Function $\phi (x)$ is defined as:
$$\phi(x) = \begin{cases} 1 & \text{ if } 0 \leq x \leq 1\\0 & \text{otherwise} \end{cases} $$
How do I find the convolution of $\phi(x)$ with itself? I tried to take the Fourier transform of $\phi(x)$ and square it, then take the inverse Fourier transform. However, in the latter step I couldn't figure out what the limits on the integral are (I think it would be integral in all space, but I'm not sure). More importantly, I don't know how to do that integral for the inverse Fourier step. Thank you!
Fourier Analysis – Convolution of a Function with Itself
convolutionfourier analysis
Related Solutions
First, since $f(\tau)=0$ outside of $[-1,1]$, we have $$ f\ast f(t)=\int_{-1}^1f(t-\tau)d\tau. $$
Now change the variable by $u=t-\tau$: $$ f\ast f(t)=\int_{t-1}^{t+1}f(u)du. $$
If $t+1\leq -1$ (ie $t \leq -2$), or if $t-1\geq 1$ (ie $t\geq 2$), then $(-1,1)\cap (t-1,t+1)=\emptyset$ so $f\ast f(t)=0$.
Let us consider the case $-2<t<2$ now.
First, if $-2<t\leq 0$, then $1+t\leq -1$ and $[t-1,t+1]\cap [-1,1]=[-1,t+1]$ so $$ f\ast f(t)=\int_{-1}^{t+1}1du=2+t. $$
Second, if $0\leq t<2$ then $1-t\geq 1$ and $[t-1,t+1]\cap [-1,1]=[t-1,1]$ so $$ f\ast f(t)=\int_{t-1}^{1}1du=2-t. $$
Use the original definition: $$\space f*f=\int_{-\infty}^{+\infty} f(y)f(x-y) dy$$ When $|y|>-1$, $f(y)=0$, otherwise it is $\frac{1}{2}$. So the integral is reduced to $$\frac{1}{2}\int_{-1}^{1} f(x-y) dy$$
Now substitute $t=x-y$,
$$\frac{1}{2}\int_{x-1}^{x+1} f(t) dt$$
It is easy to see when $x\leq -2$ or $x\geq 2$, $f(t)$ is entirely $0$, so this function is equal to $0$ in those intervals. Between $-2$ and $2$, $f(t)$ is $\frac{1}{2}$ in the green region:
In this picture, the upper boundary of the green region is $y=x+1$, the lower boundary is $x-1$. The height of the green region is the distance we want. So you can see that the integral is equal to $$(x+1)-(-1)=x+2, \quad\quad -2\leq x\leq 0\\ 1-(x-1)=2-x, \quad\quad 0\leq x\leq 2$$
And this is the required function after multiplying by $\frac{1}{4}$.
Best Answer
Simply apply the definition. So $$\phi * \phi (x) = \int_{-\infty}^{+\infty} \phi(x-y)\phi(y) \ dy$$ The integrand is $\neq 0$ only in the case that $y, x-y \in [0,1]$, i.e. $$0 \leq y \leq 1$$ $$x-1 \leq y \leq x$$
In the case that $x \leq 0$ or $x \geq 2$, the integrand must be $0$, so $\phi * \phi (x) = 0$.
In the case that $0 \leq x \leq 1$ you have
$$\phi * \phi (x) = \int_{0}^{x} \ dy = x$$
In the case that $1 \leq x \leq 2$ you have
$$\phi * \phi (x) = \int_{x-1}^{1} \ dy = 2-x$$
So
$$\phi * \phi (x) = \begin{cases} x & \text{ if } 0 \leq x \leq 1\\ 2-x & \text{ if } 1 \leq x \leq 2\\ 0 & \text{ otherwise} \end{cases} $$