0
$\begingroup$

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)} $

$\endgroup$
0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ No I can not finish it $\endgroup$ – Ralph Jul 28 '17 at 11:09
  • $\begingroup$ 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? $\endgroup$ – T. Linnell Jul 28 '17 at 11:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.