Nyquist–Shannon sampling theorem: Difference between revisions

Revision as of 21:22, 4 January 2024

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efn|group=proof|Multiplying both sides of {{EquationNote|Eq.2}} by <math>T = 1/2B</math> produces, on the left, the scaled sample values <math>(T\cdot x(nT))</math> in Poisson's formula ({{EquationNote|Eq.1}}), and, on the right, the actual formula for Fourier expansion coefficients.

}} the ''n''<supmath>n^{th}</supmath> coefficient in a Fourier-series expansion of the function <math>X(\omega),</math> taking the interval <math>-B</math> to <math>B</math> as a fundamental period. This means that the values of the samples <math>x(n/2B)</math> determine the Fourier coefficients in the series expansion of <math>X(\omega).</math>&nbsp; Thus they determine <math>X(\omega),</math> since <math>X(\omega)</math> is zero for frequencies greater than <math>B,</math> and for lower frequencies <math>X(\omega)</math> is determined if its Fourier coefficients are determined. But <math>X(\omega)</math> determines the original function <math>x(t)</math> completely, since a function is determined if its spectrum is known. Therefore the original samples determine the function <math>x(t)</math> completely.

Shannon's proof of the theorem is complete at that point, but he goes on to discuss reconstruction via [[sinc function]]s, what we now call the [[Whittaker–Shannon interpolation formula]] as discussed above. He does not derive or prove the properties of the sinc function, as the Fourier pair relationship between the [[rectangular function|rect]] (the rectangular function) and sinc functions was well known by that time.<ref>{{cite book |last1=Campbell |first1=George |last2=Foster |first2=Ronald |title=Fourier Integrals for Practical Applications |date=1942 |publisher=Bell Telephone System Laboratories |___location=New York}}</ref>

{{quote|

Let <math>x_n</math> be the ''<math>n''^{th}</math> sample. Then the function <math>x(t)</math> is represented by:

:<math display="block">x(t) = \sum_{n=-\infty}^{\infty}x_n{\sin(\pi(2Bt-n)) \over \pi(2Bt-n)}.</math>

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