit and Telephone Transmission Reference Systems. The Bell System Technical Journal, 3(3), 400
Martin, W.H. (1929) Decibel — The Name for the Transmission Unit. The Bell System Technical Journal, 8(1) ,1
David M. Howard, Jamie Angus (2017) Acoustics and Psychoacoustics. Fifth Edition. Focal Press, New York.
Don Davies, Eugene Patronis, Jr. (2006) Sound System Engineering. Third Edition. Elsevier Focal Press, Amsterdam
Bob McCarthy (2016) Sound Systems: Design and Optimization. Modern Techniques and Tools for Sound System Design and Alignment. Third Edition, Focal Press, New York.
Engineering at Alberta (Open Education Resources; Sound Power Level and Sound Pressure)
Holland, Keith R. and Newell, Philip R. (1997) Loudspeakers, Mutual Coupling and Phantom Images in Rooms. Audio Engineering Society Convention 103, September 1997

dB or not dB (pdf)

1 file(s) 1.09 MB

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Loudspeakers, Mutual Coupling and Phantom Images in Rooms (AES convention paper) (pdf)

1 file(s) 745.18 KB

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Published On: May 19th, 2024Last Updated: September 21st, 2024Categories: Audio Processing Education, Audio technologyTags: audio technology, decibels, education

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Music, composing, mixing, and mastering

What the h*ll is a decibel?

The Transmission Unit and Telephone Transmission Reference Systems (pdf)

Decibel — The Name for the Transmission Unit (pdf)

dB or not dB (pdf)

Loudspeakers, Mutual Coupling and Phantom Images in Rooms (AES convention paper) (pdf)

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