EE E6820: Speech & Audio Processing & Recognition Lecture 2: Acoustics 1 The wave equation 2 Acoustic tubes: reflections & resonance 3 Oscillations & musical acoustics 4 Spherical waves & room acoustics Dan Ellis <dpwe@ee.columbia.edu> http://www.ee.columbia.edu/~dpwe/e6820/ Columbia University Dept. of Electrical Engineering Spring 2002 E6820 SAPR - Dan Ellis L02 - Acoustics - 2002-02-04 - 1
Acoustics & sound 1 • Acoustics is the study of physical waves • (Acoustic) waves transmit energy without permanently displacing matter (e.g. ocean waves) • Same math recurs in many domains • Intuition: pulse going down a rope E6820 SAPR - Dan Ellis L02 - Acoustics - 2002-02-04 - 2
The wave equation • Consider a small section of the rope: y S φ 2 ε x φ 1 S ε • displacement is y(x), tension S, mass ·dx ⋅ ( φ 2 ) ⋅ ( φ 1 ) → F y = S sin – S sin lateral force is 2 ∂ y ⋅ ⋅ = S d x x 2 ∂ E6820 SAPR - Dan Ellis L02 - Acoustics - 2002-02-04 - 3
Wave equation (2) F = ma • Newton’s law: 2 2 ∂ y ∂ y ⋅ ⋅ ε x ⋅ S d x = d x 2 t 2 ∂ ∂ c 2 S ε ⁄ = • Call (tension to mass-per-length) hence the wave equation : 2 2 ∂ y ∂ y c 2 ⋅ = x 2 t 2 ∂ ∂ .. partial DE relating curvature and acceleration E6820 SAPR - Dan Ellis L02 - Acoustics - 2002-02-04 - 4
Solution to the wave equation ( , ) ( ) y x t = f x – ct • If then ∂ y ∂ y ( ) ⋅ ( ) = f ' x – ct = – c f ' x – ct ∂ ∂ x t 2 2 ∂ y ∂ y c 2 ( ) ⋅ ( ) = f '' x – ct = f '' x – ct x 2 t 2 ∂ ∂ ( , ) ( ) y x t = f x + ct also works for Hence, general solution: 2 2 ∂ y ∂ y c 2 ⋅ = x 2 t 2 ∂ ∂ y + x y – x ⇒ ( , ) ( ) ( ) y x t = – ct + + ct E6820 SAPR - Dan Ellis L02 - Acoustics - 2002-02-04 - 5
Solution to the wave equation (2) y + x y – x ( ) ( ) – ct + ct • and are travelling waves - shape stays constant but changes position: y y+ time 0: y- x ∆ x = c·T y+ time T: y- c is travelling wave velocity ( ∆ x / ∆ t ) • y + moves right, y – moves left • • resultant y(x) is sum of the two waves E6820 SAPR - Dan Ellis L02 - Acoustics - 2002-02-04 - 6
Wave equation solutions (3) What is the form of y + , y – ? • - any doubly-differentiable function will satisfy wave equation • Actual waveshapes dictated by boundary conditions - y(x) at t = 0 - constraints on y at particular x’s e.g. input motion y(0, t) = m(t) rigid termination y(L, t) = 0 y y(0,t) = m(t) x y + (x,t) y(L,t) = 0 E6820 SAPR - Dan Ellis L02 - Acoustics - 2002-02-04 - 7
Terminations and reflections • System constraints: - initial y(x, 0) = 0 (flat rope) - input y(0, t) = m(t) (at agent’s hand) ( → y + ) - termination y(L, t) = 0 (fixed end) - wave equation y(x,t) = y + (x - ct) + y – (x + ct) • At termination: y(L, t) = 0 → y + (L - ct) = – y – (L + ct) i.e. y + and y – are mirrored in time and amplitude around x=L → inverted reflection at termination y+ y(x,t) = y+ + y– y– x = L → simulation [travel1.m] E6820 SAPR - Dan Ellis L02 - Acoustics - 2002-02-04 - 8
Acoustic tubes 2 • Sound waves travel down acoustic tubes: pressure x - 1-dimensional; very similar to strings • Common situation: - wind instrument bores - ear canal - vocal tract E6820 SAPR - Dan Ellis L02 - Acoustics - 2002-02-04 - 9
Pressure and velocity ξ x t ( , ) • Consider air particle displacement : ξ ( x ) x ∂ξ ( , ) v x t = • Particle velocity ∂ t ( , ) ⋅ ( , ) u x t = A v x t hence volume velocity ∂ξ 1 ( , ) ⋅ p x t = – - - - • Air pressure ∂ κ x E6820 SAPR - Dan Ellis L02 - Acoustics - 2002-02-04 - 10
Wave equation for a tube • Consider elemental volume: Area dA Force p·dA Force ( p+ ∂ p/ ∂ x·dx ) ·dA x Volume dA·dx Mass ρ ·dA·dx F = ma • Newton’s law: ∂ p ∂ v ⋅ ⋅ ρ dAdx ⋅ – dx dA = ∂ ∂ t x ∂ p ∂ v ⇒ ρ t = – ∂ ∂ x 2 2 ∂ ξ ∂ ξ 1 c 2 ⋅ = c = - - - - - - - - - - - • Hence ρκ x 2 t 2 ∂ ∂ E6820 SAPR - Dan Ellis L02 - Acoustics - 2002-02-04 - 11
Acoustic tube traveling waves • Traveling waves in particle displacement: ξ + x ξ - x ξ x t ( , ) ( ) ( ) = – ct + + ct u + α ∂ ξ + α ( ) cA α ( ) = – • Call ∂ ρ c Z 0 = - - - - - - A • Then pressure: ∂ξ u + x u - x 1 ( , ) ⋅ ⋅ [ ( ) ( ) ] p x t = – - - - = Z 0 – ct + + ct κ ∂ x • Volume velocity: ∂ξ u + x u - x ( , ) ⋅ ( ) ( ) u x t = A = – ct – + ct ∂ t • (Scaled) sum and difference of traveling waves E6820 SAPR - Dan Ellis L02 - Acoustics - 2002-02-04 - 12
