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翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク convolution theorem ： ウィキペディア英語版 convolution theorem In mathematics, the convolution theorem states that under suitable conditions the Fourier transform of a convolution is the pointwise product of Fourier transforms. In other words, convolution in one domain (e.g., time domain) equals point-wise multiplication in the other domain (e.g., frequency domain). Versions of the convolution theorem are true for various Fourier-related transforms. Let $\backslash \; f$ and $\backslash \; g$ be two functions with convolution $\backslash \; f$ *g . (Note that the asterisk denotes convolution in this context, and not multiplication. The tensor product symbol $\backslash otimes$ is sometimes used instead.) Let $\backslash \; \backslash mathcal$ denote the Fourier transform operator, so $\backslash \; \backslash mathcal\backslash$ and $\backslash \; \backslash mathcal\backslash$ are the Fourier transforms of $\backslash \; f$ and $\backslash \; g$, respectively. Then : $\backslash mathcal\backslash \; =\; \backslash mathcal\backslash \; \backslash cdot\; \backslash mathcal\backslash$ where $\backslash cdot$ denotes point-wise multiplication. It also works the other way around: : $\backslash mathcal\backslash =\; \backslash mathcal\backslash$ *\mathcal\ By applying the inverse Fourier transform $\backslash mathcal^$, we can write: : $f$ *g= \mathcal^\big\\cdot\mathcal\\big\} and: : $f\; \backslash cdot\; g=\; \backslash mathcal^\backslash big\backslash$ *\mathcal\\big\} Note that the relationships above are only valid for the form of the Fourier transform shown in the Proof section below. The transform may be normalized in other ways, in which case constant scaling factors (typically $\backslash \; 2\backslash pi$ or $\backslash \; \backslash sqrt$) will appear in the relationships above. This theorem also holds for the Laplace transform, the two-sided Laplace transform and, when suitably modified, for the Mellin transform and Hartley transform (see Mellin inversion theorem). It can be extended to the Fourier transform of abstract harmonic analysis defined over locally compact abelian groups. This formulation is especially useful for implementing a numerical convolution on a computer: The standard convolution algorithm has quadratic computational complexity. With the help of the convolution theorem and the fast Fourier transform, the complexity of the convolution can be reduced to O(''n'' log ''n''). This can be exploited to construct fast multiplication algorithms. == Proof == ''The proof here is shown for a particular normalization of the Fourier transform. As mentioned above, if the transform is normalized differently, then constant scaling factors will appear in the derivation.'' Let ''f'', ''g'' belong to ''L''1(R''n''). Let $F$ be the Fourier transform of $f$ and $G$ be the Fourier transform of $g$: :$F(\backslash nu)\; =\; \backslash mathcal\backslash \; =\; \backslash int\_\; f(x)\; e^\; \backslash ,\; \backslash mathrmx$ :$G(\backslash nu)\; =\; \backslash mathcal\backslash \; =\; \backslash int\_g(x)\; e^\; \backslash ,\; \backslash mathrmx,$ where the ''dot'' between ''x'' and ''ν'' indicates the inner product of R''n''. Let $h$ be the convolution of $f$ and $g$ :$h(z)\; =\; \backslash int\backslash limits\_\; f(x)\; g(z-x)\backslash ,\; \backslash mathrm\; x.$ Now notice that :$\backslash int\backslash !\backslash !\backslash int\; |f(x)g(z-x)|\backslash ,dz\backslash ,dx=\backslash int\; |f(x)|\; \backslash int\; |g(z-x)|\backslash ,dz\backslash ,dx\; =\; \backslash int\; |f(x)|\backslash ,\backslash |g\backslash |\_1\backslash ,dx=\backslash |f\backslash |\_1\; \backslash |g\backslash |\_1.$ Hence by Fubini's theorem we have that $h\backslash in\; L^1(\backslash mathbb^n)$ so its Fourier transform $H$ is defined by the integral formula :$$ \begin H(\nu) = \mathcal\ &= \int_ h(z) e^\, dz \\ &= \int_ \int_ f(x) g(z-x)\, dx\, e^\, dz. \end Observe that $|f(x)g(z-x)e^|=|f(x)g(z-x)|$ and hence by the argument above we may apply Fubini's theorem again (i.e. interchange the order of integration): :$H(\backslash nu)\; =\; \backslash int\_\; f(x)\backslash left(\backslash int\_\; g(z-x)e^\backslash ,dz\backslash right)\backslash ,dx.$ Substitute $y=z-x$; then $dy\; =\; dz$, so: :$H(\backslash nu)\; =\; \backslash int\_\; f(x)\; \backslash left(\; \backslash int\_\; g(y)\; e^\backslash ,dy\; \backslash right)\; \backslash ,dx$ :::$=\backslash int\_\; f(x)e^\; \backslash left(\; \backslash int\_\; g(y)\; e^\backslash ,dy\; \backslash right)\; \backslash ,dx$ :::$=\backslash int\_\; f(x)e^\backslash ,dx\; \backslash int\_\; g(y)\; e^\backslash ,dy.$ These two integrals are the definitions of $F(\backslash nu)$ and $G(\backslash nu)$, so: :$H(\backslash nu)\; =\; F(\backslash nu)\; \backslash cdot\; G(\backslash nu),$ QED. 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「convolution theorem」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. 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