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hw6slns - Homework#6Solutions(Total:82points(3 3 3...

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Homework #6 Solutions (Total: 82 points)
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(3+3+3+3 = 12 points)
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(5+5 = 10 points)
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(10 points)
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Alternate diagram for part c) (parallel implementation): These zeroes satisfy the conditions for linear phase from class.
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-7 -6 -1 -4 -3 (2+3+5 = 10 points) 5. This problem can also be solved using techniques from discussion (although the above solution is much easier). The original filter H II (z) is Type II. Therefore, H II (z) has an odd number of zeros at 1, an even number of zeros at 1, and L is odd (see the Venn diagram from discussion). If this filter is multiplied by (1 z 1 ) , then that means we are adding a zero at z=1 (so the number of zeros at 1 is now odd), and the order of the new filter H(z) is L’=L+1, which is even. Looking at the Venn diagram, a filter with an odd
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number of zeros at 1, and odd number of zeros at 1, and L’ being even corresponds to a Type III filter. We know that a Type III filter is linear phase, so this concludes our proof. (10 points) 6.
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