Hearing Science SPPA 2060
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The Physics of Sound 1 The Physics of Sound Sound lies at the very center of speech communication A sound wave is both the end product ofthe speech production mechanism and the primary source of raw material used by the listener to recover the speaker39s message Because ofthe central role played by sound in speech communication it is important to have a good understanding ofhow sound is produced modi ed and measured The purpose ofthis chapter will be to review some basic principles underlying the physics of sound with a particular focus on two ideas that play an especially important role in both speech and hearing the concept ofthe spectrum and acoustic filtering The speech production mechanism is a kind of assembly line that operates by generating some relatively simple sounds consisting of various combinations of buzzes and hisses and then ltering those sounds by making anumber of ne adjustments to the tongue lips jaw and other articulators We will also see that a crucial step at the receiving end occurs when the ear breaks this complex sound into is individual frequency components in much the same way that a prism breaks white light into componenw of different optical frequencies Before getting into these ideas it is rst necessary to cover the basic principles of vibmtion and sound propagation Sound and Vibration A sound wave is an air pressure disturbance that resulw from the vibration of an object or a column ofair The two conditions that are required for the generation of a sound wave are a vibrating object and an elastic medium the most familiar ofwhich is air We will begin by describing the characteristics ofvibrating objecw and then see what happens when vibratory motion occurs in an elastic medium such as air We can begin by examining a simple vibrating object such as the one shown in Figure 31 If we set this object into vibration by tapping it from the bottom the bar will begin an upward and downward oscillation until the internal resistance ofthe bar causes the vibration to cease The graph to the right ofFigure 31 is a visual representation ofthe upward and downward motion of the bar To see how this graph is created imagine that we use a strobe light to take a series of snapshots of the bar as it vibrates up and down For each snapshot we measure the instantaneous displacement ofthe bar which is the difference between the position of the bar at the split second that the snapshot is taken and the position ofthe bar at rest The rest position of the bar is arbitrarily given a displacement of zero39 positive numbers are used for displacements above the rest position and negative numbers are used for displacements below the rest position So the rst snapshot taken just as the bar is struck will show an instantaneous displacement of zero39 the next snapshot will show a small positive displacement the next will show a somewhat larger positive displacement and so on The pattern that is traced out has a very speci c shape to it The type ofvibratory motion that is produced by a sirn le vibratory system ofthis kind is called simple harmonic motion or uniform circular motion and the pattern that is traced out in the graph is called a sine wave or a sinusoid Figure 31 A bar is xed at one and is set into vibration by tapping it from the bottom Imagine that a strobe light is used to take a series of snapshots ofthe bar as it vibrates up and down At each snapshot the instantaneous displacement ofthe bar is measured Instantaneous displacement is the distance between the rest position of the bar de ned as zero displacement and its position at any particular instant in time Positive numbers signify displacements that are above the rest position while negative numbers signify displacements that are below the rest position The vibratory pattern that is traced out when the sequence of displacements is graphed is called a sinusoid Instantaneous Dlsplacement O r l The Physics afSaund 2 Basic Terminology We are now in a position to define some of the basic terminology that applies to sinusoidal vibration periodic The vibratory pattern in Figure 31 and the waveform that is shown in the graph are examples of periodic vibration which simply means that there is a pattern that repeats itself over time cycle Cycle refers to one repetition of the pattern The instantaneous displacement waveform in Figure 31 shows four cycles or four repetitions of the pattern period Period is the time required to complete one cycle ofvibration For example if 20 cycles are completed in 1 second the period is 120th of a second s or 005 s For speech applications the most commonly used unit of measurement for period is the millisecond ms 1 ms 11000 s 0001s 10393 s A somewhat less commonly used unit is the microsecond us 1 us 11000000 s 0000001 s 10396 s freguency Frequency is defined as the number of cycles completed in one second The unit of measurement for frequency is hertz Hz and it is fully synonymous the older and more straightforward term cycles per second cps Conceptually frequency is simply the rate of vibration The most crucial function of the auditory system is to serve as a frequency analyzer 7 a system that determines how much energy is present at different signal frequencies Consequently frequency is the single most important concept in hearing science The formula for frequency is f 1t where f frequency in Hz t period in seconds So for aperiod 005 s f 1t 1005 20 Hz It is important to note that period must be represented in seconds in order to get the answer to come out in cycles per second or Hz If the period is represented in milliseconds which is very often the case the period first has to be converted from milliseconds into seconds by shifting the decimal point three places to the left For example for a period of 10 ms f 110 ms 1001s 100 Hz Similarly for a period of 100 ps f1100 us 100001s 10000 Hz The period can also be calculated if the frequency is known Since period and frequency are inversely related t 1f So for a 200 Hz frequency t 1200 0005 s 5 ms Characteristics of Simple Vibratory Systems Simple vibratory systems of this kind can differ from one another in just three dimensions frequency amplitude and phase Figure 32 shows examples of signals that differ in frequency The term amplitude is a bit different from the other terms that have been discussed thus far such as force and pressure As we saw in the last chapter terms such as force and pressure have quite specific definitions as various combinations of the basic dimensions of mass time and distance Amplitude on the other hand will be used in this text as a generic term The Physics afSaund 3 meaning quothow much quot How much what The term amplitude can be used to refer to the magnitude of displacement the magnitude of an air pressure disturbance the magnitude of a force the magnitude of power and so on In the present context the term amplitude refers to the magnitude of the displacement pattern Figure 33 shows two displacement waveforms that differ in amplitude Although the concept of amplitude is as straightforward as the two waveforms shown in the figure suggest measuring amplitude is not as simple as it might seem The reason is that the instantaneous amplitude of the waveform in this case the displacement of the object at a particular split second in time is constantly changing There are many ways to measure amplitude but a very simple method called peaktopeak amplitude will serve our purposes well enough Peakto peak amplitude is simply the difference in amplitude between the maximum positive and maximum negative peaks in the signal For example the bottom panel in Figure 33 has a peaktopeak amplitude of 10 cm and the top panel has a peaktopeak amplitude of 20 cm Figure 34 shows several signals that are identical in frequency and amplitude but differ from one another in phase The waveform labeled 0 phase would be produced if the bar were set into vibration by tapping it from the bottom The waveform labeled 180 phase would be produced if the bar were set into vibration by tapping it from the top so that the initial movement of the bar was downward rather than upward The waveforms labeled 90 phase and 270 phase would be produced if the bar were set into vibration by pulling the bar to maximum displacement and letting go beginning at maximum positive displacement for 90 phase and beginning at maximum negative displacement for 270 phase So the various vibratory patterns shown in Figure 34 are identical except with respect to phase39 that is they begin at different points in the vibratory cycle As can be seen in Figure 35 the system for representing phase in degrees treats one cycle of the waveform as a circle39 that is one cycle equals 360 For example a waveform that begins at zero displacement and shows its initial movement upward has a phase of 0 a waveform that begins at maximum positive displacement and shows its initial movement downward has a phase of 90 and so on Instantaneous Amp Instantaneous Amp Time ms Figure 32 Two vibratory patterns that differ in frequency The panel on top is higher in frequency than the panel on bottom The Physics afSaund Instantaneous Amp Instantaneous Amp Ti m e m s Figure 33 Two vibratory patterns that differ in amplitude The panel on top is higher in amplitude than the panel on bottom Phase0 Phase90 Phase180 Figure 34 Four vibratory patterns that differ in phase Shown above are vibratory pattems with phases of 0 90 180 and 270 4 The Physics of Sound 5 90 O 180 0360 Instantaneous Amplitude Time gt 270 Figure 35 The system for representing phase treats one cycle of the vibratory pattern as a circle consisting of 360 A pattern that begins at zero amplitude heading toward positive values ie heading upward is designated 00 phase a waveform that begins at maximum positive displacement and shows its initial movement downward has a phase of 90 a waveform that begins at zero and heads downward has a phase of 180 and a waveform that begins at maximum negative displacement and shows its initial movement upward has a phase of 270 The four phase angles that are shown above are just examples An infinite variety of phase angles are possible Springs and Masses We have noted that objects can vibrate at different frequencies but so far have not discussed the physical characteristics that are responsible for variations in frequency There are many factors that affect the natural vibrating frequency of an object but among the most important are the mass and stiffness of the object The effects of mass and stiffness on natural vibrating frequency can be illustrated with the simple springandmass systems shown in Figure 36 In the pair of springandmass systems to the left the masses are identical but one spring is stiffer than the other If these two springandmass systems are set into vibration the system with the stiffer spring will vibrate at a higher frequency than the system with the looser spring This effect is similar to the changes in frequency that occur when a guitarist turns the tuning key clockwise or counterclockwise to tune a guitar string by altering its stiffness1 The springandm ass systems to the right have identical springs but different masses When these systems are set into vibration the system with the greater mass will show a lower natural vibrating frequency The reason is that the larger mass shows greater inertia and consequently shows greater opposition to changes in direction Anyone who has tried to push a car out of mud or snow by rocking it back and forth knows that this is much easier with a light car than a heavy car The reason is that the more massive car shows greater opposition to changes in direction In summary the natural vibrating frequency of a springandmass system is controlled by