Question

$\text{The periodic message signal, shown in he figure below, is applied to a modulator to}\text{modulate the amplitude of a sinusodial carrier signal. If the amplitude sensitivity of the}\text{modulator is}0.25V^{-1}\text{, then the transmission efficiency achieved by the resultant modulated}\text{signal will be}$

• A. $11.11$%
• B. $20$%
• C. $33.33$%
• D. $50$%

Answer 1

The correct option is B $20$%



$\text{The tranmission efficiency of an AM signal can be given by}\eta =\frac{k_a^2{P}_m}{1+k_a^2{P}_m}\text{Here,}{k}_a=\text{amplitude sensitivity of modulator}=0.25V^{-1}{P}_m=\text{Power of the message signal}\text{For the given message signal,}{P}_m=A^2=(2{)}^2=4\text{So,}\eta =\frac{\left(0.25{)}^2\right(4)}{1+\left(0.25{)}^2\right(4)}=\frac{0.25}{1+0.25}=\frac{1}{5}=0.20(or)20$