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Comments, Quiz # 1

Comments, Quiz # 1. So far: Historical overview of speech technology - basic components/goals for systems Quick overview of pattern recognition basics Quick overview of auditory system. Next talks focus on the nature of the signal: Acoustic waves in small spaces (sources)

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Comments, Quiz # 1

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  1. Comments, Quiz # 1

  2. So far: • Historical overview of speech technology - basic components/goals for systems • Quick overview of pattern recognition basics • Quick overview of auditory system • Next talks focus on the nature of the signal: • Acoustic waves in small spaces (sources) • Acoustic waves in large spaces (rooms)

  3. Acoustic waves - a brief intro • A way to bridge from thinking about EE to thinking about acoustics: • Acoustic signals are like electrical ones, only much slower … • Pressure is like voltage • Volume velocity is like current(and impedance = Pressure/velocity) • For wave solutions, c is a lot smaller for sound (106) • To analyze, look at constrained models of common structures: strings and tubes

  4. String and tube models • Vibrating Strings – excitation • Violin – bowed or plucked • Guitar – plucked • Cello – bowed or plucked • Piano – struck • Acoustic tube – excitation • Trumpet – lip vibrations • Clarinet - reed • Human voice – glottal vibration

  5. String model assumptions • No stiffness • Constant tension S throughout • Constant mass density εthroughout • Small vertical displacement • Ignore gravity, friction

  6. Vibrating string geometry δy/δx + (δ2y/δx2) dx δy/δx F = ma Fy = S (tan Φ2 - tan Φ1 ) S (δ2y/δx2) dx= εdx δ2y/δt2 Let c = √S/ε c2 (δ2y/δx2)= δ2y/δt2

  7. String wave equation is the wave equation for transverse vibration (vibration perpendicular to wave motion direction) on a string So δ2y δ2y c2 = δx2 δt2 Where c can be derived from the properties of the medium, and is the wave propagation speed

  8. Solutions to wave equation • Solutions dependent on boundary conditions • Assume form f(t – x/c) for positive xdirection(equivalently, f(ct – x) ) • Then f(t + x/c) for negative xdirection(or, f(ct + x) ) • Sum is A f(t - x/c) + B f(t +x/c) • (or, A f(ct –x) + B f(ct + x) )

  9. Traveling -> standing waves • Let g = sin(λx – ct), q = sin(λx + ct) • sin u + sin v = 2 sin ((u+v)/2)cos((u-v)/2) • g + q = 2 sin(λx)cos(ct) • Fixed phase in x dimension, time-varying amplitude (with max fluctuation determined by position); a “standing wave” • Basic phenomenon in strings, tubes, rooms

  10. Excitation Open end x 0 L Uniform tube, source on one end, open on the other

  11. Assumptions • Plane wave propagation for frequencies below ~4 kHz;as λ increases, plane assumption is better Since c = fλ, and c ≈ 340 m/sf = 3400 Hz λ= .1m = 10 cm • No thermal conduction losses • No viscosity losses • Rigid walls • Cross-sectional area is constant

  12. Further: Using Newton’s second law, mass conservation, and assume pressure change is proportional to air density change δ2p δ2p δx2 δt2 δ2u δ2u δx2 δt2 c2 c2 − − − − = = Solving, we can show that c = speed of sound

  13. Solutions to wave eqn • Assume the form f(x – ct) [rightward wave] • 2nd time derivative is c2 times 2nd space derivative, so it works • Same for f(x + ct) [leftward wave] • For sinusoids, sum gives a standing wave (as before)

  14. Resonance in acoustic tubes • Velocity: u(x,t) = u+(x-ct) – u-(x+ct) • Pressure: p(x,t) = = Z0[ u+(x-ct) + u-(x+ct)] • Let u+(x-ct) = A ejω(t - x/c), u-(x-ct) = B ejω(t +x/c) • Assume u(0,t) = ejωt, p(L,t) = 0

  15. u(0,t) = ejωt= A ejω(t - 0/c) - B ejω(t + 0/c) Problem: Find A and B to match boundary conditions Solve for A and B (eliminate t) Now you can get equation 10.24 in text, for excitation U(ω) ejωt: p(L,t) = 0 = A ejω(t - L/c) + B ejω(t + L/c) u(x,t) = cos[ω(L-x)/c] U(ω) ejωt cos[ωL/c] Poles occur when: ω= (2n + 1)πc/2L f = (2n + 1)c/4L

  16. First 3 modes of an acoustic tube open at one end

  17. Example Human vocal tract during phonation of neutral vowel (vocal tract like open tube) – average male valuesc ≈ 340 m/s L = 17 cm, so 4L = 68 cm = .68m f1= 340/.68 = 500 Hz, f2 = 1500 Hz, f3 = 2500 HzSimilar to measured resonances

  18. Effect of losses in the tube • Upward shift in lower resonances • Poles no longer on unit circle - peak values in frequency response are finite

  19. Effect of nonuniformities in the tube • Impedance mismatches cause reflections • Can be modeled as a succession of smaller tubes • Resonances shift - hence the different formants for different speech sounds

  20. X-ray tracing and area function for /i/

  21. “small acoustics” summary • Voice, many instruments, modeled by tubes • Traveling waves in both directions yield standing waves • Standing waves correspond to resonances • Variations from the idealization give the variety of speech sounds, musical timbre

  22. Homework #2 (due next Wednesday) • Problems 9.2, 9.4, 10.1, 10.5, 14.4, 14.5 from the book (as noted in the email) • Also the problem: “Describe phase locking in the auditory nerve. Over roughly what frequencies does this take place?”

  23. Planning ahead (further) • Quiz #2, Monday March 12 • Base on chapters 13, 19-22 (mostly chapter 22), and 23, and related classes • Room acoustics, speech feature extraction, and linguistic categories. • Project proposal, due March 21

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