Lecture 2: Acoustics

EE E6820: Speech & Audio Processing & Recognition Lecture 2: Acoustics 1 The wave equation Dan Ellis & Mike Mandel Columbia University Dept. of Electrical Engineering http://www.ee.columbia.edu/ dpwe/e6820 January 29, 2009 2 Acoustic tubes: reflections & resonance 3 Oscillations & musical acoustics 4 Spherical waves & room acoustics E6820 SAPR (Ellis & Mandel) Acoustics January 29, 2009 1 / 38

Outline 1 The wave equation 2 Acoustic tubes: reflections & resonance 3 Oscillations & musical acoustics 4 Spherical waves & room acoustics E6820 SAPR (Ellis & Mandel) Acoustics January 29, 2009 2 / 38

Acoustics & sound 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 (Ellis & Mandel) Acoustics January 29, 2009 3 / 38

The wave equation Consider a small section of the rope y S x ε φ 2 φ 1 S Displacement y(x), tension S, mass ɛ dx Lateral force is F y = S sin(φ 2 ) S sin(φ 1 ) S 2 y x 2 dx E6820 SAPR (Ellis & Mandel) Acoustics January 29, 2009 4 / 38

Wave equation (2) Newton s law: F = ma S 2 y x 2 dx = ɛ dx 2 y t 2 Call c 2 = S/ɛ (tension to mass-per-length) hence, the Wave Equation: c 2 2 y x 2 = 2 y t 2... partial DE relating curvature and acceleration E6820 SAPR (Ellis & Mandel) Acoustics January 29, 2009 5 / 38

Solution to the wave equation If y(x, t) = f (x ct) (any f ( )) then y x = f (x ct) 2 y x 2 = f (x ct) also works for y(x, t) = f (x + ct) Hence, general solution: y t = cf (x ct) 2 y t 2 = c2 f (x ct) c 2 2 y x 2 = 2 y t 2 y(x, t) = y + (x ct) + y (x + ct) E6820 SAPR (Ellis & Mandel) Acoustics January 29, 2009 6 / 38

Solution to the wave equation (2) y + (x ct) and y (x + ct) are traveling waves shape stays constant but changes position time 0: y y + y - time T: x = c T x y + y - x c is traveling wave velocity ( x/ t) y + moves right, y moves left resultant y(x) is sum of the two waves E6820 SAPR (Ellis & Mandel) Acoustics January 29, 2009 7 / 38

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 e.g. y(x) at t = 0 plus constraints on y at particular xs 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 (Ellis & Mandel) Acoustics January 29, 2009 8 / 38

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 [simulation travel1.m] x = L y E6820 SAPR (Ellis & Mandel) Acoustics January 29, 2009 9 / 38

Outline 1 The wave equation 2 Acoustic tubes: reflections & resonance 3 Oscillations & musical acoustics 4 Spherical waves & room acoustics E6820 SAPR (Ellis & Mandel) Acoustics January 29, 2009 10 / 38

Acoustic tubes 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 (Ellis & Mandel) Acoustics January 29, 2009 11 / 38

Pressure and velocity Consider air particle displacement ξ(x, t) ξ(x) x Particle velocity v(x, t) = ξ t hence volume velocity u(x, t) = Av(x, t) (Relative) air pressure p(x, t) = 1 κ ξ x E6820 SAPR (Ellis & Mandel) Acoustics January 29, 2009 12 / 38

Wave equation for a tube Consider elemental volume Area da Force p da x Volume da dx Mass ρ da dx Force (p+ p / x dx) da Newton s law: F = ma p v dx da = ρ da dx x t p x = ρ v t c 2 2 ξ x 2 = 2 ξ t 2 c = 1 ρκ E6820 SAPR (Ellis & Mandel) Acoustics January 29, 2009 13 / 38

Acoustic tube traveling waves Traveling waves in particle displacement: ξ(x, t) = ξ + (x ct) + ξ (x + ct) Call u + (α) = ca α ξ+ (α), Z 0 = ρc A Then volume velocity: And pressure: u(x, t) = A ξ t = u+ (x ct) u (x + ct) p(x, t) = 1 ξ κ x = Z [ 0 u + (x ct) + u (x + ct) ] (Scaled) sum and difference of traveling waves E6820 SAPR (Ellis & Mandel) Acoustics January 29, 2009 14 / 38

