# Is it possible to simplify the convolution integral if the functions are non-zero in disjoint areas? [duplicate]

I have a function $f(x,y)$ and $h(x,y)$. $f(x,y)$ has a value of $\frac{1}{3}$ when $x$ is between $\frac{1}{3}$ and $\frac{5}{9}$, and $y$ is between $0$ and $1$. The function has a value of $0$ everywhere else. Meanwhile, $h(x,y)$ has a value of 1 when both $x$ and $y$ are between $-\frac{1}{19}$ and $\frac{1}{19}$.

These two functions are non-zero in completely disjoint regions. Is there a way to leverage this property to simplify the convolution integral?

## marked as duplicate by Dilip Sarwate, Lorem IpsumMar 23 '12 at 4:42

• The short answer is yes, as you can use the regions where you know the signal is zero to constrain the limits on the convolution integral. Instead of integrating from $-\infty$ to $\infty$, you can set the limits such that they only cover the regions where the two functions overlap as you slide them across one another. – Jason R Jan 30 '12 at 20:52
• In the other case, the function $f(x,y)$ has value $\frac{1}{3}$ for $|x| < 0.5, 0 < y < 0.4$ and $h(x,y)$ has value $1$ for $-\frac{1}{50} \leq x,y\leq \frac{1}{50}$. Are the questions really all that different? Moderators: please merge the questions. – Dilip Sarwate Mar 5 '12 at 12:14

From left to right: $f$, $h$, and $f*h$: