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 $B$.

S.H.W · Accepted Answer · 2020-07-26T12:00:10.950

0

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.

edited Jul 26 '20 at 12:00

answered Jul 26 '20 at 11:44

S.H.W

4,379

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Thank you so much !!!!! – Elena Martini Jul 28 '20 at 09:59
@S.H.W why did you take $X(j \omega)$ instead of $X(\omega)$? – Migalobe Mar 11 '22 at 17:51
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@Migalobe It's just a convention in EE. – S.H.W Mar 11 '22 at 18:36
ok, but the result is the same right? – Migalobe Mar 11 '22 at 19:17
@Migalobe Yes, different names for the same function. – S.H.W Mar 11 '22 at 19:20
@S.H.W they are different. btw, if I understood well, you took $2\pi f_0 >0$ bcuz it makes what is inside the first delta function vanish. Is that correct? – Migalobe Mar 11 '22 at 22:10

Sampling theorem condition

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