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I was learning of Fourier Transform from this book 'Scientist and Engineer Guide to DSP'. Suppose we have a digital periodic signal and there are 128 of samples of it in the time domain i.e X[0] to X[127], then on converting it to frequency domain (only real DFT is considered i.e no complex variables) why do we need only 65 points in the frequency domain, is this related to the Nyquist rate since the highest frequency in the sample can have a maximum value of 64. Are the last 64 points not needed for calculation

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  • $\begingroup$ Are your $X$'s in the time domain ? Oh, $0, \ldots, 128$ that's 129 samples. $\endgroup$ – Gilles Feb 18 '16 at 16:01
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    $\begingroup$ Real signals have a conjugate symmetric transform. $\endgroup$ – AnonSubmitter85 Feb 18 '16 at 16:07
  • $\begingroup$ @Gilles Fixed it now $\endgroup$ – Cloverr Feb 18 '16 at 16:24
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This statement is only true if your time-domain signal $x[n]$ is real. A real signal has a Fourier transform that is conjugate-symmetric. Stated differently the "negative frequencies" are just the conjugate of the corresponding "positive frequencies." Therefore, the negative frequencies add no new information, so they can be discarded from the DFT of a real signal.

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Different way but related way to answer: A real valued signal requires 128 real numbers to represent. The DFT also has 128 values, 126 of which are complex and two of which are real (DC and Nyquist). The 126 complex values are all conjugate symmetric so only half of them are independent and need to be stored. So we end up with 2 real plus 63 complex values which requires a total of 2*63+2 = 128 real numbers to be represented. Hence the information content (room to store) is exactly the same in both domains.

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