Question

The Nyquist interval corresponding to the continuous time signal $x\left(t\right)=\left[S{a}^3\right(150\pi t)\ast S{a}^2(200\pi t\left)\right]Sa\left(600\pi t\right)$ is _______ msec.

• A. 1

Answer 1

The correct option is A 1
$S{a}^3\left(150\pi t\right)\to {\omega }_{m1}=3\times 150\pi =450\pi rad/sec$

$S{a}^2\left(200\pi t\right)\to {\omega }_{m2}=2\times 200\pi =400\pi rad/sec$

for $\left(f_1\right(t{)}^{\ast }{f}_2\left(t\right))\to {\omega }_{mf}=min\{{\omega }_{m1},{\omega }_{m2}\}=400\pi rad/sec$

$Sa\left(600\pi t\right)\to {\omega }_{m_z}=600\pi rad/sec$

for $f(t{)}^{\prime }.g(t)\to {\omega }_m={\omega }_{mf}+{\omega }_{mg}$

$=400\pi +600\pi =1000\pi rad/sec$

Nyquist interval, $T_N=\frac{\pi }{{\omega }_m}=\frac{\pi }{1000\pi }=1msec$