When i take an N point DFT of a signal, it comes out to be conjugate symmetric about the point N/2 . Could someone please tell how to understand this intuitively or mathematically ?
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$\begingroup$ lacks research. Wikipedia. DFT. Properties. The Real-input DFT. Anyway, if you need to prove that, just insert a real-valued signal into the DFT formula and show that symmetrical terms are conjugate to each other. $\endgroup$– Marcus MüllerCommented Sep 4, 2017 at 18:57
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2 Answers
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First of all it is only conjugate symmetric if the input signal is Real Signal.
The reason for that is the DFT is built by Complex Exponential which can be decomposed into Sine and Cosine.
Since the Sine governs the Complex part and is $ N $ periodic one could see it creates conjugate symmetry of the DFT.
You just need to follow the definition of the DFT.
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Given strictly real input, the DFT result components should sum to something that is strictly real. If you have an FFT coefficient result with a non-zero imaginary component (representing a Sine), then the only way to get a real sum is to also have another coefficient of the same (aliased) frequency with the exact opposite (negated) non-zero imaginary component.
Thus the mirror symmetry must be conjugated.