mass and stiffness Frequency is directly proportional to stiffness STFT and inversely proportional to mass It is important to recognize that these rules apply to all objects and not just simple springandmass systems For example we will see that the frequency of vibration of the vocal folds is controlled to a very large extent by muscular forces that act to alter the mass and stiffness of the folds We will also see that the frequency analysis that is carried out by the inner ear depends to a large extent on a tuned membrane whose stiffness varies systematically from one end of the cochlea to the other Sound Propagation As was mentioned at the beginning of this chapter the generation of a sound wave requires not only vibration but also an elastic medium in which the disturbance created by that vibration can be transmitted see Box 31 bell jar experiment described in Patrick s science book nalyel writtenD To say that air is an elastic medium means that air like all other matter tends to return to its original shape after it is deformed through the application of a force he example of tuning a guitar string is imperfect since the mass of the vibrating portion of the string decreases slightly as the string is tightened This occurs because a portion of the string is wound onto the tuning key as it is tightened The Physics afSaund 6 The prototypical example of an object that exhibits this kind of restoring force is a spring To understand the mechanism underlying sound propagation it is useful to think of air as consisting of collection of particles that are connected to one another by springs with the springs representing the restoring forces associated with the elasticity of the medium Air pressure is related to particle density When a volume of air is undisturbed the individual particles of air distribute themselves moreorless evenly and the elastic forces are at their resting state A volume of air that is in this undisturbed state it is said to be at atmospheric pressure For our purposes atmospheric pressure can be defined in terms of two interrelated conditions 1 the air molecules are approximately evenly spaced and 2 the elastic forces represented by the interconnecting springs are neither compressed nor stretched beyond their resting state When a vibratory disturbance causes the air particles to crowd together ie producing an increase in particle density air pressure is higher than atmospheric and the elastic forces are in a compressed state Conversely when particle spacing is relatively large air pressure is lower than atmospheric HTIME 963996396399 V V Figure 37 Shown above is a highly schematic illustration of the chain reaction that results in the propagation of a sound wave modeled after Denes and Pinson 1963 When a vibrating object is placed in an elastic medium an air pressure disturbance is created through a chain reaction similar to that illustrated in Figure 37 As the vibrating object a tuning fork in this case moves to the right particle at which is immediately adjacent to the tuning fork is displaced to the right The elastic force generated between particles 11 and b not shown in the figure has the effect a split second later of displacing particle b to the right This disturbance will eventually reach particles c d e and so on and in each case the particles will be momentarily crowded together This crowding effect is called compression or condensation and it is characterized by dense particle spacing and consequently air pressure that is slightly higher than atmospheric pressure The propagation of the disturbance is analogous to the chain reaction that occurs when an arrangement of dominos is toppled over Figure 37 also shows that at some close distance to the left of a point of compression particle spacing will be greater than average and the elastic forces will be in a stretched state This effect is called rarefaction and it is characterized by relatively wide particle spacing and consequently air pressure that is slightly lower than atmospheric pressure The compression wave along with the rarefaction wave that immediately follows it will be propagated outward at the speed of sound The speed of sound varies depending on the average elasticity and density of the medium in which the sound is propagated but a good working figure for air is about 35000 centimeters per second or approximately 783 miles per hour Although Figure 37 gives a reasonably good idea of how sound propagation works it is misleading in two respects First the scale is inaccurate to an absurd degree a single cubic inch of air contains approximately 400 billion molecules and not the handful of particles shown in the figure Consequently the compression and rarefaction effects are statistical rather than strictly deterministic as shown in Figure 37 Second although Figure 37 makes it appear that the air pressure disturbance is propagated in a simple straight line m mm gm mad man Whammidyka m u dmzmns mm mm m m 3 ma samhubam magn z Xuxhkhdww smmnPthgII mmwa with th ammm mmm uwm lymmsm m mum b mum mnmumuamssm WMBW dun 991mm 2mm me mnmmm mymhsxem mm mm m mmma mxumwms FIVEgnu a m m m gm umquot mum mm mm humha 51m wwwmm umullyrpha am wwwmmu mm Slyempasedmm gl nmm g bhehd hm m vaVPgum mh n mm wuwfmm Mum msmawmfamukasmhldn m amsmmmvm mm mm um m gm m mum mum nvmmwms a m mmuumm u wnhexm yvhe M m msmewmfamduus mmm m af x 9236 gt 3 m x mumufhywtglumvfsammw mm mm mnmmmm dun has m rpu mmngm mum mm mum gm x m f m zzquot rd A 5 mmm Mapummmmsmmxw mprml ufsmamwnhyemduflnmmltamqm umqumyu nn z mwwmmmmmswm wmamswmmm 3 mm m mu mhmng mus Pngmlw m m swmmnnmmmmm w mm mum Whimlgwhnmmumus unwasemwnhwhlyhasessavha up mmm soume unusesz m quotmmy a man mmhw vi m awqmmm mphum memwmshwwsmk Ina12W wwwmwa mu wwwm mummm m mu sznhhugniuf m cumpm mm 3 Sumst x mm mm Wynn sunk m m Puwdk mmsLhxdun B Pummduulyuu mi mm m Kmph mmsdulmua 3 my m mummy mmmmm m B ammdmvlu vwlmydmmx nsmhmmska IRK ngm msw y Wm m m 5 syun cummpwm sm mmmnpmmmmmxmm mam useszmphxyemdlz sands wnsmufmuhm swfnq umyzmpamm Almmmunml wwwmmmmmmwm m The Physics afSaund 9 Period 10 ms Freq 100 Hz Amp 400 Phase 90 Amplitude spectrum phase spectrum 200 39 39 39 39 39 39 39 400 360 9 a 100 a 300 270 w 395 w E 43 o a 200 180 lt E lt E 100 100 90 39200 39I I I I I I I39 0 0 0 100 200 300 400 500 0 100 200 300 400 500 0 5 10 15 20 25 30 Frequency Hz Frequency Hz Period 5 ms Freq 200 Hz Amp 200 Phase 180 200 39 39 39 39 39 39 39 400 360 9 a 100 a 300 270 a 39o w E 2 r x 0 Ta 200 180 lt E n 17 lt E 100 100 90 39200 I I I I I I I39 0 100 200 300 400 500 0 100 200 300 400 500 0 5 10 15 20 25 30 Frequency Hz Frequency Hz Period 25 ms Freq 400 Hz Amp 200 Phase 270 200 39 39 39 39 39 39 39 400 360 9 a 100 a 300 270 a 39o w E 2 r x 0 39 Ta 200 180 lt E lt E 100 100 90 39200 39I I I I I I I39 39 0 100 200 300 400 500 0 100 200 300 400 500 0 5 15 25 30 Frequency Hz Frequency Hz TIme msec TIME DOMAIN FREQUENCY DOMAIN Figure 310 Time and frequency domain representations of three sinusoids The frequency domain consists of two graphs an amplitude spectrum and a phase spectrum An amplitude spectrum is a graph showing what frequencies are present with what amplitudes and a phase spectrum is a graph showing the phases of each frequency component Figure 311 shows several examples of complex periodic signals along with the amplitude spectra for these signals The time required to complete one cycle of the complex pattern is called the fundamental period This is precisely the same concept as the term period that was introduced earlier The only reason for using the term quotfundamental periodquot instead of the simpler term quotperiodquot for complex periodic signals is to differentiate the fundamental period the time required to complete one cycle of the pattern as a whole from other periods thatmay be present in the signal eg more rapid oscillations that might be observed within each cycle The symbol for fundamental period is to Fundamental frequency f0 is calculated from fundamental period using the same kind of formula that we used earlier for sinusoids fo 1to The signal in the top panel ofFigure 311 has a fundamental period of 5 ms so fo 10005 200 Hz Examination of the amplitude spectra of the signals in Figure 311 confirms that they do in fact consist of more than a single frequency In fact complex periodic signals show a very particular kind of amplitude spectrum called a harmonic spectrum A harmonic spectrum shows energy at the fundamental frequency and at Whole number multiples of the fundamental frequency For example the signal in the top panel of Figure 311 has energy present at 200 Hz 400 Hz 600 Hz 800 Hz 1000 Hz 1200 Hz and so on Each frequency component in the The Physics afSaund Figure 311 Time and frequency domain representations of three complex periodic signals Complex periodic signals have harmonic spectra with energy at the fundamental frequency f0 and at Whole number multiples of f0 f039 2 f039 3 f039 4 etc For example the signal in the upper left with a fundamental frequency of 200 Hz shows energy at 200 Hz 400 Hz 600 Hz etc In the spectra on the right amplitude is measured in arbitrary units The main point being made in this figure is the distribution of harmonic frequencies at Whole number multiples of f0 for complex periodic signals 10 The Physics afSaund 11 Q 200 D 3 1100 8 i iquot 50 a Z amp 0 0 4M 0 01001 101600 1o 20 30 4o 50 1 3 4 5 e a 9 10 t0 10 100 Hz FrequencyH Q 2A0 100 20 ii 60 80 W 3 m0 3 60 e 40 l lt 4 g yVJ 7 20 s 20 I II 39200 quotAql IA all I0 11 0 5 10 153 20 25 30 20 40 60 800 1C00120014001600 A 100 Frequency Hz no 80 20 a 5 V n 5 E 60 q 80 g ltE40 n E 60 9 20 E lt 40 I s 2012345678910 Frequency Hz 5 10 15 20 25 30 A TIMEnIaoMAI N FREdl E c gmM ng 0 160 Figure 31 39 ain representations of W i signals Unli ewmgm mls complex aperiodic signals s Eg li srsga N across the spectrum This type of spectrum is called dense or continuous These spectra have a very different appearance from the picket fence look that is associated with the discrete harmonic spectra of complex periodic signals amplitude spectrum of a complex periodic signal is called a harmonic also known as a partial The fundamental frequency in this case 200 Hz is also called the first harmonic the 400 Hz component 2 f0 is called the second harmonic the 600 Hz component 3 f0 is called the third harmonic and so on The second panel in Figure 311 shows a complex periodic signal with a fundamental period of 10 ms and consequently a fundamental frequency of 100 Hz The harmonic spectrum that is associated with this signal will therefore show energy at 100 Hz 200 Hz 300 Hz 400 Hz 500 Hz and so on The bottom panel of Figure 311 shows a complex periodic signal with a fundamental period of 25 ms a fundamental frequency of 400 Hz and harmonics at 400 800 1200 1600 and so on Notice that there two completely interchangeable ways to define the term fundamental frequency In the time domain the fundamental frequency is the number of cycles of the complex pattern that are completed in one second In the frequency domain except in the case of certain special signals the fundamental frequency is the lowest harmonic in the harmonic spectrum Also the fundamental frequency defines the harmonic spacing that is when the fundamental frequency is 100 Hz harmonics will be spaced at 100 Hz intervals ie 100 200 300 when the fundamental frequency is 125 Hz harmonics will be spaced at 125 Hz intervals ie 125 250 375 and when the fundamental frequency is 200 Hz harmonics will be spaced at 200 Hz The Physics afSaund 12 Rap on Desk 100 200 7 so 100 n V U D g 60 E 0 s lt 40 g 100 20 200 o 10 20 30 4o 50 60 70 so 90 100 1 2 3 4 5 CI ap 100 200 7 so 100 n v 39c D g 60 5 0 E 17 100 lt 40 E 20 200 o 10 20 30 4o 50 60 70 so 90 100 1 2 3 4 5 Tap on Cheek 100 200 7 80 g 100 n V U D g 60 E 0 s 17 100 lt 40 E 