Acoustic tube traveling waves (2) • Different residuals for pressure and vol. veloc.: Acoustic tube x c u+ c u- u(x,t) = u+ - u- p(x,t) = Z 0 [u+ + u-] E6820 SAPR - Dan Ellis L02 - Acoustics - 2002-02-04 - 13
Terminations in tubes • Equivalent of ‘fixed point’ for tubes? Solid wall forces hence u+ = u- u(x,t) = 0 u 0 (t) (Volume velocity input) Open end forces p(x,t) = 0 hence u+ = -u- • Open end is like fixed point for rope: reflects wave back inverted • Unlike fixed point, solid wall reflects traveling wave without inversion E6820 SAPR - Dan Ellis L02 - Acoustics - 2002-02-04 - 14
Standing waves • Consider (complex) sinusoidal input: U 0 e j ω t ( ) ⋅ u 0 t = Ke j ω t ( φ ) + • At any point, values will have form • Hence traveling waves: ( ω t φ A ) u + x j – kx + + ( ) – ct = A e ( ω t φ B ) j kx + + u - x ( ) + ct = B e ω c ⁄ k = where (spatial frequency, rad/m) λ ⁄ 2 π c ω ⁄ = c f = (wavelength ) • Pressure / vol. veloc. resultants show stationary pattern: standing waves - even when | A | ≠ | B | → simulation [sintwavemov.m] E6820 SAPR - Dan Ellis L02 - Acoustics - 2002-02-04 - 15
Standing waves (2) U 0 e j ω t pressure = 0 ( node ) kx = π vol.veloc. = max x = λ / 2 ( antinode ) For lossless termination (|u + | = |u - |), • have true nodes & antinodes • Pressure and vol. veloc. are phase shifted - in space and in time ∗ E6820 SAPR - Dan Ellis L02 - Acoustics - 2002-02-04 - 16
Transfer function U 0 e j ω t ( ) ⋅ u 0 t = • Consider tube excited by : - sinusoidal traveling waves must satisfy termination ‘boundary conditions’ - satisfied by complex constants A and B in u + x u - x ( , ) ( ) ( ) u x t = – ct + + ct ( ω t ) ( ω t ) Ae j – kx + Be j kx + = + e j ω t Ae jkx – Be jkx ⋅ ( ) = + - standing wave pattern will scale with input magnitude - point of excitation makes a big difference E6820 SAPR - Dan Ellis L02 - Acoustics - 2002-02-04 - 17
Transfer function (2) • For open-ended tube of length L excited at x = 0 U 0 e j ω t by : ( ) ω - U 0 e j ω t cos k L – x ( , ) ⋅ - - - - - - - - - - - - - - - - - - - - - - - - - - - - u x t = k = - - - - cos kL c (works at x = 0) • i.e. standing wave pattern e.g. varying L for a given ω (and hence k ): U 0 e j ω t U 0 U L U 0 e j ω t U 0 U L E6820 SAPR - Dan Ellis L02 - Acoustics - 2002-02-04 - 18
Transfer function (3) Varying ω for a given L : • ( , ) u L t 1 1 - - - - - - - - - - - - - - - - = - - - - - - - - - - - - - - - = - - - - - - - - - - - - - - - - - - - - - - - - - - - - - at x = L , ( , ) ( ω L c ⁄ ) u 0 t cos kL cos u ( L ) u (0) L ∞ u ( L ) at ω L / c = (2 r +1) π /2, r = 0,1,2... u (0) • Output vol. veloc. always larger than input ) π c )λ ( ( L = 2 r + 1 - - - - - - - = 2 r + 1 - - - • Unbounded for 2 ω 4 i.e. resonance E6820 SAPR - Dan Ellis L02 - Acoustics - 2002-02-04 - 19
Resonant modes • For lossless tube m λ ⋅ , m odd, λ wavelength, L = - - - with 4 ( ) u L is unbounded, meaning: - - - - - - - - - - - ( ) u 0 - transfer function has pole on frequency axis - energy at that frequency sustains indefinitely L = 3 · λ 1 /4 → ω 1 = 3 ω 0 L = λ 0 /4 • e.g 17.5 cm vocal tract, c = 350 m/s → ω 0 = 2 π · 500 Hz (then 1500, 2500 ...) E6820 SAPR - Dan Ellis L02 - Acoustics - 2002-02-04 - 20
Scattering junctions At abrupt change in area: • pressure must be continuous p k (x, t) = p k+1 (x, t) u+ k+1 u+ k • vol. veloc. must be continuous u- k u k (x, t) = u k+1 (x, t) u- k+1 • traveling waves u+ k , u- k , u+ k+1 , u- k+1 Area Ak Area Ak+1 will be different Solve e.g. for u - k and u + • k+1 : (generalized term.) 2r 1 + r u+ k+1 u+ k + A k+1 r = 1 - r r - 1 A k 1 + r r + 1 “Area ratio” u- k u- k+1 + 2 r + 1 E6820 SAPR - Dan Ellis L02 - Acoustics - 2002-02-04 - 21
Concatenated tube model • Vocal tract acts as a waveguide Lips x = L Lips u L ( t ) Glottis u 0 ( t ) x = 0 Glottis • Discrete approx. as varying-diameter tube: A k , L k A k+1 , L k+1 x E6820 SAPR - Dan Ellis L02 - Acoustics - 2002-02-04 - 22
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