Sampling theorem condition

Question

I have this signal $$ x(t) = \sin(2 \pi f_0 t ) $$ with $ T_c= \frac{1}{2f_0} $ but I don't know if the Nyquist condition Is verified. The condition should be $ f_c \geq B_x $ where $ B_x $ is the bilateral band. I know that $ f_c= \frac{1}{T_c} = 2f_0 $ but I don’t know how to find $B$.

Answer 1

Let $x(t) = \sin(2 \pi f_0 t )$ then $X(j\omega) = \frac{\pi}{j}[\delta(\omega - 2 \pi f_0) - \delta(\omega + 2 \pi f_0)]$. So for satisfying sampling theorem condition we should have $$\omega_s \gt 2\times2\pi f_0$$ Where $\omega_s = \frac{2\pi}{T}$ is sampling frequency. Note that it should be $\gt$ instead of $\ge$ . See here for the reason.