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Possible Duplicate:
When convolving two functions that are constants in a region and 0 everywhere else, where does the integration start?

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?

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marked as duplicate by Dilip Sarwate, Lorem Ipsum Mar 23 '12 at 4:42

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ 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. $\endgroup$ – Jason R Jan 30 '12 at 20:52
  • $\begingroup$ 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. $\endgroup$ – Dilip Sarwate Mar 5 '12 at 12:14
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Link to the closed-form solution:

wolframalpha

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

enter image description here

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  • $\begingroup$ Sorry, I don't understand. This isn't explaining my question really.. $\endgroup$ – toast Jan 31 '12 at 13:38
  • $\begingroup$ I thought you might be interested in a closed form solution since the question had explicit values for f, h and the boundaries. If that is completely irrelevant, let me know and I'll just delete my answer. $\endgroup$ – Matthias Odisio Jan 31 '12 at 14:37
  • $\begingroup$ Maybe a plot of the result would help illustrate it. $\endgroup$ – Jason R Jan 31 '12 at 15:39
  • $\begingroup$ @JasonR Thanks, I have added 3D plots. $\endgroup$ – Matthias Odisio Jan 31 '12 at 17:49

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