I know except for some special cases, aliasing is unavoidable. Assume we time-limit a function, $f(t)$, so that it is zero outside an interval say $[0,T]$ to form $y(t)$. Then, in the frequency domain, infinite frequency components are introduced.

If the system that makes this time limitation has impulse response $h(t)$, we have its Fourier representation $H(u)$. We can then calculate $Y(u)$ using convolution to obtain: $$Y(u)=F(u)*H(u)$$ Obviously, it has frequency components that are replicated to infinity. This phenomenon is because of sliding it across the original function to calculate Convolution.

Could anyone possibly let me know in what condition we can have band-limited input and output.

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    $\begingroup$ I have difficulties following which domain you mean in each sentence when you say band-limited. The standard jargon is that the function "has compact support" when it is zero-valued outside some interval. I think the question would be clearer if you said it like that for the time domain. $\endgroup$ Commented Apr 4, 2016 at 17:04
  • $\begingroup$ Do you actually mean "time-limited" $f(t)$? $\endgroup$ Commented Apr 4, 2016 at 17:21
  • $\begingroup$ @LaurentDuval, out of a finite interval, the function is zero. en.wikipedia.org/wiki/Bandlimiting $\endgroup$
    – David
    Commented Apr 4, 2016 at 17:26
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    $\begingroup$ @David Laurent is pointing out that you are confusing time-limited and band-limited signals. Please read the page you reference, specifically this part. $\endgroup$
    – Peter K.
    Commented Apr 4, 2016 at 19:20
  • $\begingroup$ The answer are correct but it seems to me that you are reaching for Wavelets. Where you can control the time and frequency resolutions; and range. You are of course limited by a form of the uncertainty principle (a rephrasing of the answers). $\endgroup$
    – rrogers
    Commented Apr 6, 2016 at 16:39

2 Answers 2


The condition is that the function $f(t)=0$, if you are talking about continuous time signals and frequencies (some of your formulations are confusing to me). In other words, a (nonzero) time-limited signal cannot be also band-limited. In other words, a function and its (continuous) Fourier transform cannot both have finite support.

The core of one proof (using complex analysis) relies on Paley-Wiener theorems: the Fourier transform of a time-limited function would be an entire function (a complex-valued function, holomorphic over the whole complex plane), that would vanish on an open interval, and hence be zero (almost) everywhere.

You can find details in Fourier transforms of compactly supported functions, and the paper (for instance) The uncertainty principle for Fourier transforms on the real line. This is a reason for the development of time-frequency analysis.

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    $\begingroup$ I would like to add that, for a slight relaxation of support requirements, we can talk about quasi compactly supported functions on time-frequency. If we only measure the support, where the function does not decay faster than any rational function, time-frequency distributions like the Gabor atom have finite support and give rise to a beautiful theory of area quantisation in the time-frequency plane. Part of the elegance of that theory comes from the fact that if we add the ignored support back in, all important results are preserved. $\endgroup$
    – Jazzmaniac
    Commented Apr 4, 2016 at 19:09
  • $\begingroup$ Indeed, the notions of essential support or "domain" concentration are more useful that standard support. Do you have a source for the definition of "quasi compactly supported"? $\endgroup$ Commented Apr 4, 2016 at 19:15
  • $\begingroup$ @Laurent, I think Jazzmaniac might refer to this: math.u-bordeaux.fr/~pjaming/exposes/2006-2010/budapest.pdf --Best. $\endgroup$
    – user20387
    Commented Apr 4, 2016 at 21:30
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    $\begingroup$ There are two very nice, relatively accessible proofs in section 6.8 of this excellent, freely-available book: afidc.ethz.ch/A_Foundation_in_Digital_Communication/Home.html $\endgroup$
    – MBaz
    Commented Apr 4, 2016 at 21:46
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    $\begingroup$ @LaurentDuval, I don't have a reference at hand right now, I will need to look it up. There are different ways to formulate a suitable condition and not all are equivalent. A good starting point is $f:R^n \to R$ is quasi (or essentially) compactly supported, iff $S(\epsilon):=\partial \{r:|f(r)|<\epsilon\}$ is compact for all $\epsilon>0$ and $\lim_{\epsilon\to +0} \epsilon |S(\epsilon)|^k = 0$ for all natural $k$. For time frequency support you would take the Wigner distribution for $f$. The tricky part is actually quantifying the essential support. $\endgroup$
    – Jazzmaniac
    Commented Apr 4, 2016 at 23:24

This is not possible for actual signals (i.e., functions of a real variable). It is possible for functions on a domain that is already compact itself, e.g. functions from the unit circle (or, as multivariate generalisations, from tori or spheres).

That is not completely hypothetical: perfectly periodic signals can be understood as functions on a compact time interval, which is however closed by the ends to be periodic.

  • $\begingroup$ well, agreed, if we call it compact carrier, you're right; I think what OP meant, though was "functions that are non-zero only within a compact interval, and zero almost everywhere", i.e. there needs to be a region from which you can take compact intervals that can be arbitrarily large compared to the carrier interval. $\endgroup$ Commented Apr 6, 2016 at 8:29

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