Question

A 360 W carrier is simultaneously modulated by two audio waves with modulation percentages of 55 and 65 respectively. The effective modulation index and the total power radiated are
1. 0.85 and 490.5 W 
2. 0.65 and 490.5 W
3. 0.85 and 450.5 W
4. 0.65 and 450.5 W

Answer 1

Correct Answer - Option 1 : 0.85 and 490.5 W 

Concept:

When a carrier is modulated by different waves having different modulation indexes, the effective modulation index is given by:

\({{\rm{μ }}_{{\rm{eff}}}} = \sqrt {{\rm{μ }}_1^2 + {\rm{μ }}_2^2}\)

Also, the total power of the modulated signal for the given effective modulation index is given by:

\({P_t} = {P_c}\left( {1 + \frac{{{μ_{eff}^2}}}{2}} \right)\)

Pc = Carrier Power

μ_eff = Effective modulation Index

Calculation:

Given μ₁ = 0.55 and μ₂ = 0.65, the effective modulation index will be:

\({{\rm{μ }}_{{\rm{eff}}}} = \sqrt{(0.55^2) + (0.65)^2}\)

μ_eff = 0.85

The total power radiated will be:

\({P_t} = {360}\left( {1 + \frac{{{(0.85^2)}}}{2}} \right)\)

P_t ≈  490.5 W