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After performing the fft transform there are 2 options to plot the spectrum

the complex (Real and imaginair) spectrum and the magnitude with the phase.
(I neglect the magnitude only for this question)
My question is why is the spectrum of magnitude with phase always single sided. and why is the complex spectrum always double sided.

second questsion is make it sense to create a double sided magnitude phase plot or an single sided complex plot?

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For real-valued sequences, the DFT is always conjugate symmetrical, i.e.

$$X[k]=X^*[N-k],\qquad k=0,1,\ldots N-1\tag{1}$$

where $N$ is the DFT length. From $(1)$ with $X[k]=X_R[k]+jX_I[k]$ it follows that the real part of $X[k]$ is symmetrical, and the imaginary part is anti-symmetrical:

$$X_R[k]=X_R[N-k]\\ X_I[k]=-X_I[N-k]$$

Furthermore, with $X[k]=M[k]e^{j\phi[k]}$ we get for magnitude and phase

$$M[k]=M[N-k]\\ \phi[k]=-\phi[N-k]$$

So for real-valued sequences, you get the above symmetries in the real and imaginary parts, and also in the magnitude and phase. So no matter if you represent the DFT in terms of real and imaginary parts, or in terms of magnitude and phase, a one-sided spectrum (i.e. for indices $0,\ldots, \lfloor N/2\rfloor$) is always sufficient, as long as the sequence is real-valued.

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  • $\begingroup$ Does this mean that when the signal is not real-valued the second part of the the spectrum(N/2+1,...,N) is needed? Even if this not exist in nature? And may second question was the other way around.. Does the half real imaginary spectrum exist? $\endgroup$ – Jan-Bert Jan 18 '16 at 21:06
  • $\begingroup$ @Jan-Bert: For a complex signal, there is generally no symmetry in the DFT, so you need all DFT values to represent the signal. I can't make sense of your second question (what is a "real imaginary spectrum"?). $\endgroup$ – Matt L. Jan 18 '16 at 21:21
  • $\begingroup$ I think i understand it at least for now $\endgroup$ – Jan-Bert Jan 23 '16 at 18:51

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