0
493views
Prove if x(n) is a real-valued sequence, then its DFT X(K) =X* (N-K).
1 Answer
written 18 months ago by |
Solution:
We know that,
$ \begin{aligned}\\ x(k) &=D F T\{x(n)\} \\\\ &=\sum_{n=0}^{N-1} x(n) W_N^{k n} ; k=0,1 \cdot N-1\\ \end{aligned}\\ $
Taking conjugates on both sides, we get,
$ X^*(k)=\sum_{n=0}^{N-1} x^*(n) w_N^{-k n}\\ $
Since $x(n)$ is real, we have $x^*(n)=x(n)$
$ \begin{aligned}\\ \therefore x^*(K) &=\sum_{n=0}^{N-1} x(n) W_N^{-k n} \\\\ …