Express a function as convolution of two function - How to get intuition for the convolution

Question

$$x\left(t\right)=\begin{cases}1-t&0\le t\le 1\\ 1+t&-1\le t\le 0\\ 0&else\end{cases}$$ How can I understand that: $$x(t)=h(t)*h(t)$$ when: $$h\left(t\right)=\begin{cases}1&-\frac{1}{2}\le t\le \frac{1}{2}\\ 0&else\end{cases}$$ I am studying to a test and can not really understand the intuition to how I see the convolution from a function. ( it was a question to find Fourier Trasnform, I can do it without convolution, but I wanted to know for me how to get intuition for it )

Answer 1

You do that using the explicit formula -

$x(t) = \int h(s)h(t-s)ds $

Since $h$ is simply an indicator function of the interval $[-1/2, 1/2]$, then $h(s)h(t-s)$ is also an indicator function! It is an indicator of the intersection between the intervals $[-1/2, 1/2]$ and $[t-1/2, t+1/2]$ (both of length $1$). This is the set ${ -1/2 \le s , t-s \le 1/2} $, which can be re-written as

${ max(-1/2, t-1/2) \le s \le min(1/2, t+1/2) }$

as long as $t-1/2 \le 1/2$ (the left endpoint of the first interval is less than the right endpoint of the second) and $-1/2 \le t+1/2$ (and vice versa) - i.e., $-1\le t\le 1$. Otherwise the intervals are disjoint and their intersection is the empty set (which its indicator's integral is 0).

If $t \le 0$, the LHS is $-1/2$, and if $t \ge 0$, the RHS is $1/2$.

To sum up, we got that the integrand is the indicator function of the interval -

$[-1/2, t+1/2]$ for $t \le 0$

$[t-1/2, 1/2]$ for $t \ge 0$

The integral of an indicator is simply the length of the interval, which is $1+t$ in the first case, and $1-t$ in the second case. To sum up, we got the exact answer you have written down.

Intuitively, I like to think of a convolution by looking at the formula directly - For each $t$, I'm multiplying $h(s)$ and $h(t-s)$, thinking of one as a "slider" of the values of $h$ starting from $0$ and going to the right, and the other as a "slider" starting from $t$ and going to the left. Looking at this graphically makes sense in my opinion.