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How to solve the Fast Fourier transform?

What is the relationship between the Fourier transform and Fourier series representation of a continuous function?

  • "a Fourier series decomposes any periodic function or periodic signal into the sum of a (possibly infinite) set of simple oscillating functions" A fourier series is a sum of discrete elements, where as the fourier transform is the decomposition over a continuous set of basis functions (integral instead of summation).  How do I convert a distribution over the continuous basis set, into a sum of discrete basis functions?

  • Answer:

    In short, fourier series is for periodic signals and fourier transform is for aperiodic signals. Fourier series is used to decompose signals into basis elements (complex exponentials) while fourier transforms are used to analyze signal in another domain (e.g. from time to frequency, or vice versa).

Aakash Prasad at Quora Visit the source

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Typically, you would only consider a Fourier series for a function on an interval---say, [−π,π][−π,π][-\pi,\pi]. Then the kth Fourier series coefficient is given by (up to a constant, it is defined differently for different disciplines, and the sign on the imaginary unit really doesn't matter as long as you are consistent) ∫π−πf(x)eikxdx=∫∞−∞f(x)eikxdx=F(k),∫−ππf(x)eikxdx=∫−∞∞f(x)eikxdx=F(k), \int_{-\pi}^{\pi} f(x) e^{i k x} dx = \int_{-\infty}^\infty f(x) e^{i k x}dx = \mathcal{F}(k), where FF\mathcal{F} is the continuous Fourier transform of f, given by F(τ)=∫∞−∞f(x)eiτxdx.F(τ)=∫−∞∞f(x)eiτxdx. \mathcal{F}(\tau) = \int_{-\infty}^\infty f(x) e^{i \tau x} dx. Similar reasoning works for periodic functions if you truncate them to one period.

Michael McCoy

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