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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!
gradstudent
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  • hint: The derivative of the ramp (vs. $t$) is a step function ( multiplied by the steepness of the ramp). – G Cab Sep 09 '16 at 16:34
  • added: the ramp is the convolution of a forward step (from $0$) with a backward step (from $t$). – G Cab Sep 09 '16 at 16:38
  • One big problem with the way that you want to use (square of the Fourier transform of the Heaviside function) is that the square of the Dirac delta is ill defined. Compare http://mathoverflow.net/q/48067/39447. This does not mean that this way is wrong, it is just very difficult to prove the equivalence of the result with that of a different approach. – bers Sep 12 '16 at 00:39
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    3 years have elapsed since you asked your question. You should validate one of the 2 good answers you have had. This is in this way that this site can live (I have been looking at your question because I wanted to give a reference in my answer to this recent question (https://math.stackexchange.com/q/3303146)). – Jean Marie Jul 25 '19 at 17:44

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In fact the convolution theorem is a bit more complicated than you think for distributions. To apply it you need to approximate your distributions with something like functions $\in L^1$. So $$1_{ x > 0} = \lim_{a \to 0^+} e^{-ax} 1_{x > 0}$$

$$x 1_{ x > 0} = \lim_{a \to 0^+} e^{-ax} 1_{x > 0} \ast e^{-ax} 1_{x > 0}$$

$$\mathcal{F}[x 1_{ x > 0}] = \lim_{a \to 0^+} \mathcal{F}[e^{-ax} 1_{x > 0} \ast e^{-ax} 1_{x > 0}]$$ $$ = \lim_{a \to 0^+} \mathcal{F}[e^{-ax} 1_{x > 0} ]^2$$ $$ = \lim_{a \to 0^+} \frac{1}{(a+2i \pi \xi)^2}$$ $$ = \lim_{a \to 0^+} \frac{-1}{4\pi^2} \frac{1}{(\xi-ia)^2}$$ $$ = \lim_{a \to 0^+} \frac{1}{4\pi^2} \frac{d^2}{d\xi^2} \log(\xi-ia)$$ $$ = \lim_{a \to 0^+} \frac{1}{4\pi^2}\frac{d^2}{d\xi^2}( -i\pi 1_{\xi < 0} +\log|\xi|)$$ $$ = \frac{i}{4\pi} \delta'(\xi)- \frac{1}{4 \pi^2 } fp(\frac{1}{\xi^2})$$ where $fp(\frac{1}{\xi^2})$ is the finite part the second derivative of $-\log |\xi|$

reuns
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"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

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

msm
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  • Thanks but I wanted to use the convolution picture. The ramp is the convolution of two Heavisides and hence FT of the ramp should be the product of the FTs of the two Heavisides. How do I use this picture to derive the FT of the ramp? – gradstudent Sep 11 '16 at 17:32
  • A ramp function can also be written as, $ (x-b)\theta (x - a)$ and one could compute the integral, $FT (x-b)\theta (x - a) = \int_a ^\infty e^{ixt} (x-b) dx$ by using an $\epsilon$ regulator as, $lim_{\epsilon \rightarrow 0} [\int_a ^\infty e^{i(x+i\epsilon)t} (x-b) dx ] $. This method does not seem to produce the $\delta'$ term. Can you kindly explain why this is being missed? – gradstudent Sep 11 '16 at 18:07
  • @gradstudent "I would otherwise like to see ... the Fourier transform of the ramp function from scratch" – msm Sep 11 '16 at 22:07
  • Could you kindly explain (1) the error in my epsilon argument and (2) the FT of the Heaviside function and (3) How the convolution argument is to be used? – gradstudent Sep 12 '16 at 13:38
  • I would mention it is the principal value of $1/f$ and the finite part of $1/f^2$, since we are considering distributions. – reuns Jul 14 '17 at 09:51
  • @gradstudent I tried an answer – reuns Nov 20 '18 at 02:33
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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.

Raghu
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