컨볼루션의 성질

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

$$ (f\ast g)'=f^{\prime}\ast g=f\ast g' $$

(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.

(b)

$$ \begin{align*} f \ast (g+h)(x) &= \int f(y) (g+h)(x-y)dy \\ &= \int f(y)\left( g(x-y) + h(x-y) \right) dy \\ &= \int f(y)g(x-y)dy + \int f(y) h(x-y)dy \\ &= f\ast g(x)+f\ast h(x) \end{align*} $$

(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)

$$ \begin{align*} a(f\ast g)(x) &= a\int f(y)g(x-y)dy \\ &= \int af(y) g(x-y)dy =af\ast g(x) \\ &=\int f(y)ag(x-y)dy=f\ast ag(x) \end{align*} $$

(e)

$$ \begin{align*} (f\ast g)'(x) &= \lim \limits_{h \to 0}\frac{f\ast g(x+h)-f\ast g(x) }{h} \\ &= \lim \limits_{h \to 0}\frac{\int f(y)g(x+h-y)dy-\int f(y)g(x-y)dy }{h} \\ &= \lim \limits_{h \to 0}\int f(y)\frac{ g(x+h-y)-g(x-y) }{h}dy \\ &=\int f(y)\lim \limits_{h \to 0}\frac{ g(x+h-y)-g(x-y) }{h}dy \\ &= \int f(y)g'(x-y)dy \\ &= f\ast g'(x) \end{align*} $$

이때 (a) 에 의해서

$$ (f\ast g)'(x)=(g\ast f)'(x)=g\ast f^{\prime}(x)=f^{\prime}\ast g(x) $$

이므로

$$ (f\ast g)'(x)=f^{\prime}\ast g(x)=f\ast g'(x) $$

(f)

$$ \begin{align*} \overline{f\ast g(x)} &= \overline{\int f(y)g(x-y)dy} \\ &= \int \overline{f(y)g(x-y)}dy \\ &= \int \overline{f(y)}\ \overline{g(x-y)}dy \\ &= \overline{f} \ast \overline{g}(x) \end{align*} $$

(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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