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Homework 1 § 11 . 7 1. Show that (a) R 0 cos xw + w sin xw 1+ w 2 dw = 0 if x < 0 π/ 2 if x = 0 πe - x if x > 0 . (b) R 0 sin w w cos xwdw = π/ 2 if 0 x < 1 π/ 4 if x = 1 0 if x > 1 . Sol. (a) We ﬁnd the Fourier integral of f ( x ) = 0 if x < 0 e - x if x > 0 . A ( w ) = 1 π Z 0 e - v cos wvdv = 1 π 1 1 + w 2 , B ( w ) = 1 π Z 0 e - v sin wvdv = 1 π w 1 + w 2 . Fourier integral of f : f ( x ) = 1 π Z 0 cos wx + w sin wx 1 + w 2 dw. Notice that f ( x ) is discontinuous only at x = 0. Z 0 cos xw + w sin xw 1 + w 2 dw = πf ( x ) = 0 if x < 0 π f (0+0)+ f (0 - 0) 2 = π/ 2 if x = 0 πf ( x ) = πe - x if x > 0 . (b) We ﬁnd the Fourier cosine integral of f ( x ) = 1 if | x | < 1 0 if | x | > 1 . A ( w ) = 2 π Z 1 0 cos wvdv = 2 π sin w w Fourier cosine integral of f : f ( x ) = 2 π Z 0 sin w w cos wxdw. Notice that f ( x ) is discontinuous only at x = 1 , - 1. Z

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## This note was uploaded on 11/08/2011 for the course EE 111 taught by Professor Kim during the Spring '11 term at Korea University.

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11.7-11.9ê³¼ì œ í’€ì´

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