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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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  • $\begingroup$ wouldn't that be called a negative associative property? $\endgroup$
    – robjohn
    Oct 16, 2013 at 20:59
  • $\begingroup$ Yes. Certainly not distributive. $\endgroup$
    – Rajesh D
    Oct 16, 2013 at 22:45

1 Answer 1

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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 $$ (f\star g)\otimes h=-f\star(g\otimes h)\tag{6} $$

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  • $\begingroup$ I forgot to mention that $f$ is an odd function. I guess it holds now. Actually $f(t )=1/t$, a kernel of Hilbert transform, I was expecting this result to prove my answer on the metric question. Please correct me if I am wrong.Thanks for the answer. $\endgroup$
    – Rajesh D
    Oct 16, 2013 at 22:40
  • $\begingroup$ @RajeshD: It would be nice to add that $f$ is odd to the question. $\endgroup$
    – robjohn
    Oct 16, 2013 at 22:52

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