The problem (taken from here) asks for possible sampling rates that will not cause aliasing in the following frequency spectrum:

enter image description here

The range of possible values after some math is given as $B_2 < f_s < 2B_1$. As an example, they provide the following image:

enter image description here

I understand the frequency spectra do not overlap, which is why we have the solution given in the key. However, I'm curious as to why frequencies not overlapping means that no aliasing occurs. The periodized version of the spectra (at the lower sampling rate) looks nothing like the original spectra, which I believe should cause a different signal than the original one to be computed. How would signal reconstruction know to avoid the 'doubling' this apparent mirroring and doubling?

  • $\begingroup$ Frequencies not overlapping means that no frequencies are modified and we can reconstruct the original signal perfectly with a proper resample to the original sampling rate. $\endgroup$
    – ZR Han
    Commented May 12, 2022 at 3:37

3 Answers 3


No aliasing means that the original information of the continuous-time signal (or higher rate discrete-time signal) is still contained unambiguously in the sampled signal. The original signal can - at least theoretically - be retrieved by means of a lowpass or bandpass filter. In the case of your example you'd need a bandpass filter with passband between $B_1$ and $B_2$. Of course, in practice you would need to make sure that there are guard bands to allow for non-ideal frequency-selective filters.

  • $\begingroup$ Thinking: Is the original information really not contained in the sampled signal when aliasing occurs? What if we have two copies with different sampling rates sufficient to resolve the ambiguity? Maybe "...is still unambiguously contained..."? $\endgroup$ Commented May 12, 2022 at 12:06
  • 1
    $\begingroup$ @DanBoschen As you know, the sampling theorem is an "if", not an "if and only if". There are cases when sampling below Nyquist is OK. However, in the general case and in the abscence of more information, the original information really is lost. $\endgroup$
    – MBaz
    Commented May 12, 2022 at 12:59
  • $\begingroup$ @MBaz I am being nitpicking about semantics. Something can be lost in the woods, but it doesn't mean that it isn't in the woods. So consider if I did have that other message that allowed us to extract the information (and someone else didn't), the information is still in their message, even though from their perspective it is "lost" due to aliasing. Does being "contained" require the ability to extract it? Not sure. $\endgroup$ Commented May 12, 2022 at 13:11
  • $\begingroup$ @DanBoschen: OK, I'm not sure if "contained" is the right word, any suggestions are of course welcome. If I have a noisy signal $y=x+n$, is the original signal $x$ "contained" in $y$? Can it be retrieved from $y$? $\endgroup$
    – Matt L.
    Commented May 12, 2022 at 13:15
  • $\begingroup$ @MattL Right, exactly my thought. We do retrieve or at least estimate x. That's why I though "unambiguously" would help. I'm interested, as such a one sentence catch-all definition of "no aliasing" would be good to have, if it could exist. $\endgroup$ Commented May 12, 2022 at 13:20

The expression $B_2 < f_s < 2B_1$. is only part of the correct answer. The full correct answer is: $${2B_2\over m+1} ≤ f_s ≤ {2B_1\over m}$$ where $m$ is any positive integer ensuring that $f_s ≥ 2(B_2-B_1)$


With any sampling rate fs, there is aliasing between frequencies. Therefore we have to read such questions with some implicit assumption. The main assumption is that that we don't care whether two frequencies alias, if we somehow know that only one of those two frequencies can occur (i.e. have non-zero energy) in the domain of possible input signals.

The typical case that's taught first is are band-limited signals with an upper limit $f_{Nyquist}$ in which case $f_s > 2*f_{Nyquist}$ is sufficient. In reconstructing, we can reconstruct the full spectrum from the sampled spectrum by setting $f>f_{Nyquist}$ to 0.

In your case, the domain of possible input signals is further restricted. Not only do frequencies above B2 carry no energy, so do frequencies below B1. Furthermore, the energy of the frequencies between B1 and B2 is given. That doesn't mean we can exactly reconstruct the signal from that - the phases are unknown.

So in this case we just require that frequencies between B1 and B2 as well as between -B2 and -B1 will alias only with frequencies for which the energy is known to be zero.


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.