This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: EE 261 The Fourier Transform and its Applications Fall 2010 Solutions to Problem Set Four 1. (10 points) Solving the wave equation An infinite string is stretched along the x-axis and is given an initial displacement described by a function f ( x ). It is then free to vibrate. The displacement u ( x,t ) at a time t > 0 and at a point x on the string is described by the wave equation 2 u t 2 = 2 u x 2 . One often includes physical constants in the equation, e.g, the speed of the wave, but these are suppressed to keep things simple. Assume that u ( x, 0) = f ( x ) and u t ( x, 0) = 0 (zero initial velocity) and use the Fourier transform to show that u ( x,t ) = 1 2 ( f ( x + t ) + f ( x- t )) . This is dAlemberts (famous) solution to the wave equation. Solution: Well take the Fourier transform on both sides of the wave equation with respect to the variable x . To simplify notation, write U ( s,t ) = F ( u ( x,t )) . We also let F ( s ) = F f ( s ) . so that U ( s, 0) = F ( u ( x, 0)) = F f ( s ) = F ( s ) . Taking the Fourier transform of the wave equation we get 2 t 2 U ( s,t ) = (2 is ) 2 U ( s,t ) or 2 t 2 U ( s,t ) + (2 s ) 2 U ( s,t ) = 0 . This is an ordinary differential equation in t . The solution has the form U ( s,t ) = Ae 2 ist + Be- 2 ist . 1 where A and B will depend on s . We find A and B from the initial conditions. First, U ( s, 0) = F ( s ) = A + B. The initial velocity is assumed to be zero, i.e., u t ( x, 0) = 0, so U t ( s, 0) = 0 , which gives 0 = A (2 is ) e + B (- 2 is ) e or 0 = A- B. Putting the two equations for A and B together, we have A = B = (1 / 2) F ( s ) . and U ( s,t ) = 1 2 F ( s ) e 2 ist + 1 2 F ( s ) e- 2 ist . Take the inverse Fourier transform (apply the modulation theorem, or compute directly): u ( x,t ) = F- 1 F ( s ) 2 e 2 ist + F ( s ) 2 e- 2 ist = 1 2 Z - F f ( s ) e 2 is ( x + t ) ds + Z - F f ( s ) e 2 is ( x- t ) ds = 1 2 ( f ( x + t ) + f ( x- t )) . 2. (20 points) Cross Correlation The cross-correlation (sometimes just called correlation) of two real-valued signals f ( t ) and g ( t ) is defined by ( f ? g )( x ) = Z - f ( y ) g ( x + y ) dy . ( f ? g )( x ) is often described as a measure of how well the values of g , when shifted by x , correlate with the values of f . It depends on x ; some shifts of g may correlate better with f than other shifts. To get a sense of this, think about when ( f ? g )( x ) is positive (and large) or negative (and large) or zero (or near zero). If, for a given x , the values f ( y ) and g ( x + y ) are tracking each other both positive or both negative then the integral will be positive and so the value ( f ? g )( x ) will be positive. The closer the match between f ( x ) and g ( x + y ) (as y varies) the larger the integral and the larger the cross-correlation....
View Full Document