Two ways of writing the Fourier transform of the mod function

Question

I feel there are two ways of writing the Fourier transform of $\vert x \vert$ and they are,

• that it is $\pm i \sqrt{2\pi} \delta'(t)$ for $x \geq0$ or $<0$
• that it is $-\sqrt{\frac{2}{\pi}} \frac{1}{t^2}$

(for frequency $t$)

What is the relationship between these two above expressions? Its not clear that they are the same!

Answer 1

Assuming $|x|=x=\text{ramp}(x)$ for $x>0$, then $|x|=\text{ramp}(x)+\text{ramp}(-x)$.

Since $$\mathcal{F}\{f(-x)\}=F(-\omega)$$ we have $$\mathcal{F}\{|x|\}=\mathcal{F}\{\text{ramp}(x)+\text{ramp}(-x)\}=G(\omega)+G(-\omega)$$ where $G(\omega)=\mathcal{F}\{\text{ramp}(x)\}=j\pi\delta'(\omega)-\frac{1}{\omega^2}$

Hence, $$\mathcal{F}\{|x|\}=-\frac{2}{\omega^2}=-\frac{1}{2\pi^2f^2}$$