2
$\begingroup$

I am trying to compute the convolution of two characteristic functions over a Cartesian mesh. First, I define my Cartesian mesh of the interval $[0,1]$ as follows

$$ x_{i} = i \Delta x, i = 0, 1, 2\\ \Delta x = \frac{1}{N}, $$ with $N$ being the number of grid points. Secondly, we can compute the convolution of two characteristic functions analytically, which gives

$$ f(x) = \mathcal{X}_{(-1/2,1/2)}(x) \star \mathcal{X}_{(-1/2,1/2)}(x) = \begin{cases} 0 & x > 1/2\;\text{or}\;x < 1/2 \\ x + 1 & x \in [-1/2,0] \\ 1-x & x \in [0,1/2] \end{cases} $$

When computing $f(x)$ over a mesh, would I calculate the characteristic function like this? $$ f(x_{i}) = \mathcal{X}_{(-1/2,1/2)}(x_{i}) \star \mathcal{X}_{(-1/2,1/2)}(x_{i}) = \begin{cases} 0 & x_{i} > 1/2\;\text{or}\;x_{i} < 1/2 \\ x_{i} + 1 & x_{i} \in [-1/2,0] \\ 1-x_{i} & x_{i} \in [0,1/2] \end{cases} $$

As a side note, I know that the convolution of these two functions provide a B-Spline.

$\endgroup$
1
  • 1
    $\begingroup$ Your formula looks correct -- it is just the evaluation of the previous one at a particular point $x_i$. As a side note, it's not useful to write $f(x)=g(x)\ast h(x)$ because on the right hand side, the $\ast$ is a non-local operation. In other words, it is not an operation that somehow operates at $g$ evaluated at $x$ and $h$ evaluated at $x$. It is better to write it as $f(x)=(g\ast h)(x)$, i.e., you are evaluating the convolution of the two functions at $x$. $\endgroup$ Dec 29, 2019 at 16:42

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.