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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 $B$.

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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.

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    $\begingroup$ Thank you so much !!!!! $\endgroup$ Jul 28, 2020 at 9:59
  • $\begingroup$ @S.H.W why did you take $X(j \omega)$ instead of $X(\omega)$? $\endgroup$
    – Migalobe
    Mar 11, 2022 at 17:51
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    $\begingroup$ @Migalobe It's just a convention in EE. $\endgroup$
    – S.H.W
    Mar 11, 2022 at 18:36
  • $\begingroup$ ok, but the result is the same right? $\endgroup$
    – Migalobe
    Mar 11, 2022 at 19:17
  • $\begingroup$ @Migalobe Yes, different names for the same function. $\endgroup$
    – S.H.W
    Mar 11, 2022 at 19:20

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