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Fourier transform
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=== Sampling the Fourier transform === {{Broader|Poisson summation formula}} The Fourier transform of an [[Absolutely integrable function|integrable]] function <math>f</math> can be sampled at regular intervals of arbitrary length <math>\tfrac{1}{P}.</math> These samples can be deduced from one cycle of a periodic function <math>f_P</math> which has [[Fourier series]] coefficients proportional to those samples by the [[Poisson summation formula]]: <math display="block">f_P(x) \triangleq \sum_{n=-\infty}^{\infty} f(x+nP) = \frac{1}{P}\sum_{k=-\infty}^{\infty} \widehat f\left(\tfrac{k}{P}\right) e^{i2\pi \frac{k}{P} x}, \quad \forall k \in \mathbb{Z}</math> The integrability of <math>f</math> ensures the periodic summation converges. Therefore, the samples <math>\widehat f\left(\tfrac{k}{P}\right)</math> can be determined by Fourier series analysis: <math display="block">\widehat f\left(\tfrac{k}{P}\right) = \int_{P} f_P(x) \cdot e^{-i2\pi \frac{k}{P} x} \,dx.</math> When <math>f(x)</math> has [[compact support]], <math>f_P(x)</math> has a finite number of terms within the interval of integration. When <math>f(x)</math> does not have compact support, numerical evaluation of <math>f_P(x)</math> requires an approximation, such as tapering <math>f(x)</math> or truncating the number of terms.
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