20 200 n o 10 20 30 4o 50 60 70 so 90 100 1 2 3 4 5 TIME msec Frequency kHz TIME DOMAIN FREQUENCY DOMAIN intervals ie 200 400 600 For some special signals this will not be the case 2 So when fo is low harmonics will be closely spaced and when fo is high harmonics will be widely spaced This is clearly seen in Figure 31 1 the signal with the lowest f0 100 Hz the middle signal shows the narrowest harmonic spacing while the signal with the highest f0 400 Hz the bottom signal shows the widest harmonic spacing There are certain characteristics of the spectra of complex periodic sounds that can be determined by making simple measurements of the time domain signal and there are certain other characteristics that require a more complex analysis For example simply by examining the signal in the bottom panel of Figure 311 we can determine that it is complex periodic ie it is periodic but not sinusoidal and therefore it will show a harmonic spectrum with energy at whole number multiples of the fundamental frequency Further by measuring the fundamental period 25 ms and converting it into fundamental frequency 400 Hz we are able to determine that the signal will have energy at 400 800 1200 1600 etc But how do we know the amplitude of each of these frequency components And how do we know the phase of each component The answer is that you cannot determine harmonic amplitudes or phases simply by inspecting the signal or by making simple measurements of the 2There are some complex periodic signals that have energy at odd multiples of the mdamental frequency only A square wave for example is a signal that ma imum nosi ive rquot 4 39 39 rquot 4 The spectrum of square wave shows energy at odd multiples of the fundamental frequency only Also a variety of simple signal processing tricks can be used to create signals with harmonics at any arbitrary set of frequencies For example it is a simple matter to create a signal with energy at 400 500 and 600 Hz only While these kinds of signals can be quite useful for conducting auditory perception experiments it remains true that most naturally occurring complex periodic signals have energy at all whole number multiples ofthe mdamental frequency The Physics afSaund 13 time domain signals with a ruler We will see soon that a technique called Fourier analysis is able to determine thaw aid m an L sgme 139 represent mm p I t t ato u Examples of nonspeech aperiodic sounds include a drummer39s cymbal or nare drum the hiss r39ia L aperiodu Inst Air Pre ape examples of transients In common dequot mplitude spectrum and the phase spectrum of any signal We will also see that the inner ears of humans other animals have developed a trick that is able to produce a neural representation that is comparable in I mI I Jam1 minim a I a I I alent to a phase ectrum This exp e spectrum is for speech and hearing applications than the phase spectrum We will return to this point later DrillHI A A I 39mmarize l a complex periodic signal is any periodic signal that is not sinusoidal 2 complex periodic ther f 1 t 4 v I thef ment lt1 l a dg produced by a are two types of aperiodic sounds 1 continuous Is no sha h I I I 2 ind static sound produced by a poorly tuned radio There ds 1 o 39 a tra 39 ts tho eir rie ur 0 on io s refl e ur 1g s amples of time domain representations and amplitude spectra for continuous aperiodic sounds The lack of in the time domain is quite evident that is unlike the periodic sounds we have seen there is no pattern AAA AAA AAA A 6 J UV UV VVmMMQ ire 3 Time and frequXch domain represlntations of three dmationgtPops clicks and the sound gun fire are iodic signals that are defined by their brief with longer d h o n complex aperiodic signals transients show Figure 314 Illustration of the principle underlying Fourier analysis The complex periodic signal shown in panel 6 was derived by pointforpoint summation of the sinusoidal signals shown in panels ad Pointforpoint summation simply means beginning at time zero ie the start of the signal and adding the instantaneous amplitude of signal a to the instantaneous amplitude of signal I at time zero then adding that sum to the instantaneous amplitude of signal 0 also at time zero then adding that sum to instantaneous amplitude of signal d at time zero The sum of instantaneous amplitudes at time zero of signals ad is the instantaneous amplitude of the composite signal 6 at time zero For example at time zero the amplitudes of sinusoids ad are 0 100 200 and 0 respectively producing a sum of 100 This agrees with the instantaneous amplitude at the very eginning of composite signal e The same summation procedure is followed for all time points A 39lic The Physics afSaund 14 FREQUENCYDOMAW 0 C B i E lt HMEDOMNN 0 200 400 600 800 Frequency Hz 5 n Fo 39 lt 4 1 I E I Time gt g n 0 200 400 600 800 Frequency Hz Figure 315 A signal enters a Fourier analyzer in the time domain and exits in the frequency domain As outputs the Fourier analyzer produces two frequencydomain representations an amplitu e spectrum that shows the amplitude of each sinusoidal component that is present in the input signal and a phase spectrum that shows the phase of each of the sinusoids The input signal can be reconstructed perfectly by summing sinusoids at frequencies amplitudes and phase that are shown in the Fourier amplitude and phase spectra using the summing method that is illustrated in Figure 314 All aperiodic sounds both continuous and transient are complex in the sense that they always consist of energy at more than one frequency The characteristic feature of aperiodic sounds in the frequency domain is a dense or continuous spectrum which stands in contrast to the harmonic spectrum that is associated with complex periodic sounds In a harmonic spectrum there is energy at the fundamental frequency followed by a gap with little or no energy followed by energy at the second harmonic followed by another gap and so on The spectra of aperiodic sounds do not share this quotpicket fencequot appearance Instead energy is smeared moreorless continuously across the spectrum The top panel in Figure 312 shows a specific type of continuous aperiodic sound called white noise By analogy to white light white noise has a at amplitude spectrum that is approximately equal amplitude at all frequencies The middle panel in Figure 312 shows the sound s and the bottom panel shows sound f Notice that the spectra for all three sounds are dense that is they do not show the quotpicket fencequot look that reveals harmonic structure As was the case for complex periodic sounds there is no way to tell how much energy there will be at different frequencies by inspecting the time domain signal or by making any simple measures with a ruler Likewise there is no simple way to determine the phase spectrum So after inspecting a timedom ain signal and determining that it is aperiodic all we know for sure is that it will have a dense spectrum rather than a harmonic spectrum Figure 313 shows time domain representations and amplitude spectra for three transients The transient in the top panel was produced by rapping on a wooden desk the second is a single clap of the hands and the third was produced by holding the mouth in position for the vowel o and tapping the cheek with an index finger Note the brief durations of the signals Also as with continuous aperiodic sounds the spectra associated with transients are dense39 that is there is no evidence of harmonic organization In speech transients occur at the instant of articulatory release for stop consonants There are also some languages such as the South African languages Zulu Hottentot and Xhosa that contain mouth clicks as part of their phonemic inventory MacKay 1986 Fourier Analysis The Physics of Sound 15 Fourier analysis is an extremely powerful tool that has widespread applications in nearly every major branch of physics and engineering The method was developed by the 19 11 century mathematician Joseph Fourier and although Fourier was studying thermal waves at the time the technique can be applied to the frequency analysis of any kind of wave Fourier s great insight was the discovery that all complex waves can be derived by adding sinusoids together so long as the sinusoids are of the appropriate frequencies amplitudes and phases For example the complex periodic signal at the bottom of Figure 314 can be derived by summing sinusoids at 100 200 300 and 400 Hz with each sinusoidal component having the amplitude and phase that is shown in the figure see the caption of Figure 314 for an explanation of what is meant by summing the sinusoidal components The assumption that all complex waves can be derived by adding sinusoids together is called Fourier39s theorem and the analysis technique that Fourier developed from this theorem is called Fourier analysis Fourier analysis is a mathematical technique thattakes a time domain signal as its input and determines 1 the amplitude of each sinusoidal component that is present in the input signal and 2 the phase of each sinusoidal component that is present in the input signal Another way of stating this is that Fourier analysis takes a time domain signal as its input and produces two frequency domain representations as output 1 an amplitude spectrum and 2 a phase spectrum The basic concept is illustrated in Figure 315 which shows a time domain signal entering the Fourier analyzer Emerging at the output of the Fourier analyzer is an amplitude spectrum a graph showing the amplitude of each sinusoid that is present in the input signal and a phase spectrum a graph showing the phase of each sinusoid that is present in the input signal The amplitude spectrum tells us that the input signal contains 1 200 Hz sinusoid with an amplitude of 100 Pa a 400 Hz sinusoid with an amplitude of 200 Pa and a 600 Hz sinusoid with an amplitude of 50 Pa Similarly the phase spectrum tells us that the 200 Hz sinusoid has a phase of 90 the 400 Hz sinusoid has a phase of 180 and the 600 Hz sinusoid has a phase of 270 If Fourier39s theorem is correct we should be able to reconstruct the input signal by summing sinusoids at 200 400 and 600 Hz using the amplitudes and phases that are shown In fact summing these three sinusoids in this way would precisely reproduce the original time domain signal that is we would get back an exact replica of our original signal and not just a rough approximation to it For our purposes it is not important to understand how Fourier analysis works The most important point is Fourier39s idea that visual appearances aside all complex waves consist of sinusoids of varying frequencies amplitudes and phases In fact Fourier analysis applies not only to periodic signals such as those shown in Figure 315 but also to noise and transients In fact the amplitude spectra of the aperiodic signals shown in Figure 313 were calculated using Fourier analysis In later chapters we will see that the auditory system is able to derive a neural representation that is roughly comparable to a Fourier amplitude spectrum However as was mentioned earlier the auditory system does not derive a representation comparable to a Fourier phase spectrum As a result listeners are very sensitive to changes in the amplitude spectrum but are relatively insensitive to changes in phase Some Additional Terminology Overtones vs Harmonics The term overtone and the term harmonic refer to the same concept39 they are just counted differently As we have seen in a harmonic series such as 100 200 300 400 etc the 100 Hz component can be referred to as either the fundamental frequency or the first harmonic the 200 Hz component is the second harmonic the 300 Hz component is the third harmonic and so on An alternative set of terminology would refer to the 100 Hz component as the fundamental frequency the 200 Hz component as