1 / 25

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)

marie
Download Presentation

Comments, Quiz # 1

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

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. 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

More Related