Question

Consider a discrete time signal \(y[n]=\frac{\sin (\pi n / 5)}{\pi n}+\cos (\pi n / 10) .\) Let's build the continuous time representation of \(y[n]\) using \(T_{s}=1 / 10\) as follows:

$$ y_{\delta}(t)=\sum_{n=-\infty}^{\infty} y[n] \delta\left(t-n T_{s}\right)=\sum_{n=-\infty}^{\infty} y[n] \delta(t-n / 10) $$

Choose the right expression for \(Y_{\delta}(j \omega)\).

(Hint: You can first sketch \(Y\left(e^{j \Omega}\right)\). Then, you can obtain the sketch of \(Y_{\delta}(j \omega)\) from \(Y\left(e^{j \Omega}\right)\). Note that in the earlier lecture on sampling, we derived that \(Y_{\delta}(j \omega)=\left.Y\left(e^{j \Omega}\right)\right|_{\Omega=\omega T_{x}}=Y\left(e^{j \omega T_{\nu}}\right)\).

In the past lectures and homework, we have practiced steps to obtain the plot of \(Y\left(e^{j \Omega}\right)\) from the plot of \(Y_{\delta}(j \omega)\) several times. You can reverse these steps to obtain the plot of \(Y_{\delta}(j \omega)\) from the plot of \(Y\left(e^{j \Omega}\right) .\) It will be easier to obtain the answer if you sketch the spectrums.)

(a) \(Y_{\delta}(j \omega)\) is \(20 \pi\)-periodic. For \(\omega \in[-10 \pi, 10 \pi], Y_{\delta}(j \omega)=\operatorname{rect}\left(\frac{\omega}{4 \pi}\right)+10 \pi \delta(\omega-\pi)+10 \pi \delta(\omega+\pi)\)

(b) \(Y_{\delta}(j \omega)\) is \(20 \pi\)-periodic. For \(\omega \in[-10 \pi, 10 \pi], Y_{\delta}(j \omega)=\operatorname{rect}\left(\frac{\omega}{4 \pi}\right)+\pi \delta(\omega-\pi)+\pi \delta(\omega+\pi)\).

(c) \(Y_{\delta}(j \omega)\) is \(20 \pi\)-periodic. For \(\omega \in[-10 \pi, 10 \pi], Y_{\delta}(j \omega)=10 \cdot \operatorname{rect}\left(\frac{\omega}{4 \pi}\right)+10 \pi \delta(\omega-\pi)+10 \pi \delta(\omega+\pi)\).

(d) \(Y_{\delta}(j \omega)\) is \(20 \pi\)-periodic. For \(\omega \in[-10 \pi, 10 \pi], Y_{\delta}(j \omega)=10 \cdot \operatorname{rect}\left(\frac{\omega}{4 \pi}\right)+10 \pi \delta(\omega-2 \pi)+10 \pi \delta(\omega+2 \pi)\)

Answer 1