Question

Find the inverse fourier transform of 6e^(4iw) * cos(2w) / 9+w2

Answer 1

Just to clarify, that's
\[\mathcal{F}^{-1}\left\{\frac{6e^{4i\omega}\cos(2\omega)}{9+\omega^2}\right\},\]
right?

Answer 2

Yes

Answer 3

Eh, it's been a while... By the definition of the inverse transform, you have
\[f(t)=\frac{1}{2\pi}\int_{-\infty}^\infty F(\omega)e^{i\omega t}~d\omega\]
Substituting the function,
\[f(t)=\frac{1}{2\pi}\int_{-\infty}^\infty \frac{6e^{4i\omega}\cos(2\omega)}{9+\omega^2}e^{i\omega t}~d\omega\]
Any idea where to go from here? I know I have a table of transforms somewhere.

Answer 4

http://www.gatsby.ucl.ac.uk/~aguez/tn1/additional/general/FourierTransformPairs.pdf

Answer 5

i got way different answers in wolfram, everyone comes up with something different lol

Answer 6

Is this the answer you're getting in WolframAlpha?
http://www.wolframalpha.com/input/?i=InverseFourierTransform%5B6*Exp%5B4*i*x%5D*Cos%5B2*x%5D%2F%289%2Bx%5E2%29%2Cx%2Ct%5D

Answer 7

this is it: http://www.wolframalpha.com/input/?i=InverseFourierTransform [6*e^%284*i*w%29*Cos[2*w]%2F%289%2Bw^2%29]

Answer 8

This might help with the integral:
\[\begin{align*}\cos(2x)&=\cos^2x-\sin^2x\\
&=\large\left(\frac{e^{ix}+e^{-ix}}{2}\right)^2-\left(\frac{e^{ix}-e^{-ix}}{2}\right)^2\\
&=\large\frac{e^{2ix}+e^{-2ix}}{2}
\end{align*}\]
So in the integral, you have
\[\large f(t)=\frac{6}{4\pi}\int_{-\infty}^\infty \frac{\left(e^{6i\omega}+e^{2i\omega}\right)}{9+\omega^2}e^{i\omega t}~d\omega\]
Split up the integral into two:
\[\large f(t)=\frac{3}{2\pi}\left[\int_{-\infty}^\infty \frac{e^{6i\omega}}{9+\omega^2}e^{i\omega t}~d\omega+\int_{-\infty}^\infty \frac{e^{2i\omega}}{9+\omega^2}e^{i\omega t}~d\omega\right]\]
Now, that table in the link you gave says
\[\large \mathcal{F}^{-1}\left\{\frac{2\alpha}{\alpha^2+\omega^2}\right\}(t)=e^{-\alpha|t|}\]
and that
\[\large \mathcal{F}^{-1}\left\{F(\omega)e^{-i\omega t_0}\right\}(t)=f(t-t_0)\]
I'll continue this in another post. Computer's starting to lag...

Answer 9

I'm going to go backwards a bit. Forget all the coefficient-factoring I did earlier. I'll work with the first integral:
\[\large f(t)=\frac{1}{2}\frac{1}{2\pi}\int_{-\infty}^\infty \frac{6e^{6i\omega}}{9+\omega^2}e^{i\omega t}~d\omega\]
I'm not sure how to deal with the coefficients without directly integrating ... but in any case, you immediately get something of the form
\[\huge e^{-3|t-6|}\]
as a result.

Similarly, the second integral gives you something of the form
\[\huge e^{-3|t-2|}\]
so both agree somewhat with Wolfram's answer.