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When I read the notes, a convolution is defined as:

$(f*g)(x) =\int_{-\infty}^{+\infty} f(\tau)g(x-\tau)\rm{d}\tau.$

What is the difference if we define a convolution integral as follows:

$(f*g)(x) =\int_{-\infty}^{+\infty} f(\tau)g(\tau-x)\rm{d}\tau.$

Bernard
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Ray
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    Unless $g$ is a even function, you are likely to get a different answer from the two integrals. The point in the first integral is that $x = \tau + (x-\tau)$ – Henry Jun 20 '18 at 23:21

1 Answers1

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The second would be $(f\ast h)(x)$ where $h(x)=g(-x)$.   So unless $g$ is symmetric around $0$, they would not be the same function.

As to the reason we use the first definition rather than the second, as Henry commented, $f\ast g(x)$ is $\int_{x=\tau+\nu} f(\tau)g(\nu)\mathsf d \tau=\int_\Bbb R f(\tau)g(x-\tau)\mathsf d \tau$

Graham Kemp
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