Nyquist–Shannon sampling theorem: Difference between revisions

Revision as of 18:01, 2 January 2024

Revision as of 23:03, 3 January 2024

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When <math>x(t)</math> is a function with a [[Fourier transform]] <math>X(f)</math>''':'''

:<math display="block">X(f)\ \triangleq\ \int_{-\infty}^{\infty} x(t) \ e^{- i 2 \pi f t} \ {\rm d}t,</math>

the [[Poisson summation formula]] indicates that the samples, <math>x(nT),</math>, of <math>x(t)</math> are sufficient to create a [[periodic summation]] of <math>X(f).</math>. The result is:

{{NumBlk|:

|<math display="block">X_s(f)\ \triangleq \sum_{k=-\infty}^{\infty} X\left(f - k f_s\right) = \sum_{n=-\infty}^{\infty} T\cdot x(nT)\ e^{-i 2\pi n T f}</math> ({{EquationRef|Eq.1}})

|{{EquationRef|Eq.1}}

}}

[[File:AliasedSpectrum.png|thumb|upright=1.8|right|<math>X(f)</math> (top blue) and <math>X_A(f)</math> (bottom blue) are continuous Fourier transforms of two {{em|different}} functions, <math>x(t)</math> and <math>x_A(t)</math> (not shown). When the functions are sampled at rate <math>f_s</math>, the images (green) are added to the original transforms (blue) when one examines the discrete-time Fourier transforms (DTFT) of the sequences. In this hypothetical example, the DTFTs are identical, which means {{em|the sampled sequences are identical}}, even though the original continuous pre-sampled functions are not. If these were audio signals, <math>x(t)</math> and <math>x_A(t)</math> might not sound the same. But their samples (taken at rate <math>f_s</math>) are identical and would lead to identical reproduced sounds; thus <math>x_A(t)</math> is an alias of <math>x(t)</math> at this sample rate.]]