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Decibels, Logarithms, Sound Pressure & Sound Power (explained)

MrPeabody

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1. Introduction

Decibels are used routinely for expressing electrical power, electrical voltage, sound power and (more commonly) sound pressure level (SPL). Each of these different physical quantities is either a form of power or else has a very specific and simple relationship with power. Let’s begin with two essential facts about decibels:

Fact #1: Decibel values are a type of “dimensionless ratio”. Two familiar examples of dimensionless ratios are Mach numbers and relative density. Mach numbers are obtained by dividing a given speed by the speed of sound, which is used as a reference. Relative density is obtained by dividing the density of a given substance by the density of another substance used as a reference, usually water. When the given quantity is divided by the reference value, the physical units of measure vanish. (Physicists regard all units of measure as “dimensions”, hence a value that is not accompanied by physical units of measure is “dimensionless”.)

Fact #2: Decibel scaling is logarithmic. Logarithmic scaling pushes large values closer together and spreads small values further apart. This effect is prominent, however the underlying rationale is not obvious and has to do with the mathematical properties of logarithms.

Whenever power, voltage, etc., is converted to a dimensionless ratio (the first step in translation to decibel), the denominator in the division operation will often be a standard reference value such as one milliWatt or one milliVolt. This isn’t always the case, and when it isn’t the case, the numerator and denominator will still agree on the physical units of measure. When decibels are used to express dynamic range (the dynamic range of the Compact Disc format for example), the dimensionless ratio is the ratio of the loudest signal that can be encoded in the format, to the quietest non-zero signal that can be encoded in the format. When decibels are used to express signal-to-noise ratios (which are always expressed in decibels), the dimensionless ratio is the ratio of the strength of the signal to the strength of the noise included in the signal.

Occasionally in this article I will need to make some point by saying something about the corresponding physical value, i.e., the value that is accompanied by physical units of measure. Whenever I use the word “physical”, this is my meaning.


2. Base-10 Logarithms

Logarithms arise from the concept of the inverse function. Not all mathematical functions have an inverse function, but if a function has an inverse function, the two functions are complementary in the sense that each reverses the effect of the other. This behavior is exemplified by the familiar calculator function ‘1/X’, which function is its own inverse.

Logarithms are defined by way of inverse functions for exponential functions. So what exactly is an exponential function? Suppose you are given some positive integer number N, and you raise the constant number 10 to the power N (treating N as the independent variable). In a nutshell, this is what an exponential function is, however we should note that even though we are mainly concerned with base 10, that in general the base of an exponential function can be any constant. No matter the base, any exponential function implicitly has an inverse, known as a logarithm function. One consequence of the abstract way in which logarithms are defined is that there isn’t any straightforward way to calculate a logarithm. Prior to the proliferation of computers and calculators, it was ordinarily done using a lookup table or else a slide rule. (Or else approximated by summing the first several terms of a series expansion of the function, or by using numerical methods that involve iteration. The algorithms used in computers typically use iterative numerical methods to obtain highly accurate results, and may use either a lookup table or a crude formula to obtain an initial approximation.)

Although generally it is no simple task to obtain the logarithm of a given number, it is trivial to obtain a logarithm when the given number (the argument) happens to be an integer power of the base of the logarithm (and when we are familiar with the base and with the integer powers of the base). It is thus trivial to know the base-10 log of numbers such as 1, 10, 100, 1000, etc., because these numbers are integer powers of the base and we recognize them as such. For example, since we know that 10^3 = 1000, we also know immediately that the base-10 logarithm of 1000 is 3 (because 3 is the power that you need to raise 10 to, to get 1000). We can easily see that:

- The base-10 logarithm of 100 is 2, because 2 is the value that you need to raise 10 to, to obtain the given number (100).

- The base-10 logarithm of 10 is 1, because 1 is the value that you need to raise 10 to, to obtain the given number (10).

It is easy to see that for numbers between 10 and 100, the base-10 logarithm will be between 1 and 2. For numbers between 100 and 1000, the base-10 logarithm will be between 2 and 3. To extend this pattern to numbers between 1 and 10, we need to make use of a basic fact from elementary algebra: any number raised to the power 0 is 1. It follows from this basic fact that regardless of the base of the logarithm, the logarithm of 1 is always 0. Thus, for numbers between 1 and 10, the base-10 logarithm will be between 0 and 1.

