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Comments, Quiz # 1. So far: Historical overview of speech technology - basic components/goals for systems Quick overview of pattern recognition basics.

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Presentation on theme: "Comments, Quiz # 1. So far: Historical overview of speech technology - basic components/goals for systems Quick overview of pattern recognition basics."— Presentation transcript:

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 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 (10 6 ) To analyze, look at constrained models of common structures: strings and tubes Acoustic waves - a brief intro

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 F = ma F y = S (tan Φ 2 - tan Φ 1 ) S (δ 2 y/δx 2 ) dx = ε dx δ 2 y/δt 2 Let c = √S/ ε c 2 (δ 2 y/δx 2 ) = δ 2 y/δt 2 δy/δx δy/δx + (δ 2 y/δx 2 ) dx

7 So δ 2 y δ 2 y δ x 2 δ t 2 c 2 is the wave equation for transverse vibration (vibration perpendicular to wave motion direction) on a string Where c can be derived from the properties of the medium, and is the wave propagation speed = String wave equation

8 Solutions dependent on boundary conditions Assume form f(t – x/c) for positive x direction (equivalently, f(ct – x) ) Then f(t + x/c) for negative x direction (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) ) Solutions to wave equation

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

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

13 Further: Using Newton’s second law, mass conservation, and assume pressure change is proportional to air density change δ 2 p δ 2 p δx 2 δt 2 c2c2 = − − δ 2 u δ 2 u δ x 2 δ t 2 c2c2 = − − Solving, we can show that c = speed of sound

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

15 Resonance in acoustic tubes Velocity: u(x,t) = u + (x-ct) – u - (x+ct) Pressure: p(x,t) = = Z 0 [ u + (x-ct) + u - (x+ct)] Let u + (x-ct) = A e jω(t - x/c ), u - (x-ct) = B e jω (t +x/c ) Assume u(0,t) = e jωt, p(L,t) = 0

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

17 First 3 modes of an acoustic tube open at one end

18 Example Human vocal tract during phonation of neutral vowel (vocal tract like open tube) – average male values c ≈ 340 m/s L = 17 cm, so 4L = 68 cm =.68m f 1 = 340/.68 = 500 Hz, f 2 = 1500 Hz, f 3 = 2500 Hz Similar to measured resonances

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

20 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

21 X-ray tracing and area function for /i/

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23 “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

24 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?”

25 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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