At a high-level Wikipedia states: "A convolution between two functions produces a third expressing how the shape of one is modified by the other."

But there are clearly many ways of combining functions to get a third one. A convolution is a specific type of such combination; one that requires reversing and shifting one of the operands and that combines them with a product and an integral to generate the output. While not very complex algebraically, the operation itself is somewhat "convoluted" (pun intended).

Why are convolutions noteworthy? What physical phenomena can be explained mathematically as a convolution?

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – David Z
    May 31, 2020 at 6:52
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    $\begingroup$ +1 for the nice question and pun $\endgroup$ May 31, 2020 at 16:26

2 Answers 2



The lesson here is that graphical intuition isn't always the best choice. For example, you can say that the intuition for derivatives is that they're slopes, and for integrals is that they're areas. But why would "slopes" be that useful in physics? I mean, besides inclined planes, you don't see that many literal slopes in a physics class. And why "areas"? We live in 3D space, so shouldn't volumes be more important?

In case that all sounds dumb, the point is that sometimes, you can actually make something less physically intuitive by explaining it visually, because usually the visual explanation is completely devoid of the dynamic context that would be present in a real physics problem.

One of the main reasons derivatives and integrals appear so often in introductory physics is that they're with respect to time, so the derivative means "a rate of change" and the integral means "an accumulation over time". This is a distinct intuition from the geometric one. The point of the geometric intuition is to help you see what the derivative and integral is given a graph, but it doesn't really help you interpret what it physically does.

Similarly, there is a complicated geometric intuition for the convolution, which can hypothetically help you eyeball what the convolution of two graphs of functions would look like. But in this case the "dynamic" intuition is much simpler.

One piece of intuition

Convolutions occur whenever you have a two-stage process where the stages combine linearly and independently.

Suppose that I kick an initially still mass on a spring at time $t = 0$, and the subsequent trajectory of the spring is $x(t)$. If I apply the same kick at time $t = 1$, then by time translational invariance, the subsequent trajectory is $x(t-1)$. Now suppose that I kick both at $t = 0$ and $t = 1$, with strengths $f(0)$ and $f(1)$. Then by linearity, the subsequent trajectory is $$f(0) x(t) + f(1) x(t-1).$$ This is like the "reversing and shifting" structure of a convolution. So more generally, if I apply a continuous force $f(t)$, then the trajectory is $$\int dt' \, f(t') x(t-t')$$ which is precisely a convolution. This is where a large fraction of the convolutions in physics and electrical engineering come from. (A large fraction of the remainder come from the fact that the Fourier transform of the convolution is the product of the Fourier transforms, and products are simple and ubiquitous.)

  • $\begingroup$ Thanks. What exactly is the product of $f(i)$ and $x(j)$ supposed to represent? I assume $x$ is location, but if $f$ is a force, what do you get as their product? Perhaps you meant "force" more figuratively (?). Just trying to fully understand that operation since it looks like you provided a great example! $\endgroup$
    – Josh
    May 30, 2020 at 18:43
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    $\begingroup$ @Josh Yeah, I was a little wonky with the dimensions to keep things simple. If you want, think of $x(t)$ as really standing for "displacement per unit impulse", then the integral has dimensions of position . $\endgroup$
    – knzhou
    May 30, 2020 at 18:55
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    $\begingroup$ @Josh There's a lot to convolutions; try a lookup: "convolution intuition". Here's a nice one with animations. To complement knzhou's take - the key is visualize correctly, rather than not at all. An important step is to identify the system as a state (e.g. image) or a process (e.g. signal), where latter is sequence set of former - and adjust viz methods and their interpretations accordingly. $\endgroup$ May 31, 2020 at 3:21

Whenever I think of a convolution I imagine a moving average: Suppose we have a function $f(x)$ and we like to calculate the moving average using the weight function $w(x)$. What we calculate is $$ \bar{f}(x_0) = \int w(x) f(x_0-x) dx $$
which is a convolution. A physical example can be found e.g. in optics: Suppose we have a structured surface, which is described by the function $f(x)$. We observe this surface using a microscope, which has a finite resolution. Thus, instead of seeing the function $f(x)$, each point of the surface is convoluted (=averaged) with the weight function $w(x)$.

Also note the close relation between the convolution and the cross-correlation function.


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