When we apply the DFT operation on a real signal $x[n]$ to get $X[k]$, then take the square magnitude of the $X[k]$, $\lvert X[k]\rvert^2$, the power spectrum is symmetrical. You can take the positive frequencies or negative frequencies as the frequency information in $X[k]$.

However this is not true for complex valued signals; the power spectrum is not symmetrical.

  • In this case, how would you determine the frequency components in original signal?
  • Can we just drop the negative frequency part?
  • $\begingroup$ Note that positive and negative frequencies are linear combinations of sine and cosine, thus, positive and negative frequencies are required to get the right phase. For complex signals, you must add the power in both frequencies to get the total power at that frequency. $\endgroup$ Commented Jan 22, 2018 at 3:52

2 Answers 2


For a real signal, content at the negative frequencies generated by using the DFT is redundant. This is due to the well-known property of real signals with respect to the Fourier transform family: their transforms are Hermitian symmetric. That is, for any real signal $x[n]$,

$$ X[k] = \sum_{n=0}^{N-1}x[n] e^{-j2\pi n k / N} = (\sum_{n=0}^{N-1}x[n] e^{-j(-2\pi n k / N)})^* = X[-k] = X[N - k] $$

So, if your input is a real signal, all of the information is in the positive frequency bins; the negative frequencies can be discarded for many applications. However, there is no such proprety for general complex signals. They can have power spectra that are asymmetric about zero frequency, so you may not discard any of the frequency bins of a complex signal's power spectrum without loss of information.

  • 1
    $\begingroup$ I think the relation between $X[k]$ and $X[N-k]$ is: $$ X[N-k] = \sum_{n=0}^{N-1}x[n] e^{-j2\pi n (N-k) / N} = \sum_{n=0}^{N-1}x[n]e^{-j2\pi N n/N} e^{j2\pi n k / N} =\sum_{n=0}^{N-1}x[n] e^{j2\pi n k / N}=(X[k])^* $$ $\endgroup$
    – Mike
    Commented Oct 17, 2011 at 3:07
  • $\begingroup$ You are correct. I left that intermediate step out; it shows how you get from effectively $X[-k]$ to $X[N-k]$. $\endgroup$
    – Jason R
    Commented Oct 17, 2011 at 12:53

In case of real we can drop e.g in practice when you use the spectrum analyzer, for real wave it is simpler to see the half because the other side is mirror. But in case of complex signal no real device answers and you have a theoretical studies so you should keep both sides.

  • $\begingroup$ if I have one complex signal composed of two close frequencies, in the PSD, i found the positive frequency side shows two peaks, while the negative side shows one peak. So, can I come up the conclusion the signal has two frequencies ? $\endgroup$
    – Mike
    Commented Oct 16, 2011 at 18:32
  • $\begingroup$ No you cannot say because the signal is complex. $\endgroup$
    – Hossein
    Commented Oct 16, 2011 at 21:35
  • 1
    $\begingroup$ what is the physical meaning for the negative frequency? $\endgroup$
    – Mike
    Commented Oct 17, 2011 at 3:12
  • 1
    $\begingroup$ @Mike dsp.stackexchange.com/questions/431/… $\endgroup$
    – GummiV
    Commented Oct 17, 2011 at 14:23
  • $\begingroup$ @Mike I believe I may have answered that... [please see] (dsp.stackexchange.com/questions/431/…) $\endgroup$
    – Spacey
    Commented Oct 19, 2011 at 9:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.