0
$\begingroup$

When we transform a complex signal into frequency signal by using a complex DFT,

The range from N/2 to N point on the X axis of the spectrum mean a negative frequency...

But i cant understand why it is..

In addition, we should not use the range over the N/2 point because of the aliasing issue. But i heard that the negative frequency region is used in case of complex dft spectrum of a complex signal.

Very confused..

Could you clearly explain it for me?

Thank you.

$\endgroup$
1
$\begingroup$

it's because the DFT is periodic with period $N$.

$$ X[k+N] = X[k] \qquad \forall k \in \mathbb{Z} $$

this periodic extension is also true about the discrete-time signal $x[n]$

$$ x[n+N] = x[n] \qquad \forall n \in \mathbb{Z} $$

doesn't matter if either time-domain or frequency-domain representations are complex or not.

that means for $-\frac{N}{2}\le k < 0 $ (the negative frequencies), then $X[k] = X[k+N]$. the explicit negative frequency component $X[k]$ has a copy $X[k+N]$, but that is in latter half of the DFT result.

If $-\frac{N}{2}\le k < 0 $, then $\frac{N}{2}\le k+N < N $.

$\endgroup$
0
$\begingroup$

They are only considered negative by convention. The issue with aliases is that you can't tell which is the "true one" because they all look the same. This is equivalent to asking what is the proper range for the inverse cosine function. By convention we say $0$ to $\pi$, but the truth is multiple values are possible.

It is also equivalent to modular math. Consider 100 stepping stones arranged in a circle and you are standing on one. Now if I tell you to take three steps counterclockwise you end up on a different stone. On the hand, if you take 97 steps in a clockwise direction you end up on the same stone. So, if I didn't see you move, did you take 3 steps counterclockwise or 97 steps clockwise? I can't tell, so by convention I'll say you took 3 counterclockwise steps.

$\endgroup$
0
$\begingroup$

If a signal is strictly real and bandlimited below Fs/2, then the DFT results above N/2 are (merely, redundantly) complex conjugates of the below N/2. This is not true for complex signals that are not strictly real (e.g. have non-zero imaginary components), or for signals that are not bandlimited to only containing spectrum below Fs/2.

If, for instance, a complex signal is bandlimited to complex spectrum between 0 and Fs (e.g. contains no negative frequencies), then the DFT result of all the bins, from 0 to N-1, will represent positive frequencies, and there will be no negative frequencies, complex conjugates or otherwise, to alias against the positive frequency data.

Another possibility is for a complex signal to be bandlimited between -Fs/2 and Fs/2, in which case the DFT result bins above N/2 will represent negative frequencies, below 0 at baseband, or below the carrier frequency in the case of complex modulation (e.g. before IQ heterodyne-ing down a strictly real higher-frequency signal that is narrowly-enough bandlimited).

$\endgroup$
  • $\begingroup$ Thank you for your answer. So, if IQ signals are received, we can compute the complex DFT. Suppose that the IQ signals have really negative frequency components ranged from 2/N to N, if we compute the complex DFT, the spectrum values between N/2 and N mean real negative frequency values which the signal really have(not mirrored version), right? Thank you. $\endgroup$ – Sinecosine Feb 19 at 13:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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