20082ee102_1_HW_8_solution_2

20082ee102_1_HW_8_solution_2 - HW8 solution, SPRING 2008,...

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HW8 solution, SPRING 2008, EE 102 1. (Prob.2.) F { cosw 0 tf ( t ) } = F { e iw 0 t + e - iw 0 t 2 f ( t ) } = (1 / 2) F { e iw 0 t f ( t ) } +(1 / 2) F { e - iw 0 t f ( t ) } = 1 2 F ( iw - iw 0 ) + 1 2 F ( iw + iw 0 ) (Prob.3)The standard Fourier relation is F ( iw ) = Z -∞ e - iwt f ( t ) dt Take derivative of the above equation d n F ( iw ) dw n = Z -∞ d n e - iwt dw n f ( t ) dt = Z -∞ e - iwt ( - it ) n f ( t ) dt So d n F ( iw ) dw n and ( - it ) n f ( t ) are Fourier transform pair. (Prob.4) Using the properties in Prob.2 above, and F { 1 } = 2 πδ ( w ), we obtain F { cosw 0 t } = πδ ( w - w 0 ) + πδ ( w + w 0 ) 2. Note F { e iw 0 t } = 2 πδ ( w - w 0 ) and f ( t ) = n = -∞ F n e inw 0 t F { f ( t ) } = Z -∞ e - iwt f ( t ) dt = Z -∞ e - iwt X n = -∞ F n e inw 0 t dt = X n = -∞ F n Z -∞ e - iwt e inw 0 t dt = X n = -∞ F n F { e inw 0 t } = X n = -∞ 2 πF n δ ( w - nw 0 ) 1
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3. (i) Note:
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20082ee102_1_HW_8_solution_2 - HW8 solution, SPRING 2008,...

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