What about the base-10 logarithm for a number between 0 and 1? To answer this question, we ask this related question: What are the values of N for which the operation 10^N will return a result in this range, between 0 and 1? You should not have much difficulty convincing yourself that in order for 10^N to return a value between 0 and 1, N must be negative. This means that the base-10 logarithm of a number between 0 and 1 will be negative.


3. Bels and decibels

Roughly a century ago, some engineers at Bell Labs began using logarithms for expressing power losses in transmission lines. The essential advantage of using a logarithmic scale for power losses was that cumulative power losses could be readily totaled using addition instead of multiplication. This is a consequence of one of the basic properties of logarithms: the log of the product of two or more numbers is equal to the sum of the logarithms of the individual numbers. The Bell Labs engineers defined the "Transmission Unit" or ‘TU’ as 10 x the base-10 log of the power loss ratio. If power suffered a loss to one-half the previous value, this was a loss of -3.01 TU. If the remaining power is again reduced by one-half, this is an additional loss of -3.01 TU, for a total loss of -6.02 TU. Of course it is trivial to multiply .5 x .5, but even so, the ability to total cumulative losses in a simple additive manner is compelling, to the point that it would not occur to most any engineer to do it any other way.

At some point those Bell Labs engineers got around to using “decibel” instead of “TU”, implicitly defining the Bel as the base-10 log of a power ratio.

To translate physical power to decibels, we divide the physical power by a reference power and then take the base-10 logarithm. We then multiply by 10, which we may think of as conversion from Bel to decibel. Let’s suppose that signal power is 500 milliWatts, and that we want to express this in terms of dBmW, i.e., referenced to 1 milliWatt. When we divide 500 mW by 1 mW, the only effect is that the physical units of measure are eliminated. The numerical value doesn’t change. Since we know that the base-10 logarithm of any number between 100 and 1000 is between 2 and 3, we know that if we take the base-10 log of 500 and then multiply by 10, the decibel value will be between 20 and 30. Since 500 is geometrically closer to 1000 than to 100 (1000 is 2x greater than 500, but 500 is 5x greater than 100), we expect that the answer will be closer to 30 dB than to 20 dB. An excellent guess would be 27 dBmW. The geometric midpoint between 100 and 1000 is the value Y such that Y/100 and 1000/Y are equal. In other words, the solution to Y^2 = 10^5, or Y = 10^(5/2). Most likely you will use a calculator to solve this, and you’ll find that the geometric midpoint between 100 and 1000 is roughly 316. If the given value had been 316 mW, the answer would have been 25 dBmW.

What if signal power had been smaller than 1 mW? For example, .5 mW? Since the ratio is less than 1, we know that the result will be negative. The base-10 logarithm of .5 is -.301, so the answer is -3.01 dBmW.

The reference value is the physical value that is mapped to 0 on the decibel scale. I know this is obvious and that the reason why is obvious, but I will note anyway that if you translate the physical reference value to decibels, you will only be taking the log of 1, and the log of 1 is 0. Further obvious yet worth noting is the fact that 0 dB never corresponds to the physical quantity 0. This would imply division by 0, which is not a mathematically meaningful concept.


4. Decibel scaling for voltage and sound pressure

It wasn’t long before the practice emerged of representing voltage in decibels. The reason that it made sense to do this is that problems routinely encountered were problems of the sort, “How much increase in voltage will be needed in order to increase load power by 250%?” Problems of this sort are routine and they lead naturally to the practice of expressing voltage in the decibel scale. There is a catch, though. The catch is that in order for it to be useful to express voltages in decibels, for solving problems of this common sort without creating a world of confusion, it is necessary that decibel scaling be applied to voltage in a manner whereby each increment of +/- 3.01 dB consistently implies a doubling or halving of power.

As most everyone knows, power consumption in a given load is proportional to the square of voltage across the load. Said another way, load voltage varies as the square root of load power. Physical quantities that are like voltage in this way are sometimes referred to as “root-power” quantities. Since load power varies as the square of load voltage, the most obvious way to maintain scaling consistency with decibel representation of power is to square the voltage ratio, thereby converting the voltage ratio to the corresponding power ratio. If voltage is increased by a factor of 2, the corresponding power ratio will be 2^2 = 4. Since the base-10 logarithm of 4 is .602, the decibel increment for doubling of voltage is +6.02. Of course this is twice the decibel increment for doubling/halving of power, which was to be expected.

