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Basic Acoustics
decibels
Sound pressure and signal levels (power and voltage) are often measured using a decibel scheme,
or dB for short. dB's are logarithmic in nature and thus tend to compress the range of signal
levels. For voltage and current gain, use the form Av’ = 20*log10 Av, and for power gain use the
form G’ = 10*log10 G. For specific signal levels, replace the gain value with the ratio of the
desired signal to the desired reference. Common references are 1 volt (dBV), .775 volts (dBu), 1
watt (dBW), and 1 milliwatt (dBm). Thus, for specifying a power in dBm, use the equation P’ =
10*log10 (P/1 mW). Positive dB values represent increases in level while negative values denote
decreases in level. To place this into a more familiar listening level context, 1 dB represents a
just discernable change in volume under good conditions for trained listeners. 3 dB is a factor of
two for power, but only represents a clearly discernable change in volume for the average
listener. To double perceived volume, a level change of 8 to 10 dB is required. Note that 10 dB
represents a factor of 10 change in power. It is clear then that the human hearing mechanism is
sort of “super logarithmic”. When dealing with sound pressure levels, 0 dB-SPL represents the
quietest sound pressure that a human with good hearing can discern. Conversational levels are
usually in the 70 dB-SPL range. Workplace values that exceed approximately 90 dB-SPL come
under OSHA control in the USA, and time limits are placed on exposures without hearing
protection. The higher the sound pressure, the shorter the allowable exposure. Excessive
exposure to high sound pressures may cause hearing damage. Generally, 120 dB-SPL is agreed to
be the “threshold of feeling” and approximately 140 dB-SPL is considered to be the “threshold of
pain”. (Different sources may quote slightly different values.)
Acoustic Waves
Sound is an acoustic pressure wave. Some device or action imparts a pressure differential in a
medium (air, water, etc.), and that pressure propagates through the medium. A classic example of
a wave is the expanding ripples resulting from a pebble cast into water. Just as there are crests
there are also valleys. This is an example of a transverse wave. In a transverse wave, the particle
motion is at right angles to the direction of the wave. Imagine a bobber on the surface of the
water. As the wave passes, the bobber moves up and down. In air, sound can be viewed as an
alternating pressure superimposed upon the existing air pressure. If you think of voltage as
electrical pressure then this is similar to the voltage that would be seen at the collector of a
transistor amplifier where the AC signal rides upon the fixed DC bias. Sound is an example of a
longitudinal wave; where the motion is in line with the direction of propagation. A good visual
example of a longitudinal wave can be demonstrated by stretching a slinky between two people
and then pushing on one end. The resulting compression can be seen to travel along the direction
of the slinky. It is important to understand that a medium is required to transmit an acoustic
pressure wave. As there is no medium, there is no such thing as sound in outer space, in spite of
what most science fiction movies portray. In space, explosions are silent. It’s like casting a
pebble onto a dry lake bed; there will be no ripples. The medium will directly affect the
transmission of the pressure wave, most notably in terms of velocity of propagation and
absorption of various frequencies. Note that the particle velocity (i.e., movement of molecules) is
considerably smaller than the wave velocity. The velocity of sound in air at sea level on a balmy
day is about 1130 feet per second. The warmer the air, the faster the velocity. In contrast, the
speed of sound in water, a denser medium, is on the order of 4750 feet per second.
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Based on the velocity, it is possible to compute the wavelength of a sound. The wavelength is
equivalent to how far the wave propagates in the time it takes to complete one full cycle. The
lower the frequency, the longer the period, and thus, the longer the wavelength. Wavelength is
given the Greek symbol Lambda . Wavelength is equal to the velocity divided by the frequency,
 = v/f. Note that the wavelength has nothing to do with the magnitude of the pressure
differential. The pressure differential is what gives rise to the sensation of loudness, while the
frequency gives rise to the sensation of pitch.
Hearing Range
Human hearing encompasses the range of approximately 20 Hz to 20 kHz for healthy young
adults. This represents three decades of range or about 10 octaves (an octave being a factor of
two1). This range will reduce as humans age, most notably in the higher frequencies. An
individual past retirement age may top out at only 10 kHz. Their overall sensitivity to sound will
most likely be reduced as well. Frequencies below the range of human hearing are called
infrasonic, although the term subsonic is sometimes used instead. Frequencies above the range of
human hearing are referred to as ultrasonic. It is interesting to note that many land mammals also
exhibit a three decade range, although it may be entirely different from a human’s. Larger
mammals, such as elephants, with their larger inner ear structures can efficiently denote
frequencies much lower than humans can. These larger structures have difficulty vibrating at
higher frequencies, so these animals cannot hear as high as humans can. Conversely, dogs and
cats can hear higher than humans can but not as low. A typical dog will sense from 40 Hz to 40
kHz while a cat may range from 50 Hz to 50 kHz. This explains the operation of the so-called
“silent” dog whistle: If the whistle is tuned to resonate just above 20 kHz, humans won’t sense it,
but fido can. Other animals may exhibit wide variations from the human norm. Insects typically
hear sounds much higher than humans are capable. Aquatic mammals such as dolphins and
specialized mammals such as bats hear well into the ultrasonic range. This extended range may
be used for communication but also for echo location (a means of sightless navigation).
It is important to remember that just because a human can’t hear something doesn’t mean that
there is no sound present. The sound may be either beyond the human hearing frequency range or
it may be too low in intensity to be noticed (either in an absolute sense or due to being masked by
a louder sound). In this regard, human hearing can be likened to a bandpass filter with a specific
noise floor. Many things that we hear do in fact produce appreciable energy beyond the range of
normal hearing. A cymbal, for example, may produce significant energy in the 20 kHz to 30 kHz
region. This can present problems during recording if the audio design engineers are not careful.
