How to calculate the Fourier transform of $\frac{1}{(t-2\pi)^2} e^{2jt} \sin^2 t$? In an exam I was asked to calculate the Fourier transform of $$\frac{1}{(t-2\pi)^2} e^{2jt} \sin^2 t$$
I've spent a lot of time trying to figure out which combination of properties use but I can't get to a good solution. 
I've tried to convert the $\sin^{2}$ to exponentials using Euler formulas and then tried to simplify all the exponentials but I don't get a result I can transform.
I also tried using properties frequency shift and inverse but it doesn't work for me.
I would be grateful if someone could tell me some tips on how to solve the exercise.
 A: Observe that $\sin(t)=\sin(t-2\pi)$ and then
$$
\mathrm e^{j2t}\frac{\sin^2(t)}{(t-2\pi)^2}=\mathrm e^{j2t}\left(\frac{\sin(t-2\pi)}{t-2\pi}\right)^2=\mathrm e^{j2t}\mathrm{sinc}^2\left(t-2\pi\right)
$$
where $\mathrm{sinc}(x)=\frac{\sin x}{ x}$.
Now using the Fourier transform defined as $\mathcal F\{f(x)\}=\hat{f}(\xi)=\int _{-\infty }^{\infty }f(x)e^{-2\pi jx\xi }\,\mathrm dx$ we have
$$\mathcal F\{\mathrm{sinc}^2(t)\}=\pi\,\mathsf{Tri}(\pi\xi)$$
where $\mathsf{Tri}(x)$ is the triangular function
and using the shift property in time and frequency $$\mathcal F\{f(x-a)\mathrm e^{jbx}\}=\mathrm e^{-2\pi ja\xi}\hat f\left(\xi-\frac{b}{2\pi}\right)$$
we find, putting $a=2\pi$ and $b=2$
$$
\mathcal F\left\{\mathrm e^{j2t}\mathrm{sinc}^2\left(t-2\pi\right)\right\}=\pi\,\mathsf{Tri}(\pi\xi-1)\mathrm e^{-j4\pi^2\xi}
$$
A: First rewrite
$$e^{2jt} \sin^2 t 
= e^{2jt} \frac{1}{(2j)^2} \left(e^{jt} - e^{-jt}\right)^2 
= e^{2jt} \frac{1}{-4} \left(e^{2jt} - 2 + e^{-2jt}\right) 
= \frac{1}{4} \left( - e^{4jt} + 2e^{2jt} - 1 \right)$$
using the approach you mentioned. You'll get a weighted sum of three delta pulses. On the other hand, the term
$$\frac{1}{t^2}$$
has a known transform, look it up in a table. Just apply the time shift property to obtain the transform of 
$$\frac{1}{(t-2\pi)^2} \ .$$
Your expression is the product of the two discussed terms, so its transform is the frequency-domain convolution of their individual transforms. Convolution with a sum of delta pulses is very easy. Then you're done.
