Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In Wikipedia, there is Shannon's proof on Nyquist-Shannon sampling theorem. ( http://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem#Shannon.27s_original_proof )

The original proof presented by Shannon is elegant and quite brief, but it offers less intuitive insight into the subtleties of aliasing, both unintentional and intentional. Quoting Shannon's original paper, which uses ''f'' for the function, ''F'' for the spectrum, and ''W'' for the bandwidth limit:Let $\scriptstyle F(\omega)$ be the spectrum of $\scriptstyle f(t).$ Then $f(t)= {1 \over 2\pi} \int_{-\infty}^{\infty} F(\omega) e^{i\omega t}\;{\rm d}\omega \ = {1 \over 2\pi} \int_{-2\pi W}^{2\pi W} F(\omega) e^{i\omega t}\;{\rm d}\omega \ $ since $\scriptstyle F(\omega)$ is assumed to be zero outside the band ''W''. If we let $t = {n \over {2W}}\,$ where ''n'' is any positive or negative integer, we obtain $f \left({n \over {2W}} \right) = {1 \over 2\pi} \int_{-2\pi W}^{2\pi W} F(\omega) e^{i\omega {n \over {2W}}}\;{\rm d}\omega.$

What I am not getting is where $\infty$ is replaced with $2\pi W$. Why is it substituted like this? Also, why is $t$ substitued with $n \over 2W$?


share|improve this question
add comment

1 Answer

up vote 0 down vote accepted

Well, have a look at the statement of the theorem - it assumes that the signal is band-limited i.e. it has finite frequency content, so the first integral over $(-\infty,\infty)$ reduces to a finite integral over $[-2\pi W,2\pi W]$ because the signal contains no frequencies larger than $W$. Intuitively, you need this assumption because if we have infinite frequency content, then your signal may vary arbitrarily fast - in which case you might need to sample your signal at an arbitrarily high frequency to obtain a perfect reconstruction.

Finally, substituting $t$ with $\frac{n}{2W}$ is actually a trick to see that the LHS ($f(\frac{n}{2W})$ i.e. $f$ sampled at a rate $\ge 2W$, twice the largest frequency in the signal) actually gives you the Fourier coefficients of the signal, which is the RHS (note that by assumption, we only need angular frequencies in the range $[-2\pi W,2\pi W]$, corresponding to frequencies in the range $[-W,W]$). And since the Fourier transform (if it exists) uniquely determines the signal, we conclude that it must be possible to reconstruct the signal - and Shannon's interpolating formula gives an explicit way to do this.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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