Fourier transform of $g(t)=\begin{cases}1-|t|&|t|\leq1,\\~~~~0&\text{everywhere else}\end{cases}$

Question

I'm having trouble on how I should work with this particular Fourier transform:

$$ g(t)= \begin{cases} 1-|t| & |t|\leq1,\\ ~~~~0 &\text{everywhere else} \end{cases} $$

I understand the integral notation, but I can't seem to reason that any of the Fourier pairs known to me can be used to simplify it.

Thank you in advance!

Answer 1

Let $f$ denote the characteristic function of the interval $[-1/2, 1/2]$. You can easily check that $g$ is the convolution of $f$ with itself: $$g(x) = \int_{-1/2}^{1/2}f(y)f(x-y)\ dy = \int_{-1/2}^{1/2}f(x-y)\ dy$$ Consequently, the Fourier transform of $g$ is the product of the Fourier transform of $f$ with itself: $$\mathcal Fg = (\mathcal Ff)^2$$ It is straightforward to compute $\mathcal Ff$: $$\begin{aligned} \mathcal Ff(u) &= \int_{-1/2}^{1/2}f(x)e^{-iux}\ dx \\ &= \int_{-1/2}^{1/2}e^{-iux}\ dx \\ &= \frac{1}{-iu}(e^{-iu/2} - e^{iu/2}) \\ &= \frac{-2i\sin(u/2)}{-iu} \\ &= \frac{\sin(u/2)}{u/2} \end{aligned}$$ Consequently, $$\mathcal Fg(u) = (\mathcal Ff(u))^2 = \left(\frac{\sin(u/2)}{u/2}\right)^2$$

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Alternatively, you can calculate the Fourier transform directly: $$\begin{aligned} \mathcal Fg(u) &= \int_{-1}^{1}g(x) e^{-iux}\ dx \\ &= \int_{-1}^{1}g(x)\cos(ux)\ dx \qquad \text{since $g\cdot \sin$ is odd} \\ &= 2\int_0^{1}g(x)\cos(ux)\ dx \qquad \text{since $g\cdot \cos$ is even} \\ &= 2\int_0^1(1-x)\cos(ux)\ dx \\ &= 2\int_0^1\cos(ux)\ dx - 2\int_0^1 x\cos(ux)\ dx \\ \end{aligned}$$ The first integral is easy: $$2\int_0^1\cos(ux)\ dx = \frac{2}{u}(\sin(u) - \sin(0)) = \frac{2\sin(u)}{u}$$ The second integral can be evaluated using integration by parts to yield $$2\int_0^1x\cos(ux)\ dx = \frac{2(u\sin(u)+\cos(u)-1)}{u^2}$$ Therefore, $$\begin{aligned} \mathcal Fg(u) &= \frac{2\sin(u)}{u} - \frac{2(u\sin(u)+\cos(u)-1)}{u^2} \\ &= \frac{2(1 - \cos(u))}{u^2} \\ &= \frac{2(2\sin^2(u/2))}{u^2} \\ &= \frac{\sin^2(u/2)}{(u/2)^2} \\ &= \left(\frac{\sin(u/2)}{u/2}\right)^2 \end{aligned}$$ The first proof is quite a bit less work! Exploit properties such as convolution $\mapsto$ product when you can.

Answer 2

You can split absolute value and work with function $$ g(t)= \begin{cases} 1+t & -1\leq t<0 ,\\ 1-t & 0\leq t<1 . \end{cases} $$ for the Fourier transform of $g(x)$ we see by definition \begin{align} {\cal F}(g) &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-i w t}g(t)\,dt \\ &= \frac{1}{\sqrt{2\pi}}\left(\int_{-1}^0 (1+t) e^{-i w t}\, dt + \int_0^1 (1+t) e^{-i w t}\, dt\right) \\ &= \frac{i w+1-e^{i w}}{\sqrt{2\pi}w^2}-\frac{iw-1+e^{-iw}}{\sqrt{2\pi}w^2} \\ &= \dfrac{2}{\sqrt{2\pi}}\dfrac{1-\cos w}{w^2} \end{align}