Ain’t so convoluted
What’s a convolution, you ask? If you have two functions and , the convolution of f and g, is defined as
where u is the dummy variable of integration.
The notation is also used to denote the convolution of f and g, although I’ll be using the former notation. Note that the convolution of f and g is itself a function of x. That is,
Here’s one way to visualize the convolution. If you have two functions f and g with respect to the dummy variable, what the convolution actually does is,
STEP 1: “Flips” one of the functions about the y-axis. For example, to
STEP 2: Shifts it by an amount x. ie, to
STEP 3: “Slides” along the u-axis by keeping u fixed, and allowing x to vary all the way from to . The value of the convolution at some point , is
So, if you consider the convolution integral, the value of the product , and hence is zero when the two functions do not intersect. However, when the two functions do intersect, the value of the convolution at that point will be the integral of the product over the entire overlapping region (where the product is non-zero), and this value is simply the area of the overlapping region.
You can see detailed animated illustrations of this idea at Wolfram’s MathWorld.
Another useful way of thinking about the convolution of two functions, is by the concept of a functional. We can consider as a functional of the function . That is, for every given function , there will be a corresponding value of . So, to calculate the value of at some point , we still need to know the entire function .
There are useful properties associated with convolution. For example, convolutions are
So, it doesn’t matter which function you flip.
The convolution theorem is immensely useful to calculate fourier transforms and find fourier pairs. It states that, if and ,
It’s an interesting result that the fourier transform of the convolution of two functions, is the product of the corresponding fourier transforms of the individual functions. This means that a convolution in the normal domain, becomes a product in the fourier domain and vice-versa
It’s actually easy to prove this useful result. For, if the convolution of two functions is given as
Then the fourier transform of h(x) is
Since x is a dummy variable, set , while holding u constant, so that the integral reduces to
We can apply this theorem to find various convolutions. Here’s an example.
(i) Convolution with a delta function
We’re now interested in the convolution of f with a shifted delta function,
To find this, let’s apply the convolution theorem. We know that and . Therefore
And taking the inverse fourier transform of , we find that is . Notice that f(x) is shifted by an amount a.
Finally, I’ll discuss one more property I’ll need to use before I move on to a part of what I’m currently working on, which is the application of fourier analysis in interferometry and diffraction. This property is the derivative of fourier transforms.
which you can easily check this, by use of the Leibniz rule. This property of fourier transforms is useful in solving linear PDE’s.