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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

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.

Ask a homework question - tutors are online