Question

The Nyquist sampling rate of the signal

$x\left(t\right)=sinc{}^2\left({10}^{4}t\right)\sin c^2\left({10}^{5}t\right)is$

• A. 240
• B. 210
• C. 220
• D. 230

Answer 1

The correct option is C 220

$\sin c\left(f\right)\stackrel{CTFT}{⟷}rect\left(f\right)$

$\sin c\left({10}^{4}t\right)\stackrel{CTFT}{⟷}rect\left(\frac{f}{{10}^4}\right)$

$\sin c\left({10}^{4}t\right)\stackrel{CTFT}{⟷}rect$

$\sin c^2\left({10}^{4}t\right)\stackrel{CTFT}{⟷}rect$

$\sin c^2\left({10}^{5}t\right)\stackrel{CTFT}{⟷}rect$



$x\left(t\right)\stackrel{CTFT}{⟷}$

$f_{max}=100+10=110kHz$

$f_{m(max)=2f_{max}=220kHz}$