ACOUSTIC THEORY OF SPEECH PRODUCTION ... - Ling.cam.ac.uk

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8 of 20 Paper 9 etc: AcThSpProd Figure 9: Review: Relationship between wavelength and frequency A B 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.11 time (s) Wavelength = distance (cm) travelled in one cycle Thus, wavelength depends on speed of sound (referred to as c) In air, c = 34000 cm/s Wavelength = c/frequency Thus, higher frequencies (e.g. A) have shorter wavelengths than lower frequencies (e.g. B).) 6.1 Boundary conditions When the wavelength of a sound is the right length to fulfill these boundary conditions, the sound will keep being reflected by each end of the tube (it will keep bouncing back and forth along it) and it will be amplified because it takes less energy to move the air particles. A frequency at which this amplification arises is a resonance frequency. You can think of it as happening when the boundary conditions are such that the particular frequency keeps bouncing back and forth along the tube, reinforcing itself as it goes, rather than causing random fluctuation or cancelling itself out as it goes back and forth. More technically, reinforcement rather than random fluctuation or cancellation takes place when the boundary conditions cause pressure and velocity waves of the same frequency to become 90° rather than 0° out of phase. The boundary conditions necessary for standing waves to occur depend on the state of the tube's ends. closed end: pressure maximum; open end: pressure minimum L = length of tube i.e., L when the end is closed, pressure must be at a maximum (velocity at a minimum); when the end is open, pressure must be at a minimum (velocity at a maximum). For a tube of uniform cross-sectional area that is closed at one end and open at the other, the lowest frequency that meets these boundary conditions has a wavelength 4 times the length of the tube: i.e. one quarter of one period of this particular sine wave fits into the tube such that it meets the boundary conditions of having a pressure maximum at the closed end, and a pressure minimum at the open end. relative amplitude of pressure +1 0 -1 L amplitude envelope of lowest standing wave of pressure, which is a ¼ wavelength of a sinusoidal sound wave. Only the upper (positive) half of the envelope need be shown, since the upper and lower parts are symmetrical. M4_0708_AcousticTheorySpeechProduction_07-8.doc
Paper 9 etc: M4_AcThSpProd 9 of 20 The formula for the lowest resonance of such a tube is F1 = c where c = speed of sound (c. 34,000 cm/s in air); 4L L = length of tube (16-17 cm for a man; c. 14 cm for a woman.) in speech, L represents distance from glottis; glottis is modelled as closed; lips are modelled as open Successively higher resonances occur at odd-number integer multiples: F2 = 3c , F3 = 5c & so on. 4L 4L Work out that this must be so if the above statements are right. Check your reasoning against Figure 10. Figure 10. Standing waves of velocity for a tube of uniform cross-sectional area, closed at one end, open at the other. Draw the standing wave of pressure in. (Maximum at the closed end, minium at the open end.) |U(x)| = volume velocity (absolute value) SWP = standing wave pattern (not “of pressure” in these pictures) A tube that is closed at one end and open at the other is called a “quarter-wave resonator”. An average man’s vocal tract is 17 cm long, so for [´] (which is modelled as a single unconstricted tube), his formant frequencies are: F1 : 34,000 = 500 Hz F2: 3(500) = 1500 Hz F3: 5(500) = 2500 Hz 4 x 17 M4_0708_AcousticTheorySpeechProduction_07-8.doc
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