Nyquist–Shannon sampling theorem: Difference between revisions

Revision as of 22:18, 30 May 2024

Revision as of 22:25, 30 May 2024

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: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:

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