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Properties of Convolution 📂Fourier Analysis

Properties of Convolution

Theorem

Convolution satisfies the following properties.

(a) Commutative Law

f∗g=g∗f f \ast g = g \ast f

(b) Distributive Law

f∗(g+h)=f∗g+f∗h f \ast (g+h) = f \ast g + f \ast h

(c) Associative Law

f∗(g∗h)=(f∗g)∗h f \ast (g \ast h) = (f \ast g) \ast h

(d) Scalar Multiplication Associative Law

a(f∗g)=(af∗g)=(f∗ag) a(f \ast g)=(af\ast g)=(f\ast ag)

(e) Differentiation

(f∗g)′=f′∗g=f∗g′ (f\ast g)^{\prime}=f^{\prime}\ast g=f\ast g^{\prime}

(f) Complex Conjugate

f∗g‾=f‾∗g‾ \overline{f\ast g}=\overline{f} \ast \overline{g}

(g) Dirac Delta Function

f∗δ=f f \ast \delta =f

Proof

(a)

f∗g(x)=∫−∞∞f(y)g(x−y)dy=∫∞−∞f(x−z)g(z)(−dz)=∫−∞∞f(x−z)g(z)dz=∫−∞∞g(z)f(x−z)dz=g∗f(x) \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*}

In the second equality, it was substituted with x−y=zx-y=z.

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(b)

f∗(g+h)(x)=∫f(y)(g+h)(x−y)dy=∫f(y)(g(x−y)+h(x−y))dy=∫f(y)g(x−y)dy+∫f(y)h(x−y)dy=f∗g(x)+f∗h(x) \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)

f∗(g∗h)(x)=∫yf(y)g∗h(x−y)dy=∫yf(y)h∗g(x−y)dy=∫yf(y)∫zh(z)g(x−y−z)dzdy=∫zh(z)∫yf(y)g(x−z−y)dydz=∫zh(z)f∗g(x−z)dz=h∗(f∗g)(x)=(f∗g)∗h(x) \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)

a(f∗g)(x)=a∫f(y)g(x−y)dy=∫af(y)g(x−y)dy=af∗g(x)=∫f(y)ag(x−y)dy=f∗ag(x) \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)

(f∗g)′(x)=lim⁡h→0f∗g(x+h)−f∗g(x)h=lim⁡h→0∫f(y)g(x+h−y)dy−∫f(y)g(x−y)dyh=lim⁡h→0∫f(y)g(x+h−y)−g(x−y)hdy=∫f(y)lim⁡h→0g(x+h−y)−g(x−y)hdy=∫f(y)g′(x−y)dy=f∗g′(x) \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*}

At this point, due to (a),

(f∗g)′(x)=(g∗f)′(x)=g∗f′(x)=f′∗g(x) (f\ast g)^{\prime}(x)=(g\ast f)^{\prime}(x)=g\ast f^{\prime}(x)=f^{\prime}\ast g(x)

Therefore,

(f∗g)′(x)=f′∗g(x)=f∗g′(x) (f\ast g)^{\prime}(x)=f^{\prime}\ast g(x)=f\ast g^{\prime}(x)

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(f)

f∗g(x)‾=∫f(y)g(x−y)dy‾=∫f(y)g(x−y)‾dy=∫f(y)‾ g(x−y)‾dy=f‾∗g‾(x) \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)

f∗δ(x)=∫f(y)δ(x−y)dy=f(x)∫δ(x−y)dy=f(x) \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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