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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