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We have two functions of time $f(t)$ and $g(t)$, for which convolution and correlation are defined as following:

Convolution: $(f(t)\ast g(t))(\tau) = \int_{-\infty}^\infty{f(t)g(\tau-t)dt}$

Correlation: $(f(t)\star g(t))(\tau) = \int_{-\infty}^\infty{f^\ast(t)g(\tau+t)dt}$

In the english wikipedia and in other sources I found that the following relationship should hold:

$(f(t)\star g(t))(\tau) = (f^\ast(-t)\ast g(t))(\tau)$

Is this correct? If so, how can i prove this? Usually, i would try substitution, but how to change the $g(\tau+t)$ to $g(\tau-t)$?

Timitry
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2 Answers2

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I figured out the answer while writing down the question. Here it is:

$(f^\ast(-t)\ast g(t))(\tau) = \int_{-\infty}^\infty{f^\ast(-t)g(\tau-t)dt} = -\int_{\infty}^{-\infty}{f^\ast(t)g(\tau+t)dt} = \int_{-\infty}^{\infty}{f^\ast(t)g(\tau+t)dt} = (f(t)\star g(t))(\tau)$

In the second step, the substitution $t\to -t$ took place.

Timitry
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We have -
Convolution - $ f(t) * g(t) = \int_{-\infty}^{\infty}f(\tau) . g(t-\tau) d\tau$
Correlation - $ R_{fg}(\tau) = \int_{-\infty}^{\infty}f(t) . g(t-\tau) dt$

Now, $$ f(t) * g(t) = \int_{-\infty}^{\infty}f(\tau) . g(t-\tau) d\tau$$
Replace $\tau$ by $u$-
$$ f(t) * g(t) = \int_{-\infty}^{\infty}f(u) . g(t-u) du$$ Replace $t$ by $k$-
$$f(t) * g(t) = \int_{-\infty}^{\infty}f(u) . g(k-u) du = \int_{-\infty}^{\infty}f(t) . g(k-t) dt = \int_{-\infty}^{\infty}f(t) . g(-(t-k)) dt = f(t) \star g(-t)$$ which is the correlation between $f(t)$ and $g(-t)$. This can be seen as - $$f(t) \star g(-t) = \int_{-\infty}^{\infty}f(t) . g(-(t-\tau)) dt$$
So, $(f(t)⋆g(-t))(\tau)=(f(t)∗g(t))(\tau)$.
Similarly, we can prove that $(f(t)⋆g(t))(\tau)=(f(t)∗g(-t))(\tau)$.

Swapnil Agarwal
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