the rst overtone the 300 Hz component as the second overtone and so on Use of the term overtone tends to be favored by those interested in musical acoustics while most other acousticians tend to use the term harmonic Octaves vs Harmonics An octave refers to a doubling of frequency So if we begin at 100 Hz the next octave up would 200 Hz the next would be 400 Hz the next would be 800 Hz and so on Note that this is quite different from a harmonic progression A harmonic progression beginning at 300 Hz would be 300 600 900 1200 1500 etc while an octave progression would be 300 600 1200 2400 4800 etc There is something auditorilly natural about octave spacing and octaves play a very important role in the organization of musical scales For example on a piano keyboard middle A A5 is 440 Hz A above middle A A5 is 880 Hz A7 is 1760 and so on See Box 32 Wavelength The concept of wavelength is best illustrated with an example given by Small 1973 Small asks us to imagine dipping a finger repeatedly into a puddle of water at a perfectly regular interval Each time the finger hits the water a wave is propagated outward and we would see a pattern formed consisting of a series of concentric The Physics afSaund 16 circles see Figure 316 Wavelength is simply the distance between the adjacent waves Precisely the same concept can be applielilh lw Simply the distance betV iEQWePnEIFe the next or one raref c 39 ve andt e next r more rally the distance b tween any tw resio ding p 39 in adjacent waves 12 wsl avselem mm to be made abo mmaggn iWavalem relationship between frequency and wavelength Using the puddle example imagine that we begin by dipping our 39 e for c ulating wavelen tim form a from 39 ior high scho Figure 316 Wavelength is a measure of the distance between the crest of one cycle of a wave and the crest of the next cycle or trough to trough or in fact the distance between any two corresponding points in the wave Wavelength and frequency are related to one another Because the wave has only a short time to travel from one cycle to the next high frequencies produce short wavelengths Conversely because of the longer travel times low frequencies produce long wavelengths Ike Physics afSaund 17 70 Vowel a f0 100 Hz 70 Vowel I f0 150 Hz a 60 60 C g 50 g 50 g 40 40 E 30 E 30 20 20 10 10 0 C 0 1 2 3 0 1 2 3 Frequency kHz Frequency kHz 70 Vowel a f0 200 Hz 70 Vowel u f0 150 Hz b 60 60 d g 50 g 50 g 40 g 40 CL CL E 30 E 30 20 20 10 10 l C C 0 1 2 3 0 1 2 3 Frequency kHz Frequency kHz Figure 317 A spectrum envelope is an imaginary smooth line drawn to enclose an amplitude spectrum Panels 11 and b show the spectra of two signals the vowel with different fundamental frequencies note the differences in harmonic spacing but very similar spectrum envelopes Panels 0 and d show the spectra of two signals with different spectrum envelopes the vowels i and u in this case but the same fundamental frequencies ie the same harmonic spacing vowel quality ie Whether a vowel sounds like i vs a vs u etc For example panels a andb in Figure 317 show the vowel produced at two different fundamental frequencies We know that the fundamental frequencies are different because one spectrum shows Wide harmonic spacing and the other shows narrow harmonic spacing The fact that the two vowels are heard as a despite the difference in fundamental frequency can be attributed to fact that these two signals have similar spectrum envelopes Panels 0 and d in Figure 317 show the spectra of two signals with different spectrum envelopes but the same fundamental frequency ie with the same harmonic spacing As we Will see in the chapter on auditory perception differences in fundamental frequency are perceived as differences in pitch So for signals a and b in Figure 317 the listener will hear the same vowel produced at two different pitches Conversely for signals c and d in Figure 317 the listener will hear two different vowels produced at the same pitch We will return to the concept of spectrum envelope in the chapter on auditory perception Amplitude Envelope The term amplitude envelope refers to an imaginary smooth line that is drawn on top of a time domain signal Figure 318 shows sinusoids that are identical except for their amplitude envelopes It can be seen that the different amplitude envelopes re ect differences in the way the sounds are turned on and off For example panel a shows a signal that is turned on abruptly and turned off abruptly panel I shows a signal that is turned on gradually and turned off abruptly and so on Differences in amplitude envelope have an important effect on the quality of a sound As we Will see in the chapter on auditory perception amplitude envelope along with spectrum envelope discussed above is another physical parameter that affects timbre or sound quality For example piano players know that a given note will sound different depending on Whether or not the damping pedal is used Similarly notes played on a stringed instrument such as a violin or cello will sound different depending on Whether the note is plucked or bowed In both cases the underlying acoustic difference is amplitude envelope The Physics afSaund 18 Signals Differing in Amplitude Envelope nnnnmmnmmmmmnnmmnnmn a umumummuiuumummmmuHm Inst Air Pres Inst Air Pres Inst Air Pres Time gt Figure 318 Amplitude envelope is an imaginary smooth line drawn to enclose a timedomain signal This feature describes how a sound is turned on and turned off39 for example whether the sound is turned on abruptly and turned off abruptly panel 1 turned on gradually and turned off abruptly panel 1 turned on abruptly and turned off gradually panel c or turned on and off gradually panel d Acoustic Filters As will be seen in subsequent chapters acoustic filtering plays a central role in the processing of sound by the inner ear The human vocal tract also serves as an acoustic filter that modifies and shapes the simple sounds that are created by the larynx and other articulators For this reason it is quite important to understand how acoustic filters work In the most general sense the term filter refers to a device or system that is selective about the kinds of things that are allowed to pass through versus the kinds of things that are blocked An oil filter for example is designed to allow oil to pass through while blocking particles of dirt Of special interest to speech and hearing science are frequency selective filters These are devices that allow some frequencies to pass through while blocking or attenuating other frequencies The term attenuate means to weaken or reduce in amplitude A simple example of a frequency selective filter from the world of optics is a pair of tinted sunglasses A piece of white paper that is viewed through red tinted sunglasses will appear red Since the original piece of paper is white and since we know that white light consists of all of the visible optical frequencies mixed in equal amounts the reason that the paper appears red through the red tinted glasses is that optical frequencies other than those corresponding to red are being blocked or attenuated by the optical filter As a result it is primarily the red light that is being allowed to pass through Starting at the lowest optical frequency and going to the highest light will appear red orange yellow green blue indigo and violet A graph called a frequency response curve is used to describe how a frequency selective filter will behave A frequency response curve is a graph showing how energy at different frequencies will be affected by the filter Specifically a frequency response curve plots a variable called quotgainquot as a function of variations in the frequency of the input signal Gain is the amount of amplification provided by the filter at different signal frequencies Gains are interpreted as amplitude multipliers for example suppose that the gain of a filter at 100 Hz is 13 If a 100 Hz sinusoid enters the filter measuring 10 uPa the amplitude at the output of the filter at 100 Hz will measure 13 uPa The Physics ofSotmd 19 Lowpass Filter Highpass Filter Bandpass Filter 4 4 4 E E E e e e o o o R o y G B l v R o v G B l v R o y G B l v law Heq High Heq law Heq High Heq law Heq High Heq U N ENCY FREQUENCY Figure 3719 q r L r39 39 39 39 39 mucnin optical energy mouth optical unmeii in blocking energy at higher andlower frequencies 10 uPax 1313 uPaTL n 39 that the effect ofthe lter will be to atten signal at 100 Hz will measure 5 uPa the ller i e a 39 39 39 39 39 q 1 meaning uate the signal For example ifthe gain at 100 Hz is 05 a 10 upa input at the output ofthe lter When the filter gain is 10 the signal is unaffected by 10 L Figure 319 shows frequency response curves for several optical filters Panel a shows a frequency response em 39 39 39 l L p 39 39 39 39 athe Lu r signal rquot quot L in h hi h In This is L L 39 39 39 39 p This is an r aniiei 39 Pane L L r39 39 uuei thi lm will an 39 lg A 39 uuiu therefore appear violet 39 39 39 Panel r a mm in n 39 39 components UJui he A 39 umu appea tee This is calleda bandpass filter 4 alone To see how the 1 r 1 39 in fart L 39 lll um The wine glass will 1 p 39 A hnmn in Figure 120 we amplitude for the lter For L r r L39 r39 will 39 glass is 500 Hz We now ask the singer to produce a low frequency signal say 50 Hz since this frequency is quite L 39 u 39 39 in W ite 10W et uu ei auu uu ei 39 n lel dlibd m an 39 rue ia q r 39 We would MPMmUXm an x we m g y gig f i k MzLu Hmzznnnmmufhmv um mmmslsulmw mum ampmmpmm Nahum mm m mm mm Mm sun mmamm mummy m m gas In mm mm mum vi mmm unuhswlmqm Wnsmhmmwmnmme WW mum furyme unwewmdnsuvh mmumynrpmsa mm m bm pusfmx mummy mel nuuuy lumm mwmmm Zlms ewmynrpmsum farm Munmmaumww a mum Vmu mm mm B uf m39whnse ewmnrpmsa mm mm Med N mu m muw My Musmmmms Nam ubww sun um ubImssfmxuulPsses m mmqu ax mu m m um mm m m ufmawhnse mmm mm mm m The Physics afSaund 21 Lowpass Filters with Different Highpass Filters with Different Cutoff Frequencies Cutoff Frequencies 10 1 0 08 08 E 06 E 06 U U D 04 D 04 02 02 00 00 0 1000 2000 3000 4000 0 1000 2000 3000 4000 Frequency Hz Frequency Hz Figure 321 Lowpass and highpass filters differing in cutoff frequency Bandpass Filters Differing in Bandwidth lt Narrow Band Filter Wide Band Filter Gain 1000 2000 3000 4000 Figure 322 Frequency response curves for two bandpass filters with identical center frequencies but different bandwidths Both filters pass a band of energy centered around 2000 Hz but the narrow band filter is more selective than the Wide band filter that is gain decreases at a higher rate above and below the center frequency for the narrow band filter than for the Wide band filter example the tuning dial on a radio controls the center frequency of a narrow bandpass filter that allows a single radio channel to pass through While blocking channels at all other frequencies The hum an vocal tract is an example of a variable filter of the most spectacular sort For example 1 during the occlusion interval that occurs in the production of a sound like b the vocal tract behaves like a lowpass filter 2 in the articulatory posture for sounds like s and sh the vocal tract behaves like a highpass filter and 3 in the production of vowels the vocal tract behaves like a series of bandpass filters connected to one another and the center frequencies of these filters can be adjusted by changing the positions of the tongue lips and jaw To a very great extent the production of speech involves making adjustments to the articulators that have the effect of setting the vocal tract filter in differ modes to produce the desired sound quality We Will have much more to say about this in later chapters The Physics afSaund 22 Frequency Response Curves vs Amplitude Spectra It is not uncommon for students to confuse a frequency response curve with an amplitude spectrum The