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I ran into a problem, which wants to find the convolution $\delta (3-t) * \delta (t-2)$ and I am stuck. How can I approach it?

Andy Walls
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Mitsos
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    Hint: $f(t)*\delta(t-a) = f(t-a)$ – Andy Walls Oct 13 '18 at 01:56
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    Here's a non-rigorous but intuitively useful way to think about it. If two independent real-valued random variables have probability density functions $f$ and $g$, then the sum of those random variables has density function $fg$. The delta function $\delta(t-a)$ can be (non-rigorously) considered the density function of a random variable that takes the value $a$ with probability $1$. So $\delta(t-a)\delta(t-b)$ "should" be $\delta(t-(a+b))$. – Andreas Blass Oct 13 '18 at 02:14
  • @AndyWalls So because δ is an even distribution then δ(3-t) = δ(t-3). And δ(t-2) = δ(t-3-(-1)) so if x = t-3 we have δ(x)*δ(x-(-1)) . And because of your hint we have δ(x+1)=δ(t-2)? Because from Andreas' comment the answer is δ(t-5) – Mitsos Oct 13 '18 at 03:17
  • I though it was a somewhat straightforward $\delta(3-t)*\delta(t-2)=\delta(3-(t-2))=\delta(5-t)$. Conceptually, convolving a target function with a shifted delta function just shifts the origin of the target function to the position of the impulse. – Andy Walls Oct 13 '18 at 08:46

2 Answers2

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It helps to look at the definition of a convolution: Given two functions $f$ and $g$ (assumed from $\mathbb R$ to $\mathbb R$), the convolution $f*g$ is defined as $$(f*g) (t) = \int\limits_{-\infty}^\infty f(t-y) g(y) \,\text d y\,.$$

In your case, $f(t)=\delta(3-t)$ and $g(t)=\delta(t-2)$, so we have $$(f*g) (t) =\int \delta(3-(t-y)) \,\delta(y-2) \,\text{d} y\,.$$ (You can switch which function you call $f$ and which one $g$ without changing the result, since the convolution is symmetric.) , Now the essential property of the $\delta$ funcion is that $\int f(y) \delta(y-a)\,\text d y=f(a)$, so the result is $$(f*g) (t) =\int \delta(3-(t-y)) \,\delta(y-2) \,\text{d} y =\delta(3-t+y)\big|_{y=2}=\delta(5-t)\,.$$

Toffomat
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You could do it using the Laplace transform and the convolution theorem for Laplace transforms. The Laplace transform of a Dirac delta is $$\mathcal{L}(\delta(t-a)) = e^{-as}$$ and the convolution theorem states that $\mathcal{L} ((f*g)(t)) = \mathcal{L}(f(t))\mathcal{L}(g(t))$, so you can multiply the Laplace transforms of your deltas and then take the inverse. There is likely a more direct method though.

Kaleb R.
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