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$$H(\Omega)=\begin{cases}\exp(-j \pi/2) ,\;&\Omega >0 \\ \exp(j\pi/2) ,\;&\Omega<0.\end{cases}$$

How can I find the inverse fourier transform of this $(h(t))$ using fourier properties?

Fourier Transform Used: $$X(\Omega)\equiv\int_{-\infty}^{\infty}x(t)\,e^{-j\Omega t}\,dt.$$

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    $\begingroup$ Is this a piece-wise defined function? $\endgroup$ – Adrian Keister Nov 19 '18 at 16:00
  • $\begingroup$ @AdrianKeister yeah i had an issue with the format $\endgroup$ – Prestyy Nov 19 '18 at 16:01
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    $\begingroup$ Ok, there's a better way to typeset: use the cases environment. $\endgroup$ – Adrian Keister Nov 19 '18 at 16:03
  • $\begingroup$ Which definition of the Fourier Transform and Fourier Inverse Transform are you using? (There are at least three, maybe more.) $\endgroup$ – Adrian Keister Nov 19 '18 at 16:04
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    $\begingroup$ @Prestyy: I'm sorry, but that's not what I was asking. You need to type up the exact formula for the Fourier Transform that you're using, and add that to your question body. There are at least three different ways of handling the constants that multiply the integrals in the definitions. $\endgroup$ – Adrian Keister Nov 19 '18 at 16:12
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First, to simplify notation a bit, we notice that, since $e^{j\pi/2}=j,$ it follows that $$H(\Omega)=-j\operatorname{sgn}(\Omega). $$ In looking at the wiki table of Fourier Transforms, in the non-unitary, angular frequency column (which corresponds to the definition you're using), we see that the tricky part here is to note that $$\mathcal{F}^{-1}(-j\pi\operatorname{sgn}(\Omega))=\frac1t,$$ so that $$\mathcal{F}^{-1}(-j\operatorname{sgn}(\Omega))=\frac{1}{\pi t},$$ since the Inverse Fourier Transform is linear. However, this does not take into account the situation in which $t=0$. Technically, your $H(\Omega)$ is undefined there, unless you made a typo in your definition and one of the inequalities is non-strict.

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