Midterm_Slns_Fall_08

# Midterm_Slns_Fall_08 - 1 Midterm Solution 1 a N 1 X[k = 1 N...

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Midterm Solution 1) a) X [ k ] = N - 1 X n =0 g [ n ] h [ n ] e - J 2 π N nk = 1 N N - 1 X n =0 N - 1 X m =0 G [ m ] e J 2 π N mn h [ n ] e - J 2 π N nk = 1 N N - 1 X m =0 G [ m ] N - 1 X n =0 h [ n ] e - J 2 π N n ( k - m ) = 1 N N - 1 X m =0 G [ m ] N - 1 X n =0 h [ n ] e - J 2 π N n<k - m> N = 1 N N - 1 X m =0 G [ m ] H [ < k - m > N ] b) g [ n ] ⇐⇒ G [ k ] = G ( e k ) ± ± ω k = 2 π N k g 1 [ n ] ⇐⇒ G 1 [ k ] = G [2 k ] = G ( e k ) ± ± ω k = 2 π N/ 2 k By frequency sampling theorem, g 1 [ n ] = X r = -∞ g [ n + r N 2 ] = g [ n ] + g [ n + N 2 ] ,n = 0 ,.., N 2 - 1 2) a) x [( - n ) 12 ] = { 1 , 0 , 0 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 } x [( n - 4) 10 ] = { 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 } b) h [( - n ) 12 ] = { 1 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 1 , 1 , 1 } x [ n ] = { 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 0 , 0 } y [0] = 11 X n =0 x [ n ] h [( - n ) 12 ] = 2 h [(1 - n ) 12 ] = { 1 , 1 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 0 , 1 , 1 } x [ n ] = { 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 1 , 0 , 0 } y [1] = 11 X n =0 x [ n ] h

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Midterm_Slns_Fall_08 - 1 Midterm Solution 1 a N 1 X[k = 1 N...

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