SolutionsIX

# SolutionsIX - ECE 3250 HOMEWORK IX SOLUTIONS Fall 2009 1(a...

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Unformatted text preview: ECE 3250 HOMEWORK IX SOLUTIONS Fall 2009 1. (a) By definition, x 1 * x 2 ( n ) = ∞ X k =-∞ x 1 ( k ) x 2 ( n- k ) = 1023 X k =0 x 1 ( k ) x 2 ( n- k ) = if n < P n k =0 x 1 ( k ) x 2 ( n- k ) if n ≥ . The sum in the last line has at most the single nonzero term x 1 (0) x 2 (0) if n = 0, at most two nonzero terms if n = 1, and so on up until n = 1023, where it has at most 1023 nonzero terms. It also has a max of 1023 nonzero terms for n = 1024, then 1022 nonzero terms for n = 1025, 1021 nonzero terms for n = 1026, and so on up until n = 2047, where its only possibly nonzero term is x 1 (1023) x 2 (1024). Each nonzero term in the sum entails one multiplication, so the computation takes a nominal 1 + 2 + 3 + ··· + 1023 + 1023 + 1022 + 1021 + ··· + 1 multiplications, which is the same as 2 1023 X k =0 k ! = 2 × 1023 × 1024 2 = 512 . 5 × 2 11 . This is a worst-case result because some of the possibly nonzero terms in the sum might actually be zero. (b) To compute the convolution using N-point DFTs, you need to regard x 1 and x 2 as N-point signals, where N is the “length” of x 1 * x 2 , and we know that length is N = 1024 + 1025- 1 = 2048, which, conveniently, is a power of 2, namely 2 11 . Using the FFT to compute an N-point DFT or inverse-DFT requires N + N log 2 N = 12 × 2 11 multiplications. To compute the x 1 * x 2 , you need to – compute the N-point DFTs of x 1 and x 2 , each regarded as an N-point signal, which takes a nominal 24 × 2 11 multiplications; – multiply these DFTs together, which takes N = 2 11 multiplications; – take the inverse N-point DFT of the result of the previous step, which takes a nominal 12 × 2 11 multiplications....
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SolutionsIX - ECE 3250 HOMEWORK IX SOLUTIONS Fall 2009 1(a...

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