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Find the inverse Fourier Transform of

$$ { F(\omega)=\frac{1}{2\pi(a+j\omega)^2} \ } $$ using the convolution theorem. Hint: the Fourier Transform of $e^{-at} u(t)=\frac{1}{\sqrt{2\pi}(a+j\omega)} $

Ralph
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The convolution theorem gives us: $$\mathcal{F}^{-1}[\mathcal{F}(f)\cdot\mathcal{F}(g)]= f*g$$

From the hint: $$F(\omega) = \mathcal{F}(e^{-at}u(t))\cdot\mathcal{F}(e^{-at}u(t))$$ and you seek $\mathcal{F}^{-1}[F(\omega)]$.

Can you finish it?

Tez LaCoyle
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  • No I can not finish it – Ralph Jul 28 '17 at 11:09
  • Use the convolution theorem to get: $\mathcal{F}^{-1}[F(\omega)] = \mathcal{F}^{-1}[\mathcal{F}(e^{-at}u(t))\cdot\mathcal{F}(e^{-at}u(t))] = (e^{-at}u(t))*(e^{-at}u(t))$ And the convolution is then: $\int_{-\infty}^{\infty}e^{-a(t-\tau)}u(t-\tau)e^{-a\tau}u(\tau)d\tau$

    Does this help?

     – Tez LaCoyle Jul 28 '17 at 11:42