Question

If $x\left(t\right)$ is represented as given below and $h\left(t\right)=Sa\left(\frac{9\pi }{2}t\right)$ then the Nyquist rate for output $y\left(t\right)$ in $Hz$ will be________

• A. 3

Answer 1

The correct option is A 3
Time period for x(t), $T_0=2sec$

$\omega =\frac{2\pi }{T_0}=\frac{2\pi }{2}=\pi rad/sec$

$x\left(t\right)=dc+oddharmonicsofcosterms$

$=a_0+a_{1}cos{\omega }_{0}t+a_{3}cos3{\omega }_{0}t+....$

Frequency component available in input : $0,\pi ,3\pi ,5\pi ,....$

$h\left(t\right)=Sa\left(4.5\pi t\right)$




LPF cut off frequency,

${\omega }_c=4.5\pi$

Frequency component present in $y\left(t\right):0,\pi ,3\pi ,...$

upto 4.5 $\pi$ only 3 frequency components : $0,\pi ,3\pi$

So, ${\omega }_{es}{\mid }_{y\left(t\right)}=3\pi rad/sec$

Nyquist rate of output,

${\omega }_{ny}{\mid }_{y\left(t\right)}=2{\omega }_m=6\pi rad/sec$

$So,f_{\pi y}{\mid }_{y\left(t\right)}=\frac{6\pi }{2\pi }=3Hz$