The source/filter theory of speech acousticsSound sources -vocal fold vibration (complex periodic sound)frication (turbulent noise)stop release burst (impulse source)Filter - vocal tract resonancesvowel acousticsvoice source – complex periodic wavevocal tract filter – resonant tube[ ә ] Fn = (2n-1)c/ 4L[ a ] ??[ i ] ??A sequence of glottal pulses – one pulse every 6.6 msso the glottal frequency is 150 Hz (150 pulses per second)The “glottal frequency” is also called the fundamental frequency of voicing and is abbreviated F0.The spectrum of the sequence of glottal pulses.It has a series of peaks – these are called “harmonics”Higher frequency harmonics generally have lower amplitude than lower frequency harmonics.The frequency of the lowest harmonic (the first harmonic)is equivalent to the fundamental frequency of the voice.1st harmonic = F0 = 150 Hz (in this example)The higher harmonics are integer multipules of the fundamental – thus the 10th harmonic is at 1500 Hz.This is the spectrum of the voice SOURCE. It hasyet to be filtered by a vocal tract!Review schwa acoustics.- Vocal tract is a uniform tube (same diameter for the length of the tube)- This tube has certain resonant frequencies- We can calculate the resonances given the lengthof the vocal tract (assume 17.5 cm for now)and the speed of sound (assume 35,000 cm/s)- Fn = (2n-1)c/4LF1 = 500 HzF2 = 1500 HzF3 = 2500 Hz....The filtered voice source - [ ә ]Same set of harmonics as in the unfiltered sourcenow harmonics at 500, 1500 and 2500 Hz are louder.500 1500 2500Not all vowels sound like schwa.Today we’ll look at the “tube models” approach tomodeling vowel differencesNext time we’ll look at the “perturbation theory” approach.A tube model of the vowel [ɑ] (low back unrounded)narrow back tubewide front tubeA tube model of the vowel [ɑ] (low back unrounded)fn = (2n-1)c/4lbfn = (2n-1)c/4lfA nomogram of the two tube model: resonant frequencies for different back tube lengths.A nomogram of the two tube model.If the back tube is 3 cm long, the front tube is 13 cm long.F1 = c/(4*13) = 673 HzF2 = 3c /(4*13) = 2019 HzF3 = c/ (4*3) = 2917 HzA nomogram of the two tube model.In the vowel [ a ] the front and back tubes are about equal. F1 ≅ F2 = c/(4*8) = 1093 HzɑsF2F1Prediction of the tube model is basically correct – we find F1 and F2in [ ɑ ] are very close to each other infrequency.A tube model for the vowel [ i ]back tubefront tubeconstrictionA tube model for the vowel [ i ]fn = nc/2lbfn = (2n-1)c/4lfconstrictionA tube model for the vowel [ i ]A Helmholtz resonatorf = c/2π √ (Ac/Ablblc)[i] with constriction 11 cm from the glottisF1 = 300 Hz (Helmholtz)F2 = 1900 Hz = c/(2*9) lb=9 cmF3 = 2200 Hz = c / (4*4) lf = 4 cmThis tube model prediction is approximately correct for [ i ].F3 = 2500 HzF2 = 1900 HzF1 = 400
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