Sampling a signal with halved frequency

Question

I have this signal $ x(t) = sin ( 2 \pi f_0 t ) $ , with $ T_c = \frac{1}{f_0} $ and I have to find $ x_\delta (t) $ and the reconstruction signal. first I found that $ f_c > 2 f_s $ so the Nyquist condition isn’t verified and there’s aliasing. After, I found that $ x_ \delta (t) = 0 $, because $\sin(2\pi)=0$. Now I know that the reconstruction signal $ x_{rec} (t) $ is the inverse Fourier transform of $ X(f) $. I found that $ X(f) = \frac{i}{2} \delta (f + f_c) - \frac{i}{2} \delta ( f - f_c ) $ but my book write that’s $0$

Answer 1

Let $x(t) = \sin(2\pi f_0 t)$ and sampling period be $T_s = \frac{1}{f_0}$. So we have $$p(t) = \sum_{n = -\infty}^{+\infty}\delta(t-nT_s)$$Where $p(t)$ is sampling function. Multiplying $x(t)$ and $p(t)$ gives us $$x_p(t) = \sum_{n = -\infty}^{+\infty}x(nT_s)\delta(t-nT_s)$$ In this case $x(nT_s) = \sin(2\pi n) = 0$ for all $n \in \mathbb{N}$. So $x_p(t)$ is identically zero which implies $X_p(j\omega) = 0$ for all $\omega \in \mathbb{R}$: $$X_p(j\omega) = \int_{-\infty}^{+\infty}x_p(t)e^{-j\omega t}dt = \int_{-\infty}^{+\infty}0 \ dt = 0$$Therefore the reconstructed signal using lowpass filter will be identically zero.