Calculate the convolution: $\frac{\sin(4t)}{\pi t}*( \cos(t)+\cos(6t) )$ using Fourier transform

Question

$$\frac{\sin(4t)}{\pi t}*[ \cos(t)+\cos(6t) ]$$

We know,

$$\mathscr{F}\{\cos(t)\}=\frac{1}{2} \delta\left(f-\frac{1}{2 \pi}\right)+\frac{1}{2} \delta\left(f+\frac{1}{2 \pi}\right)$$

and

$$\mathscr{F}\{\cos(6t)\}=\frac{1}{2} \delta\left(f-\frac{3}{ \pi}\right)+\frac{1}{2} \delta\left(f+\frac{3}{ \pi}\right)$$

Furthermore, using duality, \begin{align} & \mathscr{F}\left\{ \text{rect}\left(\frac{t}{T}\right) \right\}= \frac{\sin(\pi f T)}{\pi f } \\ \implies & \mathscr{F}\left\{ \frac{\sin(4t)}{\pi t} \right\} = \text{rect}\left(-f \frac{\pi}{4}\right) \end{align}

Is it correct?

How can I continue the exercise?

Thanks

Answer 1

$$x_1(t)=\frac{\sin 4t}{\pi t}=\frac{4}{\pi}Sinc(\frac{4}{\pi}t)\\x_2(t)=\cos t+\cos 6t\to \\ X_1(i\omega)=rect(\frac{\pi}{4}\omega) \\ X_2(i\omega)=\pi(\delta(\omega-1)+\delta(\omega+1)+\delta(\omega-6)+\delta(\omega+6))\to \\ Y(i\omega)=X_1(i\omega)X_2(i\omega)=\pi(\delta(\omega-1)+\delta(\omega+1))\to y(t)=x_1(t)*x_2(t)=\cos t$$

Answer 2

I presume using the Convolution Theorem since it is specified "using the Fourier transform" in the title:

$\mathscr{F}\{{f*g}\}=\mathscr{F}\{{f}\}\mathscr{F}\{{g}\}$

$\mathscr{F}\{{fg}\}=\mathscr{F}\{{f}\}*\mathscr{F}\{{g}\}$ (where $f$ and $g$ are any two functions)