Acoustic traveling waves (2) Different resultants for pressure and volume velocity: c c x Acoustic tube u + u - u(x,t) = u + - u - p(x,t) = Z 0 [u + + u - ] Volume velocity Pressure E6820 SAPR (Ellis & Mandel) Acoustics January 29, 2009 15 / 38

Terminations in tubes Equivalent of fixed point for tubes? Solid wall forces u(x,t) = 0 hence u + = u - 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 (Ellis & Mandel) Acoustics January 29, 2009 16 / 38

Standing waves Consider (complex) sinusoidal input u 0 (t) = U 0 e jωt Pressure/volume must have form Ke j(ωt+φ) Hence traveling waves: u + (x ct) = A e j( kx+ωt+φ A) u (x + ct) = B e j(kx+ωt+φ B) where k = ω/c (spatial frequency, rad/m) (wavelength λ = c/f = 2πc/ω) Pressure and volume velocity resultants show stationary pattern: standing waves even when A B [simulation sintwavemov.m] E6820 SAPR (Ellis & Mandel) Acoustics January 29, 2009 17 / 38

Standing waves (2) U 0 e jωt kx = π x = λ / 2 pressure = 0 (node) vol.veloc. = max (antinode) For lossless termination ( u + = u ), have true nodes and antinodes Pressure and volume velocity are phase shifted in space and in time E6820 SAPR (Ellis & Mandel) Acoustics January 29, 2009 18 / 38

Transfer function Consider tube excited by u 0 (t) = U 0 e jωt sinusoidal traveling waves must satisfy termination boundary conditions satisfied by complex constants A and B in u(x, t) = u + (x ct) + u (x + ct) = Ae j( kx+ωt) + Be j(kx+ωt) = e jωt (Ae jkx + Be jkx ) standing wave pattern will scale with input magnitude point of excitation makes a big difference... E6820 SAPR (Ellis & Mandel) Acoustics January 29, 2009 19 / 38

Transfer function (2) For open-ended tube of length L excited at x = 0 by U 0 e jωt jωt cos k(l x) u(x, t) = U 0 e cos kl k = ω c (matches at x = 0, maximum at x = L) i.e. standing wave pattern e.g. varying L for a given ω (and hence k): U 0 e jωt U 0 U L U 0 e jωt U 0 U L magnitude of U L depends on L (and ω) E6820 SAPR (Ellis & Mandel) Acoustics January 29, 2009 20 / 38

Transfer function (3) Varying ω for a given L, i.e. at x = L U L u(l, t) = U 0 u(0, t) = 1 cos kl = 1 cos(ωl/c) u(l) u(0) L u(l) at ωl/c = (2r+1)π/2, r = 0,1,2... u(0) Output volume velocity always larger than input Unbounded for L = (2r + 1) πc 2ω = (2r + 1) λ 4 i.e. resonance (amplitude grows without bound) E6820 SAPR (Ellis & Mandel) Acoustics January 29, 2009 21 / 38

Resonant modes For lossless tube with L = m λ 4, m odd, λ wavelength 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 = 3ω 0 compare to time domain... e.g. 17.5 cm vocal tract, c = 350 m/s ω 0 = 2π 500 Hz (then 1500, 2500,... ) L = λ 0/ 4 E6820 SAPR (Ellis & Mandel) Acoustics January 29, 2009 22 / 38

Scattering junctions u + k u - k Area A k Area A k+1 u + k+1 u - k+1 At abrupt change in area: pressure must be continuous p k (x, t) = p k+1 (x, t) vol. veloc. must be continuous u k (x, t) = u k+1 (x, t) traveling waves u + k, u- k, u+ k+1, u- k+1 will be different Solve e.g. for u k and u+ k+1 : (generalized term) u + k u - k 1 - r 1 + r + 2r 1 + r 2 r + 1 + r - 1 r + 1 u + k+1 u - k+1 r = A k+1 A k Area ratio E6820 SAPR (Ellis & Mandel) Acoustics January 29, 2009 23 / 38

Concatenated tube model Vocal tract acts as a waveguide Lips x = L Lips u L (t) Glottis u 0 (t) x = 0 Glottis Discrete approximation as varying-diameter tube A k, L k x A k+1, L k+1 E6820 SAPR (Ellis & Mandel) Acoustics January 29, 2009 24 / 38

