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I have read about Fourier transformation that real signals are "mirrored" in the real and negative halves of the Fourier transform because of the nature of the Fourier transform. For example, if we have the real signal $x$ whose length is $N$, its $X = FFT(x)$ should be conjugate symmetric. That means the $X^*[m] = X[N-m]$.

That is ok, and I have read it clearly HERE

My question: assuming you have a real signal $x = [-1, 1, -1, 1];$ its $X = FFT(x) = [0, 0, -4 , 0]$ So, in that case, we cannot say that $X^*[m] = X[N-m]$. Why ?

Is what I said above is right ? in which cases is right?

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  • $\begingroup$ For an intuitive understanding of this property check out my article: dsprelated.com/showarticle/768.php The DFT is equivalent to a center of mass calculation of the signal wrapped around the unit circle varying number of times. The $k$ case "wraps forward" and the $N-k$ case is the mirror image which "wraps backward". Since the "mirror" is the real axis, mirror images mean complex conjugates. $\endgroup$ – Cedron Dawg Jun 17 at 12:11
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We can! In your example, $N=4$, and the DFT is real-valued, so you get $X[k]=X[4-k]$, and that's true:

$$\begin{align}k=0:\;X[0]&=X[4]=0\\k=1:\;X[1]&=X[3]=0\\k=2:\;X[2]&=X[2]=-4\end{align}$$

Note that by definition the DFT is $N$-periodic, so $X[N]=X[0]$.

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  • $\begingroup$ Is that correct for any real signal $x$ ? $\endgroup$ – Gze Jun 17 at 11:23
  • $\begingroup$ @Gze: You mean $X^*[k]=X[N-k]$? Yes, that's a basic property of the DFT of real-valued signals. $\endgroup$ – Matt L. Jun 17 at 11:39
  • $\begingroup$ The real take-away from this is the factoid that in the length-N DFT (N even) of a real-valued sequence $x[\cdot]$, $X\left[\frac N2\right]$ is a real number, not a complex number (with nonzero imaginary part).. $\endgroup$ – Dilip Sarwate Jun 17 at 21:25
  • $\begingroup$ Yes, I am thinking now in the OFDM system, when using BPSK modulation, it means that half of bandwidth is wasted because the transmitted signal is located in the first half which is similar to that one in second half of bandwidth !! right? $\endgroup$ – Gze Jun 18 at 9:35

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