Editing Nyquist–Shannon sampling theorem (section)

==Shannon's original proof==
Poisson shows that the Fourier series in {{EquationNote|Eq.1}} produces the periodic summation of <math>X(f)</math>, regardless of <math>f_s</math> and <math>B</math>. Shannon, however, only derives the series coefficients for the case <math>f_s=2B</math>. Virtually quoting Shannon's original paper:

:Let <math>X(\omega)</math> be the spectrum of <math>x(t).</math>&nbsp; Then

::<math>x(t) = {1 \over 2\pi} \int_{-\infty}^{\infty} X(\omega) e^{i\omega t}\;{\rm d}\omega = {1 \over 2\pi} \int_{-2\pi B}^{2\pi B} X(\omega) e^{i\omega t}\;{\rm d}\omega,</math>

:because <math>X(\omega)</math> is assumed to be zero outside the band <math>\left|\tfrac{\omega}{2\pi}\right| < B.</math>&nbsp; If we let <math>t = \tfrac{n}{2B},</math> where <math>n</math> is any positive or negative integer, we obtain:

{{Equation box 1|title=
|indent=: |cellpadding= 0 |border= 0 |background colour=white
|equation = {{NumBlk|:|
<math>x \left(\tfrac{n}{2B} \right) = {1 \over 2\pi} \int_{-2\pi B}^{2\pi B} X(\omega) e^{i\omega {n \over {2B}}}\;{\rm d}\omega.</math> &nbsp; &nbsp;
|{{EquationRef|Eq.2}}}}
}}

:On the left are values of <math>x(t)</math> at the sampling points. The integral on the right will be recognized as essentially{{

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 <math>n^{th}</math> 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>

{{blockquote|
Let <math>x_n</math> be the <math>n^{th}</math> sample. Then the function <math>x(t)</math> is represented by:

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

As in the other proof, the existence of the Fourier transform of the original signal is assumed, so the proof does not say whether the sampling theorem extends to bandlimited stationary random processes.

===Notes===
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