Fourier Transform of 1-|x|

Question

This should be really simple, I'm baffled by the fact that I just cannot get it to work. I'm either overthinking it, or just tired.

I'd like to compute the Fourier Transform of the following function:

$$ f(x) = \begin{cases} 1-|x|, & \text{If $|x|\leq 1$} \\ 0, & \text{If $|x|>1$} \end{cases}$$

I've tried the "obvious" approach of splitting the function into 2 integrals. The F.T. of 1 would be the delta function times $\sqrt{2\pi}$. I also know the F.T. of $|x|$. The answer however is supposed to look like this:

$({sin(\pi k)\over{\pi k}})^2$

Now that I think of it, it's actually the sinc function squared. Which I think would be obtained if we use convolution on 2 rectangular functions. I probably know how to do it now, but I've typed everything out so far, and it's 3am. Heading to bed.

Edit: Convolution of 2 rectangular functions is a triangular function if I'm not mistaken. I sketched out f(x) and it indeed looked like a triangle. I think I just suck at functions involving absolute values, never found a good way to deal with them. I'll try the approach of splitting the integral into 2 sets of boundaries later.

Answer 1

To verify your intuition (without doing integral "now"): By $\text{Rect(t)}*\text{Rect}(t)=\text{Tri(t)},$ $$\mathscr{F}(\text{Tri(t)})=\mathscr{F}(\text{Rect(t)}*\text{Rect}(t))=\mathscr{F}(\text{Rect}(t))\cdot\mathscr{F}(\text{Rect}(t))=\text{sinc}(f)\cdot\text{sinc}(f)=\text{sinc}^2(f)$$

The second equality follows from the property of Fourier Transform. Therefore you're correct.

Answer 2

\begin{align} \hat f(k) &= \int_{\mathbb R}e^{-2\pi ikx}f(x)\ \mathsf dx\\ &= \int_{-1}^0 (1+x)e^{-2i\pi kx}\ \mathsf dx + \int_0^1 (1-x)e^{-2i\pi kx}\ \mathsf dx\\ &= \frac{1+2 i \pi k-e^{2 i \pi k}}{4 \pi ^2 k^2} + \frac{1-2 i \pi k-e^{-2 i \pi k}}{4 \pi ^2 k^2}\\ &= \frac{2-e^{2 i \pi k}-e^{-2 i \pi k}}{4 \pi ^2 k^2}\\ &= \left(\frac{\left(e^{i\pi k}- e^{-i\pi k} \right)/2i}{\pi k}\right)^2\\ &= \left(\frac{\sin\pi k}{\pi k}\right)^2. \end{align}