Does negative distributive property of convolution over cross correlation holds?

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Let $\star$ denote convolution binary operation and $\otimes$ denote cross correlation binary operation between two functions.

Let $f,g,h$ be functions. Does this negative distribution property holds? $$f\star ( g\otimes h) = - (f \star g) \otimes h$$

Edit : It is given that $f$ is an odd function.

Definitions :

$$f\star g = \int \limits_{-\infty}^\infty f(x)g(\tau-x)dx$$ and $$f\otimes g = \int \limits_{-\infty}^\infty f(x)g(x+\tau)dx$$

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Let $$ \bar{h}(x)=h(-x)\tag{1} $$ then $$ \begin{align} g\otimes h(x) &=\int g(t)h(t+x)\,\mathrm{d}t\\ &=\int g(-t)h(-t+x)\,\mathrm{d}t\\ &=\int\bar{g}(t)h(x-t)\,\mathrm{d}t\\ &=\bar{g}\star h(x)\tag{2} \end{align} $$ Furthermore $$ \begin{align} g\otimes h(x) &=\int g(t)h(t+x)\,\mathrm{d}t\\ &=\int g(t-x)h(t)\,\mathrm{d}t\\ &=h\otimes g(-x)\\ &=\overline{h\otimes g}(x)\tag{3} \end{align} $$ and $$ \begin{align} \overline{f\star g}(x) &=f\star g(-x)\\ &=\int f(t)g(-x-t)\,\mathrm{d}t\\ &=\int\bar{f}(-t)\bar{g}(x+t)\,\mathrm{d}t\\ &=\int\bar{f}(t)\bar{g}(x-t)\,\mathrm{d}t\\ &=\bar{f}\star\bar{g}(x)\tag{4} \end{align} $$ Therefore, $$ \begin{align} (f\star g)\otimes h &=\overline{f\star g}\star h\\ &=\bar{f}\star\bar{g}\star h\\ &=\bar{f}\star(g\otimes h)\tag{5} \end{align} $$ Since $f$ is an odd function, $(5)$ becomes $$ (f\star g)\otimes h=-f\star(g\otimes h)\tag{6} $$