I tried solving the following exercise but I am not sure about the result. Could you help me?
The signal $x(t)= 12\text{sinc}^2(4t) + 4 \cos(4 \pi t)$ is sampled with a rate $f_c = 3$. It then becomes $$y[n]=y(nT_c)= 12\text{sinc}^2\left(\frac{4}{3} n\right) + \cos\left(\frac{4 \pi}{3} n\right) = y_{1s}(t) + y_{2s}(t) $$. What is the Fourier transform of $y[n]$? Sampling an aperiodic signal in the time domain brings us a continuous and periodic spectrum in the frequency domain, so $$Y_{1s}(f) = f_c \cdot \sum_{k=-\infty}^{+\infty} {Y_1(f) * \delta(f-kf_c)}= 9 \cdot \sum_{k=-\infty}^{+\infty} {\text{tr}\left(\frac{f-3k}{4}\right)}$$
Now, $y_{2}[n]=4 \cos(\frac{4}{3} \pi n) $ is a periodic signal: to obtain its Fourier transform do I use the same method as before, getting $$Y_{2s}(f)= 6 \cdot \sum_{k=-\infty}^{+\infty} {\delta(f-2-3k) + \delta(f+2-3k)}$$ ? Is it correct?
If I then decide to send $Y(f)$ in input to a low pass filter with frequency response $H_{LP}(f)= \frac{1}{f_c} \cdot \text{rect}(\frac{f}{f_c})$, I should have $Z(f)=Y(f) \cdot H_{LP}(f)= 3 tr(\frac{f}{4}) \cdot \text{rect}(\frac{f}{3}) + 2[\delta(f-1) + \delta(f +1)] $. Now I have to get $z[n]$ from this. We know that $\displaystyle\mathscr{F}[x[n]]= \sum_{n=-\infty}^{+\infty}{x[n] \cdot e^{-j2 \pi fT_c n}}$ and $x[n]= T_c \cdot \int_{\frac{-f_c}{2}}^{\frac{f_c}{2}} {X(f) \cdot e^{j2 \pi fT_c n}}df$ . $$Z(f)=3 \cdot \text{triangle}\left(\frac{f}{4}\right) + 2[\delta(f-1) + \delta(f+1)]$$ , so $$z[n]=12\text{sinc}^2(4n)*\text{sinc}(3n)+ \frac{4}{3}\cos(2 \pi n)$$. Is there any mistake?