How does one derive the Fourier transform of the Ramp function?

Question

One approach could have been to see that the Ramp function ( http://mathworld.wolfram.com/RampFunction.html ) is the convolution of $2$ Heavisides (at $0$). Hence its Fourier transform should have been the product of the Fourier transforms of Heavisides. The Fourier transform of the Heaviside (http://mathworld.wolfram.com/HeavisideStepFunction.html) is, $\frac{1}{2} [\delta(t) - \frac{1}{\pi t} ] $. But its not clear to me as to how its square is the Fourier transform of the Ramp at $0$ which is $\frac{i}{4\pi} \delta'(t) - \frac{1}{4\pi^2 t^2} $

• I would otherwise like to see a reference (or if someone can type in!) which derives the Fourier transform of the ramp function from scratch!

Answer 1

"Frequency derivative" is a property of Fourier transform which is: $$\mathcal{F}\{x(f(x)\}=j\frac{d}{d\omega}F(\omega)$$

Plug $f(x)=u(x)$ (i.e. heaviside function) whose FT is $F(\omega)=\pi\delta(\omega)-\frac{j}{\omega}$.

Since $\text{ramp}(x)=xu(x)$ we get

$$\mathcal{F}\{\text{ramp}(x)\}=j\frac{d}{d\omega}\left(\pi\delta(\omega)-\frac{j}{\omega}\right)=j\pi\delta'(\omega)-\frac{1}{\omega^2}$$

If you want to represent it versus $f$, since $\omega=2\pi f$ it becomes

$$\mathcal{F}\{\text{ramp}(x)\}=(j\pi)\frac{1}{(2\pi)^2}\delta'(f)-\frac{1}{4\pi^2f^2}=\frac{j}{4\pi}\delta'(f)-\frac{1}{4\pi^2f^2}$$

Answer 2

If I am right Fourier analysis is applicable only to bounded or finite integral or stable system (Dirichlet Conditions). And ramp function is not a bounded signal. So we can't apply Fourier transform for ramp function.