Question

The Nyquist sampling rate for the signal $m\left(t\right)=\frac{\sin \left(600\pi t\right).\cos \left(800\pi t\right)}{\pi t}$ is ________kHz.

• A. 1.4

Answer 1

The correct option is A 1.4
Given signal,
$m\left(t\right)=\frac{\sin \left(600\pi t\right).\cos \left(800\pi t\right)}{\pi t}=\frac{\sin (600\pi t}{\pi t}.\cos \left(800\pi t\right)$

$\text{Let}m\left(t\right)=x_1\left(t\right).x_2\left(t\right)$

$\text{where,}{x}_1\left(t\right)=\frac{\sin \left(600\pi t\right)}{\pi t}$

$x_2\left(t\right)=\cos \left(800\pi t\right)$

$x_2\left(t\right)=\frac{e^{j800\pi t}+e^{-j800\pi t}}{2}$

$x_1\left(t\right).x_2\left(t\right)\stackrel{F.T}{⟷}{X}_1\left(f\right)\ast X_2\left(f\right)$

$\text{where,}{X}_2\left(f\right)=\delta (f-400)+\delta (f+400)$

$\therefore \text{Nyquist rate,}{f}_N=2f_{\text{max}}=2\times 700Hz$

$\therefore f_N=1.4\text{kHz}$