Question

A modulator and a demodulator of a DSB-SC system are cascaded as shown in the figure below.

The message signal m(t) is band limited to $“f_m”$ and $fm<<f_c$. If the average power of the signal m(t) is 80 W, then the average power of the signal $y\left(t\right)$ is _____ W.

• A. 5

Answer 1

The correct option is A 5

From the figure given in the question, we get,

$s\left(t\right)=m\left(t\right)sin(2\pi f_{c}t+{30}^{\circ })$

$x\left(t\right)=m\left(t\right)sin(2\pi f_{c}t+{30}^{\circ })cos\left(2\pi f_{c}t\right)$

$=m\left(t\right)sin(2\pi f_{c}t+{30}^{\circ })sin(2\pi f_{c}t+{90}^{\circ })$

$=\frac{1}{2}m\left(t\right)\left[cos\right({60}^{\circ })–cos(4\pi f_{c}t+{120}^{\circ }\left)\right]$

$=\frac{cos\left({60}^{\circ }\right)}{2}m\left(t\right)-\frac{m\left(t\right)}{2}cos(4\pi f_{c}t+{120}^{\circ })$

After passing x(t) through LPF, we get,

$y\left(t\right)=\frac{cos\left({60}^{\circ }\right)}{2}m\left(t\right)=\frac{m\left(t\right)}{4}$

If the power of the signal m(t) is $P_m$, then the power of the signal y(t) can be given as

$P_y={\left(\frac{1}{4}\right)}^2{P}_m=\frac{P_m}{16}=\frac{80}{16}W=5W$

$\because \text{Given that,}{P}_m=80W$