HW_8_sol

# HW_8_sol - HW 8 ECE 2704 Due 7.4.6 Find the inverse Fourier...

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DKL:11/10/06 HW 8 1/2 HW 8 ECE 2704 Due 11-10-06 7.4.6 Find the inverse Fourier transform of each of the Fourier transforms shown in Figures P7.4.6ac. |H( ! )| ! 1 " #" \$ H( ! ) ! -t d " #" Figure P7.4.6c (a) (5) Find a function describing H ( ! ) = H ( ! ) e " H ( ! ) . (b) (10) Set up the integral for finding the inverse Fourier transform using the definition. (c) (10) Evaluate the integral to find the inverse Fourier transform. Solution The inverse Fourier transform is found using Definition 7.4.1: h ( t ) = 1 2 ! H ( " ) e j " t d " #\$ \$ % In Section 14.6 we will use these functions to define ideal filters. (a) H LP ( ! ) = 1 e " jt d ! # ! 2 \$ % & ( ) * (b) h LP ( t ) = 1 2 ! H ( " ) e j " t d " #\$ \$ % = 1 2 ! 1 e # jt d " ( ) e j " t d " #& & % (c) h LP ( t ) = 1 2 ! 1 e " jt d # ( ) e j # t d # "\$ \$ % = 1 2 ! 1 e j ( t " t d ) # d # "\$ \$ % = 1 2 ! 1 j ( t " t d ) e j ( t " t d ) # & ( ) * + "\$ \$ = 1 ! 1 ( t " t d ) e j ( t " t d ) \$ " e " j ( t " t d ) \$ ( ) 2 j = \$ ! sin ( t " t d ) \$ ( ) ( t " t d ) \$ = \$ ! Sa ( t " t d ) \$ ( )

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DKL:11/10/06 HW 8 2/2 7.5.1 (10) Find the Fourier transform of each of the following signals using the Table and properties of the Fourier transform. x ( t ) = ! ( t + t 0 ) + ! ( t " t 0 ) Solution This problem using the properties of the Fourier transform in Table 7.6.2 and the Fourier
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Unformatted text preview: This problem using the properties of the Fourier transform in Table 7.6.2 and the Fourier transform pairs in Table 7.6.3. (x) Using the Time Shift Property X ( ! ) = 1 e j ! t + 1 e " j ! t = 2 e j ! t + e " j ! t 2 # \$ % & ’ ( = 2cos( ! t ) 7.5.3 (10) Find the inverse Fourier transform of the following functions. X ( ! ) = Sa(4 !" 5) + Sa(4 ! + 5) Solution This problem using the properties of the Fourier transform in Table 7.6.2 and the Fourier transform pairs in Table 7.6.3. (i) X ( ! ) = 2 1 2 Sa 4 ! " 5 4 # \$ % & ’ ( ) * + ,-. + Sa 4 ! + 5 4 # \$ % & ’ ( ) * + ,-. / 1 2 3 4 Sa 4 ! [ ] = 1 8 " t 8 # \$ % & ’ ( x ( t ) = 1 4 ! t 8 " # \$ % & ’ cos 5 4 t " # \$ % & ’...
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• Spring '08
• DJStilwell
• inverse Fourier transform

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