axis labels are rather similar an amplitude spectrum plots amplitude on the y axis and frequency on the x axis while a frequency response curve plots gain on the y axis and frequency on the x axis The apparent similarities are deceiving however since a frequency response curve and an amplitude spectrum display very different kinds of information The difference is that an amplitude spectrum describes a sound while a frequency response curve describes a ller For any given sound wave an amplitude spectrum tells us what frequencies are present with what amplitudes A frequency response curve on the other hand describes a filter and for that filter it tells us what frequencies will be allowed to pass through and what frequencies will be attenuated Keeping these two ideas separate will be quite important for understanding the key role played by filters in both hearing and speech science Resonance The concept of resonance has been alluded to on several occasions but has not been formally defined The term resonance is used in two different but very closely related ways The term resonance refers to l the phenomenon of forced vibration and 2 natural vibrating frequency also resonant frequency or resonance frequency To gain an appreciation for both uses of this term imagine the following experiment We begin with two identical tuning forks each tuned to 435 Hz Tuning fork A is set into vibration and placed one centimeter from tuning fork B but not touching it If we now hold tuning fork B to a healthy ear we will find that it is producing a 435 Hz tone that is faint but quite audible despite the fact that it was not struck and did not come into physical contact with tuning fork A The A quot for this quot ti 4 t r t 4 quot A is that the sound wave generated by tuning fork A forces tuning fork B into vibration that is the series of compression and rarefaction waves will alternately push and pull the tuning fork resulting in vibration at the frequency being generated by tuning fork A The phenomenon of forced vibration is not restricted to this quotactionatadistance case The same effect can be demonstrated by placing a vibrating tuning fork in contact with a desk or some other hard surface The intensity of the signal will increase dramatically because the tuning fork is forcing the desk to vibrate resulting in a larger volume of air being compressed and rarefied 3 Returning to our original tuning fork experiment suppose that we repeat this test using two mismatched tuning forks for example tuning fork A with a natural frequency of 256 Hz and tuning fork B with a natural vibrating frequency of 435 Hz If we repeat the experiment 7 setting tuning fork A into vibration and holding it one centimeter from tuning fork B 7 we will find that tuning fork B does not produce an audible tone The reason is that forced vibration is most efficient when the frequency of the driving force is closest to the natural vibration frequency of the obj ect that is being forced to vibrate Another way to think about this is that tuning fork B in these experiments is behaving like a filter that is being driven by the signal produced by tuning fork A Tuning forks in fact behave like rather narrow bandpass filters In the experiment with matched tuning forks the filter was being driven by a signal frequency corresponding to the peak in the filter39s frequency response curve Consequently the filter produced a great deal of energy at its output In the experiment with mismatched tuning forks the filter is being driven by a signal that is remote from the peak in the filter39s frequency response curve producing a low amplitude output signal To summarize resonance refers to the ability of one vibrating system to force another system into vibration Further the amplitude of this forced vibration will be greater as the frequency of the driving force approaches the natural vibrating frequency resonance of the system that is being forced into vibration Cavity Resonators An airfilled cavity exhibits frequency selective properties and should be considered a filter in precisely the way that the tuning forks and wine glasses mentioned above are filters The human vocal tract is an airfilled cavity that behaves like a filter whose frequency response curve varies depending on the positions of the articulators Tuning forks and other simple filters have a single resonant frequency Note that we will be using the terms quotnatural vibrating frequencyquot and quotresonant frequencyquot interchangeably Cavity resonators on the other hand can have an infinite number of resonant frequencies 3The increase in intensity that would occur as the tuning fork is placed in contact with a hard surface does not mean that additional energy is created The increase in intensity would be offset by a decrease in the duration ofthe tone so the total amount of energy would not increase relative to a freely vibrating tuning fork The Physics afSaund 23 1 0 500 1500 2500 3500 4500 08 g 06 o 0 4 175 cm Uniform Tube 02 00 1 0 4375 13125 21875 30625 39375 08 g 06 o 0 4 20 cm Uniform Tube 02 00 1 0 5833 17500 29167 40833 52250 08 g 06 o 0 4 15 cm Uniform Tube 02 00 0 1000 2000 3000 4000 5000 Frequency Hz Figure 323 Frequency response curves for three uniform tubes open at one end and closed at the other These kinds of tubes have an infinite number of resonances at odd multiples of the lowest resonance As the figure shows shortening the tube shifts all resonances to higher frequencies while lengthening the tube shifts all resonances to lower frequencies A simple but very important cavity resonator is the uniform tube This is a tube whose crosssectional area is the same uniform at all points along its length A simple water glass is an example of a uniform tube The method for determining the resonant frequency pattern for a uniform tube will vary depending on whether the tube is closed at both ends open at both ends or closed at just one end The configuration that is most directly applicable to problems in speech and hearing is the uniform tube that is closed at one end and open at the other end The ear canal for example is approximately uniform in crosssectional area and is closed medially by the ear drum and open laterally Also in certain configurations the vocal tract is approximately uniform in crosssectional area and is effectively closed from below by the vocal folds and open at the lips The resonant frequencies for a uniform tube closed at one end are determined by its length The lowest resonant frequency F1 for this kind of tube is given by F1 c4L where c the speed of sound L the length of the tube For example for a 175 cm tube F1 c4L 3500070 500 Hz This tube will also have an infinite number of higher frequency resonances at addmultiples of the lowest resonance F1F1l 500Hz The Physics afSaund 24 F2 F1 3 1500 Hz F3 F1 5 2500 Hz F4 F1 7 3500 Hz The frequency response curve for this tube for frequencies below 4000 Hz is shown in the solid curve in Figure 323 Notice that the frequency response curve shows peaks at 500 1500 2500 and 3500 Hz and valleys in between these peaks The frequency response curve in fact looks like a number of bandpass filters connected in series with one another It is important to appreciate that what we have calculated here is a series of natural vibrating frequencies of a tube What this means is that the tube will respond best to forced vibration if the tube is driven by signals with frequencies at or near 500 Hz 1500 Hz 2500 Hz and so on Also the resonant frequencies that were just calculated should not be confused with harmonics Harmonics are frequency components that are present in the amplitude spectra of complex periodic sounds resonant frequencies are peaks in the frequency response curve of filters We next need to see what will happen to the resonant frequency pattern of the tube when the tube length changes If the tube is lengthened to 20 cm F1 c4L 3500080 4375 Hz F2 F13 13125 Hz F3 F1 5 21875 Hz F4 F1 7 30625 Hz It can be seen that lengthening the tube from 175 cm to 20 cm has the effect of shifting all of the resonant frequencies downward see Figure 323 Similarly shortening the tube has the effect of shifting all of the resonant frequencies upward For example the resonant frequency pattern for a 15 cm tube would be F1 c4L 3500060 5833 Hz F2 F1 3 1750 Hz F3 F1 5 29167 Hz F4 F1 7 40833 Hz The general rule is quite simple all else being equal long tubes have low resonant frequencies and short tubes have high resonant frequencies This can be demonstrated easily by blowing into bottles of various lengths The longer bottles will produce lower tones than shorter bottles This effect is also demonstrated every time a water glass is filled The increase in the frequency of the sound that is produced as the glass is filled occurs because the resonating cavity becomes shorter and shorter as more air is displaced by water This simple rule will be quite useful For example it can be applied directly to the differences that are observed in the acoustic properties of speech produced by men women and children who have vocal tracts that are quite different in length Resonant Frequencies and Formant Frequencies The term quotresonant frequencyquot refers to natural vibrating frequency or equivalently to a peak in a frequency response curve For reasons that are entirely historical if the filter that is being described happens to be a human vocal tract the term formant frequency is generally used So one typically refers to the formant equencies of the vocal tract but to the resonant frequencies of a plastic tube the body of a guitar the diaphragm of a loudspeaker or most any other type of filter other than the vocal tract This is unfortunate since it is possible to get the mistaken idea that formant frequencies and resonant frequencies are different sorts of things The two terms are in fact fully synonymous The Decibel Scale The final topic that we need to address in this chapter is the representation of signal amplitude using the decibel scale The decibel scale is a powerful and immensely flexible scale for representing the amplitude of a sound wave The scale can sometimes cause students difficulty because it differs from most other measurement scales in not just one but two ways Most of the measurement scales with which we are familiar are absolute and linear The decibel The Physics of Sound 25 scale however is relative rather than absolute and logarithmic rather than linear Neither of these characteristics is terribly complicated but in combination they can make the decibel scale appear far more obscure than it is We wil examine these features one at a time and then see how they are put together in building the decibel scale Linear vs Logarithmic Measurement Scales Most measurement scales are linear To say that a measurement scale is linear means that it is based on equal additive distances This is such a common feature of measurement scales that we do not give it much thought For example on a centigrade or Fahrenheit scale for measuring temperature going from a temperature of 90 to a temperature of 91 involves adding one 1 One rather obvious consequence of this simple additivity rule is that the difference in temperature between 10 and 11 is the same as the difference in temperature between 90 and 91 However there are scales for which this additivity rule does not apply One of the best known examples is the Richter scale that is used for measuring seismic intensity The difference in seismic intensity between Richter values of 40 and 50 50 and 60 60 and 70 is not some constant amount of seismic intensity but rather a constant multiple Specifically a 70 on the Richter scale indicates an earthquake that is 10 times greater in intensity than an earthquake that measures 60 on the Richter scale Similarly an 80 on the