Question

A DSB-SC mdoulated signal is given as $\varphi \left(t\right)=m\left(t\right)\cos \left(2\pi f_{c}t\right)$ is multiplied with local carrier$s\left(t\right)=\cos ({\omega }_{C}t+\theta )$ and the output is passed through at LPF with a bandwith equal to the bandwidth of message signal. If the power of the message signal is given as $P_m$ then the power of the, LPF signal at the outputs equal to

• A. $\frac{P_m\cos \theta }{2}$
• B. $\frac{P_m\cos \theta }{2}$
• C. $\frac{P_m\cos \theta }{4}$
• D. $\frac{P_m{\cos }^2\theta }{4}$

Answer 1

The correct option is D

$\frac{P_m{\cos }^2\theta }{4}$

The signal $x\left(t\right)=m\left(t\right)\cos \left({\omega }_{c}t\right)\cos ({\omega }_{c}t+\theta )$

$=\frac{1}{2}m\left(t\right)\cos \theta +\frac{1}{2}m\left(t\right)\cos (2{\omega }_{c}t+\theta )$
After passing the signal through a low pass filter, we have

$y\left(t\right)=\frac{1}{2}m\left(t\right)\cos \theta$

Thus, the power of the output signal $P_y$ is equal to

$P_y=\frac{1}{4}{P}_m{\cos }^2\theta$