Convolution of a function with itself using substitution.

Question

f(t) = \begin{cases} 1, & \text{0<=t<=1} \\ 0, & \text{otherwise} \end{cases}

I know how to find the convolution when there are 2 functions but how do I find the convolution of a function with itself? The textbook claims that substitution is a method that can be useful but I don't understand how the boundaries would work with substitution.

EDIT: Convolution is from negative infinity to infinity. convolution function

Answer 1

By definition, $$ f\star f(t) =:g(t)= \int_0^t f(s)f(t-s)\,\mathsf ds = \int_0^t \mathsf 1_{(0,1)}(s)\mathsf 1_{(0,1)}(t-s)\,\mathsf ds. $$ If $t\in[0,1)$, then $t-1<s<t$ and $0<s<1$ imply $0<s<t$, so we compute $$ g(t) = \int_0^t \mathsf ds = t. $$ If $t\in[1,2]$, then similarly we compute $$ g(t) = \int_{t-1}^1 \mathsf ds = 2-t. $$ Hence, $$g(t) = t\cdot\mathsf 1_{[0,1)}(t) + (2-t)\cdot\mathsf 1_{[1,2]}. $$

Answer 2

$f=1_{[0,1]}$.

$f(t-x) = 1_{[0,1]}(t-x) = 1_{[t-1,t]}(x)$.

$(f * f)(t) = \int f(x) f(t-x) dx = \int 1_{[0,1]}(x) 1_{[t-1,t]}(x) dx = \max(0, t-|t-1|)$.

Answer 3

Imagine the following experience :

Consider a stripe of paper that has a square hole punched into it. Do a duplicate of this stripe. Slide one on the other in front of a bulb.

The amount of light that flows through the superposition of the two stripes is exactly the result ("tent" function).

(the holes modelize the characteristic function of interval $(0,1)$ and the sliding the convolution).