Richter scale is 10 times greater in intensity than a 70 Whenever jumping from one scale value to the next involves multiplying by a constant rather than adding a constant the scale is called logarithmic The multiplicative constant need not be 10 See Box 32 for an example of a logarithmic scale 7 an octave progression 7 that uses 2 as the constant Another way of making the same point is to note that the values along the Richter scale are exponents rather than ordinary numbers for example a Richter value of 6 indicates a seismic intensity of 10 a Richter value of 7 indicates a seismic intensity of 107 etc The Richter values can of course just as well be referred to as powers or logarithms since both of these terms are synonyms for exponent The decibel scale is an example of a logarithmic scale meaning that it is based on equal multiples rather than equal additive distances Absolute vs Relative Measurement Scales A simple example of a relative measurement scale is the Mach scale that is used by rocket scientists to measure speed The Mach scale measures speed not in absolute terms but in relation to the speed of sound For example a missile at Mach 20 is traveling at twice the speed of sound while a missile at Mach 09 is traveling at 90 of the speed of sound So the Mach scale does not represent a measured speed Sm in absolute terms but rather represents a measured speed in relation to a reference speed SmSQ The reference that is used for the Mach scale is the speed of sound so a measured absolute speed can be converted to a relative speed on the Mach scale by simple division For example taking 783 mph as the speed of sound 1200 mph 1200783 Mach 153 The decibel scale also exploits this relative measurement scheme The decibel scale does not represent a measured intensity 1quot in absolute terms but rather represents the ratio of a measured intensity to a reference intensity TmL The decibel scale is trickier than the Mach scale in one important respect For the Mach scale the reference is always the speed of sound but for the decibel scale many different references can be used In explaining how the decibel scale works we will begin with the commonly used intensity reference of 103912 wm2 watts per square meter which is approximately the intensity that is required for an average normal hearing listener to barely detect a 1000 Hz pure tone So for our initial pass through the decibel scale 103912 wm2 will serve as 1 and will perform the same function that the speed of sound does for the Mach scale Table 31 lists several sounds that cover a very broad range of intensities The second column shows the measured intensities of those sounds and the third column shows the ratio of those intensities to our reference intensity Whispered speech for example measures approximately 10398 wmz which is 10000 times more intense than the reference intensity 10 103912 104 10000 The main point to be made about column 3 is that the ratios become very large very soon Even a moderately intense sound like conversational speech is 1000000 times more intense than the reference intensity The awkwardness of dealing with these very large ratios has a very simple solution Column 4 shows the ratios written in exponential notation and column 5 simplifies the situation even further by recording the exponent only The term exponent and the term logarithm are synonymous so the measurement scheme that is expressed by the numbers in column 5 can be summarized as follows 1 divide a measured intensity by a reference intensity in this case 103912 wmz 2 take the logarithm of this ratio ie write the number in exponential notation and keep the exponent only This method in fact is a completely legitimate way to represent signal intensity The unit of measure is called the bel after AG Bell and the formula is The Physics afSaund 26 bel loglo LnL whereLn a measured intensity a reference intensity Table 31 Sound intensities and intensity ratios showing how the decibel scale is created Column 2 shows the measured intensities Tm of several sounds Column 3 shows the ratio of these intensities to a reference intensity of 103912 wmz Column 4 shows the ratio written in exponential notation while column 5 shows the exponent only The last column shows the intensity ratio expressed in decibels which is simply the logarithm of the intensity ratio multiplied by 10 Measured Ratio Ratio in Exponent Decibel Sound Intensity Im LnL Exp Not loglo 10 x loglo Threshold 103912 Vizin2 1 100 0 0 1 kHz Whisper 10398 wmz 10000 104 4 40 Conversational 106 wmz 1000000 106 6 60 Speech City Traffic 10394 wmz 100000000 108 8 80 Rock amp Roll 10392 wmz 10000000000 1010 10 100 Jet Engine 100 wmz 1000000000000 1012 12 120 Legitimate or not the bel finds its sole application in textbooks attempting to explain the decibel For reasons that are purely historical the loglo of the intensity ratio is multiplied by 10 changing bel into the decibel dB As shown in the last column of Table 31 this has the very simple effect of turning 4 bels into 40 decibels 8 bels into 80 decibels etc The formula for the decibel then is dBlL lO logm LnL where Ln ameasured intensity L areference intensity The designation quotILquot stands for intensity level and it indicates that the underlying measurements are of sound intensity and not sound pressure As will be seen below a different version of this formula is needed if sound pressure measurements are used The multiplication by 10 in the dBlL formula is a simple operation but it can sometimes have the unfortunate effect of making the formula appear more obscure that it is The decibel values that are calculated however should be readily interpretable For example 30 dBlL means 3 factors of 10 more intense than L 60 dBlL means 6 factors of 10 more intense than L and 90 dBlL means 9 factors of 10 more intense than L Deriving a Pressure Version of the dB Formula In a simple world we would be finished with the decibel scale The problem is that the formula is based on measurements of sound intensity but as a purely practical matter sound intensity is difficult to measure Sound pressure on the other hand is quite easy to measure An ordinary microphone for example is a pressure sensitive device The problem then is that the decibel is defined in terms of intensity measurements but the measurements The Physics afSaund 27 that are actually used will nearly always be measures of sound pressure This problem can be addressed since there is a predictable relationship between intensity I and pressure E intensity is proportional to pressure squared 10C EZ Knowing this relationship allows us to create a completely equivalent version of the decibel formula that will work when sound pressure measurements are used instead of sound intensity measurements All we need to do is substitute squared pressure measurements in place of the intensity measurements dBlL lO loglo lml intensity version of formula stpL 10 logio ElmE2 pressure version of formula The designation quot SPLquot stands for sound pressure level and it indicates that measures of sound pressure have been used and not measures of sound intensity Although the stpL formula shown here will work fine it will almost never be seen in this form The reason is that the formula is algebraically rearranged so that the squaring operation is not needed The algebra is shown below 1 dBlL lO loglo lml the intensity version of the formula 2 stpL 10 logio ElmE2 measures of E2 replace measures ofl because 1 0C E2 3 stpL 10 logm Emmi aim2 402 4 stpL lO 2 logio EmE this is the only tricky step log ab b log a 5 dBSPL 2010g10EmE 2 lO 20 With the possible exception of the fourth step4 the algebra is straightforward but the details of the derivation are less important than the following general points 1 The decibel formula is defined in terms of intensity ratios The basic formula is dBlL lO loglo lmlp 2 While sound intensity is difficult to measure sound pressure is easy to measure It is therefore necessary to derive a version of the decibel formula that works when measures of sound pressure are used instead of sound intensity 3 The derivation of the pressure version of the formula is based entirely on the fact that intensity is proportional to pressure squared 1 0C E2 This allows measures of E2 to replace measures of l turning dBlL 10 lo 10 lml into stpL 10 loglo ElmE A few algebra tricks are applied to turn this formula into the more aesthetically pleasing final version stpL 2010gio EmE 4 The two versions of the formula are fully equivalent to one another see Box 33 This last point about the equivalence of the intensity and sound pressure versions of the formula is explained in some detail in Box 33 but the basic point is quite simple The pressure version of the dB formula was derived from the intensity version of the formula through algebraic manipulations based on this relationship 1 0C E2 The whole 4 Step 4 is the only tricky part of derivation The reason it works is that squaring a number and then taking a log is the same as taking the log rst and then multiplying the log by 2 For example note that the two calculations below produce the same result log 1u1002 log lu10000 4 square rst then take the log log In 1002 log iu100 x 2 2 x 2 4 take the log then multiply by 2 The Physics afSaund 28 BOX 32 HARMONICS OCTAVES LINEAR SCALES AND LOGARITHMIC SCALES As we will see when the decibel scale is introduced there is an important distinction to be made between linear scales which are quite common and logarithmic scales which are less common but quite important This distinction can be illustrated by examining the difference between a harmonic progression and an octave progression Notice that in a harmonic progression the spacing between the harmonics is always the same39 that is the difference between H1 and H2 is the same as the difference between H2 and H3 and so on This is because increases in frequency between one harmonic and the next involve adding a constant with the constant being the fundamental frequency For example H1 500 H2 1000 add 500 H3 1500 add 500 HA 2000 add 500 To get from one scale value to another on an octave progression involves multiplying by a constant rather than adding a constant For example an octave progression starting at 500 Hz looks like this 01 Oz 1000 multiply by 2 03 2000 multiply by 2 O4 4000 multiply by 2 As a result of the fact that we are multiplying by a constant rather than adding a constant the spacing is no longer even ie the spacing between 01 and Oz is 500 Hz the spacing between Oz and 03 is 1000 Hz and so on The point to be made of this is that there are two fundamentally different kinds of scales 1 scales like harmonic progressions that are created by adding a constant which are by far the more common and 2 scales like octave progressions that are created by multiplying by a constant Scales that are created by adding a constant are called linear scales while scales that are created by multiplying by a constant are called logarithmic scales Note that for an octave progression the multiplier happens to be 2 meaning that progressing from one frequency to an octave above that frequency involves multiplication by 2 However a logarithmic scale can be built using any multiplier We will return to the distinction between linear and logarithmic scales when we talk about the decibel scale and there we will see that a logarithmic scale is built around multiplication by a constant value of 10 rather than 2 The Physics afSaund 29 point of algebra of course is to keep the expression on the left equal to the expression on the right The simple and useful point that emerges from this is this If an intensity meter shows that a given sound measures 60 dBlL for example a pressure meter will show that the same sound measures exactly 60 dBSPL This may seem counterintuitive