How to get output from High Pass filter on this signal? I'm trying to solve this exercise:


  Consider a High-Pass filter with a cutoff frequency of $W=2\pi$
  
  Calculate the output signal if the input signal is $x(t)=\displaystyle \frac{\sin(4\pi t)}{\pi t}$


The fourier transform for the input signal is:

$X(\omega)= \left\{ \begin{array}{lcc}
             1 &amp;   \text{if}  &amp; -4\pi \lt \omega \lt 4\pi \\[2ex]
             0 &amp;  \text{otherwis}e \\
             \end{array}
   \right.$

Now, the output signal is :
$Y(\omega)=H(\omega)X(\omega)$

$Y(\omega)= \left\{ \begin{array}{lcc}
             1 &amp;   \text{if}  &amp; 2\pi \le |\omega| \le 4\pi \\[2ex]0 &amp;  \text{otherwise} \\
             \end{array}
   \right.$

I can get here, but the solution to the exercise converts this function with inverse fourier transform, giving the result of $x(t)=2\displaystyle \frac{\sin(\pi t)}{\pi t}\cos(3\pi t)$.

I don't know how did they get to that conclusion, I have tried to search for it, but i don't know how to look for that.

I hope that the question is clear. Sorry if I'm lacking some details, it's my first time asking this type of questions.
What you get as $x(\omega)$ is two $\mathrm{sinc}(\pi t) = \displaystyle\frac{\sin(\pi t)}{\pi t}$ shifted by $+\omega_0$ and $-\omega_0$ in frequency ($\omega_0 = 3 \pi$). By the frequency shift property of the FT:

$$ 
\mathcal F\left\{x(t) \cdot e^{i\omega_0 t}\right\} = X\left(e^{j(\omega - \omega_0)}\right)
$$

Then, remember $e^{i\omega t} = \cos(\omega t) + i \cdot \sin(\omega t)$. Therefore:

$$
x(\omega) = \sqcap_{2\pi} (\omega - \omega_0) + \sqcap_{2\pi} (\omega + \omega_0)
$$

Doing inverse:
$$
\begin{align}
x(t) &amp;= \mathcal F^{-1}\left\{x(\omega)\right\} \\
 &amp;=\mathrm{sinc} (\pi t) \cdot \big(\cos(\omega_0 t) + i \cdot \sin(\omega_0 t)\big) +  \mathrm{sinc} (\pi t) \cdot \big(\cos(-\omega_0 t) + i \cdot \sin(-\omega_0 t)\big) \\
&amp;= 2 \cdot \mathrm{sinc} (\pi t)\cdot \cos(\omega_0 t) \\
&amp;= 2 \cdot \frac{\sin(\pi t)}{\pi t}\cdot \cos(3 \pi t)
\end{align}
$$

since $\cos(-x)=\cos(x)$ and $\sin(-x)=-\sin(x)$