Let be $v(k)$ the DFT of the $u(n)$ sequence where $0\leq n,k \leq N-1$. Why If $u$ is a real sequence then a DFT is the vector $N\times 1$
$$v(0),\left\{{Re(v(k)),k=0,...,N/2-1}\right\},\left\{{Im(v(k)),k=0,...,N/2-1}\right\}, v(N/2)?$$
pdta: I know that $v^*(N/2+k)=v(K/2-k)$. i.e the symetric plot for DFT of $u(n)$ real.
Reference: pag. 144 Anil K Jain Fundamentals of Digital Image Processing.
