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I want to convolve two triangles. The equation satisfied by one triangle is

$f(y) = \begin{cases} y + 1 & −1 < y < 0\\ \\ 1 − y & 0 \leq y < 1 \end{cases}.$

So, the overall duration of a triangle is $\;-1\;$ to $\;1$.

What I have found is that the convolution of a triangle with itself is a $\operatorname{sinc}$ function but does one obtain this $\operatorname{sinc}$ from the above equation?

Secondly, what should be the limits of integral for calculating convolution?

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    It might be worthwhile asking the moderators to migrate this question to dsp.SE. With regard to your question about the limits on the integral for calculating convolutions, there is not a single integral that you have to compute, but different integrals depending on your choice for the _argument_ of the _function_ that is the _result_ of the convolution. For a _detailed exposition_ of how to go about convolving two signals, see [my answer to a similar question](http://dsp.stackexchange.com/a/1405/235) on dsp.SE2012-12-22

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We have to compute $g(x):=\int_{-\infty}^\infty f(x-t)\ f(t)\ dt=\int_{-1}^0 f(x-t)\ (1+t)\ dt + \int_0^1 f(x-t)\ (1-t)\ dt$ for given $x\in{\mathbb R}$. Because of the nonanalytic way the function $f$ is defined this is a tedious matter. It's pretty obvious that $g(-x)=g(x)$, and that $g(x)=0$ when $x>2$.

Let an $x\in{\mathbb R}$ be fixed. Then $f(x-t)=\cases{ 0&$(tx+1)$\cr}\quad.$

We distinguish the cases (i): $\ 0< x< 1$, and (ii): $\ 1< x<2$. (The cases $x\in\{0,1,2\}$ will follow by continuity of $g$.)

In case (i) we have $-1 and therefore $g(x)=\int_{x-1}^0 (1-x+t)(1+t)\ dt +\int_0^x (1-x+t)(1-t)\ dt +\int_x^1 (1+x-t)(1-t)\ dt\ ,$ which Mathematica computes to $g(x)={2\over3}-x^2+{x^3\over2}\ .$ In case (ii) we have $0. Therefore the integral $\int_{-1}^0\ldots$ vanishes, and we are left with $g(x)=\int_{x-1}^1 (1-x+t)(1-t)\ dt={4\over3}-2x+x^2-{x^3\over6}\ .$

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Use the formula of convolution from here http://en.wikipedia.org/wiki/Convolution the limits of your integral will be from $-2$ to $2$. For bounded signals in some interval $[a,b]$, the result of the convolution will double the range from $b-a$ to $2(b-a)$.

Your range is $1-(-1)=2$, and you can expect to double it to $2*2=4$ and how is it related to defining the limits of the integral?

The resul of the convolution is not a $sinc$ function. $sinc$ function is not positive everywhere but the result of a convolution of two non-negative functions is always non-negative everywhere.

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    you will not write a single function in the integral. Your function says that the function is $y+1$, if -1 and $1-y$ if 0. Depending on the limits you must use either of them. Suggestion: try to plot the function on a paper.2012-12-22