soln5 - EECS 451 1a. ck = ck = 1 6 3 6 SOLUTIONS TO PROBLEM...

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EECS 451 SOLUTIONS TO PROBLEM SET #5 1a. c k = 1 6 5 n =0 x ( n ) e j 2 πnk/ 6 = 1 6 (3+2 e j 2 πk 6 +1 e j 4 πk 6 +0 e j 6 πk 6 +1 e j 8 πk 6 +2 e j 10 πk 6 ) c k = 3 6 + 4 6 cos( πk/ 3) + 2 6 cos(2 πk/ 3). c o = 9 6 ; c 1 = c 5 = 4 6 ; c 2 = c 4 = 0; c 3 = 1 6 . 1b. Time domain: 1 6 5 n =0 | x ( n ) | 2 = 1 6 (3 2 + 2 2 + 1 2 + 0 2 + 1 2 + 2 2 ) = 19 6 . Freq domain: 5 k =0 | c k | 2 = 1 6 2 (9 2 + 4 2 + 0 2 + 1 2 + 0 2 + 4 2 ) = 19 6 2a. c k = cos 4 x ( n ) = 7 k =0 cos 4 cos 2 πkn 8 = 4 δ ( n - 1) + 4 δ ( n + 1). c k = sin 3 4 x ( n ) = j 7 k =0 sin 3 4 sin 2 πkn 8 = j 4 δ ( n - 3) - j 4 δ ( n + 3). Hence x ( n )=sum of the above two signals, repeated with period=8. 2c. x ( n ) = 4 k = 3 c k e j 2 πnk 8 = 2+1 e j 2 πn 8 + 1 2 e j 4 πn 8 + 1 4 e j 6 πn 8 + 1 4 e j 10 πn 8 + 1 2 e j 12 πn 8 +1 e j 14 πn 8 x ( n ) = 2 + 2 cos πn 4 + 1 cos πn 2 + 1 2 cos 3 πn 4 . Real, as expected. 3a.
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This note was uploaded on 12/10/2009 for the course EECS 451 taught by Professor Andrewyagle during the Spring '08 term at University of Michigan.

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soln5 - EECS 451 1a. ck = ck = 1 6 3 6 SOLUTIONS TO PROBLEM...

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