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Unformatted text preview: R. Balan Homework #6 Solutions MATH 464 1. Note the function f is the convolution between the box function and the unit Gaussian function , that is: f ( x ) = Z  ( u ) ( x u ) du , ( u ) = 1 ,  u  < 1 2 ,  u  > 1 2 , ( u ) = e  u  2 Thus the Fourier transform is the product of the Fourier transforms of and , respectively: F ( s ) = sinc ( s ) e  s  2 = sinc ( s ) ( s ) 2. Note f 1 = 1 2 df dx The Fourier transforms of f and f 1 are: F ( s ) = e 2 s 2 respectively F 1 ( s ) = isF ( s ) = is e 2 s 2 . Let g = f * f and g 1 = f * f 1 . Then their Fourier transforms are: G ( s ) = F ( s ) 2 = e 2 2 s 2 G 1 ( s ) = F ( s ) F 1 ( s ) = is 2 e 2 2 s 2 Next we need to compute the inverse Fourier transforms. Again apply the dilation and the derivative rules, and get: g ( x ) = r 2 e x 2 / 2 g 1 ( x ) = 1 2 dg dx = x 2 r 2 e x 2 / 2 3....
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This note was uploaded on 05/05/2010 for the course MATH 464 taught by Professor Staff during the Spring '08 term at Maryland.
 Spring '08
 staff
 Math

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