Question

If X(k) is the DFT of x(n) which is defined as x(n)=x1(n)+jx2(n), 0≤ n≤ N-1, then what is the DFT of x1(n)?

(a) \(\frac{1}{2} [X*(k)+X*(N-k)]\)

(b) \(\frac{1}{2} [X*(k)-X*(N-k)]\)

(c) \(\frac{1}{2j} [X*(k)-X*(N-k)]\)

(d) \(\frac{1}{2j} [X*(k)+X*(N-k)]\)

I got this question during an internship interview.

My question is taken from Applications of FFT Algorithms topic in section DFT Efficient Computation – Fast Fourier Transform Algorithms of Digital Signal Processing

Answer 1

Right option is (a) \(\frac{1}{2} [X*(k)+X*(N-k)]\)

The explanation: We know that if x(n)=x1(n)+jx2(n) then x1(n)=\(\frac{x(n)+x*(n)}{2}\)

On applying DFT on both sides of the above equation, we get

X1(k)=\(\frac{1}{2} {DFT[x(n)]+DFT[x*(n)]}\)

We know that if X(k) is the DFT of x(n), the DFT[x*(n)]=X*(N-k)

=>X1(k)=\(\frac{1}{2} [X*(k)+X*(N-k)]\).