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# soln6 - R Balan Homework#6 Solutions MATH 464 1 Note the...

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R. Balan Homework #6 Solutions MATH 464 1. Note the function f is the convolution between the box function Π and the unit Gaussian function γ , that is: f ( x ) = Z -∞ Π( u ) γ ( x - u ) du , Π( u ) = 1 , | u | < 1 2 0 , | u | > 1 2 , γ ( u ) = e - π | u | 2 Thus the Fourier transform is the product of the Fourier transforms of Π and γ , respectively: F ( s ) = sinc ( s ) e - π | s | 2 = sinc ( s ) γ ( s ) 2. Note f 1 = - 1 2 df 0 dx The Fourier transforms of f 0 and f 1 are: F 0 ( s ) = πe - π 2 s 2 respectively F 1 ( s ) = - πisF 0 ( s ) = - isπ πe - π 2 s 2 . Let g 0 = f 0 * f 0 and g 1 = f 0 * f 1 . Then their Fourier transforms are: G 0 ( s ) = F 0 ( s ) 2 = πe - 2 π 2 s 2 G 1 ( s ) = F 0 ( s ) F 1 ( s ) = - isπ 2 e - 2 π 2 s 2 Next we need to compute the inverse Fourier transforms. Again apply the dilation and the derivative rules, and get: g 0 ( x ) = r π 2 e - x 2 / 2 g 1 ( x ) = - 1 2 dg 0 dx = x 2 r π 2 e - x 2 / 2 3. (Preferred Solution) Let F denote the Fourier transform of f . Define g : R C , g ( u ) = f ( x - u ). The Fourier transform of g is G : R C given by G ( s ) = e - 2 πisx F ( s ) Then the Parseval identity (which requires R -∞ | f ( x ) | 2 dx < ) implies: Z -∞ f ( u ) f ( x - u

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