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# soln6 - EECS 216 SOLUTIONS TO PROBLEM SET#6 Winter 2008 1...

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EECS 216 SOLUTIONS TO PROBLEM SET #6 Winter 2008 1. #4.5 Throughout F{ x ( t ) } = X ( ω ) = rect (( ω 1) / 2) = braceleftBig 1 for 0 ω 2 0 otherwise . Note: X ( ω ) negationslash = X ( ω ) * so x ( t ) is NOT real–DON’T use conjugate symmetry! 1a. F{ x ( t ) } = X ( ω ) = rect (( ω + 1) / 2). 1b. F{ tx ( t ) } = j dX = ( ω ) ( ω 2). 1c. F{ x ( t + 1) } = X ( ω ) e = e rect (( ω 1) / 2). 1f. F{ dx dt } = jωX ( ω ) = for 0 ω 2 and 0 otherwise. 1g. F{ t dx dt } = j d [ jωX ( ω )] = 2 δ ( ω 2) rect (( ω 1) / 2). 1i. F{ x ( t ) e - j 2 t } = X ( ω + 2) = rect (( ω + 1) / 2)=answer to (a). 2a. #4.14 X ( ω ) = F{ e - at u ( t ) } = 1 + a , H ( ω ) = F{ sin (2 t ) πt } = rect ( ω/ 4)=0 outside | ω | < 2. Y ( ω ) = H ( ω ) X ( ω ) = rect ( ω/ 4) + a . 2b. Input energy= integraltext -∞ | x ( t ) | 2 dt = integraltext 0 e - 2 at dt = 1 2 a . Output energy= 1 2 π integraltext 2 - 2 | 1 + a | 2 = 1 2 π integraltext 2 - 2 1 ω 2 + a 2 = 1 tan - 1 2 a . So: 1 tan - 1 2 a = 1 2 1 2 a tan - 1 2 a = 4 a = π 4 a = 2. 3. #4.24 X ( ω ) = braceleftbigg 2 π for | ω | < ω 1 π for ω 1 < | ω | < ω 2 and 0 otherwise. 3a. Y ( ω ) = 2 π for | ω | < ω f y ( t ) = 2 π sin( ω f t ) πt = 2 sin( ω f t ) t . 3b. Y ( ω ) = braceleftbigg 2 π for | ω | < ω 1 π for ω 1 < | ω | < ω f y ( t ) = sin( ω 1 t ) t + sin( ω f t ) t 3c. Since H ( ω ) passes all of x ( t ), y ( t ) = x ( t ) = sin( ω 1 t ) t + sin( ω 2 t ) t 4. #4.19 Mult. by p ( t ) modulates x ( t ): shifts spectrum up and down by ω 0 and halves. Filtering by H 1 ( ω ) doesn’t affect this if ω B > ω m ; clips this if ω B < ω m .
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