Consider two sequences x[n] and h[n], and let y[n] denote their linear convolution, y[n] = x[n]...

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Consider two sequences x[n] and h[n], and let y[n] denote their linear convolution, y[n] = x[n] *h[n]. Assume that x[n] is zero outside the interval 21 ≤ n ≤ 31, and h[n] is zero outside the interval 18 ≤ n ≤ 31. (a) The signal y[n] will be zero outside of an interval N₁ ≤ n ≤ N₂. Determine the values of N₁ and N₂. (b) We compute the 32-point DFTs of and x₁ [n] = h₁ [n]: = 0, (x[n], 0, (h[n], n = 0, 1,..., 20 n = 21, 22, ..., 31 n = 0, 1, ..., 17 18, 19, 31 n= ") (i.e., the zero samples at the beginning of each sequence are included). Then, we form the product Y₁[k] = X₁[k]H₁[k]. If we define y₁ [n] to be the 32-point inverse DFT of Y₁[k], how is y₁ [n] related to the linear convolution y[n]? Write an equation expressing y₁ [n] in terms of y[n] for 0 ≤ n ≤ 31. (c) Suppose that you are free to choose the DFT length N in part (b) so that the sequences are also zero-padded at their ends. What is the minimum value of N so that y₁ [n] = y[n] for 0 ≤ n ≤ N - 1? Consider two sequences x[n] and h[n], and let y[n] denote their linear convolution, y[n] = x[n] *h[n]. Assume that x[n] is zero outside the interval 21 ≤ n ≤ 31, and h[n] is zero outside the interval 18 ≤ n ≤ 31. (a) The signal y[n] will be zero outside of an interval N₁ ≤ n ≤ N₂. Determine the values of N₁ and N₂. (b) We compute the 32-point DFTs of and x₁ [n] = h₁ [n]: = 0, (x[n], 0, (h[n], n = 0, 1,..., 20 n = 21, 22, ..., 31 n = 0, 1, ..., 17 18, 19, 31 n= ") (i.e., the zero samples at the beginning of each sequence are included). Then, we form the product Y₁[k] = X₁[k]H₁[k]. If we define y₁ [n] to be the 32-point inverse DFT of Y₁[k], how is y₁ [n] related to the linear convolution y[n]? Write an equation expressing y₁ [n] in terms of y[n] for 0 ≤ n ≤ 31. (c) Suppose that you are free to choose the DFT length N in part (b) so that the sequences are also zero-padded at their ends. What is the minimum value of N so that y₁ [n] = y[n] for 0 ≤ n ≤ N - 1?

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SOLUTION The linear Convolution output is yu x Then according to the Given intervals hn 21 3 ... View the full answer