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Unformatted text preview: 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 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 dω = jδ ( ω ) − jδ ( ω − 2). 1c. F{ x ( t + 1) } = X ( ω ) e jω = e jω rect (( ω − 1) / 2). 1f. F{ dx dt } = jωX ( ω ) = jω for 0 ≤ ω ≤ 2 and 0 otherwise. 1g. F{ t dx dt } = j d dω [ 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 jω + a , H ( ω ) = F{ sin (2 t ) πt } = rect ( ω/ 4)=0 outside  ω  < 2. Y ( ω ) = H ( ω ) X ( ω ) = rect ( ω/ 4) jω + a . 2b. Input energy= integraltext ∞∞  x ( t )  2 dt = integraltext ∞ e 2 at dt = 1 2 a ....
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This note was uploaded on 04/04/2008 for the course EECS 216 taught by Professor Yagle during the Winter '08 term at University of Michigan.
 Winter '08
 Yagle
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