If we split the 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) and F1(k) and F2(k) are the N/2 point DFTs of f1(k) and f2(k) respectively, then what is the N/2 point DFT X(k) of x(n)?a)F1(k)+F2(k)b)F1(k)- WNkF2(k)c)F1(k)+WNkNkF2(k)d)None of the mentionedCorrect answer is option 'C'. Can you explain this answer?

Question

Answer

From the question, it is given that
f₁(n)=x(2n)
f₂(n)=x(2n+1) ,n=0,1,2…N/2-1

X(k) = F₁(k)+WNk_N^k F₂(k)

Answer

Introduction:
In this question, we are given a data sequence x(n) of length N and we need to split it into two sequences f1(n) and f2(n) corresponding to the even and odd numbered samples of x(n). We also have N/2 point DFTs of f1(n) and f2(n) denoted as F1(k) and F2(k) respectively. We need to find the N/2 point DFT X(k) of x(n).

Explanation:
To solve this problem, let's first understand the concept of splitting a data sequence and its DFT.

When we split a data sequence x(n) into even and odd numbered samples, we get two sequences f1(n) and f2(n). The even numbered samples are given by f1(n) = x(2n) and the odd numbered samples are given by f2(n) = x(2n+1).

The N/2 point DFT of f1(n) is denoted as F1(k), and it is defined as:

F1(k) = Σ[n=0 to N/2-1] f1(n) * exp(-j2πkn/(N/2))

Similarly, the N/2 point DFT of f2(n) is denoted as F2(k), and it is defined as:

F2(k) = Σ[n=0 to N/2-1] f2(n) * exp(-j2πkn/(N/2))

Now, let's find the N/2 point DFT X(k) of x(n) using F1(k) and F2(k).

Solution:
To find the N/2 point DFT X(k) of x(n), we can write it in terms of f1(n) and f2(n) as follows:

X(k) = Σ[n=0 to N-1] x(n) * exp(-j2πkn/N)
= Σ[n=0 to N/2-1] x(2n) * exp(-j2πk(2n)/N) + Σ[n=0 to N/2-1] x(2n+1) * exp(-j2πk(2n+1)/N)
= Σ[n=0 to N/2-1] f1(n) * exp(-j2πkn/(N/2)) + Σ[n=0 to N/2-1] f2(n) * exp(-j2πk(2n+1)/N)

Now, let's compare this expression with the definition of F1(k) and F2(k):

X(k) = F1(k) + Σ[n=0 to N/2-1] f2(n) * exp(-j2πk(2n+1)/N)

We can rewrite the second term on the right-hand side using the identity exp(-j2πk(2n+1)/N) = exp(-j2πkn/N) * exp(-j2πk/N), where exp(-j2πk/N) is denoted as WNk:

X(k) = F1(k) + Σ[n=0 to N/2-1] f2(n) * exp(-j2πkn/N) * exp

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