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Prove if x(n) is a real-valued sequence, then its DFT X(K) =X* (N-K).

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Prove if x(n) is a real-valued sequence, then its DFT X(K) =X* (N-K).

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written 2.0 years 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} \\\\ \Rightarrow x^{-*}(K) &\left.=\sum_{n=0}^{N-1} x(n) W_N^{-k n} \cdot W_N^{N n} \text { (since } W_N^{N-1}\right) \\\\ &=\sum_{n=0}^{N-1} x(n) W_N(N-k)_n \\\\ &=x(N-K) . \end{aligned}\\ $

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