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SolutionsIII_1

SolutionsIII_1 - ECE 3250 HOMEWORK III SOLUTIONS Fall 2009...

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Unformatted text preview: ECE 3250 HOMEWORK III SOLUTIONS Fall 2009 1. Note that h has finite duration. (a) By Criterion 1, D h = F Z because h has finite duration. (b) S ( x ) = Conv h ( x ), so for every n ∈ Z , S ( x )( n ) = ∞ X k =-∞ h ( k ) x ( n- k ) = 6 X k =0 3(5- k ) = 3(1- 5- 7 ) 1- 1 / 5 = 15 4 (5- 7- 1); . Note that S ( x ) is a constant signal. (c) S ( x ) = Conv h ( x ), so for every n ∈ Z , S ( x )( n ) = ∞ X k =-∞ h ( k ) x ( n- k ) = 6 X k =0 5- k u ( n- k ) = 8 < : if n < P n k =0 5- k if 0 ≤ n < 7 P 6 k =0 5- k if n ≥ 7 = 8 > < > : if n < 1- 5- ( n +1) 1- 1 / 5 if 0 ≤ n < 7 1- 5- 7 1- 1 / 5 if n ≥ 7 = 8 < : if n < 5 4 (5- ( n +1)- 1) if 0 ≤ n < 7 5 4 (5- 7- 1) if n ≥ 7 . (d) Since x has duration 13, we can find N 1 so that x ( n ) = 0 when n < N 1 and x ( n ) = 0 when n > N 1 + 12. Now S ( x ) = Conv h ( x ), so for each n ∈ Z S ( x )( n ) = ∞ X k =-∞ h ( k ) x ( n- k ) = 6 X k =0 5- k x ( n- k ) . The sum features terms x ( n ), x ( n- 1), . . . , x ( n- 6). All these are zero if n < N 1 . Similarly, all are zero if n- 6 > N 1 + 12, which is the same as n > N 1 + 18. It follows that S ( x )( n ) = 0 when n < N 1 and S ( x )( n ) = 0 when n > N 1 + 18. The maximum “duration interval” for S ( x ) is therefore N 1 ≤ n ≤ N 1 + 18, so the maximum possible duration for S ( x ) is 19. (It’s easy to find x so that S ( x ) has duration exactly 19. Just suppose x ( n ) = 1 on its “duration interval.”) 2. (a) I just wanted you to think about what the averager does. What you need to do is find a signal x so that the average value of x ( k ) over any k-interval of length 5 is equal to 11. A reasonable way to do this is to find 5 numbers that average to 11 and just let x be the signal that cycles periodically through those numbers....
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SolutionsIII_1 - ECE 3250 HOMEWORK III SOLUTIONS Fall 2009...

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