Faltung

2009-03-17

Im R³:

$(f*g)(R):=∫_{rεR³} f(R-r)⋅g(r) dV_r$

Substituiere:

$u:=R-r=\begin{pmatrix} R_1-x \\ R_2-y \\ R_3-z \end{pmatrix}$

$u=\begin{pmatrix} u_1 \\ u_2 \\ u_3 \end{pmatrix}$

$(f*g)(R)=∫_{-∞}^{∞}∫_{-∞}^{∞}∫_{-∞}^{∞} f(R-r)⋅g(r) dx dy dz$

x	u₁
-∞	+∞
+∞	-∞

y	u₂
-∞	+∞
+∞	-∞

z	u₃
-∞	+∞
+∞	-∞

$(f*g)(R)=∫_{-∞}^{∞}∫_{-∞}^{∞}∫_{-∞}^{∞} f(u)⋅g(R-u)⋅(-du_1)⋅(-du_2)⋅(-du_3)$

$(f*g)(R)=-∫_{-∞}^{∞}∫_{-∞}^{∞}∫_{-∞}^{∞} f(u)⋅g(R-u)⋅du_1⋅du_2⋅du_3$

$(f*g)(R)=-∫_{-∞}^{∞}∫_{-∞}^{∞}∫_{-∞}^{∞} f(u)⋅g(R-u)⋅dV_u$

$(f*g)(x):=∫_{-∞}^{∞} f(ξ)⋅g(x-ξ) dξ$

$f(v):=\left\{\begin{array}{cl} 1, & |v|<1\\÷{1}{(v-2)^4}, & \mbox{sonst} \end{array}\right.$

$(f*g)(x)=∫_{-∞}^{1} ÷{1}{(ξ-2)^4}⋅(x-ξ+1)²dξ+∫_{-1}^{1} (x-ξ+1)²dξ+∫_{1}^{∞} ÷{1}{(ξ-2)^4}⋅(x-ξ+1)²dξ$

Substituiere:

$(f*g)(x)=-∫_{∞}^{x+1} ÷{1}{(x-u-2)^4}⋅(u+1)²du+∫_{x+1}^{x-1}⋅(u+1)²⋅(-du)+∫_{x-1}^{-∞} ÷{1}{(x-u-2)^4}⋅(u+1)²⋅(-du)$

$(f*g)(x)=∫_{-∞}^{x-1} ÷{(u+1)²}{(x-u-2)^4}du+∫_{x-1}^{x+1}⋅(u+1)²du+∫_{x+1}^{∞} ÷{(u+1)²}{(x-u-2)^4}du$

Author: Ewa Weinmüller

Last modification on: Sat, 04 May 2024 .