HWSol_Ch4 - Homework Solutions Assignment 5 4.21(a The...

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Homework Solutions Assignment 5 4.21 (a) The given signal is e - at cos( w 0 t ) u ( t )=0 . 5 e - at e jw 0 t u ( t )+0 . 5 e - at e - jw 0 t u ( t ) Use the Table 4.2, X ( jw )= 1 2( a - jw 0 + jw ) - 1 2( a - jw 0 - jw ) (c) Directly use the definition of Fourier Transform in (4.9) X ( jw )= 2sin w w + sin w π - w - sin w π + w (g) Directly use the definition of Fourier Transform in (4.9) X ( jw )= 2 j w bracketleftbigg cos2 w - sin w w bracketrightbigg 4.22 (b) x ( t )= 1 2 e - jπ/ 3 δ ( t - 4)+ 1 2 e jπ/ 3 δ ( t +4) (d) x ( t )= 2 j π sin t + 3 π cos(2 πt ) 4.30 (a) For w ( t ) w ( t )=cos t F T W ( jw )= π [ δ ( w - 1)+ δ ( w +1) and
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Unformatted text preview: FT ↔ G ( jw ) = 1 2 π [ X ( jw ) ⋆ W ( jw )] Then, G ( jw ) = 1 2 X ( j ( w-1)) + 1 2 X ( j ( w + 1)) We know that, G ( jw ) is rectangular window with duration 4, and it is a sum of two shifted version of X ( jw ) . Then, X ( jw ) is also rectangular window function with duration 2. Then, x ( t ) = 2 sin t πt 1...
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  • Fall '14
  • Cos, Short-time Fourier transform

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