Let’s now add a third essential fact to the two essential facts we’ve already covered:

Fact #3: Decibels can be used to represent root-power quantities such as voltage, and when so, translations between the physical quantity and decibels are done in a way such that the decibel increment +/- 6.02 corresponds to the doubling/halving of the root-power quantity.

Let’s now look at another one of the important properties of logarithms, sometimes referred to as the “exponent rule”:

log( X^N ) = N x log( X )

The exponent rule gives us an alternative to squaring the voltage ratio. Instead of squaring the voltage ratio, we can simply use 20 for the multiplier, in place of 10. It is for this reason that in formulas used to express root-power quantities in decibels, the multiplier is 20. Whenever you encounter one of these formulas and the multiplier is 20, you can infer that the physical quantity is a root-power quantity.

Because +/- 3.01 dB always implies doubling/halving of power, the reference value that corresponds to 0 dB is the only difference between decibel representation of voltage and decibel representation of power. In many applied contexts, the reference value is immaterial, because all that truly matters are the relative, geometric differences among different physical values, i.e., the physical ratios. And with decibel scaling, the geometric differences among physical values (the physical ratios) are transformed into arithmetic differences. You can interpret a signal-to-noise ratio of 50 dB in terms of either physical voltage or physical power, and in neither case will the reference value be the least bit relevant.

In frequency response plots (either the gain plot for an amplifier or the sensitivity plot for a loudspeaker), the primary concern is with the deviations from a flat response, which is to say, the arithmetic differences in decibels for the peaks (and dips) vs. the nominal value (the mean value over the audible spectrum). The reference value is again immaterial. There is a traditional practice of treating +/- 3dB as having particular significance in interpreting the severity of the peaks and dips. As you know, this is the increment for which root-power physical quantities (voltage and sound pressure) either increase by a factor of 1.41 or decrease by a factor of .707 (respectively, the square root of 2 and the square root of 1/2).

The reference value becomes relevant when the concern is specifically with loudspeaker sensitivity. For loudspeaker sensitivity given in SPL, the physical interpretation of the 0 dB point is the threshold of sound detection (at 1 kHz) by a person with normal, undamaged hearing. Thus, if the sensitivity of a speaker is given as 85.0 dB, this means that (at the stipulated input voltage and the stipulated distance from the speaker) the physical sound pressure produced by the loudspeaker will be 17,800 times greater than physical sound pressure at the threshold of human hearing. (To obtain this number, do this: 10^(85/20).) The accepted value for the threshold of sound detection is .00002 Pa. (Pa is short for Pascal, the unit of measure usually used for sound pressure in acoustics.) Thus, at the stipulated input voltage and at the stipulated distance, the physical sound pressure produced by the speaker will be 17,800 x 2 x 10^-5 = .356 Pa. This is less than .001% of ambient air pressure!

(The stipulated distance for specification of loudspeaker sensitivity is usually 1 meter, and the stipulated input voltage is usually 2.83 volts. 2.83 volts is the voltage implied for 1 watt of power dissipation in an 8 ohm resistive load.)

Consider now the dynamic range for 16-bit linear quantization. The smallest non-zero value that can be encoded is obviously 1, and the largest value is 2^16 -1. The reason for subtracting 1 is that one of the 2^16 unique values is 0. The ratio will be (2^16 -1)/1 = 65535. Since the encoded information represents voltage, we take the base-10 log and multiply by 20. We get 96.3 dB. In common practice, the number of bits is simply multiplied by 6.02. This is actually an approximation, albeit a highly accurate approximation when the number of bits is great enough to be useful in linear quantization of audio. The reason this is an approximation is that the expression ‘N x 6.02’ is equivalent to ‘N x 20 x Log(2)’ and thus equivalent to ‘20 x Log(2^N)’. The expressions ‘20 x Log(2^N)’ and ‘20 x Log(2^N -1)’ are not equivalent expressions, which is particularly evident when they are both evaluated for N smaller than about 5. As N becomes greater, the -1 rapidly diminishes in significance relative to 2^N, and the two expressions rapidly converge. It has become common practice to express improvements in dynamic range and in signal-to-noise ratio in terms of the equivalent quantity of bits. An improvement of 6.02 dB is deemed a one-bit improvement, an improvement of 12.04 dB is deemed a two-bit improvement, etc.