Concatenated tube resonances Concatenated tubes scattering junctions lattice filter u + k + e -jωτ 1 + e -jωτ 2 + u - k + e -jωτ 1 + e -jωτ 2 + + + + + e -jω2τ 1 + e -jω2τ 2 + Can solve for transfer function all-pole 1 5 0-1 -0.5 0 0.5 1 Approximate vowel synthesis from resonances [sound example: ah ee oo] E6820 SAPR (Ellis & Mandel) Acoustics January 29, 2009 25 / 38

Outline 1 The wave equation 2 Acoustic tubes: reflections & resonance 3 Oscillations & musical acoustics 4 Spherical waves & room acoustics E6820 SAPR (Ellis & Mandel) Acoustics January 29, 2009 26 / 38

Oscillations & musical acoustics Pitch (periodicity) is essence of music why? why music? Different kinds of oscillators simple harmonic motion (tuning fork) relaxation oscillator (voice) string traveling wave (plucked/struck/bowed) air column (nonlinear energy element) E6820 SAPR (Ellis & Mandel) Acoustics January 29, 2009 27 / 38

Simple harmonic motion Basic mechanical oscillation ẍ = ω 2 x x = A cos(ωt + φ) Spring + mass (+ damper) F = kx m ζ x e.g. tuning fork Not great for music fundamental (cos ωt) only relatively low energy ω 2 = k m E6820 SAPR (Ellis & Mandel) Acoustics January 29, 2009 28 / 38

Relaxation oscillator Multi-state process one state builds up potential (e.g. pressure) switch to second (release) state revert to first state, etc. e.g. vocal folds: p u http://www.youtube.com/watch?v=ajbcjiyhfky Oscillation period depends on force (tension) easy to change hard to keep stable less used in music E6820 SAPR (Ellis & Mandel) Acoustics January 29, 2009 29 / 38

Ringing string e.g. our original rope example tension S mass/length ε Many musical instruments guitar (plucked) piano (struck) violin (bowed) Control period (pitch): change length (fretting) change tension (tuning piano) change mass (piano strings) Influence of excitation... [pluck1a.m] L ω 2 = π2 S L 2 ε E6820 SAPR (Ellis & Mandel) Acoustics January 29, 2009 30 / 38

Wind tube Resonant tube + energy input energy nonlinear element ω = acoustic waveguide π c 2 L e.g. clarinet lip pressure keeps reed closed reflected pressure wave opens reed reinforced pressure wave passes through finger holds determine first reflection effective waveguide length scattering junction (tonehole) (quarter wavelength) E6820 SAPR (Ellis & Mandel) Acoustics January 29, 2009 31 / 38

Outline 1 The wave equation 2 Acoustic tubes: reflections & resonance 3 Oscillations & musical acoustics 4 Spherical waves & room acoustics E6820 SAPR (Ellis & Mandel) Acoustics January 29, 2009 32 / 38

Room acoustics Sound in free air expands spherically: Spherical wave equation: radius r 2 p r 2 + 2 p r r = 1 2 p c 2 t 2 solved by p(r, t) = P 0 r e j(ωt kr) Energy p 2 falls as 1 r 2 E6820 SAPR (Ellis & Mandel) Acoustics January 29, 2009 33 / 38

Effect of rooms (1): Images Ideal reflections are like multiple sources: virtual (image) sources reflected path source listener Early echoes in room impulse response: h room (t) direct path early echos t actual reflections may be h r (t), not δ(t) E6820 SAPR (Ellis & Mandel) Acoustics January 29, 2009 34 / 38

Effect of rooms (2): Modes Regularly-spaced echoes behave like acoustic tubes Real rooms have lots of modes! dense, sustained echoes in impulse response complex pattern of peaks in frequency response E6820 SAPR (Ellis & Mandel) Acoustics January 29, 2009 35 / 38

Reverberation Exponential decay of reflections: h room (t) ~e -t /T Frequency-dependent greater absorption at high frequencies faster decay Size-dependent larger rooms longer delays slower decay Sabine s equation: RT 60 = 0.049V Sᾱ Time constant varies with size, absorption t E6820 SAPR (Ellis & Mandel) Acoustics January 29, 2009 36 / 38

Summary Traveling waves Acoustic tubes & resonances Musical acoustics & periodicity Room acoustics & reverberation Parting thought Musical bottles E6820 SAPR (Ellis & Mandel) Acoustics January 29, 2009 37 / 38

References E6820 SAPR (Ellis & Mandel) Acoustics January 29, 2009 38 / 38