The act of hearing is really one of interpretation by the brain. Further details on human hearing
are given later in the course.
1
The term octave is borrowed from Western music. The Latin root oct refers to the fact that there are eight
tones in a Western musical scale, do-re-mi-fa-so-la-ti-do, where do to do is a doubling of frequency. (That’s
Western as in Western Europe, not Country & Western!)
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Physical Interactions
The medium itself may absorb some acoustical energy and this will result in a fall-off of
amplitude. Generally, however, the reduction in amplitude that people are familiar with is caused
by the fact that the pressure wave is a spherical wave-front, meaning that it is growing in two
dimensions. Thus, if the distance is doubled, the same energy must be spread over four times the
initial area producing a fall off in intensity. This is a square-law phenomenon: Intensity falls off
as the square of the distance ratio. Thus, if the sound source is moved 10 times distant, the
received intensity will be 100 times smaller. The foregoing assumes that there are no interactions
with objects or boundaries. This is called the free-field or free-space case. In practical
applications, outdoor sound is often considered free-space. This is not entirely accurate in the
most strict sense as there is some interaction with the ground surface.
Things are a little different in enclosed spaces or with sizable obstructions (we consider
something “sizable” if it’s on the order of the wavelength involved). When a wave hits an
obstruction, one of two things can happen: A portion may be reflected, and a portion may be
either absorbed or diffracted around the object. Frequency (wavelength) will play a major role in
this. Some objects may be quite absorptive at some frequencies but reflective at others. If the
object is large and non-transmissive, a sound “shadow” will exist, meaning that the intensity will
be lower immediately behind the object. The sound will diffract (bend) around the object and
thus at a considerable distance from the object there will be no discernable change in intensity in
line with the obstruction. Again, note that “large” means with respect to the wavelength, so low
frequencies bend around obstructions more easily than do high frequencies. In the case of
reflected energy, it may be widely dispersed if the surface is irregular or it may reflect off in a
single direction if the surface is regular (again, with regard to the wavelength involved). The later
case gives rise to the familiar echo phenomenon. If sound is directed at and perpendicular to a
large flat surface some distance away, a sizable portion of the energy will be reflected directly
back to the sound source. Due to the finite velocity of propagation, the sound will returned
delayed in time.
If the reflected sound can hit other objects, further reflections and absorptions will occur. This
leads to the case of an enclosed space (room). It is possible to consider the sound source as
producing an infinite number of “rays” of sound into the room. Each ray will hit obstructions
(walls, furniture, etc.) and be partially absorbed and partially reflected. Each reflection will in
turn hit other objects, and so on. With each collision the energy of the ray is reduced to do partial
absorption. Eventually, these rays reach a listener in the room. The listener now hears the direct
initial sound plus a series of delayed reflected rays. The more a ray is reflected before it reaches
the listener, the longer it takes to arrive and the lower its intensity. This can be considered to be a
series of very tightly packed echoes of both increasing density and decreasing amplitude. As they
are so closely spaced in time, individual echoes are not noticed, but rather the echoes tend to
smear together. This is called reverberation. Reverberation, or reverb for short, is what gives
rooms their acoustic signature or ambience. One measure of reverb is the time it takes for the
echoes to die away (specifically, 60 dB down). This is called reverb time or RT 60. Other
important characteristics include the relative absorption at various frequencies, the density of the
echoes, and the amount of time elapsed before the first echo appears. These factors are what
make the difference between a theatre, a living room, and an auditorium. Large spaces with little
absorption tend to have long reverb times. If RT60 is too long, it can hinder speech intelligibility,
although it might make for a pleasant concert. A room is said to be “dead” if the reverb time is
relatively short and there is considerable high frequency absorption. A “live” room, in contrast,
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tends to have less high frequency absorption and a longer reverb time. A carpeted living room
with upholstered furniture and drapes is “dead” while a formal dinning room with hardwood
floors and bare wooden furniture is relatively “live”. Bathrooms are usually “live” as well. The
ultimate “dead” room is an anechoic chamber. This is a room with virtually no reverb and is used
to test items such as loudspeakers. There is further discussion of reverb and echo in the later part
of the course.
Example Problems
1. Q: Express a power of 10 watts in dBm. A: P’=10*log10 (P/1 mW), so P’ = 10*log10 (10 W /1
mW), P’ = 40 dBm.
2. Q: You see a flash of lightening and hear the corresponding thunder crash 3 seconds later.
How far away was the lighten strike? A: Assuming standard temperature and pressure, the
speed of sound is about 1130 feet per second. The speed of light is fast enough to ignore in
this case. As there was a 3 second delay, the strike must have been 3*1130, or 3390 feet
away.
3. Q: Determine the wavelength of a 1 kHz tone in air under standard conditions. A:  = v/f, so
 = 1130 fps/ 1 kHz,  = 1.13 feet.
4. Q: A sound source is measured to be 90 dB-SPL at 100 meters in free-field (i.e., not in a
room). Ignoring atmospheric absorption, what should the level be at 200 meters, assuming a
spherical wavefront? A: This represents a doubling of distance. Intensity falls off as the
square of distance change, so the intensity drop is a factor of 22, or 4. In other words, the
intensity is ¼ of the 100 meter value. G’=10*log10 G, so G’=10*log10 ¼. G’=-6dB.
Therefore, the signal is 6 dB from the prior value. Thus the 200 meter intensity should be 90
dB-SPL minus 6 dB, or 84 dB-SPL. This is a good rule of thumb to remember for free-field
calculations: A doubling of distance produces a 6 dB drop in level.
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