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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 2.4 years ago by |
Solution:
We know that,
x(k)=DFT{x(n)}=N−1∑n=0x(n)WknN;k=0,1⋅N−1
Taking conjugates on both sides, we get,
X∗(k)=∑N−1n=0x∗(n)w−knN
Since x(n) is real, we have x∗(n)=x(n)
∴x∗(K)=N−1∑n=0x(n)W−knN⇒x−∗(K)=N−1∑n=0x(n)W−knN⋅WNnN (since WN−1N)=N−1∑n=0x(n)WN(N−k)n=x(N−K).