SolutionsII

SolutionsII - ECE 3250 HOMEWORK II SOLUTIONS Fall 2009 Note...

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ECE 3250 HOMEWORK II SOLUTIONS Fall 2009 Note: I’ve expressed the solutions to the problems involving calculating convolutions in particular ways that match up well with some stuff we’ll be doing later in the course. Your answers might not look exactly like mine, but perhaps if you check you’ll find that your answers are the same as mine but in a slightly different form. 1. I’ll recount the sequence of steps and then justify each one. x 1 * x 2 ( n ) = X k = -∞ x 1 ( k ) x 2 ( n - k ) = X k = -∞ u ( k )3 - ( n - k ) u ( n - k ) = X k =0 3 - ( n - k ) u ( n - k ) =  P n k =0 3 - ( n - k ) if n 0 0 if n < 0 . The first line holds by definition of convolution. The second line results from plugging in for x 1 and x 2 . The third line holds because u ( k ) = 0 for k < 0 and u ( k ) = 1 for k 0. The last line holds because u ( n - k ) = 1 when k n and u ( n - k ) = 0 when k > n . In particular, when n < 0, u ( n - k ) = 0 for all k in the range of summation 0 k < . You can simplify the sum on the last line using the geometric-series thing as follows: n X k =0 3 - ( n - k ) = 3 - n n X k =0 3 k = 3 - n (1 - 3 n +1 ) 1 - 3 = 3 2 - 1 2 3 - n . Accordingly, x 1 * x 2 ( n ) = 3 2 - 1 2 3 - n if n 0 0 if n < 0 , which is the same as x 1 * x 2 ( n ) = 3 2 - 1 2 3 - n « u ( n ) for all n Z . I’ve written the answer in a specific form for a reason that won’t become obvious until we discuss z -transforms. 2. I’ll proceed as in the previous problem. x 1 * x 2 ( n ) = X k = -∞ x 1 ( k ) x 2 ( n - k ) = X k = -∞ 3 k u ( k )7 n - k u ( n - k ) = X k =0 3 k 7 n - k u ( n - k ) =  P n k =0 7 n (3 / 7) k if n 0 0 if n < 0 . The first line holds by definition of convolution. The second line results from plugging in for x 1 and x 2 . The third line holds because u ( k ) = 0 for k < 0 and u ( k ) = 1 for k 0. 1
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2 The last line holds because u ( n - k ) = 1 when k n and u ( n - k ) = 0 when k > n . In particular, when n < 0, u ( n - k ) = 0 for all k in the range of summation 0 k < . You can simplify the sum on the last line using the geometric-series thing as follows: n X k =0 7 n (3 / 7) k = 7 n n X k =0 (3 / 7) k = 7 n (1 - (3 / 7) n +1 ) 1 - (3 / 7) = 7 4 7 n - 3 4 3 n . Accordingly, x 1 * x 2 ( n ) = 7 4 7 n - 3 4 3 n if n 0 0 if n < 0 , which is the same as x 1 * x 2 ( n ) = 7 4 7 n - 3 4 3 n « u ( n ) for all n Z . I’ve written the answer in a specific form for a reason that won’t become obvious until we discuss z -transforms. 3. I’ll proceed as in the previous problems. x
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SolutionsII - ECE 3250 HOMEWORK II SOLUTIONS Fall 2009 Note...

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