Question

If x(n) is real and even, then what is the DFT of x(n)?

(a) \(\sum_{n=0}^{N-1} x(n) sin⁡\frac{2πkn}{N}\)

(b) \(\sum_{n=0}^{N-1} x(n) cos⁡\frac{2πkn}{N}\)

(c) -j\(\sum_{n=0}^{N-1} x(n) sin⁡\frac{2πkn}{N}\)

(d) None of the mentioned

I have been asked this question in an interview for internship.

My question comes from Properties of DFT in section Discrete Fourier Transform – Properties and Applications of Digital Signal Processing

Answer 1

Right answer is (b) \(\sum_{n=0}^{N-1} x(n) cos⁡\frac{2πkn}{N}\)

Easiest explanation: Given x(n) is real and even, that is x(n)=x(N-n)

We know that XI(k)=0. Hence the DFT reduces to

X(k)=\(\sum_{n=0}^{N-1} x(n) cos⁡\frac{2πkn}{N}\) ;0 ≤ k ≤ N-1