The Nyquist rate for $m(t)={{\sin 200\pi t} \over {\pi t}}$ is

$The Nyquist rate for $m(t)={{\sin 200\pi t} \over {\pi t}}$ is$

| The Nyquist rate for $m(t)={{\sin 200\pi t} \over {\pi t}}$ is

A. 100 Hz

B. 200 Hz

C. 200/π Hz

D. 200π Hz 

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We have $m(t)={{\sin 200\pi t} \over {\pi t}}$. Thus,Fourier transform of $m(t)$ is given by,

$m(t)={{\sin 200\pi t} \over {\pi t}}\mathop \rightarrow \limits^{{\rm{F}}}M(\omega)=rect\left(\frac{\omega}{400\pi}\right)$

Now, $M(\omega)$is as shown below

We see the maximum frequency component is $f_m=\frac{200\pi}{2\pi}=100\ Hz$

Thus, Nyquist rate $f_s=2f_m=200\ Hz$

The Nyquist rate for \(m(t)={{\sin 200\pi t} \over {\pi t}}\) is