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Hoi, I want to have the inverse fourier transform $\mathcal{F}^{-1}(\frac{1}{1+s^2})$.

So I thought about using some properties of fourier-transform. But knowing the answer I must make some sort of mistakes in my reasoning, but i dont understand what im doing wrong:

I know the answer is : $$ce^{-|x|}$$ and according to wolframalpha $c= \sqrt{\pi/2}$.

But i got this: Some calculations give:

$$\frac{1}{1+s^2} = \frac{1}{1-is}\cdot \frac{1}{1 +is} = \mathcal{F}(H(t)e^{-t})\cdot \mathcal{F}(H(-t)e^{t}) = \mathcal{F}[H(t)e^{-t}\ast H(-t)e^{t}] $$

That is according some properties of Fourer transform: $F(g \ast f) = F(g)F(f)$

So that would then imply the answer is $$H(t)e^{-t}\ast H(-t)e^{t}$$

But that doesnt give me the right answer...what is my big error. I get calculating this convolution: $\frac{1}{2}e^{-x}$

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    $\begingroup$ Is it $ \frac{1}{1+s^2} $ or $ \frac{1}{(1+s)^2} $? $\endgroup$ Commented Apr 1, 2013 at 13:34
  • $\begingroup$ $1/(1+s^2)$. Sorry. Otherwise none of what I said wouldve made sense :P $\endgroup$
    – DinkyDoe
    Commented Apr 1, 2013 at 13:47
  • $\begingroup$ You can use residue theorem. $\endgroup$ Commented Apr 1, 2013 at 18:56
  • $\begingroup$ for what exactly do i use it then. In line of the reasoning above I wouldnt need it, so what is my error? $\endgroup$
    – DinkyDoe
    Commented Apr 1, 2013 at 19:30
  • $\begingroup$ The answer should be $ \frac{1}{2}e^{-t}H(t) + \frac{1}{2}e^{t}H(-t).$ $\endgroup$ Commented Apr 2, 2013 at 14:15

5 Answers 5

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The Solution can be easily obtained by using Leibniz differentiation under integral sign, and obtaining a differential equation. Solving the D.E gives an equation with constants $C_1$ and $C_2$, or here they are equal hence $C$. This $C$ is same as the "c" obtained from wolframalpha ,

$Ce^{-\left | x \right |}$ , giving $C= \sqrt{\frac{\pi}{2}}$ , hence $$ F^{-1}\left(\frac{1}{1+s^2}\right) = \sqrt{\frac{\pi}{2}}e^{-\left | x \right |}\, . $$

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  • $\begingroup$ Thnx for the edit #zerothehero $\endgroup$
    – Vivek Adi
    Commented Dec 17, 2017 at 17:23
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In your calculation of the convolution I'm sure that at some point you get a $\sqrt{x}^2$, and you're saying that's equal to $x$, when it's actually $|x|$, that is what that answer is telling you. It's a very common mistake, check you're calculations agains to see if that is the problem.

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  • $\begingroup$ ok, do u see a problem with my reasoning here then? Cause im almost sure the convolution itself is the right answer. But see under here, what mistake do I make then... $\endgroup$
    – DinkyDoe
    Commented Apr 1, 2013 at 12:38
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There should be a mistake in your computation of convolution, because the map $t\mapsto H(t)e^{-t}\star H(-t)e^{t}$ is even (convolution is commutative).

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  • $\begingroup$ Im sorry: meant to say $H(t)e^{-t}\star H(-t)e^{t}$ $\endgroup$
    – DinkyDoe
    Commented Apr 1, 2013 at 12:28
  • $\begingroup$ So, there should be a mistake. I got that. But is my reasoning correct to conclude that $H(t)e^{-t}\star H(-t)e^{t}$ is the right answer? And if so...see under here. What am I misunderstanding here? $\endgroup$
    – DinkyDoe
    Commented Apr 1, 2013 at 12:57
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In your Calculation, x shouldn't be the lower border in the integral. It must be H(x)*x. Because, if x < 0, the integral is cut, then the lower border is 0. And in the end it is (x-2H(x)*x) = -|x| ... :)

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It has been a while, but since no answer was given, I decided to explain the mistake. The problem is indeed with the convolution.

You must take into consideration the cases $x>0$ and $x<0$. For the first case, the convolution has $$H(x)e^{-x}\ast H(-x)e^x =\frac{e^{-x}}{2} = \frac{e^{-|x|}}{2},$$ as an answer, while for the case $x<0$ you have $$H(x)e^{-x}\ast H(-x)e^x =\frac{e^x}{2} = \frac{e^{-(-x)}}{2} = \frac{e^{-|x|}}{2}.$$

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