Assume a (continuous) band-limited signal $f$, that is, a signal for which $F(s) = 0$ for all $\lvert s \rvert > p / 2$. If the signal is sampled with frequency $p$, we can reconstruct it by cutting off the frequency band corresponding to one period of its spectrum:

$$ \mathcal{F}^{-1}\left( \Pi_p(\mathcal{F}f \star \hbox{Ш}_p) \right), $$


  • $\mathcal F$ denotes the Fourier transform;
  • $\Pi_p$ denotes the cut-off function of the $\left[-p/2,+p/2\right]$ frequency interval;
  • $\hbox{Ш}_p$ denotes a train of Dirac functions spaced $p$ apart.

(This notation is borrowed Brad Osgood's lecture notes on The Fourier Transform and its Applications, see for example Chapter 5.)

If the signal is undersampled (with frequency $q < p$), we can recover an alias—another signal that agrees with the original signal $f$ at the sample values, but not necessarily at the rest of the points. The alias can be obtained by simply cutting off at the sampling frequency $q$:

$$ \mathcal{F}^{-1}\left( \Pi_q(\mathcal{F}f \star \hbox{Ш}_q) \right). $$

However, in the case of undersampling another possibility would be cutting off at the maximum frequency $p$ of the signal (in case we knew it):

$$ \mathcal{F}^{-1}\left( \Pi_p(\mathcal{F}f \star \hbox{Ш}_q) \right). $$

Graphically, the two variants would look as follows—cutting off at the sampling frequency $q$ (in brown) and cutting off at the signal's frequency $p$ (in green):

enter image description here

As far as I could tell this latter approach (of cutting off at the signal's maximum frequency) doesn't guarantee that we obtain the same values at the sample points of $f$ (so it's not really an "alias" in the precise sense of the word?!). However, I've seen this approach presented in a textbook (Digital Image Processing by Gonzalez and Woods, Figure 4.9 in the third edition of the book), so my question is: when would cutting off at the signal's frequency make sense in the case of an undersampled signal? Does this variant give a tighter approximation of the original signal?!

N.B.: As you probably may tell, I'm not well versed in signal processing, so please forgive my misuse of terminology and do correct my mistakes. Thank you!

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    $\begingroup$ Maybe related $\endgroup$ Dec 28, 2022 at 10:32
  • $\begingroup$ Thanks for the pointer @OverLordGoldDragon! I'm currently lacking the knowledge of the discrete Fourier transform to properly understand your question (but I'll revisit it once I get to grips with the material). And the answer to your question, while it treats the continuous case, I don't think it answers my question: in my case I was wondering whether using the frequencies past half the sampling frequency will be beneficial for the reconstruction. $\endgroup$ Dec 29, 2022 at 12:00

1 Answer 1


I think you're misunderstanding the meaning of the term "aliasing" as it's used here.

It gets complicated if I try to express this as a general case, but let's take the case where you have a sine wave of unknown phase at $0.4 f_s$, and another one at $0.6 f_s$.

After sampling, both of these sine waves appear at $0.4 f_s$. Actually, because sampling takes frequencies on the entire real number line and maps them to an interval that is exactly $1 f_s$ long, these sine waves both appear to be at $0.4 f_s$, $0.6 f_s$, $-0.4 f_s$, etc. -- basically $\pm 0.4 f_s + kf_s\, \forall\ k \in \mathbb I$.

The fact is that after aliasing you cannot tell which interval the original signal component came from -- that information is lost. That's where the "alias" comes from -- signal A is pretending to be signal B.

As for how to deal with an undersampled signal after aliasing, what you do depends on the character of the original signal, and what cost of the various errors in the undersampled signal may be.

Sometimes it may be best to just live with the signal as sampled.

Sometimes it may be best to filter out all possible aliased frequencies.

Sometimes it may be best to widen your filter, and accept some aliasing as the cost of getting more un-aliased content.

Sometimes it may be best to chase anyone wearing a pinstriped shirt out of your TV studio, before the aliasing happens.

  • $\begingroup$ Thank you very much for your answer and thanks for confirming that indeed the suitability of the reconstruction method will depend on multiple factors! I think my question was related to your third example "widen your filter and accept some aliasing," but what I had in mind was maybe even a more extreme case: extend the filter past half the sampling frequency, up to the maximum frequency of the signal (assuming this information is provided). $\endgroup$ Dec 29, 2022 at 11:37
  • $\begingroup$ For your example that would mean to include both the $±0.4 f_s$ frequencies and also the $±0.6 f_s$ frequencies, when reconstructing the second signal (which has frequency $0.6f_s$). Would that ever make sense? On the one hand including frequencies past the sampling frequency will induce noisy frequencies, but on the other hand we do know that the frequency of the signal is greater than what we can reconstruct—so how could we best make use of this extra information? $\endgroup$ Dec 29, 2022 at 11:38
  • $\begingroup$ And a follow-up question on the definition of a signal aliasing another. You say that signal $g$ "is pretending to be" signal $f$, but isn't this in line with what I've said in the original post: a signal $g$ is an alias of a (sampled) signal $f$ if it agrees with $f$ at the sample values, but not necessarily at the rest of the points? Otherwise, what does "pretending" mean? $\endgroup$ Dec 29, 2022 at 11:46

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