# soln7 - R Balan Homework#7 Solutions MATH 464 1 Method 1 2...

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R. Balan Homework #7 Solutions MATH 464 1 . Method 1 Let g ( x ) = e - 2 πiσx f ( x ). Note G ( s ) = F ( s + σ ) and thus G ( s ) = 0 for | s | > σ . Hence g B 2 σ (that is, g is σ -band limited). Shannon’s sampling formula yields g ( x ) = X n g ( nT ) sinc ( x - nT T But g ( nT ) = e - πin f ( nT ) = ( - 1) n f ( nT ) (recall = 1 / 2). Thus f ( x ) = e 2 πiσx g ( x ) = X n ( - 1) n f ( nT ) e 2 πiσx sinc ( x - nT T ) Method 2 Use: f ( x ) = Z -∞ F ( s ) e 2 πisx ds = Z 2 σ 0 e 2 πisx F ( s ) ds But F ( s ) = X n c n e 2 πins/ (2 σ ) = X n c n e 2 πinTs Now c n = 1 2 σ Z 2 σ 0 F ( s ) e - 2 πins/ (2 σ ) ds = Tf ( - nT ) Thus f ( x ) = X n Tf ( - nT ) Z 2 σ 0 e 2 πisx +2 πisnT ds = X n Tf ( nT ) Z 2 σ 0 e 2 πis ( x - nT ) ds = = X n Tf ( nT ) 1 2 πi ( x - nT ) ± e 2 πi 2 σ ( x - nT ) - 1 ² = X n f ( nT ) e πi x - nT T 2 πi x - nT T ± e πi x - nT T - e - πi x - nT T ² = = X n f ( nT ) e - πin e 2 πiσx sinc ³ x - nT T ´ = X n f ( nT )( - 1) n e 2 πiσx sinc ³ x - nT T ´ 2 . First note

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soln7 - R Balan Homework#7 Solutions MATH 464 1 Method 1 2...

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