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What would be the correct way to write down the convolution in "star" notation for these two functions; $h(t)$ and $\delta(t-x)$. $\delta$ is the Dirac delta function. The integral notation should be

$$ \int_{-\infty}^{\infty}h(t-\tau)\delta(\tau-x)d\tau$$

It feels a little awkward with the delta function in there.

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  • $\begingroup$ As far as I see, this evaluates to $ h(t - x) $, and this is the simplest way you can write it. Am I missing something? $\endgroup$
    – derpy
    Apr 29, 2014 at 16:48
  • $\begingroup$ That is true but I just wanted to know how to write down the notation for the convolution with a dirac delta function not centered at $0$. $\endgroup$ Apr 29, 2014 at 16:49
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    $\begingroup$ Perhaps $ h \star (\delta \circ T_x) $, where $ T_x $ is the translation by $ x $ (that is, $ T_x : \mathbb R \to \mathbb R $, defined by $ \tau \mapsto \tau - x $)? $\endgroup$
    – derpy
    Apr 29, 2014 at 16:50
  • $\begingroup$ Yea that makes sense. I guess there isn't a real convention then. Thanks $\endgroup$ Apr 29, 2014 at 16:53

2 Answers 2

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How about $h* \delta(\cdot - x)$?

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Maybe not as rigorous as the previous comments, but probably a bit nicer in notation: You could define an ensemble of functions $\delta_\tau(t):= \delta(t-\tau)$ and then write the convolution as $h*\delta_\tau$.

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