due to the differences in the formulas but see Box 33 for the explanation The equivalence of the two versions of the dB formula greatly simplifies the interpretation of sound levels that are expressed in decibels References The reference that is used for the Mach scale is always the speed of sound One of the virtues of the decibel scale is that any reference can be used as long as it is clearly specified The only reference that has been mentioned so far is 103912 wmz which is roughly the audibility threshold for a 1000 Hz pure tone This is a standard reference intensity and unless otherwise stated it should be assumed that this is used when a signal level is reported in dBlL The standard reference that is used for stpL is 20 Pa so when a signal level is reported in stpL it should be assumed that this reference is used unless otherwise stated Many references besides these two standard references can be used For example suppose that a speech signal is presented to a listener at an average level of 3500 Pa in the presence of a noise signal whose average soun pressure is 1400 Pa The speechtonoise ratio SN can be represented on a decibel scale using the level of the speech as Em and the level of the noise as E stn 20 logio Em E 20 logio 35001400 2010g10 25 20 039794 796 dB To take one more example assume that a voice patient prior to treatment produces sustained vowels that average 2300 Pa Following treatment the average sound pressures increase to 8890 Pa The improvement in sound pressure posttreatment relative to pretreatment can be represented on a decibel scale dBImproVement 20 10gm EpostEpre 20 logio 88902300 20 loglo 386522 20 058717 1174 dB A final example can be used to make the point that the decibel scale can be used to represent intensity ratios for any type of energy not just sound Bright sunlight has a luminance measuring 100000 cdm2 candela per square meter Light from a barely visible star on the other hand has a luminance measuring 00001 cdmz We can now ask how much more luminous bright sunlight is in relation to barely visible star light and the dB scale can be used to represent this value Since the underlying physical quanities here are measures of electromagnetic intensity we want the intensity version of the formula rather than the pressure version 13 10 10gm IsunlightIstarlight 10 logio 10000000001 10log10 105104 10 loglo 109 division is done by subtracting exponents 5 7 4 9 10 9 90 dB 5The standard pressure reference for stpL is sometimes given as 00002 dynescm2 rather than 20 uPa These two sound pressures are identical however in exactly the same sense that 4 quarts and 1 gallon are identical Likewise the standard reference for dBquot is often given as 103916 wcm2 instead of103912 wmz These two intensities are also identical The Physics of Sound 30 The fact that we are measuring light rather than sound makes no difference a decibel is 10 logro lml or equivalently 20 lOgro EmE regardless of whether the energy comes from sound light electrical current or any other type of energy dB Hearing Level dBHL The dB Hearing Level dBHL scale was developed specifically for testing hearing sensitivity for pure tones of different frequencies The soundlevel dials on clinical audiometers6 for example are calibrated in dBHL rather than stpL To understand the motivation for the dBHL scale examine Figure 324 which shows the sound level in stpL required for the average normalhearing listener to barely detect pure tones at frequencies between 125 and 8000 Hz This is called the audibility curve and the simple but very important point to notice about this graph is that the curve is not a flat line39 that is the ear is clearly more sensitive at some frequencies than others The differences in sensitivity are quite large in some cases For example the average normalhearing listener will barely detect a 1000 Hz pure tone at 7 stpL but at 125 Hz the sound level needs to be cranked all the way up to 45 stpL an increase in intensity of nearly 4000 1 Now suppose we were to test puretone sensitivity using an audiometer that is calibrated in stpL Imagine that a listener barely detects a 1000 Hz pure tone at 25 stpL Does this listener have a hearing loss and if so how large The only way to answer this question is to consult the data in Figure 324 which shows that the threshold of audibility for the average normal hearing listener at 1000 Hz is 7 stpL This means that the hypothetical listener in this example has a hearing loss of 257 18 dB Suppose further that the same listener detects a 250 Hz tone at 30 dBSPL The table in Figure 324 shows that normal hearing sensitivity at 250 Hz is 255 dBSPL meaning that the listener has slightly better than normal hearing at this frequency As a final example imagine that this listener barely detects a 500 Hz tone at 30 dBSPL Since the table shows that normal hearing sensitivity at 500 Hz is 115 stpL the listener has a hearing loss of 300115 185 dB The simple point to be made about these examples is that with an audiometer dial that is calibrated in stpL it is not possible to determine whether a listener has a hearing loss or to measure the size of that loss without doing some arithmetic involving the normative data in Figure 324 The dBHL scale however provides a simple solution to this problem that avoids this arithmetic entirely The solution involves calibrating the audiom eter in such a way that when the level dial is set to 0 dBm sound level is set to the threshold of audibility for the average normalhearing listener for that signal frequency For example when the level dial is set to 0 dBHL at 125 Hz the level of tone will be 45 stpL 7 the threshold of audibility for the average normal hearing listener at this frequency Now if a listener barely detects the 125 Hz tone at 0 dBHL no arithmetic is needed39 the listener has normal hearing at this frequency Further if the listener barely detects this 125 Hz tone at 40 dBHL for example the listener must have a 40 dB loss at this frequency 7 and again it is not necessary to consult the data in Figure 324 Similarly when the level dial is set to 0 dBHL at 250 Hz the level of the tone will be 255 stpL which is the audibility threshold at 250 Hz If this tone is barely detected at 0 dBm the listener has normal hearing at this frequency However if the tone is not heard until the dial is increased to 50 dB dBHL for example the listener has a 50 dB hearing loss at this frequency The same system is used for all signal frequencies in all cases the 0 dBHL reference is not a fixed number as it is for dBSPL a constant value of 20 uPa no matter what the signal frequency is or dBlL a constant value of 103912 wattsm2 again independent of signal frequency but rather a family of numbers In each case the reference for the dBHL scale is the threshold of audibility for an average normalhearing listener at a particular signal frequency What this means is that values in dBHL are a xed distance above the audibility curve although they may be very different levels in stpL For illustration Figure 325 shows the audibility curve the filled symbols and above that in the unfilled symbols a collection ofvalues that all measure 30 dBm Although the sound levels on the 30 dBHL curve vary considerably in stpL ie measured using 20 uPa as the reference every data point on this curve is a constant 3 factors of 10 or 30 dB above the audibility curve The value of 30 dB in this figure is just an example All values in dBHL and stpL are interpreted in the same way 50 stpL means that the signal being measured is 100000 times ie 5 factors of 10 more intense than the fixed reference of 20 uPa independent of frequency 50 dBm on the other hand means that the signal being measured is 100000 times again 5 factors of 10 more intense than a tone that is barely audible to a normalhearing listener at that signal frequency Similarly 20 stpL means that the signal is 20 dB 2 factors of 10 more intense than the fixed reference of 20 uPa while 20 dBHL means that the signal is 20 dB again 2 factors of 10 above the audibility curve 6A clinical audiometer is an instrument with among other things one dial that controls puretone frequency and another dial that controls the intensity of the tone The listener is asked to raise a hand when the tone is barely audible The Physim of Sound 31 Summary The decibel is apowerful care Jul 39 39 amplitude quotquot 39 39 1 similar to the Mach scale it represents signal level not in absolute terms but as a measured level divided by a reference level and 2 like the Richter scale the dB scale is logarithmic rather than linear meaning that it is based quot quot 39 39 umn equal 4 quot 439 quot quot 39b is defined interms of intensity ratios for practical reasons measures of sound pressure are far more common than measures of sound intensity Consequently a version of the decibel formula was derived that makes use of pressure ratios rather than intensity ratios The derivation was based on the fact that intensity is proportional to pressure squared The two versions of the decibel formula dBquot 10 log m ImI and stpL 20 log m EmE are fully equivalent meaning thatif quot D L L quot quot 9 rLT quot scale quot the speed of sound as a reference any number ofreferences can be used with the decibel scale The standard reference for the dBquot scale is 10quot2 wm2 and the standard reference for the stpL scale is 20 pPa However any level can be used as a reference as long as it is specified The dBHL scale widely used in audiological assessment was developed specifically for measuring sensitivity to pure tones of difference frequencies The reference that is used for the dBHL scale is the threshold of audibility at a particular signal frequency for the average normalhearing listener Sound levels in stpL an BHL 39 quot 39 quot f For example 39 stpL is 4 factors of 10 ie 40 dB greater than the fixed SPL reference of 20 pPa while a pure tone measuring 40 39s factors of 10 again 40 dB greater than a tone ofthat same frequency that is barely audible to an average normalhearing listener Freggencx Threshold 125 45 0 250 25 5 500 115 750 80 1000 70 1500 65 2000 90 3000 100 4000 95 6000 155 8000 130 Figure 3r24 The threshold of audibility for the average normalhearing listener for pure tones varying between 125 and 8000 Hz The audibility threshold is the sound level in stpL that is required for a listener to barer detect L in L a tone Values on L39 L L a thers In particular er and 0 the ear is more sensitive in a range ofmidfrequencies between about 1000 and 4000 Hz than it is at low higher frequencies The complex shape ofthis curve provides the underlying motivation for the dBHL scale See text for details The Physim of Sound 32 7 At Values on the 30 dBML Curve are a 70 Constant 3 Factors of 10 in Intensity ie 30 dB above the Audibiligz Curve 60 30 dB 5 l u 50 30 dB soida r E 40 7 1 g 3 30 dB l SOLE 30 dB 30dB U 20 J 30 dB j m 0 dBm 0 J 0113 20 Pa 125 250 500 1000 2000 4000 8000 Frequency Hz gure 3725 The lower function is the audibility curve 7 the sound level in stpL that is required for an average normal hearing listener to barely detect pure tones of different frequencies The upper unction shows sound levels for a set of tones that all measure 30 dB L These tones vary quite a bit in stpL ie relative to the constant value of 20 uPa but in all cases the tones are a constant 3 factors of 10 in intensity ie 30 dB above the curve The Physics of Sound 33 BOX 33 THE EQUIVALENCE OF THE INTENSITY AND PRESSURE VERSIONS OF THE DECIBEL FORMULA One fact about the two versions of the dB formula that is not always well understood is that the dBIL and dBSpL formulas are fully equivalent By quotfully equivalent we mean the following suppose that a sound intensity meter is used to measure the level of some sound and we find that this