Convolution of a step function with itself

Question

I have a step function that states $f(x) = 1$ for $|x| < 1$ and 0 everywhere else. So far I've found this to be $$\int_{-\infty}^{\infty}\, dy\, f(y)\,f(x-y)$$ and from boundaries, $$-1 < y <1 \\ x-1 < y < x+1$$ Which now gives the function $$1\cdot\int_{-1}^{1}\, dy\, f(x-y) = \int_{t \,= \,x-1}^{t\, = \,x+1}f(t)\,dt$$ where $t = x-y$. From here I've found that when $x < -2$ or $x + 2$ that $f(t) = 0$ and between these boundaries $f(t) = 1$. I'm stuck with what to do from here, however. I'm unsure how to use the boundary conditions and find a final answer here.

The question also asks for the Fourier transform of this convolution. Is this straightforward or is there I trick to it that I need to be aware of? Thanks.

Answer 1

First of all, the convolution of the function $f$ with itself is given by

$$\begin{align} (f*f)(x)&=\int_{-\infty}^\infty f(y)f(x-y)\,dy\\\\ &=\begin{cases} 0&,x\le -2\\\\ \int_{-1}^{x+1}(1)\,dx=(x+2)&,-2<x<0\\\\ \int_{x-1}^1(1)\,dx=2-x&,x<2\\\\ 0&,x\ge 2 \end{cases} \end{align}$$

We can write $(f*f)(x)=2-|x|$ for $|x|<2$ and $0$ otherwise.

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The Fourier Transform of $f(x)$ is

$$\begin{align} \mathscr{F}\{f\}(k)&=\int_{-\infty}^\infty f(x)e^{ikx}\,dx\\\\ &=\int_{-1}^1 e^{ikx}\,dx\\\\ &=2\frac{\sin(k)}{k} \end{align}$$

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Finally, invoking the Convolution Theorem yields

$$\bbox[5px,border:2px solid #C0A000]{\mathscr{F}\{f*f\}(k)=\frac{4\sin^2(k)}{k^2}}$$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Lets $\ds{\,\mc{F}\pars{x} \equiv \int_{-\infty}^{\infty}\mrm{f}\pars{y}\mrm{f}\pars{x - y}\,\dd y = \int_{-\infty}^{\infty}\hat{\mc{F}}\pars{k}\expo{\ic kx}{\dd k \over 2\pi}}$ such that

\begin{align} \hat{\mc{F}}\pars{k} & = \int_{-\infty}^{\infty}\mc{F}\pars{x}\expo{-\ic kx}\dd x = \int_{-\infty}^{\infty}\bracks{% \int_{-\infty}^{\infty}\mrm{f}\pars{y}\mrm{f}\pars{x - y}\,\dd y} \expo{-\ic kx}\dd x \\[5mm] & = \int_{-\infty}^{\infty}\bracks{% \int_{-\infty}^{\infty}\mrm{f}\pars{y}\mrm{f}\pars{x}\,\dd y} \expo{-\ic k\pars{x + y}}\dd x = \bracks{\int_{-\infty}^{\infty}\mrm{f}\pars{y}\expo{-\ic ky}\dd y} \bracks{\int_{-\infty}^{\infty}\mrm{f}\pars{x}\expo{-\ic kx}\dd x} \\[5mm] & = \hat{\mrm{f}}^{\,2}\pars{k} = \pars{\int_{-1}^{1}\expo{-\ic kx}\dd x}^{2} = 4\pars{\int_{0}^{1}\cos\pars{kx}\,\dd x}^{2} = \bbx{\ds{4\sin^{2}\pars{k} \over k^{2}}} \end{align}