컨볼루션의 성질
정리
컨볼루션은 다음과 같은 성질을 만족한다.
(a) 교환법칙
$$ f \ast g = g \ast f $$
(b) 분배법칙
$$ f \ast (g+h) = f \ast g + f \ast h $$
(c) 결합법칙
$$ f \ast (g \ast h) = (f \ast g) \ast h $$
(d) 스칼라곱의 결합 법칙
$$ a(f \ast g)=(af\ast g)=(f\ast ag) $$
(e) 미분
$$ (f\ast g)^{\prime}=f^{\prime}\ast g=f\ast g^{\prime} $$
(f) 켤레 복소수
$$ \overline{f\ast g}=\overline{f} \ast \overline{g} $$
(g) 디랙 델타 함수
$$ f \ast \delta =f $$
증명
(a)
$$ \begin{align*} f\ast g(x)&=\int _{-\infty} ^{\infty} f(y)g(x-y)dy \\ &=\int_{\infty} ^{-\infty} f(x-z)g(z)(-dz) \\ &=\int_{-\infty} ^{\infty} f(x-z)g(z)dz \\ &=\int_{-\infty} ^{\infty} g(z)f(x-z)dz \\ &=g\ast f(x) \end{align*} $$
두번째 등호에서 $x-y=z$로 치환하였다.
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(b)
$$ \begin{align*} f \ast (g+h)(x) &= \int f(y) (g+h)(x-y)dy \\ &= \int f(y)\left( g(x-y) + h(x-y) \right) dy \\ &= \int f(y)g(x-y)dy + \int f(y) h(x-y)dy \\ &= f\ast g(x)+f\ast h(x) \end{align*} $$
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(c)
$$ \begin{align*} f \ast (g\ast h)(x) &= \int_{y} f(y) g\ast h(x-y)dy \\ &= \int_{y} f(y) h\ast g(x-y)dy \\ &= \int_{y} f(y) \int_{z} h(z)g(x-y-z)dzdy \\ &= \int_{z} h(z) \int_{y}f(y)g(x-z-y)dydz \\ &= \int_{z} h(z) f\ast g(x-z) dz \\ &=h \ast (f\ast g)(x) \\ &= (f\ast g)\ast h(x) \end{align*} $$
(d)
$$ \begin{align*} a(f\ast g)(x) &= a\int f(y)g(x-y)dy \\ &= \int af(y) g(x-y)dy =af\ast g(x) \\ &=\int f(y)ag(x-y)dy=f\ast ag(x) \end{align*} $$ \ast \eta
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(e)
$$ \begin{align*} (f\ast g)^{\prime}(x) &= \lim \limits_{h \to 0}\frac{f\ast g(x+h)-f\ast g(x) }{h} \\ &= \lim \limits_{h \to 0}\frac{\int f(y)g(x+h-y)dy-\int f(y)g(x-y)dy }{h} \\ &= \lim \limits_{h \to 0}\int f(y)\frac{ g(x+h-y)-g(x-y) }{h}dy \\ &=\int f(y)\lim \limits_{h \to 0}\frac{ g(x+h-y)-g(x-y) }{h}dy \\ &= \int f(y)g^{\prime}(x-y)dy \\ &= f\ast g^{\prime}(x) \end{align*} $$
이때 (a) 에 의해서
$$ (f\ast g)^{\prime}(x)=(g\ast f)^{\prime}(x)=g\ast f^{\prime}(x)=f^{\prime}\ast g(x) $$
이므로
$$ (f\ast g)^{\prime}(x)=f^{\prime}\ast g(x)=f\ast g^{\prime}(x) $$
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(f)
$$ \begin{align*} \overline{f\ast g(x)} &= \overline{\int f(y)g(x-y)dy} \\ &= \int \overline{f(y)g(x-y)}dy \\ &= \int \overline{f(y)}\ \overline{g(x-y)}dy \\ &= \overline{f} \ast \overline{g}(x) \end{align*} $$
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(g)
$$ \begin{align*} f \ast \delta (x) &= \int f(y)\delta (x-y)dy \\ &= f(x)\int \delta (x-y)dy \\ &= f(x) \end{align*} $$
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