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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 xaxis 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 crosscorrelation (sometimes just called correlation) of two realvalued 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 crosscorrelation....
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