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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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Solution:

We know that,

x(k)=DFT{x(n)}=N1n=0x(n)WknN;k=0,1N1

Taking conjugates on both sides, we get,

X(k)=N1n=0x(n)wknN

Since x(n) is real, we have x(n)=x(n)

x(K)=N1n=0x(n)WknNx(K)=N1n=0x(n)WknNWNnN (since WN1N)=N1n=0x(n)WN(Nk)n=x(NK).

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