Question

If X(k) is the N/2 point DFT of the sequence x(n), then what is the value of X(k+N/2)?

(a) F1(k)+F2(k)

(b) F1(k)-WN^k F2(k)

(c) F1(k)+WN^k F2(k)

(d) None of the mentioned

This question was addressed to me during an online exam.

This intriguing question comes from Efficient Computation of DFT FFT Algorithms in division DFT Efficient Computation – Fast Fourier Transform Algorithms of Digital Signal Processing

Answer 1

The correct option is (b) F1(k)-WN^k F2(k)

For explanation I would say: We know that, X(k) = F1(k)+WN^k F2(k)

We know that F1(k) and F2(k) are periodic, with period N/2, we have F1(k+N/2) = F1(k) and F2(k+N/2)= F2(k). In addition, the factor WN^k+N/2 = -WN^k.

Thus we get, X(k+N/2)= F1(k)- WN^k F2(k).