# Inverse Fourier Transform

How can you find the inverse Fourier transform or $f(w) = \frac{\exp(-iw)}{2+iw}$ ?

I started out by using $f(t)= \frac{1}{2\pi} \int f(w)\exp(jwt)\delta$ but I'm not sure how to go about it..

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 What's the "delta" in your integral? – Matt N. Apr 11 '12 at 9:07

Substitute in the given expression. Then we are trying to calculate $$f(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{1}{2 + i\omega} e^{i\omega (t-1)} d\omega.$$

We can shift time to simplify the integrand: $$f(t+1) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{1}{2 + i\omega} e^{i\omega t} d\omega.$$

Now we flip through our favourite book of Fourier transforms and notice the following $$f(t) = e^{-at} H(t); \quad \hat{f}(\omega) = \frac{1}{a + i\omega}.$$

Here $H(t)$ is the Heaviside step function. Excellent! Then clearly $$f(t+1) = e^{-2t}H(t).$$

And as a final step we undo the shift we introduced earlier, $$f(t) = e^{-2t-2}H(t-1).$$

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For the "just the godd*** answer" method, the $exp(-iw)$ just means a delay in the time domain, and then you can probably lookup the inverse transform of $\frac{1}{2+iw}$ (I guess it would be some kind of exponential).

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