soln9 - R. Balan Homework #9 Solutions MATH 464 I . a . f (...

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Unformatted text preview: R. Balan Homework #9 Solutions MATH 464 I . a . f ( x ) = cos ( x )( x/ ) . Note: f ( x ) = 1 2 ( e ix + e- ix ) x Its Fourier transform is F ( s ) = 1 2 Z - e- 2 isx ( e ix + e- ix )( x ) dx = 1 2 Z - e- 2 i ( s- 1 / 2 ) x ( x ) dx + 1 2 Z - e- 2 i ( s +1 / 2 ) x ( x ) dx Change the integration variable y = x , x = y , dx = dy F ( s ) = 2 Z - e- 2 i ( s- 1 / 2) y ( y ) dy + 2 Z - e- 2 i ( s +1 / 2) y ( y ) dy = 2 sinc ( s- 1 2 ) + 2 sinc ( s + s 2 ) b . f ( x ) = (2 x + 1)(2 x- 1) . Note the support of f 1 ( x ) = (2 x + 1) is the interval [- 1 , 0] , whereas the support of f 2 ( x ) = (2 x- 1) is [0 , 1] . Hence the function f ( x ) is identically zero: f ( x ) = 0 . Thus F ( s ) = 0 . c . f ( x ) = e- 2 | x | sinc ( x ) . Note f ( x ) = f 1 ( x ) f 2 ( x ) where f 1 ( x ) = e- 2 | x | and f 2 ( x ) = sinc ( x ) . Hence F ( s ) = F 1 * F 2 ( s ) = Z - F 1 ( s- t )...
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soln9 - R. Balan Homework #9 Solutions MATH 464 I . a . f (...

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