Reconstructing an undersampled signal by cutting off at the signal's maximum frequency

Question

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:

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

where

• $\mathcal F$ 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):

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!

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.