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## Homework Statement

Compute the convolution ##(f*h)(t)## where

$$f(t) = \left\{\begin{matrix}1, \ \ for \ \ |t|<1 \\ 0, \ \ \ \ otherwise \end{matrix}\right.$$

and

$$h(t) = \left\{\begin{matrix}2|t|-1, \ \ for \ \ |t|<1/2 \\ 0, \ \ \ \ otherwise \end{matrix}\right.$$

## Homework Equations

Convolution integral: ##(f*h)(t)= \int^\infty_{-\infty} f(\tau) h(t-\tau) d \tau##

## The Attempt at a Solution

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Here I first made a sketch of the two functions:

f(t) is non-zero on [-1, 1] and g(t) is non-zero on [-0.5, 0.5], so adding these two we find that (f ∗ g)(t) should be nonzero on [-1.5, 1.5]. The list of change points of (f ∗ g)(t) is {-1.5, -0.5, 0.5, 1.5}. For values t that are outside this range the convolution is zero.

So, do I need to calculate the convolution for 3 separate cases?

I mean ##-1.5 \leq t \leq -0.5##, ##-0.5 \leq t \leq 0.5##, and ##0.5 \leq t \leq 1.5##?

Here is my attempt:

**First case:**we want to integrate from -1.5 to -0.5 using the formula given above (##(f*g)(t)=\int^{0.5}_{-1.5} f(\tau) h(t-\tau) d \tau##). ##h(t)## is zero. But for ##f(t)## the function is zero on the interval ##(-1.5, -1)##, whereas it is equal to 1 from ##(-1, -0.5)##, so what value do I need to be using for the function?

Any help is greatly appreciated.