Question

What is the DFT of the four point sequence x(n)={0,1,2,3}?

(a) {6,-2+2j-2,-2-2j}

(b) {6,-2-2j,2,-2+2j}

(c) {6,-2+2j,-2,-2-2j}

(d) {6,-2-2j,-2,-2+2j}

I got this question at a job interview.

Origin of the question is Frequency Domain Sampling DFT in division Discrete Fourier Transform – Properties and Applications of Digital Signal Processing

Answer 1

Right option is (c) {6,-2+2j,-2,-2-2j}

For explanation I would say: The first step is to determine the matrix W4. By exploiting the periodicity property of W4 and the symmetry property

WN^k+N/2=-WN^k

The matrix W4 may be expressed as

W4=\(\begin{bmatrix}W_4^0& W_4^0& W_4^0& W_4^1\\W_4^0& W_4^0& W_4^2& W_4^3\\W_4^0& W_4^2& W_4^0& W_4^3\\W_4^4& W_4^6& W_4^6& W_4^9\end{bmatrix}=\begin{bmatrix}W_4^0& W_4^0& W_4^0& W_4^1\\W_4^0& W_4^0& W_4^2& W_4^3\\W_4^0& W_4^2& W_4^0& W_4^3\\W_4^0& W_4^2& W_4^2& W_4^1\end{bmatrix}\)  

=\(\begin{bmatrix}1&1&1&1\\1&-j&-1&j\\1&-1&1&-1\\1&j&-1&-j\end{bmatrix}\)

Then X4=W4.x4=\(\begin{bmatrix}6\\ -2+2j\\ -2\\-2-2j\end{bmatrix}\)