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..
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..
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, $f(t) = e^{-2t-2}H(t-1).$
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).