The expression for N point DFT is given as

The expression for XR(k) is given as

Let us consider the computation of the N=2v point DFT by the divide and conquer approach. We select M=N/2 and L=2. This selection results in a split of N point data sequence into two N/2 point data sequences f1(n) and f2(n) corresponding to the even numbered and odd numbered samples of x(n), respectively, that is

If we consider the mapping n=Ml+m, then it leads to an arrangement in which the first row consists of the first M elements of x(n), the second row consists of the next M elements of x(n), and so on.

The decimation of the data sequence should be repeated again and again until the resulting sequences are reduced to one point sequences. For N=2v, this decimation can be performed v=log2N times. Thus the total number of complex additions is reduced to Nlog2N.

We know that if N=LM, then WNmqL = WN/Lmq = WMmq.

We observe that the direct computation of F1(k) requires (N/2)2 complex multiplications. The same applies to the computation of F2(k). Furthermore, there are N/2 additional complex multiplications required to compute WNk. Hence it requires N(N+1)/2 complex multiplications to compute X(k).

In decimation-in-time FFT algorithm, the input is taken in bit reversal order and the output is obtained in the order.

From the question, it is given that

The formula for calculating N point DFT is given as