F(fg)(ω)=1/√2πF(f)*F(g)
合成積のFourier変換
と
F(f)(ω)=F^{-1}(f)(-ω)
を使う
$ \mathcal F(f)*\mathcal F(g)= \mathcal F^{-1}(\mathcal F(\mathcal F(f)*\mathcal F(g)))
$ = \mathcal F^{-1}(\sqrt{2\pi}\mathcal F^2(f)\mathcal F^2(g))
$ \because
合成積のFourier変換
$ =\mathcal F^{-1}(\sqrt{2\pi}f(-x)g(-x))
$ \because
F(F(f))(ω)=f(-ω)
$ =\sqrt{2\pi}\mathcal F(fg)
$ \because
F(f)(ω)=F^{-1}(f)(-ω)
#2025-05-26
14:19:15
#2025-05-19
14:40:02