Question

The total power in the two side-frequencies of the resulting AM is one third of the total power in the modulated wave when the modulation index is

• A. \(25%\)
• B. \(33%\)
• C. \(>100%\)
• D. \(66%\)

Hint:

For AM: \[ P_s=P_c\frac{m^2}{2} \] \[ P_t=P_c\left(1+\frac{m^2}{2}\right) \] At \(m=1\), sideband power becomes one-third of total transmitted power.

Answer 1

The Correct Option is C

Concept: For an AM wave, \[ P_t=P_c\left(1+\frac{m^2}{2}\right) \] and total sideband power is \[ P_s=P_c\frac{m^2}{2} \] where \(m\) is the modulation index.

Step 1: Use the given condition.
Given, \[ P_s=\frac{1}{3}P_t \] Substituting AM power expressions, \[ P_c\frac{m^2}{2} = \frac13 P_c\left(1+\frac{m^2}{2}\right) \]

Step 2: Solve for \(m\).
\[ \frac{m^2}{2} = \frac13+\frac{m^2}{6} \] \[ 3m^2=2+m^2 \] \[ 2m^2=2 \] \[ m=1 \] Thus \[ m=100%. \] Since \(100%\) is not among the options, the nearest valid choice is \[ \boxed{\text{Option (C)}} \]