# Digital Representation of Audio Information EE599-020 Audio Signals and Systems Wave Basics Kevin D. Donohue Electrical and Computer Engineering University of Kentucky Vibrating String o 2 y2 y1 y 1 o x Given an elastic string is displaced in the y-direction, let o be the mass per unit length, assume tension o is constant and independent of position, angles (1 and 2) are small, motion is limited to x-y plane, and element x of the string is only displaced in the y-direction. Derive the equation

governing the motion of the string. Vibrating String Differential forces in x and y direction from tension o are given by Fx o (cos( 2 ) cos(1 )) Fy o (sin( 2 ) sin(1 )) For small , Fx is negligible and sin()= = dy/dx . Therefore Newtons second law for y-direction results in: o x o 2 y t 2 2 y y y o 2 1 x x t 2 1 y 2 y1 x

x 0 x x o lim t 2 o 2 y o x 2 2 y 1 2 y 2 y t 2 c 2 x 2 DAlemberts Solution The general solution of the partial differential equation:

is: 2 y 1 2 y 0 x 2 c 2 t 2 y( x, t ) Af ( x ct ) Bf ( x ct ) where f(x-ct) represents a wave traveling in the forward (or positive) x-direction, and f(x+ct) represents a wave traveling in the negative x-direction. Standing Waves Consider the first harmonic component of f(x,t) given by: 2 f1 ( x, t ) A1 cos ( x ct ) A1 cosx t where is called the angular wave number related to space and is referred to angular frequency related to time.

Consider a sinusoidal forward and backward wave in complex notation: f1 ( x, t ) .5 A1 exp j (x t ) f1 ( x, t ) .5 A1 exp j (x t ) determine the real part of the summation of the forward and backward wave. Sketch and discuss the properties of the resulting (standing) wave and its higher harmonics. Acoustic Tubes For the transverse waves of an elastic string the motion of the string was governed by the restoring force and mass/momentum. For gasses or fluids in a tube, the pressure p(x,t) and volume displacement u(x,t) are directly considered (analogous to voltage and current). The resulting equations for these quantities are: 2 p 2 p c x 2 t 2 2u 2u c

x 2 t 2 2 where c 2 1 (density compressibility ) Acoustic Tubes The solution is the same as that of the transverse wave on a string: u ( x, t ) Af ( x ct ) Bf ( x ct ) p( x, t )Z o Af ( x ct ) Bf ( x ct ) where Zo is the acoustic impedance of the gas given by: Zo c A with being the density, c being the propagation speed, and A being the cross-sectional area of the tube. Newtons second law

for volume mass results in: p u x A t Tube Resonance The general solution can be expressed: An cos( nt ) Bn sin( nt )Cn cos( n x) Dn sin( n x) u( x, t ) n 1 Initial conditions determine time amplitude and phase of standing time wave oscillation. Boundary conditions will determine the resonance frequency. If a tube is closed at both ends, the displacement must be zero: u (0, t )0 C n cos( n 0) Dn sin( n 0) u ( L, t ) 0 C n cos( n L ) Dn sin( n L ) If a tube is closed at one end and open at the other, the velocity must be zero: u(0, t )0 C n cos( n 0) Dn sin( n 0) u( x, t ) 0 Cn n sin( n L ) Dn n cos( n L )

x x L Tube Resonance The boundary conditions for a tube closed at both ends, reduces solution to: n nc u ( x, t ) An Dn cos(nt ) Bn Dn sin(nt ) sin n x where n , n L L n 1 The boundary conditions for a tube is closed at one end and open at the other, reduces u ( x, t ) A Dsolution cos( tto: ) B D sin( t ) sin x n n

n n n n n n 1 wher e (2n 1) (2n 1)c n , n 2 L 2

L Sketch standing waves in the tubes for the above cases. Homework (1) a) Electronically generate a harmonic for a recorded guitar note, where a node is placed at the center of the string. b) Repeat part (a) for when the node is placed at one third the length of the string. Extra credit (2 points), record your voice saying (or singing) an a sound (the a sound in the word about) for 2 seconds sampled at 8000 Hz. Try to perform an analogous harmonic operation as in part (a). Email me the script/function and the original data file.