Question

A DSB-SC signal is generated by using the modulator shown below.

$\text{Here m(t) is a message signal which is band-limited to}{f}_m\text{. The bandpass filter is ideal with unity passband gain over a bandwidth of}2f_m\text{centered around}{f}_c\text{. The carrier generator generates}{\cos }^3\left(2\pi f_{c}t\right),\text{but not}\cos \left(2\pi f_{c}t\right)\text{. If the average power of the message signal m(t) is 8 W, then the average power of the output modulated signal y(t) will be\_\_\_\_\_\_\_W.}$

• A. 2.25

Answer 1

The correct option is A 2.25
$x\left(t\right)=m\left(t\right){\cos }^3\left(2\pi f_{c}t\right)$
$=m\left(t\right)\left[\frac{3}{4}\cos \right(2\pi f_{c}t)+\frac{1}{4}\cos (6\pi f_{c}t\left)\right]$
$=\frac{3}{4}m\left(t\right)\cos \left(2\pi f_{c}t\right)+\frac{1}{4}m\left(t\right)\cos \left(6\pi f_{c}t\right)$
After passing through the BPF, we get,
$y\left(t\right)=\frac{3}{4}m\left(t\right)\cos \left(2\pi f_{c}t\right)=\frac{3}{4}m\left(t\right)\cos \left({\omega }_{c}t\right)$
$=\frac{3}{4}m\left(t\right)\left(\frac{e^{j{\omega }_{c}t+e^{-j{\omega }_{c}t}}}{2}\right)=\frac{3}{8}\left[m\right(t)e^{j{\omega }_{c}t}+m\left(t\right)e^{-j{\omega }_{c}t}]$
The average power of the signal y(t) will be,

$P_y={\left(\frac{3}{8}\right)}^2[P_m+P_m]=\frac{9}{32}{P}_m$
$\text{Given that,}{P}_m=8W$
$\text{So,}{P}_y=\frac{9}{32}\times 8W=2.25W$