![]() ![]() The interval at which the DTFT is sampled is the reciprocal of the duration of the input sequence. In mathematics, the discrete Fourier transform ( DFT) converts a finite sequence of equally-spaced samples of a function into a same-length sequence of equally-spaced samples of the discrete-time Fourier transform (DTFT), which is a complex-valued function of frequency. Its similarities to the original transform, S(f), and its relative computational ease are often the motivation for computing a DFT sequence. The respective formulas are (a) the Fourier series integral and (b) the DFT summation. The spectral sequences at (a) upper right and (b) lower right are respectively computed from (a) one cycle of the periodic summation of s(t) and (b) one cycle of the periodic summation of the s(nT) sequence. Fig 2: Depiction of a Fourier transform (upper left) and its periodic summation (DTFT) in the lower left corner. The FFT algorithm computes one cycle of the DFT and its inverse is one cycle of the DFT inverse. The inverse DFT (top) is a periodic summation of the original samples. Right column: The DFT (bottom) computes discrete samples of the continuous DTFT. Its Fourier transform (bottom) is a periodic summation ( DTFT) of the original transform. Center-right column: Original function is discretized (multiplied by a Dirac comb) (top). The inverse transform is a sum of sinusoids called Fourier series. Fourier transform (bottom) is zero except at discrete points. Center-left column: Periodic summation of the original function (top). Left column: A continuous function (top) and its Fourier transform (bottom). Now, here's the fun part: you can show that the spacial distribution of protons in the nucleus is the Fourier transform of the momentum distribution.Īnd bingo, a measurement of the size of the nucleus.ĭo it with a polarized target and you can get info on the shape as well.Fig 1: Relationship between the (continuous) Fourier transform and the discrete Fourier transform. (This is what nuclear physicists call "quasi-elastic scattering".) Now, if (1) we are shooting a beam of electrons at a stationary target, (2) we have a precision measurement of the momenta of the incident and scattered electrons and the ejected proton, (3) we are willing to neglect excitation energy of the remnant nucleus, and (4) we assume that $p$ mostly did not interact with the remnant after being scattered, we know the momentum of the proton inside the nucleus at the time it was struck.Ĭollect enough statistics on this and we have sampled the proton momentum distribution of the nucleus. Where $A$ represents that target nucleus and $B$ the remnant after we bounce a proton out. What we do is scatter things off of the component parts of the nucleus. Now, electron microscopy can just about provide vague picture of a medium or large atom as a out-of-focus ball, but there is no hope of employing that technique to something orders of magnitude smaller. What is the shape and size of a atomic nucleus?įrom Rutherford we learned that the nucleus is rather a lot smaller than the atom as a whole. Given that leftaroundabout and vonjd have addressed the fundamental place of the Fourier transform in the formalism, let me talk a little about an experimental application.
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