Evaluate Fouries transform using properties

Question

$$x(t)=t\ \left[\frac{\sin(t)}{\pi t}\right]^2$$ How can I find the Fourier transform of the above signal without direct integral evaluation(using Fourier Transform properties)
The answer will be$$X(jw)=\begin{cases} j/2\pi, &-2\le w \le 0 \\ -j/2\pi , & 0\le w \le 2 \\ 0, & \text{otherwise} \end{cases}$$

Answer 1

You need the following correspondences:

$$\mathcal{F}\{f^2(t)\}=\frac{1}{2\pi}F(\omega)*F(\omega)\\ \mathcal{F}\{tf(t)\}=j\frac{dF(\omega)}{d\omega}$$

where $F(\omega)$ is the Fourier transform of $f(t)$, and '$*$' denotes convolution. The Fourier transform of $\sin(t)/\pi t$ is a rectangle with height 1 for $|\omega|<1$. If you convolve it with itself, you get a triangle between $\omega=-2$ and $\omega=2$ with maximum value 2. Dividing by $2\pi$, taking the derivative, and multiplying by $j$ gives the final result.

Answer 2

The Fourier transform of a product is the convolution between the Fourier transforms of the factors. Since: $$t\left(\frac{\sin t}{\pi t}\right)^2 = \frac{\sin(t)}{\pi}\cdot\frac{\sin(t)}{\pi t},$$ the Fourier transform of $\frac{\sin t}{t}$ is a multiple of $\mathbb{1}_{[-1,1]}$ and the Fourier transform of $\sin t$ is the difference of two Dirac deltas, the claim follows.