sound is 1000 times more intense than the standard intensity reference of 103912 wmz The sound would then measure 30 dBlL 10 loglo 1000 10 3 30 dBlL Now suppose that we put the sound intensity meter away and use a sound pressure meter to measure the same sound You might think that the sound would measure 60 stpL since now we are multiplying by 20 instead of 10 but the trick is that the ratio is no longer 1000 Recall that intensity is proportional to pressure squared which means that pressure is proportional to the square root of intensity This means that if the intensity ratio is 1000 the pressure ratio must be the square root of 1000 or 316 So the formula now becomes 20 log 316 20 15 30 stpL which is exactly what we obtained originally It will always work out this way if a sound measures 50 dBlL that same sound will measure 50 dBSPL Table 32 might help to make this more clear The first column shows an intensity ratio the second column shows the corresponding pressure ratio this is always the square root of the intensity ratio the third column shows the dBlL value 10 log of the intensity ratio and the fourth column shows stpL value 20 log of the pressure ratio As you can see they are always the same Table 32 Intensity ratios equivalent pressure ratios dBlL values and stpL values showing the equivalence of the intensity and pressure versions of the dB formula Intensity Pressure dBlL stpL atio Ratio 10 loglo lmI 20 logio EmE 10 316 1000 1000 20 447 1301 1301 40 632 1602 1602 50 707 1699 1699 60 775 1778 1778 70 837 1845 1845 80 894 1903 1903 90 949 1954 1954 100 1000 2000 2000 200 1414 2301 2301 300 1732 2477 2477 400 2000 2602 2602 500 2236 2699 2699 1000 3162 3000 3000 The Physics afSaund 34 Studv quot Pthical A m e 1 Explain the basic processes that are involved in the propagation of a sound wave 2 Draw time and frequencydomain representations of simple periodic complex periodic complex aperiodic and transient sounds 3 Draw time and frequencydom ain representations of two complex periodic sounds with different fundamental frequencies 4 Draw timedomain representations of two simple periodic sounds with the same frequency and phase but different amplitudes 5 Draw timedomain representations of two simple periodic sounds with the same frequency and different amplitudes but different phases 6 Draw amplitude spectra of two sounds with the same fundamental frequencies but different spectrum envelopes 7 Draw amplitude spectra of two sounds with different fundamental frequencies but similar spectrum envelopes 8 Calculate signal frequencies for sinusoids with the following values a period 034 s b period 2 s c period 10 ms d period 2 ms e wavelength 20 cm f wavelength 100 cm Answers a f 1034 294 Hz b f 12 05 Hz c f 1001 100 Hz df1002 500 Hz e f cWL speed of soundwavelength 3500020 1750 Hz f f cWL speed of soundwavelength 35000100 350 Hz 9 Calculate the three lowest resonant frequencies of the following uniform tubes that are closed at one end and open at the other en a 10 cm b 30 cm c 40 cm Answers a wavelength of lowest resonance 40 cm 10 x 4 f 3500040 875 R1 875 R1 frequency of resonance number 1 R2 2625 R3 4375 b wavelength of lowest resonance 120 cm 30 x 4 f 35000120 2917 The Physics of Sound 35 R1 2917 R2 8750 R3 14583 c wavelength of lowest resonance 160 cm 40 x 4 f 35000160 21875 R1 21875 R2 65625 R3 109375 10 Show what the frequencyresponse curves look like for the tubes in the problem above 11 A complex periodic signal has a fundamental period of 4 msec What is the fundamental frequency of the signal At what frequencies would we expect to find energy 12 How are the terms octave and harmonic different 13 Give examples of the following kinds of graphs being sure to label both axes a amplitude spectrum b phase spectrum c frequencyresponse curve d tim edom ain representation 4 Give a brief explanation of the basic idea behind Fourier analysis What is the input to Fourier analysis and what kind of outputs does it produce Lquot Draw and label frequencyresponse curves for lowpass highpass and bandpass filters O What parameters control the frequency of vibration of a spring and mass system l Draw the time domain representation of one cycle of a sinusoid as variations in instantaneous air pressure over time and one cycle of that same sinusoid as variations in instantaneous velocity over time so How if at all are the terms resonant frequency and harmonic different O How if at all are the terms resonant frequency and formant different N O A harmonic is a peak in a a frequency response curve b an amplitude spectrum or c either a frequency response curve or an amplitude spectrum N A resonance is a peak in a a frequency response curve b an amplitude spectrum or c either a frequency response curve or an amplitude spectrum N N A formant is a peak in a a frequency response curve b an amplitude spectrum or c either a frequency response curve or an amplitude spectrum N L A frequency response curve describes a N 4 An amplitude spectrum describes a The Physics ofSound 36 Freguency Response Problems Frequency Response Curve Output Spectrum Gain Requency Requency Requency Assume that a signal With the amplitude spectIan show at the le is modi ed by a lter With the frequency response curve show in the middle Show What the output spectIan would look like Frequency Response Curve Output Spectrum Gain H 125 ZS 375 SW 625 7i X75 l Requency Requency Requency T he Physics of Sound 37 Answer to Freguency Response Problems Input Spectrum Frequency Response Curve Output Spectrum Gain F wmwnrv nemmm F wmwnrv Input Spectrum Frequency Response Curve Output Spectrum Gain Reqnency Requency Requency H N 9 gt V 0 gt1 00 H N H E H 4 H V H 0 H gt1 H so 60 dBHL at 1000 Hz means The Physics afSaund 38 Decibel Study Questions What reference is used for the dBlL scale What reference is used for the dBSPL scale What reference is used for the dBHL scale What reference is used for the dBSL scale A listener barely detects a 125 Hz pure tone at 55 stpL Does this listener have a hearing loss at 125 Hz and if so what is the size of the hearing loss A listener barely detects a 1000 Hz pure tone at 55 stpL Does this listener have a hearing loss at 1000 Hz and if so what is the size of the hearing loss A listener barely detects a 125 Hz pure tone at 55 dBm Does this listener have a hearing loss at 125 Hz and if so what is the size of the hearing loss A listener barely detects a 1000 Hz pure tone at 55 dBm Does this listener have a hearing loss at 1000 Hz and if so what is the size of the hearing loss 60 dBSPL at 1000 Hz means more intense than 60 dBlL at 1000 Hz means more intense than more intense than The reference that is used for the stpL scale is a anumber b asentence 1f the answer to the question above is a number give the number if it s a sentence give the sentence The reference that is used for the dBHL scale is a anumber b asentence 1f the answer to the question above is a number give the number if it s a sentence give the sentence A specific individual has a 70 dB hearing loss in the left ear at 1000 Hz A 90 dBHL 1000 Hz tone that is presented to this listener s left ear would measure SL A sound measures 42 dBlL On the stpL scale that same sound will measure a 84 stpL because with the stpL formula we are now are multiplying the ratio by 20 instead of 10 b 42 stpL because the two versions of the formula are equivalent A sound measures 60 dBlL a The measured intensity 1M must therefore be imes greater than the reference intensity 1R b What would the pressure ratio EMER be for this same sound 0 Do the arithmetic to show what this sound would measure in stpL The Physics of Sound 39 19 A 39 mm a greaferthjd n 39 39 quot IR 03 to hqu mes mH N K this same sound c 39 stpL 0 stpL and b 3000 Hz 20 alBHL the grid lines on they axls are spaced at 2 dB interv l 20 On the graph below put a mark at a 3000 Hz 2 a 5 Frequency Threshold in Hz in dB SPL 125 450 250 255 500 115 750 80 1000 70 1500 55 2000 90 3000 100 4000 95 5000 155 8000 130 H N 9 gt V 0 gt1 00 O H O H H H N H E H 4 H Lquot H 0 H gt1 H 00 The Physics afSaund 40 Answers to Dec1bel Stud uestlons 103912 wattsm2 20 Pa or equivalently 00002 dynescmz The threshold of audibility for an average normalhearing listener at a particular signal frequency 3 The threshold of audibility for a particular listener at a particular signal frequency Consulting the attached figure and table showing the audibility curve for average normalhearing listeners we find that the threshold of audibility at 125 Hz is 45 stpL A listener who barely detected a 125 Hz tone at 55 stpL would therefore have hearing loss of 554510 dB39 that is the hearing sensitivity of this listener would be 10 dB worse than normal Consulting the attached figure and table showing the audibility curve for average normalhearing listeners we find that the threshold of audibility at 1000 Hz is 7 stpL A listener who barely detected a 1000 Hz tone at 55 stpL would therefore have a hearing loss of 55748 dB39 that is the hearing sensitivity of this listener would be 48 dB worse than normal The reference for dBHL is the audibility threshold so this listener would have a 55 dB hearing loss at 125 Hz There is no need to consult the table The reference for dBHL is the audibility threshold so this listener would have a 55 dB hearing loss at 1000 Hz There is no need to consult the table 6 factors of 10 ie 1000000 times more intense than 20 uPa 6 factors of 10 ie 1000000 times more intense than 103912 wattsm2 6 factors of 10 ie 1000000 times more intense than a 1000 Hz tone that is barely audible to an average norm alhearing listener a number 20 Pa a sentence The threshold of audibility for an average normalhearing listener at a particular signal frequency 20 dBSL The reference for the dBSL SLsensation level is the threshold of audibility for a specific listener So what we want to know here very simply is where this 90 dBHL tone is in relation to this particular listener s threshold This listener has a 70 dB hearing loss at this frequency so the 90 dBHL tone which would be 90 dB above a normalhearing listener s threshold is only 20 dB above this particular listener s threshold 42 stpL The pressure version of the formula was derived from the intensity version through algebraic manipulations so they have to be equivalent to one another The next problem was designed to illustrate how this can be the case a 1000000 times 6 factors of 10 more intense than 1R b If the intensity ratio is 1000000 the pressure ratio has to be the square root of 1000000 which is 1000 c stpL 20 log 1000 20 39 3 60 stpL This is exactly what we got for the same sound measured in dBlL It will always be the same If a sound measures 60 that same sound will measure 60 dBamp The Physics of Sound 41 19 a 10000 times 4 factors of10 more intense than IR b Ifthe intensity ratio is 10000 the pressure ratio has N the square root of 10000 which is 100 c stpL 20 log 100 2039 2 40 stpL This is exactly what we got for the same sound measured in dBquot It will always be the same Ifa sound measures 40 dBquot thazmme sound will measure 40 dB pL See below The lower ofthe two marks is 20 dB 2 factors of 10 above the constant reference line of 20 1135 The higher ofthe two marks is 20 dB also 2 factors of 10 above the survey line which is the threshold of audibility for the average normalhearing listener Time msec 0 100 200 300 400 500 Frequency d 400 Hz Phase o 100 200 300 400 500 5 Sum of aid Frequency 2 Give the amplimde Speclrum and me phase speclrum for the complex periodic signal a mago 71qu m a 100 Hz 18 apmndmv I I 0 100 200 300 400 500 Frequency c 300 Hz d 400 Hz 359 0 100 200 300 400 500 8 Sum of a4 Frequency Time msec Answer to Fourier Analysis Problem In Physc ofSound The Physics afSaund 44
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