Consider the following procedure?

(a) Form the sequence v[n] = x₂[2n] where x₂[n] is given by Eq. (8.166). This yields?

v[n] = x[2n]? ? ? ? ? ? ? ? ? n = 0, 1 ?. N/2 ? 1

v[N ? 1 ? n] = x[2n + 1],? ? ? ? ? ? ? ? ? ? ? n =0, 1?. N/2 ? 1.

(b) Compute V[k], the N-point DFT of v[n].?

Demonstrate that the following is true:?

Note that this algorithm uses N-point rather than 2N-point DFTs as required in Eq. (8.167). In addition, since v[n] is a real sequence, we can exploit even and odd symmetries to do the computation of V[k] in one N/4-point complex DFT

n = 0, 1.....2N 1. x2[n] = x[((n))2N] + x[((-n 1))2N] = x2[n]. (8.166) %3D k = 0. 1.. . ..2N X2[k] = X[k] + X*[k]e2nk/(2N). (8.167) 1. Part B k = 0, 1..., N-1. X2(k] = 2Rele/2ak/(4N)V[k]t. N-1 ak(4n +1) = 2v[n]cos k = 0, 1. .... N - 1. N-1 ak(2n + 1) = 2x{n]cos k = 0, 1...., N-1. %3! 2N na0