# soln9 - R Balan Homework#9 Solutions MATH 464 I a f x = cos...

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Unformatted text preview: R. Balan Homework #9 Solutions MATH 464 I . a . f ( x ) = cos ( x )Π( x/π ) . Note: f ( x ) = 1 2 ( e ix + e- ix ) Π x π Its Fourier transform is F ( s ) = 1 2 Z ∞-∞ e- 2 πisx ( e ix + e- ix )Π( x π ) dx = 1 2 Z ∞-∞ e- 2 πi ( s- 1 / 2 π ) x Π( x π ) dx + 1 2 Z ∞-∞ e- 2 πi ( s +1 / 2 π ) x Π( x π ) dx Change the integration variable y = x π , x = πy , dx = πdy F ( s ) = π 2 Z ∞-∞ e- 2 πi ( πs- 1 / 2) y Π( y ) dy + π 2 Z ∞-∞ e- 2 πi ( πs +1 / 2) y Π( y ) dy = Π 2 sinc ( πs- 1 2 ) + π 2 sinc ( πs + s 2 ) b . f ( x ) = Λ(2 x + 1)Λ(2 x- 1) . Note the support of f 1 ( x ) = Λ(2 x + 1) is the interval [- 1 , 0] , whereas the support of f 2 ( x ) = Λ(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 π | x | sinc ( x ) . Note f ( x ) = f 1 ( x ) f 2 ( x ) where f 1 ( x ) = e- 2 π | 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 x = cos...

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