Question

A message signal $m\left(t\right)$ and the correspondiing DSB-SC signal $s\left(t\right)$ are given below:
$m\left(t\right)=\sin c\left(1000t\right)\sin c^2\left(2000t\right)$
$s\left(t\right)=m\left(t\right)\cos \left(2\pi f_{c}t\right)$
The carrier signal $f_c=2MHz$. If $\sin c\left(t\right)=\frac{\sin \left(\pi t\right)}{\pi t}$, then the bondwidth of the DSBSC, signal will be _____ kHz.

• A. 5

Answer 1

The correct option is A 5
$sinc\left(t\right)\stackrel{CTFT}{⟷}rect\left(f\right)$

$sinc\left(1000t\right)\stackrel{CTFT}{⟷}\frac{1}{1000}rect\left(\frac{f}{1000}\right)$

$sinc\left(2000t\right)\stackrel{CTFT}{⟷}\frac{1}{2000}rect\left(\frac{f}{2000}\right)$

$sin{c}^2\left(2000t\right)\stackrel{CTFT}{⟷}$

$sinc\left(1000t\right)sin{c}^2\left(2000t\right)\stackrel{CTFT}{⟷}$


So, the maximum frequency of the message signal is $f_{m\left(max\right)}=2.5kHz$
$Hence,(BW{)}_{DSB-SC}=2f_{m\left(max\right)}=2(2.5)=5kHz$.