Let’s now consider how to translate a distortion measurement (or distortion+noise) from a percentage to decibels (and back). The only thing you need to know, beyond what we’ve already covered, is that when distortion or distortion+noise is expressed as a percentage, the physical ratio implied by the percentage is a voltage or sound pressure ratio, which means that the multiplier is 20. By way of example, let us suppose that a THD figure is 10%. Since 10% implies a physical ratio of .1, we take the base-10 log of .1 and multiply by 20. The answer is -20 dB. If we do the same for 1%, we get -40 dB. If we do the same for .1%, we get -60 dB. The pattern is evident. 100% distortion (or 100% distortion+noise) is equivalent to 0 dB, and each factor-of-ten reduction in the percentage corresponds to a decibel step of -20 dB. Thus:

10% corresponds to -20 dB
1% corresponds to -40 dB
.1% corresponds to -60 dB
.01% corresponds to -80 dB
.001% corresponds to -100 dB
.0001% corresponds to -120 dB.


5. Sound pressure and sound power

As previously mentioned, the physical unit used for sound pressure is the Pascal, abbreviated Pa, and defined to be 1 Newton per square meter. To put this unit of measure into perspective, atmospheric pressure at altitude fifty meters above sea level is typically about 100,000 Pa. (If 100,000 N/m^2 seems a very large value for atmospheric pressure, the reason is likely twofold: ambient air pressure is probably greater than you realize, and the Newton is probably a smaller unit of force than you realize. Earth’s gravitational force acting on a person with average adult mass of 62 kg is 608 N. The force due to atmospheric pressure, acting on one square meter of flat surface at 50 meters above sea level, is 165 times greater.)

Practical scenarios where we need to translate between physical sound pressure and sound pressure expressed in decibels (SPL) are not common. The reason is that this translation is already done for us, either by the sound meter or by the software that is interpreting the voltage from a microphone. Engineers involved in the design of a sound meter or similar application software need to know that sound pressure is a root-power quantity, and also that the reference value used for SPL is .00002 Pa. As previously mentioned, this reference value for sound pressure in decibels is the accepted value for the threshold of sound detection at 1 kHz, for a young person with undamaged hearing. In engineering and scientific notation this is written 20 uPa or 2x10^-5 Pa.

Whereas sound pressure is defined for point locations within the medium surrounding a sound source, sound power is ordinarily (and correctly) interpreted as the instantaneous rate at which acoustic energy is emitted from a sound source, without regards to directional distribution. Sound power may also be interpreted as the energy flux for the propagation of sound across a virtual membrane within the medium surrounding the source. While this is a theoretically valid interpretation of sound power, it isn’t apparent how much practical value it has.

By intuition we know that sound power and sound pressure must be meaningfully related. Yet, this relationship is not practically able to be described using analytical formulas except in cases where the radiation pattern conforms to a simple mathematical model. The simplest sort of radiation pattern that is useful to this end is unobstructed spherical radiation from a point source. For this convenient radiation pattern, the relationship between physical sound pressure and physical sound power is:

Physical_pressure = [ Density x C x Physical_power / (4 x pi x R^2) ] ^(1/2)

Density is the density of air, C is the speed of sound, and R is the radius of the sphere you choose, i.e., the distance from the source to the detection point. Physical_pressure is the pressure at the surface of the chosen sphere. The speed of sound has to be corrected for air temperature. Commonly used values for the density (at 20 degrees C) and for the speed of sound are 1.20 kg/m^3 and 343 m/s respectively.

It is of course necessary to use consistent units of measure, in order to get a meaningful, useful result. The best way to do this is to express power in Watts, pressure in Pa, the air density in kg/m^3, C in meter/s, and R in meters. As previously noted, the Pa is defined as one Newton of force per square meter. The Watt is defined as one Joule of work per second, and the Joule is defined as the work performed with one Newton of force acting through a distance of one meter. Both the Pa and the Watt involve the Newton, which, being a unit of force, has base units of mass multiplied by acceleration, i.e., mass multiplied by distance per second^2. The Newton is defined as the force associated with 1 kg mass at 1 m/s^2 acceleration.

Note that in this formula, it is apparent that physical sound pressure obeys a proportionality with the square root of physical sound power. Thus, sound pressure is a root-power quantity. Note further that the denominator in the integrand expression is simply the surface area of the chosen sphere (4 x pi x R^2), which varies as the square of the radius. It is thus apparent that physical pressure varies as the inverse of the radius. With each doubling of distance, physical pressure reduces by half, which is a change of -6.02 dB.

