Attenuation of Waves

When a wave travels through a medium, its intensity diminishes with distance.  In idealized materials, the wave amplitude is only reduced by the spreading of the wave. Natural materials, however, all produce an effect which further weakens the wave. This further weakening results from scattering and absorption. Scattering is the reflection of the wave in directions other than its original direction of propagation.  Absorption is the conversion of the wave energy to other forms of energy.  The combined effect of scattering and absorption is called attenuation. 

The amplitude change of a decaying/attenuating plane wave can be expressed as:

$$A=A_{0}e^{-\alpha z}$$

In this expression A₀ is the unattenuated amplitude of the propagating wave at some location. The amplitude A is the reduced amplitude after the wave has traveled a distance z from that initial location. The quantity $\alpha$ is the attenuation coefficient of the wave traveling in the z-direction. The dimensions of $\alpha$are nepers/length, where a neper is a dimensionless quantity. The term e is the exponential (or Napier's constant) which is equal to approximately 2.71828.

The units of the attenuation value in Nepers per meter (Np/m) can be converted to decibels/length by dividing by 0.1151. Decibels is a more common unit when relating the amplitudes of two signals. More information about decibels is provided below.

The Decibel

The decibel (dB) is one tenth of a Bel, which is a unit of measure that was developed by engineers at Bell Telephone Laboratories and named for Alexander Graham Bell. The dB is a logarithmic unit that describes a ratio of two measurements. The basic equation that describes the difference in decibels between two measurements is:

$$\Delta X(dB)=10log\frac{X_{2}}{X_{1}}$$

where: delta X is the difference in some quantity expressed in decibels, X₁ and X₂ are two different measured values of X, and the log is to base 10. 

Why is the dB unit used?

Use of dB units allows ratios of various sizes to be described using easy to work with numbers. For example, consider the information in the table.

Ratio between Measurement 1 and 2	Equation	dB
1/2	dB = 10 log (1/2)	-3 dB
1	dB = 10 log (1)	0 dB
2	dB = 10 log (2)	3 dB
10	dB = 10 log (10)	10 dB
100	dB = 10 log (100)	20 dB
1,000	dB = 10 log (1000)	30 dB
10,000	dB = 10 log (10000)	40 dB
100,000	dB = 10 log (100000)	50 dB
1,000,000	dB = 10 log (1000000)	60 dB
10,000,000	dB = 10 log (10000000)	70 dB
100,000,000	dB = 10 log (100000000)	80 dB
1,000,000,000	dB = 10 log (1000000000)	90 dB

From this table it can be seen that ratios from one up to ten billion can be represented with a single or double digit number. Ease to work with numbers was particularly important in the days before the advent of the calculator or computer. The focus of this discussion is on using the dB in measuring sound levels, but it is also widely used when measuring power, pressure, voltage and a number of other things.

Use of the dB in Sound Measurements

Sound intensity is defined as the sound power per unit area perpendicular to the wave. Units are typically in watts/m² or watts/cm². For sound intensity, the dB equation becomes:

$$\Delta I(dB)=10\log\frac{I_{2}}{I_{1}}$$

However, the power or intensity of sound is generally not measured directly. Since sound consists of pressure waves, one of the easiest ways to quantify sound is to measure variations in pressure (i.e. the amplitude of the pressure wave). When making ultrasound measurements, a transducer is used, which is basically a small microphone. Transducers like most other microphones produced a voltage that is approximately proportionally to the sound pressure (P). The power carried by a traveling wave is proportional to the square of the amplitude. Therefore, the equation used to quantify a difference in sound intensity based on a measured difference in sound pressure becomes:

$$\Delta I(dB)=10\log\frac{I_{2}}{I_{1}}=10\log\frac{P_{2}^{2}}{P_{1}^{2}}=20\log\frac{P_{2}}{P_{1}}$$ (The factor of 2 is added to the equation because the logarithm of the square of a quantity is equal to 2 times the logarithm of the quantity.)

Since transducers and microphones produce a voltage that is proportional to the sound pressure, the equation could also be written as:

$$\Delta I(dB)=20\log\frac{V_{2}}{V_{1}}$$ where delta I is the change in sound intensity incident on the transducer and V₁ and V₂ are two different transducer output voltages.

Revising the table to reflect the relationship between the ratio of the measured sound pressure and the change in intensity expressed in dB produces:

Ratio between Measurement 1 and 2	Equation	dB
1/2	dB = 20 lag (1/2)	-6 dB
1	dB = 20 lag (1)	0 dB
2	dB = 20 lag (2)	6 dB
10	dB = 20 lag (10)	20 dB
100	dB = 20 lag (100)	40 dB
1,000	dB = 20 lag (1000)	60 dB
10,000	dB = 20 lag (10000)	80 dB
100,000	dB = 20 lag (100000)	100 dB
1,000,000	dB = 20 lag (1000000)	120 dB
10,000,000	dB = 20 lag (10000000)	140 dB
100,000,000	dB = 20 lag (100000000)	160 dB
1,000,000,000	dB = 20 lag (1000000000)	180 dB

From the table it can be seen that 6 dB equates to a doubling of the sound pressure. Alternately, reducing the sound pressure by 2, results in a – 6 dB change in intensity.

"Absolute" Intensity Levels

Whenever the decibel unit is used, it always represents the ratio of two values. Therefore, in order to relate different intensities it is necessary to choose a standard reference level. The reference sound pressure (corresponding to a sound pressure level of 0 dB) commonly used is that at the threshold of human hearing, which is conventionally taken to be 2×10 −5 Newton per square meter, or 20 micropascals (20 μPa). To avoid confusion with other decibel measures, the term dB(SPL) is used. Similarly for EM waves the quantities dBW and dBm have been defined where the reference is 1 Watt and 1 milliwatt, respectively.