Q. An amplitude modulated wave is modulated to 50%. What is the saving in power if carrier as well as one of the side bands are suppressed
A) 70%
B) 65.4%
C) 94.4%
D) 25.5%
Explanation
(c) $P_{sb} = P_{c}\left( \frac{m_{a}}{2} \right)^{2} = P_{c}\frac{(0.5)^{2}}{4} = 0.0625\ P_{c}$
Also $P = P_{c}\ \left( 1 + \frac{m_{a}^{2}}{2} \right) = P_{c}\left( 1 + \frac{(0.5)^{2}}{2} \right) = 1.125P_{c}$
$\therefore\%$ saving $= \frac{(1.125P_{c} - 0.0625P_{c})}{1.125P_{c}} \times 100 = 94.4\%$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.