If we calculate the physical sound pressure for 1 Watt of sound power and for R = 1 m, we get 5.72 Pa. Translating this to decibels (referenced to 20 uPa), we get 109 dB (SPL). There are two things to keep in mind here. First, the physical power value, 1 Watt, is acoustic power. The power delivered by the amplifier to the speaker includes the power dissipated as heat from the voice coil; this wasted power is on the order of a hundred times greater, depending on the efficiency of the speaker. Second, this applies only to unobstructed spherical radiation.

This analysis provides us with a useful understanding of how physical sound pressure and physical sound power are related. Although, if we were doing this sort of thing for any practical purpose, we would likely prefer a formula where pressure and power are expressed in decibels. Before I discuss this formula, I will add one more thing to the previous formula. In the numerator of the integrand, alongside density and C and physical power, you can factor in a small integer designated Q (it has nothing to do with damping of resonance). For the spherical radiation case, Q is 1, which is why I didn’t previously need to mention it. For half-spherical space where the source is located against an infinitely large plane, Q is 2. For quarter-spherical space where the source is located at the intersection of two planes at right angles to each other, Q is 4. And if this space is divided in half once again (the corner of a room), Q is 8. Now for the other formula:

Lp = Lw - |10 x Log(Q/(4 x pi x R^2)) |

Lp is sound pressure in decibels, referenced to 20 uPa. Lw is sound power in decibels, referenced to 1 picoWatt. Note the use of absolute value. Note also that this decibel formula gives the appearance that sound pressure and sound power are related in an additive fashion. This deceptive appearance is an artifact of the use of logarithmic expression for two quantities. If this were the only formula you examined, you would likely have difficulty making sense of the fact that sound pressure is a root-power quantity.

An exercise that will now be useful will be to convert the physical pressure and power values we got a few paragraphs back (when examining the previous formula) to decibel representation and then apply this formula, just to see what happens. We already did the translation for sound pressure: 109 dB SPL. For sound power in decibels, we need to convert 1 Watt to picoWatt. Since the pico prefix implies 10^-12, and since we take the base-10 log of the reciprocal of this number (because we divide 1 Watt by this number), the base-10 log value is 12, which we multiply by 10 to get 120 dBpW. That is, sound power in decibels is 120 dB. Since this is 11 dB greater than the SPL value, the expression in the absolute value operator should evaluate to either +11 dB or -11dB (for R = 1 m and Q = 1), assuming that both formulas are correct and that I didn’t make any booboos. So let’s see, 4 x pi is 12.57, and if I take the reciprocal of this and then the base-10 log, I get -1.099, and then after multiplying by 10 and rounding off to two digits and taking the absolute value, I get, wait for it, 11 dB.

The absolute value operator was obviously needed in order for this to come out correct. This raises the question of the circumstances under which the expression in the absolute value operator will be negative. Clearly it will be negative when the base-10 log returns a negative value, which it will do when its argument is less than 1. For Q = 1, this will occur when the expression ‘4 x pi x R^2’ is greater than 1, which is to say, when R^2 is > 1/(4 x pi). In other words, when R is greater than .2821 m.


6. More formulas, anyone?

If you are converting electrical power to decibels, you take the ratio, then the log, then multiply by 10. Going the other direction, you first divide by 10, then you do 10^X where X is what you got from the division. This gives you the physical, geometric ratio. If the physical quantity with which you are concerned is voltage or sound pressure, then you simply use 20 instead of 10 for the multiplier (not in place of 10 in 10^X). Other than this, all you really need to remember are the reference values for sound pressure (20 uPa) and for sound power (1 pWatt).

One formula that I think has particular value is the formula by which you can find the net sound pressure (SPL) at a frequency where multiple drivers are contributing. This formula assumes that the individual sources are mutually in phase, such that there is no destructive interference, acoustically.

net_sound_pressure = 20 x Log( 10^(A/20) + 10^(B/20) + 10^(C/20) +…)

The A, B, C, etc., are the partial contributions in decibels. For example, suppose the nominal sensitivity of a 2-way speaker is 85 dB, and that at the crossover point, the individual drivers are each at -6 dB relative to the nominal value. We get:

net_sound_pressure = 20 x Log( 10^(79/20) + 10^(79/20))
= 20 x Log( 2 x 10^(79/20))
= 20 x Log( 2 x 10^3.95)
= 20 x Log( 2 x 8912.5)
= 20 x Log(17825)
= 20 x 4.25
= 85 dB
 
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